aa r X i v : . [ m a t h . L O ] M a r Dp-finite fields IV: the rank 2 picture
Will JohnsonMarch 23, 2020
Abstract
We investigate fields of characteristic 0 and dp-rank 2. While we do not obtain aclassification, we prove that any unstable field of characteristic 0 and dp-rank 2 admitsa unique definable V-topology. If this statement could be generalized to higher ranks,we would obtain the expected classification of fields of finite dp-rank.We obtain the unique definable V-topology by investigating the “canonical topol-ogy” defined in [8]. Contrary to the expectations of [9], the canonical topology neednot be a V-topology. However, we are able to characterize the canonical topology (onfields of dp-rank 2 and characteristic 0) in terms of differential valued fields.This differential valued structure is obtained through a partial classification of , a sort of “generalized valuation” that arises naturally in fields of finite rank.Additionally, we give an example of a dp-rank 2 expansion of ACVF with a definableset of full rank and empty interior. This example interferes with certain strategies forproving the henselianity conjecture.
NIP structures play a central role in modern model theory, and so it would be desirable toclassify the NIP theories of fields. NIP can be characterized via dp-rank: a structure M isNIP iff dp-rk( M ) < ∞ . For an overview of NIP and dp-rank, see ([12], Chapters 2 and 4).From the point of view of dp-rank, the natural first step is fields of dp-rank 1 (dp-minimal fields). Dp-minimal fields were successfully classified in ([7], Chapter 9). The hopeis to generalize this proof to the next simplest case—fields of finite dp-rank (dp-finite fields). The present paper continues [8], [9], [10], which made some partial progress on the classi-fication of dp-finite fields. The overall strategy is to prove the dp-finite case of the Shelahconjecture:
Conjecture 1.1 (Shelah conjecture, dp-finite case) . Let K be a dp-finite field. Then one ofthe following holds: K is finite • K is algebraically closed • K is real closed • K admits a non-trivial henselian valuation. Modulo this conjecture, the classification of dp-finite fields is known ([3], Theorem 3.11 ).Let ( K, + , · , . . . ) be a dp-finite field, possibly with extra structure. If K is stable, then K must be algebraically closed or finite ([5], Proposition 7.2). Assume K is unstable. In [8]and [9], we constructed a field topology on K characterized by the fact that the followingfamily is a neighborhood basis of 0: { X − X : X ⊆ K, X is definable , dp-rk( X ) = dp-rk( K ) } . Here X − Y denotes the set of differences { x − y : x ∈ X, y ∈ Y } , rather than the set difference X \ Y .We call this topology the canonical topology . In a monster model K (cid:23) K , define the K -infinitesimals to be the intersection of all K -definable basic neighborhoods: J K = \ { X − X : X ⊆ K , X is K -definable , dp-rk( X ) = dp-rk( K ) } . Using the infinitesimals, we proved the Shelah conjecture for dp-finite fields of positivecharacteristic in ([8], Corollary 11.4).In [9], we sketched a strategy for attacking fields of characteristic 0. Say that K is valu-ation type if the canonical topology is a V-topology. (See [11] for a reference on topologicalfields and V-topologies.) We conjectured Conjecture 1.2 (Valuation conjecture) . If K is an unstable dp-finite field, then K is valu-ation type. Modulo this conjecture, we proved the Shelah conjecture. We also gave a seeminglyweaker criterion which implies the valuation conjecture:
Fact 1.3 (Theorem 8.11 in [9]) . If the K -infinitesimals J K contain a non-zero ideal of amulti-valuation ring on K , then K is valuation type. Here, a multi-valuation ring on K means a finite intersection of valuation rings on K . The classification in [3] is for strongly dependent fields, but the proof specializes to the case of dp-finitefields. The fields appearing in the conjectured classification of strongly dependent fields are all dp-finite ([3],Proposition 3.9). .2 Main results for dp-finite fields In the present paper, we investigate unstable fields of dp-rank 2 and characteristic 0. Wefind a counterexample to the valuation conjecture:
Theorem 1.4.
There is a valued field ( K, O ) and a subset R ⊆ K such that • The structure ( K, + , · , O , R ) has dp-rank 2. • The set R has full rank dp-rk( R ) = 2 , but has empty interior with respect to thevaluation topology. • The canonical topology is not a V-topology.
The counterexample does not contradict the Shelah conjecture, or the expected classifi-cation of dp-finite fields and valued fields. In fact, ( K, O ) | = ACVF , .In spite of the counterexample, we are able to prove the following, statement, whichwould imply the Shelah conjecture if generalized to higher ranks: Theorem 1.5.
Let K be an unstable field of characteristic 0 and dp-rank 2. • The canonical topology on K is definable, i.e., there is a uniformly definable basis ofopens. • There is a unique definable non-trivial V-topology on K . Additionally, we can give a rather explicit description of the canonical topology, in thecases where it is not a V-topology.
Theorem 1.6.
Let K be a sufficiently resplendent unstable field of characteristic 0 anddp-rank 2. Suppose the canonical topology on K is not a V-topology. Then there exists avaluation val : K → Γ and a derivation δ : K → K such that the following sets form a basisfor the canonical topology: B a,b,γ := { x ∈ K : val( x − a ) > γ and val( δx − b ) > γ } . Moreover, every B a,b,γ is non-empty. Non-emptiness of the B a,b,γ expresses some degree of independence between the derivationand the valuation. We call this sort of field topology a DV-topology . We investigate DV-topologies in § K ) = 0, we only use char( K ) = 2.In fact, we prove the valuation conjecture in odd characteristic: Theorem 1.7.
Let K be an unstable field of dp-rank 2, with char( K ) > . Then the canonicaltopology on K is a V-topology. .3 Inflators Theorem 1.5 is obtained through an algebraic analysis of 2-inflators. An inflator is somesort of “generalized valuation” that occurs naturally when attempting to prove the valuationconjecture. The basic properties of inflators were investigated in [10].The main point of [10] was that the infinitesimals J K are “governed” by an r -inflator, forsome r ≤ dp-rk( K ). Fact 1.8.
Let K be a sufficiently saturated unstable dp-finite field. Then there are smallmodels k (cid:22) K (cid:22) K and a malleable k -linear r -inflator ς on K such that • r ≤ dp-rk( K ) . • The group J K of K -infinitesimals is an ideal in the fundamental ring R ς of ς . • If ς ′ is any mutation of ς , then there is a small model K ′ (cid:23) K such that J K ′ is an idealin R ς ′ . For definitions of inflators, the fundamental ring, malleability, and mutation, see Defini-tions 4.1, 5.8, 5.31, and 10.2 (respectively) in [10]. We verify Fact 1.8 in § ς is weakly multi-valuation type ([10], Definition 5.27) if its fundamental ring R ς contains a non-zero ideal of a multi-valuation ring. Because of Fact 1.8, we can focus ourattention on 2-inflators ς with the following properties:1. ς is malleable.2. No mutation of ς is weakly multi-valuation type. Otherwise, some K ′ ≡ K would havevaluation type, implying the same for K .3. The underlying field K has characteristic 0. The classification of dp-finite fields isalready known in positive characteristic.Sections 2–5 carry out an algebraic analysis of inflators satisfying these assumptions. Theoriginal hope was to rule out these “wicked” 2-inflators. Instead, we get a rather explicitalgebraic description. Theorem 1.9.
Let ( K, O , m ) be a valued field of characteristic 0, and k be a small subfieldon which the valuation is trivial. Let ∂ : O → K/ m be a k -linear derivation. Suppose thatfor every x ∈ K/ m , the set { y ∈ O : ∂y = x } is dense in O . Let R = { x ∈ O : ∂x ∈ O / m } I = { x ∈ m : ∂x ∈ m / m = 0 } . Then R is a subring of K and I is an ideal in R , and the quotient R/I is isomorphic to k [ ε ] = k [ ε ] / ( ε ) , where k is the residue field of O . There is a 2-inflator ς n : Sub K ( K n ) → Sub k (( R/I ) n ) V ( V ∩ R n + I n ) /I n .
4e call these diffeovaluation inflators and study them in §
8. Actually, the definition isa little more general, allowing K/ m to be replaced with a “mock K/ m .”Up to mutation, diffeovaluation inflators account for all the “wicked” 2-inflators: Theorem 1.10.
Let K be a field of characteristic 0. Let ς be a malleable 2-inflator on K .Suppose that no mutation of ς is weakly multi-valuation type. Then some mutation of ς is adiffeovaluation inflator. Using this explicit characterization, we prove Theorem 1.5. The characterization stronglyhints at how to build the example in Theorem 1.4; we check the details in § A “field” may contain additional structure beyond the pure field structure. We will use bold K for sufficiently saturated and resplendent fields.Unlike [8], [9], and [10], we will write the K -infinitesimals as J K , not I K , to avoid conflictwith the fundamental ideal I of a 2-inflator.We will use the following definitions and facts from [10]: • Directories (Definition 2.1) and the characterization of semisimple directories (Theo-rem 2.7). • Inflators and equivalence of inflators (Definitions 4.1, 4.2). • The basic inflator calculus of § • The fundamental ring and ideal, and the generalized residue map (Proposition 5.7,Definition 5.8). • Tame and wild elements (Lemma 5.22 and Definition 5.23). • The notions of (weakly) multi-valuation type (Definitions 5.26, 5.27), and the charac-terization in terms of tame and wild elements (Proposition 5.25). • Mutation (Theorem 10.1, Definition 10.2), transitivity of mutation (Proposition 10.5),and commutativity of mutation (Remark 10.7). • Malleability (Definition 5.31), and its preservation under mutation (Proposition 10.13)Unlike [10], we will use k for the small ground field, rather than K . Our inflators will be k -linear. All rings will be k -algebras, and all fields will extend k .Over the course of § ς : Dir K ( K ) → Dir S ( M ), satisfyingthe following assumptions:1. The characteristic of K (or its subfield k ) is not 2.2. ς is malleable 5. No mutation of ς is weakly multi-valuation type.For § ς is isotypic: if we write M as N ⊕ N ′ with N, N ′ simple, then N ∼ = N ′ .We call (1-3) the “Weak Assumptions” and (1-4) the “Strong Assumptions.” Remark . Until Lemma 5.20, we will use only the following weaker form of (3): nomutation of ς is multi-valuation type.During our analysis, we will define a number of sets, rings, and functions. We includethe following list as a reference: • The 2-inflator will be ς : Dir K ( K ) → Dir S ( M ) , where M is a semisimple S -module of length 2. Beginning in § S will be a skew field k , and following § k will be commutative. • k [ ε ] will denote the ring of dual numbers k [ ε ] / ( ε ). • R and I will denote the fundamental ring and ideal of ς . The generalized residue mapwill be c res : R → End S ( M ), or later c res : R → End k ( M ). In §
3, we will arrange for M = k [ ε ], and show that c res factors through k [ ε ] ∼ = End k [ ε ] ( k [ ε ]) ⊆ End k ( k [ ε ]) = End k ( M ) . Beginning in §
4, we will therefore view c res as a map c res : R → k [ ε ] . • In § A will denote the image of c res in End k ( M ), and A will denote the k -algebragenerated by A . In Lemma 3.7 and Proposition 3.11, we will see A = A ∼ = k [ ε ] . • p and Q will denote the sets p = { x ∈ R : c res( x ) ∈ kε } Q = { x ∈ R : c res( x ) ∈ k } , where k [ ε ] = k ⊕ kε . Then p will be an ideal in R , and Q will be a subring of R . • O will denote the integral closure of R . In Corollary 4.5, we will see that O is avaluation ring. In Proposition 4.9, we will see that the residue field of O is k . We willlet val : K × → Γres :
O → k c res( x ) = s + tε = ⇒ res( x ) = s, for x ∈ R and s, t ∈ k . • In §
5, we will construct an O -module D and a Q -linear derivation ∂ : O → D , as wellas a valuation val : D → Γ ≤ ∪ { + ∞} . ( D is essentially K/ m ). Before constructing D and ∂ , we will define a mapval ∂ : O → Γ ≤ ∪ { + ∞} in Definition 5.12. Later, val ∂ ( x ) will turn out to be val( ∂x ). For the duration of § ς : Dir K ( K ) → Dir S ( M )will be a k -linear 2-inflator, and the following Weak Assumptions will be in place: • The characteristic of K (or its subfield k ) is not 2. • ς is malleable ([10], Definition 5.31). • No mutation of ς is weakly of multi-valuation type ([10], Definition 5.27) Remark . If ς satisfies the Weak Assumptions, then so does any mutation ς ′ , by ([10],Propositions 10.5 and 10.13).Since M is semisimple of length 2, we can write it as an internal direct sum M = A ⊕ B, with A, B simple. We say that ς is isotypic if A ∼ = B . This depends only on the isomorphismclass of the directory Dir S ( M ).By Theorem 2.7 in [10], we can assume that we are in one of two cases: • S = k and M = k , for some division algebra k over k . • S = k × k and M = k ⊕ k , for two division algebras k , k over k .The first case is isotypic, and the second case is non-isotypic.7 .1 The degeneracy subspace For any α ∈ K , let Θ α denote the lineΘ α := K · (1 , α ) = { ( x, αx ) : x ∈ K } . For any ϕ ∈ End S ( M ), let Θ ϕ denote the graph of ϕ , i.e.,Θ ϕ = { ( x, ϕ ( X )) : x ∈ M } . Recall from ([10], Definition 5.8) that the fundamental ring of ς is the set R = { α ∈ K | ∃ ϕ ∈ End S ( M ) : ς (Θ α ) = Θ ϕ } . This is a subring of K .For every α ∈ K , one of two things occurs, by Lemma 5.22 in [10]. • The fundamental ring R contains all but at most two of the numbers { α } ∪ (cid:26) α − c : c ∈ k (cid:27) . (1) • The fundamental ring R contains none of the numbers in (1).In the first case, α is called tame , and in the second case α is called wild ([10], Definition 5.23).The fact that ς does not have multi-valuation type implies that there is at least one wild α ∈ K ([10], Proposition 5.25). Lemma 2.2.
There is an S -submodule A ⊆ M of length 1 such that for every wild α , ς (Θ α ) = A ⊕ A ⊆ M ⊕ M. Proof.
Suppose α is wild. Then α / ∈ R , so ς (Θ α ) is not the graph of an endomorphism.Counting lengths, this implies 0 ⊕ A ⊆ ς (Θ α ) for some length-1 submodule A ⊆ M . Similarly, α − / ∈ R implies that 0 ⊕ A ′ ⊆ ς (Θ α − )for some length-1 submodule A ′ ⊆ M . Or equivalently, by permutation invariance, A ′ ⊕ ⊆ ς (Θ α ) . As ς (Θ α ) has length 2, we must have ς (Θ α ) = A ′ ⊕ A. We claim that A = A ′ . Suppose otherwise. Then M is an internal direct sum of A and A ′ .By GL ( k )-equivariance, ς (Θ / ( α +1) ) = ς ( { ( αx + x, x ) : x ∈ K } )= { ( y + x, x ) : x ∈ A ′ , y ∈ A } . / ( α + 1) ∈ R and α is tame, acontradiction.Thus, for any wild α there is some length-1 submodule A α ⊆ M such that ς (Θ α ) = A α ⊕ A α . It remains to show that A α doesn’t depend on α . Suppose for the sake ofcontradiction that α, β are wild and A α = A β . Then ς ( { ( x, αx, y ) : x, y ∈ K } ) = A α ⊕ A α ⊕ Mς ( { ( w, y, βy ) : w, y ∈ K } ) = M ⊕ A β ⊕ A β ς ( { ( x, αx, αβx ) : x ∈ K } ) = A α ⊕ ⊕ A β using Lemma 5.2.1 in [10] and the fact that A α ∩ A β = 0. Then ς ( { ( x, αx, αβx ) : x ∈ K } ) = A α ⊕ ⊕ A β ς ( { (0 , y,
0) : y ∈ K } ) = 0 ⊕ M ⊕ ς ( { ( x, y, αβx ) : x, y ∈ K } ) = A α ⊕ M ⊕ A β , using Lemma 5.2.2 in [10]. But a symmetric argument shows ς ( { ( x, y, βαx ) : x, y ∈ K } ) = A β ⊕ M ⊕ A α As αβ = βα , it follows that A α ⊕ M ⊕ A β = A β ⊕ M ⊕ A α , and A α = A β , a contradiction.We call A = A α the degeneracy subspace of M . Corollary 2.3.
Under the Weak Assumptions, ς has a mutation which is isotypic.Proof. Take wild α , and let A ⊆ M be the degeneracy subspace, so that ς ( K · (1 , α )) = A ⊕ A. By definition of mutation, the mutation along K · (1 , α ) is of the form ς ′ : Dir K ( K ) → Dir S ( M ′ ) , where M ′ = ς ( K · (1 , α )) = A ⊕ A . The S -module M ′ is isotypic.9 The explicit formula for ς For the duration of § ς : Dir K ( K ) → Dir S ( M )will be a k -linear 2-inflator, and the following Strong Assumptions will be in place: • char( k ) = char( K ) = 2 • ς is malleable • No mutation of ς is weakly of multi-valuation type. • ς is isotypic, i.e., if we write M as a direct sum of two simple S -modules A and B ,then A ∼ = B .Isotypy is the new assumption, not present in the Weak Assumptions of §
2. Isotypy impliesthat Dir S ( M ) ∼ = Dir k ( k )for some division k -algebra k . Therefore, we may assume S = k and M is a two-dimensional k -vector space. Remark . As in Remark 2.1, the Strong Assumptions are preserved under mutations. Forisotypy, note that any mutation of ς will have the form ς ′ : Dir K ( K ) → Dir k ( M ′ )for some k -module M ′ . The fact that k is a division ring ensures that M ′ is isotypic. k is commutative Proposition 3.2.
The division ring k is commutative (a field).Proof. Changing coordinates on k , we may assume that the degeneracy subspace is 0 ⊕ k .If α ∈ K is wild, then ς ( { ( x, αx ) : x ∈ K } ) = (0 ⊕ k ) ⊕ (0 ⊕ k ) . (2)We write elements of M n = ( k ) n as tuples ( a , b ; a , b ; . . . ; a n , b n ).Let a, b be two non-commuting elements of k . By malleability, we can find α, β ∈ K × such that ς ( { ( x, αx ) : x ∈ K } ) ⊇ { ( t, ta,
0) : t ∈ k } ς ( { ( x, βx ) : x ∈ K } ) ⊇ { ( t, tb,
0) : t ∈ k } α nor β can be wild, by equation (2). By GL ( k )-equivariance, ς ( { ( x, αx ) : x ∈ K } ) ⊇ { ( t, ta,
0) : t ∈ k } ς ( { ( x, ( α − x ) : x ∈ K } ) ⊇ { ( t, t ( a − ,
0) : t ∈ k } ς ( { ( x, α − x ) : x ∈ K } ) ⊇ { ( t, ta − ,
0) : t ∈ k } ς ( { ( x, ( α − − x ) : x ∈ K } ) ⊇ { ( t, t ( a − − ,
0) : t ∈ k } And one of α, /α, / ( α −
1) is in R . So, replacing a with one of { a, a − , ( a − − } , we mayassume that α ∈ R . Similarly, we may assume β ∈ R . Then ς ( { ( x, αx ) : x ∈ K } ) ς ( { ( x, βx ) : x ∈ K } )are graphs of endomorphisms ϕ A , ϕ B : k → k . Because( t, ta, ∈ Γ A ( t, tb, ∈ Γ B it follows that ϕ A ( t,
0) = ( ta,
0) and ϕ B ( t,
0) = ( tb, ϕ A and ϕ B do not commute.But this is impossible, as ϕ A , ϕ B both lie in the image of the homomorphism R → End k ( k ),and R is commutative.Thus k is a field extending k . A Let A be the image of R in End k ( k ) = M ( k ). This is a commutative k -algebra. Lemma 3.3.
The algebra A is not contained in the center of M ( k ) , i.e., A contains amatrix not of the form (cid:18) λ λ (cid:19) .Proof. Changing coordinates on M ∼ = k , we may assume that 0 ⊕ k is the degeneracysubspace, and so ς ( { ( x, αx ) : x ∈ K } ) = (0 ⊕ k ) ⊕ (0 ⊕ k ) . (3)for any wild α . By malleability, choose α ∈ K such that ς ( { ( x, αx ) : x ∈ K } ) ⊇ { ( t,
0; 0 , t ) : t ∈ k } ∋ (1 ,
0; 0 , . Then α cannot be wild. By GL ( k )-equivariance, we have(1 ,
0; 0 , ∈ ς ( { ( x, αx ) : x ∈ K } )(0 ,
1; 1 , ∈ ς ( { ( x, α − x ) : x ∈ K } )(1 ,
1; 1 , ∈ ς ( { ( x, ( α + 1) − ) : x ∈ K } ) .
11y tameness of α , one of the right-hand-sides is the graph of some endomorphism ϕ ∈ End k ( k ). By definition, ϕ ∈ A . But no central matrix (cid:18) λ λ (cid:19) can map (1 ,
0) to (0 ,
1) ormap (0 ,
1) to (1 ,
0) or map (1 ,
1) to (1 , A be the k -subalgebra of M ( k ) generated by A . It is a commutative k -algebra.Note that M = k is naturally an A -module. Proposition 3.4.
For any V ⊆ K n , the specialization ς n ( V ) is an A -submodule of M n .Proof. We already know that ς n ( V ) is a k -submodule, so it remains to show that ς n ( V ) isclosed under multiplication by A . Let a be an element of R , specializing to ϕ ∈ A . Then ς n ( { ( ~x, a~x ) : ~x ∈ K n } ) = { ( x , . . . , x n , ϕx , . . . , ϕx n ) : ~x ∈ M n } ς n ( { ( ~x, ~y ) : ~x ∈ V, ~y ∈ K n } ) = { ( x , . . . , x n , y , . . . , y n ) : ~x ∈ ς n ( V ) , ~y ∈ M n } ς n ( { ( ~x, ~y ) : ~x ∈ K n , ~y ∈ V } ) = { ( x , . . . , x n , y , . . . , y n ) : ~x ∈ M n , ~y ∈ ς n ( V ) } ς n ( { ( ~x, a~x ) : ~x ∈ V } ) = { ( x , x , . . . , x n , ϕx , . . . , ϕx n ) : ~x ∈ ς n ( V ) } . The first line holds because a specializes to ϕ , using compatibility with ⊕ and permutations.The second and third lines hold by compatibility with ⊕ . The fourth line holds by intersectingthe first and second lines (using Lemma 5.2.1 in [10]). Now V is a K -submodule of K n , andhence an R -submodule. Therefore { ( ~x, a~x ) : ~x ∈ V } ⊆ { ( ~x, ~y ) : ~x ∈ K n , ~y ∈ V } . As ς n is order-preserving, { ( x , x , . . . , x n , ϕx , . . . , ϕx n ) : ~x ∈ ς n ( V ) }⊆ { ( x , . . . , x n , y , . . . , y n ) : ~x ∈ M n , ~y ∈ ς n ( V ) } , which implies that ς n ( V ) is closed under multiplication by ϕ .Proposition 3.4 says that the inflator ς : Dir K ( K ) → Dir k ( M ) factors through Dir A ( M ) ⊆ Dir k ( M ). Corollary 3.5.
The degeneracy subspace X ⊆ M is an A -submodule of M .Proof. If α ∈ K is wild, then ς ( K · (1 , α )) = X ⊕ X, and so X ⊕ X is an A -submodule of M ⊕ M . This implies X is an A -submodule of M . Lemma 3.6.
The k -algebra A is isomorphic to one of the following • k × k • k [ ε ] = k [ ε ] / ( ε ) . oreover, M is a free A -module of rank 1.Proof. View A as a commutative k -subalgebra of M ( k ). Take µ ∈ A non-central. Changingthe identification M ∼ = k , we may assume we are in one of three cases: • µ = (cid:18) a b (cid:19) for some a = b in k . • µ = (cid:18) a b a (cid:19) for some a, b ∈ k with b = 0. • µ = (cid:18) a b (cid:19) for some monic irreducible quadratic polynomial x − bx − a ∈ k [ x ].The degeneracy subspace is a one-dimensional subspace of k , preserved by µ , and so µ hasan eigenvector. This rules out the third case.In the first case, the k -subalgebra generated by µ is A ′ = (cid:26)(cid:18) x y (cid:19) : x, y ∈ k (cid:27) ∼ = k × k. Then
A ⊇ A ′ . In particular, A contains the matrix µ ′ = (cid:18) (cid:19) , and A lies in the centralizerof µ ′ . By inspection, this centralizer is A ′ . Thus A = A ′ ∼ = k × k . Also, the vector (1 , k as an A ′ -module.In the second case, the k -subalgebra generated by µ is A ′′ = (cid:26)(cid:18) x y x (cid:19) : x, y ∈ k (cid:27) ∼ = k [ ε ] . Then
A ⊇ A ′′ . In particular, A contains the matrix µ ′′ = (cid:18) (cid:19) , and A lies in thecentralizer of µ ′′ . By inspection, this centralizer is A ′′ . Thus A = A ′′ ∼ = k [ ε ]. Also, thevector (1 ,
0) freely generates k as an A ′′ -module. Lemma 3.7.
The two algebras A and A are equal.Proof. Changing M up to isomorphism, we may assume M = A . Let µ be any element of A ; we will show µ ∈ A . By malleability, there is a line L ⊆ K such that ς ( L ) ⊇ k · (1 , µ ) . But ς ( L ) is an A -submodule of M (Proposition 3.4), and the A -submodule generated by(1 , µ ) is { ( x, µx ) : x ∈ A} . Thus ς ( L ) ⊇ { ( x, µx ) : x ∈ A} . k -modules, and so equality holds. Let a be the slope of L .Evidently, a = ∞ , or else ς ( L ) would be 0 ⊕ A . So a ∈ R and a specializes to µ . Therefore µ ∈ A .In particular, the natural homomorphism R ։ A ֒ → A is surjective. If I is the fundamental ideal (the kernel of R → End k ( M )), then R/I isisomorphic to k × k or to k [ ε ]. Proposition 3.8.
Fix any A -module isomorphism M ∼ = A . Then the specialization maps ς n : Sub K ( K n ) → Sub k ( A n ) are given by the formula ς n ( V ) = { ( c res( x ) , . . . , c res( x n )) : ( x , . . . , x n ) ∈ V ∩ R n } where c res is the generalized residue map R ։ R/I ∼ = A .Proof. First suppose that ( a , . . . , a n ) ∈ R n ∩ V , and a i specializes to b i ∈ A for each i . Thenthe line L = { ( x, a x, . . . , a n x ) : x ∈ K } is contained in K ⊕ V , and so ς n +1 ( L ) = { ( x, b x, . . . , b n x ) : x ∈ A} ⊆ ς n +1 ( K ⊕ V ) = A ⊕ ς n ( V ) . In particular, (1 , b , . . . , b n ) ∈ A ⊕ ς n ( V ), and so( c res( a ) , . . . , c res( a n )) = ( b , . . . , b n ) ∈ ς n ( V ) . We have seen ς n ( V ) ⊇ { ( c res( x ) , . . . , c res( x n )) : ( x , . . . , x n ) ∈ V ∩ R n } Conversely, suppose that ( b , . . . , b n ) ∈ ς n ( V ). Then(1 , b , . . . , b n ) ∈ A ⊕ ς n ( V ) = ς n +1 ( K ⊕ V ) . By malleability, there is a line L ⊆ K ⊕ V such that ς n +1 ( L ) ⊇ k · (1 , b , . . . , b n ) ∋ (1 , b , . . . , b n ) . The left hand side is an A -module, so ς n +1 ( L ) ⊇ A · (1 , b , . . . , b n ) . Both sides have length two over k , so equality holds. Now L must be the graph of a K -linearfunction K → K n ; otherwise L ⊆ ⊕ K n and ς n +1 ( L ) would be contained in 0 ⊕ A n , whichis visibly false. Thus L = K · (1 , a , . . . , a n )14or some a i ∈ K . So ς n +1 ( { ( x, a x, . . . , a n x ) : x ∈ K } ) = { ( x, b x, . . . , b n x ) : x ∈ A} . Joining this with ς n +1 (0 ⊕ K i − ⊕ ⊕ K n − i ) = 0 ⊕ A i − ⊕ ⊕ A n − i , we obtain ς n +1 ( { ( x, y , . . . , y i − , a i x, y i +1 , . . . , y n ) : x, y , . . . , y n ∈ K } )= { ( x, y , . . . , y i − , b i x, y i +1 , . . . , y n ) : x, y , . . . , y n ∈ A} . By permutation invariance and ⊕ -compatibility, we see ς ( { ( x, a i x ) : x ∈ K } ) = { ( x, b i x ) : x ∈ A} . So each a i is in R , and c res( a i ) = b i . The fact that L ⊆ K ⊕ V implies that ~a ∈ V . So wehave shown that ς n ( V ) ⊆ { ( c res( x ) , . . . , c res( x n )) : ( x , . . . , x n ) ∈ V ∩ R n } . Corollary 3.9. K = Frac( R ) .Proof. Let α be any element of K × . By Proposition 3.8, ς ( { ( x, αx ) : x ∈ K } ) = { ( c res( x ) , c res( y )) : x, y ∈ R, y/x = α } . The fact that ς ( { ( x, αx ) : x ∈ K } ) = 0 implies that there exist non-trivial x, y ∈ R suchthat y/x = α . Let I denote the fundamental ideal ([10], Definition 5.8), i.e., the kernel of the generalizedresidue map c res : R → A . Remark . Every maximal ideal of R comes from a maximal ideal of the artinian ring R/I ∼ = A . Indeed, this follows from the fact that I ⊆ J ac ( R ) ([10], Proposition 5.7.4).For example, if A ∼ = k × k , then R has two maximal ideals. Proposition 3.11.
The algebra A is isomorphic to k [ ε ] .Proof. Otherwise,
A ∼ = k × k . Let p , p : R → k be the two maps such that c res( x ) = ( p ( x ) , p ( x )) ∈ k × k ∼ = A . R are the kernels of p and p .The degeneracy subspace of ς is some rank 1 submodule of A , necessarily k × × k .Without loss of generality, it is 0 × k .Take some wild element a ∈ K . By Proposition 3.8 and the definition of the degeneracysubspace, ς ( K · (1 , a )) = { ( c res( x ) , c res( y )) : x, y ∈ R, y/x = a } = { ( p ( x ) , p ( x ); p ( y ) , p ( y )) : x, y ∈ R, y/x = a } = { (0 , t ; 0 , s ) : t, s ∈ k } . Therefore, we can find x, y, x ′ , y ′ ∈ R such that( p ( x ) , p ( x )) = (0 , p ( y ) , p ( y )) = (0 , y = ax ( p ( x ′ ) , p ( x ′ )) = (0 , p ( y ′ ) , p ( y ′ )) = (0 , y ′ = ax ′ . Then xy ′ = yx ′ , and we obtain a contradiction:1 = p ( y ) p ( x ′ ) = p ( yx ′ ) = p ( xy ′ ) = p ( x ) p ( y ′ ) = 0 . Corollary 3.12.
The fundamental ring R is a local ring.Proof. Its maximal ideals come from k [ ε ], which is a local ring.The following corollary will be useful later: Corollary 3.13. If ς : Dir K ( K ) → Dir k ( k ) satisfies the Strong Assumptions, and a ∈ K specializes to a matrix µ = (cid:18) b cd e (cid:19) , then µ must have a repeated eigenvalue, and so ( b + e ) = (Tr( µ )) = 4 det( µ ) = 4( be − cd ) . Proof.
The element a lies in R and its image in A is µ . If µ is central, the identity is clear.Otherwise, the proof of Lemma 3.6 shows that µ must be conjugate to a matrix of the form (cid:18) x y x (cid:19) or (cid:18) x y (cid:19) . In the second case, A ∼ = k × k , contradicting Proposition 3.11. So wemay assume µ = (cid:18) x y x (cid:19) , and then the desired identity is clear.Now that we have identified A , we can specify the degeneracy locus: Lemma 3.14.
Under any isomorphism of A -modules M ∼ = A , the degeneracy subspace isthe principal ideal kε = ( ε ) ⊳ k [ ε ] ∼ = A . roof. There are only three k [ ε ]-submodules of k [ ε ], and kε is the only one having dimension1 over k .We summarize the picture in the following Theorem. Theorem 3.15.
Let ς be an isotypic malleable k -linear 2-inflator on a field K with char( K ) =2 . Suppose that no mutation of ς is weakly of multi-valuation type. Then there are • A field k extending k . • A subring R ⊆ K . • An ideal I ⊳ R • An isomorphism of k -algebras R/I ∼ = k [ ε ] := k [ ε ] / ( ε ) such that ς is isomorphic to ς : Dir K ( K ) → Dir k ( k [ ε ]) ς n ( V ) = { ( c res( x ) , . . . , c res( x n )) : ~x ∈ V ∩ R n } where c res is the quotient map R ։ R/I ∼ = k [ ε ] . Moreover, • R is the fundamental ring of ς , I is the fundamental ideal, and c res is the generalizedresidue map (in the sense of Definition 5.8 in [10]). • R is a local ring, whose unique maximal ideal m is the pullback of k · ε along c res . • Frac( R ) = K . Continue the Strong Assumptions of §
3. In light of Theorem 3.15, we assume that ς has theform ς : Dir K ( K ) → Dir k ( k [ ε ]) ς n : Sub K ( K n ) → Sub k ( k [ ε ] n ) V
7→ { ( c res( x ) , . . . , c res( x n )) : ~x ∈ V ∩ R n } where c res : R ։ k [ ε ] is the generalized residue map. The fundamental ideal I ⊳ R is thekernel of c res. 17 .1 Finitely generated ideals Let Q be the set of a ∈ R such that c res( a ) lies in k , i.e., c res( a ) = x + 0 ε for some x ∈ k . The set Q is a k -subalgebra of R . Note that I is an ideal in Q . Moreover,if a ∈ Q \ I , then c res( a ) = x + 0 ε for some non-zero x , and so a ∈ R × . Then c res( a − ) = ( x + 0 ε ) − = x − + 0 ε, so that a − ∈ Q . Thus Q is a local ring and I is its maximal ideal. Lemma 4.1.
Let a, b, c be three elements of K . Then there exist x, y, z ∈ Q such that ax + by + cz = 0 , and at least one of x, y, z is 1. The same holds for R instead of Q .Proof. Consider the vector space V = { ( x, y, z ) ∈ K : ax + by + cz = 0 } . Then ς ( V ) = { ( c res( x ) , c res( y ) , c res( z )) : ( x, y, z ) ∈ R and ax + by + cz = 0 } . Also, dim k ( ς ( V )) = 2 · dim K ( V ) = 4 . Counting dimensions, ς ( V ) must have non-trivial intersection with the subspace k = { ( s + 0 ε, t + 0 ε, u + 0 ε ) : s, t, u ∈ k } ⊆ k [ ε ] . Therefore, there exist x, y, z ∈ R such that c res( x ) = s + 0 ε c res( y ) = t + 0 ε c res( z ) = u + 0 ε for some s, t, u ∈ k , not all zero. Then x, y, z ∈ Q and at least one of the three is in Q × . If x ∈ Q × , we may replace x, y, z with x/x, y/x, z/x , and arrange for x = 1. This handles thecase of Q , and the case of R follows as R ⊇ Q .We shall return to the ring Q in § Corollary 4.2.
Any finitely-generated R -submodule of K is generated by at most two ele-ments. In particular, any ideal of R is generated by at most two elements.Proof. It suffices to consider the case of three generators: R · a + R · b + R · c ≤ K . Thenthe lemma implies that one of a, b, c is in the R -submodule generated by the other two.18 .2 The integral closure of R We let p denote the unique maximal ideal of R , i.e., the set of x ∈ R such that c res( x ) hasthe form 0 + tε for some t ∈ k . Because R is local, R × = R \ p . Note that p is the pullbackof the principal ideal ( ε ) = kε ⊳ k [ ε ] along c res( − ). Lemma 4.3.
Suppose α ∈ K satisfies a monic quadratic equation over R : α + bα + c = 0 for some b, c ∈ R . If α is wild, then b − c ∈ p .Proof. Let b , b , c , c ∈ k be such that c res( b ) = b + b ε c res( c ) = c + c ε. We must show b = 4 c . Let L be the line K · (1 , α ). By Lemma 3.14, ς ( L ) = ς ( { ( x, αx ) : x ∈ K } ) = kε ⊕ kε = { ( sε, tε ) : s, t ∈ k } . It follows that ς ( { ( x, αx, − cx, − bαx ) : x ∈ K } ) = { ( sε, tε, − c sε, − b tε ) : s, t ∈ k } ς ( { ( x, αx, − bαx − cx ) : x ∈ K } ) = { ( sε, tε, ( − b t − c s ) ε ) : s, t ∈ k } ς ( { ( x, αx, αx, α x ) : x ∈ K } ) = { ( sε, tε, tε, ( − b t − c s ) ε ) : s, t ∈ k } Here, we are using the identities c res( b ) · ( sε ) = ( b s ) ε c res( c ) · ( sε ) = ( c s ) εα = − bα − c. Now let ς ′ : Dir K ( K ) → Dir k ( M ′ ) be the mutation along L . Then M ′ = ς ( L ) = { ( sε, tε ) : s, t ∈ k } ∼ = k , and ς ′ ( L ) = ς ( { ( x, αx ; αx, α x ) : x ∈ K } ) = { ( s, t ; t, ( − b t − c s )) : s, t ∈ k } . Note that ς ′ ( L ) is the graph of the k -linear map k → k (4)( s, t ) ( t, − b t − c s ) (5)Therefore α lies in the fundamental ring R ′ of ς ′ . By Remark 3.1, ς ′ continues to satisfy theStrong Assumptions, and then by Corollary 3.13, the linear map (5) must have a repeatedeigenvalue. Therefore b = (cid:18) Tr (cid:18) − c − b (cid:19)(cid:19) = 4 det (cid:18) − c − b (cid:19) = 4 c . emma 4.4. Let α be an element of K × . Then α or α − is integral over R , and in fact oneof α or α − satisfies a monic polynomial equation of degree d , where d = ( α tame α wild.Proof. Let α be an element of K . First suppose α is tame. Then there is some b ∈ k suchthat the number α ′ := 1 / ( α − b ) lies in R . Because R is a k -algebra, it contains bα ′ + 1.Because R is a local k -algebra, at least one of α ′ and bα ′ + 1 is invertible. Therefore at leastone of the following lies in R : α = bα ′ + 1 α ′ /α = α ′ bα ′ + 1 . Next suppose α is wild. By Lemma 4.1, there are x, y, z ∈ R such that x + yα + zα = 0 , and at least one of x, y, z is in R × . If z is invertible, then α + ( y/z ) α + ( x/z ) = 0 , so α is integral over R . Similarly, if x is invertible, then 1 /α is integral over R . So we mayassume y is invertible and x, z are not. Then0 ≡ x ≡ z y (mod p ) . Let β = α − α , so that α = β − β +1 . Then x + yα + zα = 0( β + 1) x + ( β − β + 1) y + ( β − z = 0( β + 2 β + 1) x + ( β − y + ( β − β + 1) z = 0( x + y + z ) β + (2 x − z ) β + ( x + z − y ) = 0 β + 2 x − zx + z + y β + x + z − yx + z + y = 0where in the final line we have used the fact that ( R, p ) is a local ring and x + z + y ≡ y p ) . Note also that 2 x − zx + z + y ≡ y (mod p ) x + z − yx + z + y ≡ − yy (mod p ) . β + b ′ β + c ′ = 0 for some b ′ , c ′ ∈ R with b ′ ≡ c ′ ≡ −
1. Because β and α arerelated by a fractional linear transformation over k , we know that β is wild. As we are notin characteristic 2, ( b ′ ) = 0
6≡ − ≡ c ′ (mod p ) , contradicting Lemma 4.3. Corollary 4.5.
Let O denote the integral closure of R (in K ). Then O is a valuation ringon K . O Let m denote the maximal ideal of the valuation ring O . Lemma 4.6.
The intersection m ∩ R is exactly the prime ideal p .Proof. Let α be an element of R . First suppose α / ∈ p . Then α − ∈ R ⊆ O , so α / ∈ m .Conversely, suppose α ∈ p but α / ∈ m . Then α − ∈ O , so there exist c , c , . . . , c n − ∈ R such that α − n + c n − α − n + · · · + c α − + c = 0 , or equivalently − c n − α + c n − α + · · · + c α n − + c α n . But the right-hand side is in p and the left hand side is not, a contradiction. Corollary 4.7.
The valuation ring O is non-trivial.Proof. By surjectivity of c res : R → k [ ε ], we can find x ∈ R with c res( x ) = ε . Then x = 0,but x ∈ p . Therefore m = 0 and O 6 = K . Lemma 4.8. If α ∈ O , then α is tame if and only if α ∈ R .Proof. If α ∈ R then α is tame by definition. Conversely, suppose α is tame. By Lemma 4.4,one of α or 1 /α is in R . If α ∈ R , we are done. Otherwise, 1 /α ∈ R and α / ∈ R , so 1 /α is anon-invertible element of R . Then 1 /α ∈ p ⊆ m , so α has negative valuation, contradictingthe assumption that α ∈ O . Proposition 4.9.
The induced map k ∼ = R/ p = R/ ( R ∩ m ) ֒ → O / m is onto, hence an isomorphism. In particular, O has residue field isomorphic to k . roof. We must show that for every x ∈ O there exists y ∈ R such that x ≡ y (mod m ) . If x ∈ R we can take y = x , so we may assume x / ∈ R . Then x is wild by Lemma 4.8. If x ∈ m we can take y = 0. So we may assume that x − ∈ O . By Lemma 4.4, at least one of x or x − satisfies a monic quadratic polynomial equation over R .First suppose it is x . Then x + c x + c = 0 (6)for some c , c ∈ R . By Lemma 4.3, c − c ∈ p . As we are not in characteristic 2, we may rewrite (6) as (cid:16) x + c (cid:17) = (cid:18) c − c (cid:19) . The right hand side is in p ⊆ m , so it has positive valuation. Therefore x + c / x + c ∈ m . Thus we can take y = − c / x − satisfies a monic quadratic polynomial equation over R : x − + c x − + c = 0 . The same argument shows that x − ≡ b (mod m ) for some b ∈ R . Then x ≡ b − (mod m ),and x / ∈ m = ⇒ b / ∈ m = ⇒ b / ∈ p = ⇒ b ∈ R × = ⇒ b − ∈ R, so we can take y = b − .We let res : O → k denote the natural residue map. Note that if a ∈ R and c res( a ) = x + yε, then res( a ) = x . Lemma 4.10.
Let α be an element of K . Suppose α / ∈ O , and α − / ∈ p . Let ς ′ denote themutation along K · (1 , α − ) , let R ′ denote the fundamental ring of ς ′ , and let p ′ denote themaximal ideal of R ′ . Then α − ∈ p ′ . roof. First note that α − ∈ m ⊆ O . If α − is tame, then α − ∈ R by Lemma 4.8, and then α − ∈ p by Lemma 4.6. This contradicts the assumptions, and so α − and α are wild. ByLemma 4.4, one of α and α − satisfies a monic quadratic polynomial equation over R . Since α does not lie in the integral closure O of R , it must be α − that satisfies the equation: α − = bα − + c. Then b + 4 c ∈ p by Lemma 4.3. We claim that c ∈ p . Otherwise, c ∈ R × , and c − = c − bα + α , contradicting the fact that α is not integral over R . So b + 4 c ≡ p ) c ≡ p )and therefore b ∈ p as well. Let β, γ be c res( b ) and c res( c ), respectively. The fact that b, c ∈ p implies that β, γ ∈ kε , and therefore β, γ annihilate kε .As α − is wild and kε is the degeneracy subspace, ς ( { ( x, α − x ) : x ∈ K } ) = { ( sε, tε ) : s, t ∈ k } . By the usual inflator calculus, one sees that ς ( { ( x, α − x, α − x ) : x ∈ K } ) = ς ( { ( x, α − x, bα − x + cx ) : x ∈ K } )= { ( sε, tε, βtε + γsε ) : s, t ∈ k } = { ( sε, tε,
0) : s, t ∈ k } . Now let ς ′ be the mutation of σ along K · (1 , α − ). Then ς ′ ( { ( x, α − x ) : x ∈ K } ) = ς ( { ( x, α − x ; α − x, α − x ) : x ∈ K } )= { ( s, t ; t,
0) : s, t ∈ k } . Thus α − specializes to the endomorphism( s, t ) ( t, , and so α − ∈ R ′ . This endomorphism fails to be invertible, so α − ∈ p ′ . Lemma 4.11.
Let ς ′ be a mutation of ς , with fundamental ring R ′ . Let p ′ be the maximalideal of R ′ . Then p ⊆ p ′ .Proof. Suppose ς ′ is the mutation along L = K · ( a , . . . , a n ). Then ς ′ is a mapDir K ( K ) → Dir k ( M ′ ) , M ′ = ς n ( L ) ⊆ ( k [ ε ]) n . By Proposition 3.4, M ′ is a k [ ε ]-submodule of ( k [ ε ]) n .Take b ∈ p . Then c res ς ( b ) = sε for some s ∈ k . This means that b ∈ R specializes (withrespect to ς ) to the endomorphism k [ ε ] → k [ ε ] z sεz. With respect to the mutation ς ′ , the element b specializes to the endomorphism M ′ → M ′ ~v sε~v, by Lemma 10.3 in [10]. This map is not onto, by Nakayama’s lemma (over the Noetherianring k [ ε ]). Therefore b ∈ R ′ , but b − / ∈ R ′ , implying that b ∈ p ′ . Proposition 4.12. If ς ′ is a mutation of ς , with fundamental ring R ′ , then R ′ ⊆ O . Con-sequently, R ′ has the same integral closure as R .Proof. Let b be an element of R ′ that is not in O . First suppose that b − ∈ p . ThenLemma 4.11 implies b − ∈ p ′ . Therefore b / ∈ R ′ , a contradiction.Next suppose that b − / ∈ p . Let τ and τ ′ be the mutations of ς and ς ′ along K · (1 , b − ).By Lemma 4.10, b − ∈ p τ . By commutativity of mutation (Remark 10.7 in [10]), τ ′ is amutation of τ . By Lemma 4.11, b − ∈ p τ ⊆ p τ ′ . This implies b / ∈ R τ ′ . But b ∈ R ′ = R ς ′ ⊆ R τ ′ , a contradiction. Corollary 4.13.
The integral closure O is the limiting ring R ∞ of [10], Definition 10.9.Proof. The limiting ring R ∞ is integrally closed, so R ∞ ⊇ O . On the other hand R ∞ is aunion of rings R ′ obtained by mutation. Proposition 4.12 says R ′ ⊆ O . Thus R ∞ ⊆ O . Continue the Strong Assumptions of § Remark . Over the next few sections, we will carry out a number of convoluted calcula-tions. The motivated reader may wish to keep two running examples in mind: • The diffeovaluation inflators of § • The “endless mutation” example of § .1 Double mutation lemma The idea of the next few lemmas is that we can calculate the residue res( r ) of an element r by passing to a mutation where r becomes tame; Proposition 4.12 ensures that the valuationdoes not change in the mutation. Lemma 5.2.
Let r be an element of K , let q be an element of k , and let L be a line (aone-dimensional subspace) in K n . Suppose that every element of ς n ( { ( ~x, r~x ) : ~x ∈ L } ) is of the form ( ~x, q~x ) . Then val( r ) ≥ and res( r ) = q .Proof. Let M ′ = ς n ( L ). Then ς n ( { ( ~x, r~x ) : ~x ∈ L } ) ⊆ { ( ~x, q~x ) : ~x ∈ k [ ε ] n } ς n ( { ( ~x, r~x ) : ~x ∈ L } ) ⊆ ς n ( L ⊕ L ) = M ′ ⊕ M ′ ς n ( { ( ~x, r~x ) : ~x ∈ L } ) ⊆ { ( ~x, q~x ) : ~x ∈ M ′ } . The first line is by assumption, the second line is by order-preservation and ⊕ -compatibility,and the third line follows by intersecting the first two lines. Counting lengths, equality musthold in the second line. Let ς ′ : Dir K ( K ) → Dir k ( M ′ )be the mutation of ς along L . Let R ′ , p ′ , I ′ , O ′ , m ′ denote the analogues of R, p , I, O , m forthe mutation ς ′ . Then ς ′ ( K · (1 , r )) = ς n ( { ( ~x, r~x ) : ~x ∈ L } ) = { ( ~x, q~x ) : ~x ∈ M ′ } . It follows that r specializes with respect to ς ′ to the endomorphism M ′ → M ′ x qx. Thus r ∈ R ′ ⊆ O ′ . Choose some p ∈ R such that c res( p ) = q = q + 0 ε . By Lemma 10.3 in[10], the element p is also in R ′ , and also specializes to this endomorphism. Therefore r − p ∈ I ′ ⊆ p ′ ⊆ m ′ . By Proposition 4.12, O = O ′ and m = m ′ , implying that r ∈ O and r − p ∈ m . Thereforeval( r ) ≥ r ) = res( p ) = q . Lemma 5.3.
Suppose a ∈ K has val( a ) > , and suppose ( u, v ) ∈ ς ( K · (1 , a )) for some u, v ∈ k [ ε ] with v = 0 . Then there is a ′ ∈ K such that ( k · v ) ⊕ ( k · v ) = ς ( K · (1 , a ′ )) . roof. If a is wild, then ς ( K · (1 , a )) = kε ⊕ kε by Lemmas 2.2 and 3.14. Then v ∈ kε , and( k · v ) ⊕ ( k · v ) = kε ⊕ kε. So we may take a ′ = a .Otherwise, a is tame, and so a ∈ R ∩ m = p by Lemmas 4.6 and 4.8. Then c res( a ) = bε for some b ∈ k , and( u, v ) ∈ ς ( K · (1 , a )) = { ( x + yε, ( bε )( x + yε )) : x, y ∈ k } = { ( x + yε, xb ) ε ) : x, y ∈ k } . Thus v ∈ kε . Then we can take a ′ to be any wild element, and( k · v ) ⊕ ( k · v ) = kε ⊕ kε = ς ( K · (1 , a ′ )) . Lemma 5.4 (Double mutation lemma) . Let a, r be elements of K , with val( a ) > . Supposethat ( s, t, u, qu ) ∈ ς ( K · (1 , r, a, ar )) for some s, t, u ∈ k [ ε ] and q ∈ k with u nonzero. Then val( r ) ≥ and res( r ) = q .Proof. Let ς ′ : Dir K ( K ) → Dir k ( M ′ ) be the mutation of ς along K · (1 , r ), where M ′ = ς ( M ′ ) ⊆ ( k [ ε ]) . Then ς ′ ( K · (1 , a )) = ς ( K · (1 , r, a, ar )) ∋ ( s, t, u, qu ) . By Lemma 5.3 applied to ς ′ , there is some a ′ ∈ K such that ς ′ ( K · (1 , a ′ )) = { ( xu, xqu ; yu, yqu ) : x, y ∈ k } . Equivalently, then ς ( K · (1 , r, a ′ , a ′ r )) = { ( xu, xqu, yu, yqu ) : x, y ∈ k } ς ( K · (1 , a ′ , r, a ′ r )) = { ( xu, yu, xqu, yqu ) : x, y ∈ k } . By Lemma 5.2 applied to the line L = K · (1 , a ′ ), it follows that val( r ) ≥ r ) = q . Recall from § Q is the subring Q = { x ∈ R : c res( x ) ∈ k } , where we view k as a subset of k [ ε ] in the natural way. Definition 5.5. If a ∈ O , a neutralizer is an a † ∈ Q such that aa † ∈ R \ Q .26eutralizers need not be unique. Lemma 5.6. If a ∈ O \ Q , then a has a neutralizer.Proof. First suppose a ∈ R \ Q . Then 1 is a neutralizer.Next suppose a ∈ O \ R . By Lemma 4.8, the element a is wild. Then ς ( K · (1 , a )) = kε ⊕ kε. On the other hand, by Proposition 3.8, ς ( K · (1 , a )) = { ( c res( x ) , c res( ax )) : x ∈ R, ax ∈ R } . Therefore, there is some a † ∈ R such that aa † ∈ R , and c res( a † ) = 0 c res( aa † ) = ε. Then a † ∈ Q and aa † ∈ R \ Q . Lemma 5.7.
Let a be a wild element. • If b ∈ R and ab ∈ R , then c res( b ) = pε c res( ab ) = qε for some p, q ∈ k . • If a † is a neutralizer of a , then c res( a † ) = 0 c res( aa † ) = qε. for some nonzero q ∈ k .Proof. Because a is wild and kε is the degeneracy subspace, { ( c res( x ) , c res( ax )) : x ∈ R, ax ∈ R } = ς ( K · (1 , a )) = kε ⊕ kε. The fact that b, ab ∈ R thus implies that c res( b ) and c res( ab ) lie in kε , and so c res( b ) = 0 + pε c res( ab ) = 0 + qε, for some p, q ∈ k . When b is a neutralizer a † , we must have p = 0 and q = 0, because a † ∈ Q and aa † / ∈ Q . 27 emma 5.8. Let a be a wild element with val( a ) > , and let a † be a neutralizer of a .Suppose b ∈ R and ab ∈ R . Then val( b ) ≥ val( a † ) . Moreover, ab ∈ Q ⇐⇒ val( b ) > val( a † ) . Proof.
By Lemma 5.7, c res( a † ) = 0 c res( b ) = pε c res( aa † ) = sε c res( ab ) = qε, for some p, q, s ∈ k with s = 0. Note ab ∈ Q ⇐⇒ q = 0. It suffices to show thatval( b/a † ) ≥ b/a † ) = q/s .By the inflator calculus, ς ( K · (1 , a † , b, aa † , ab )) = ς ( { ( x, a † x, bx, aa † x, abx ) : x ∈ K } )= { ( x, , ( pε ) x, ( sε ) x, ( qε ) y : x ∈ k [ ε ] } = ( k [ ε ]) · (1 , , pε, sε, qε ) , and therefore (0 , pε, sε, qε ) ∈ ς ( K · ( a † , b, aa † , ab )) = ς ( K · (1 , b/a † , a, ab/a † )) . By the Double Mutation Lemma 5.4 with r = b/a † , it follows that val( b/a † ) ≥ b/a † ) = q/s . Lemma 5.9.
Let a be a wild element with val( a ) > , and let a † be a neutralizer. Suppose b ∈ R and ab / ∈ R . Then val( b ) < val( a † ) .Proof. Note that b ∈ R ⊆ O and a ∈ m ⊆ O , so ab ∈ O . The fact that ab / ∈ R then impliesthat ab is wild, by Lemma 4.8. Also, ab / ∈ Q , as Q ⊆ R . By Lemma 5.6, ab has a neutralizer( ab ) † . By Lemma 5.7, c res(( ab ) † ) = 0 c res( ab ( ab ) † ) = sε, for some non-zero s ∈ k . Let c = b ( ab ) † . Then c ∈ R (because b ∈ R and ( ab ) † ∈ Q ⊆ R ).Also ac = ( ab )( ab ) † ∈ R \ Q. By Lemma 5.8, ac / ∈ Q = ⇒ val( c ) = val( a † ) . Then val( a † ) = val( c ) = val( b ) + val(( ab ) † ) > val( b ) , because res(( ab ) † ) = 0. 28emmas 5.8 and 5.9 combine to yield the following: Lemma 5.10.
Let a be a wild element with val( a ) > , and let a † be a neutralizer of a .Suppose b ∈ R . ab ∈ R ⇐⇒ val( b ) ≥ val( a † ) ab ∈ Q ⇐⇒ val( b ) > val( a † ) . Next, we weaken the assumption on a , allowing val( a ) = 0: Lemma 5.11.
Let a be a wild element with val( a ) ≥ , and let a † be a neutralizer of a .Suppose b ∈ R . Then ab ∈ R ⇐⇒ val( b ) ≥ val( a † ) . If b ∈ Q ⊆ R , then ab ∈ Q ⇐⇒ val( b ) > val( a † ) . Proof.
Let res( a ) = γ , and choose c ∈ R with c res( c ) = γ + 0 ε . Then c ∈ Q and res( c ) = γ .Let a ′ = a − c . Then res( a ′ ) = res( a ) − res( c ) = γ − γ = 0. So val( a ′ ) >
0. Because a / ∈ R and c ∈ R , we have a ′ = a − c / ∈ R , and so a ′ is wild by Lemma 4.8. Moreover, a ′ a † = aa † − ca † ∈ ( R \ Q ) − Q = R \ Q. Therefore a † is a neutralizer of a ′ , and we can apply Lemma 5.10 to a ′ , a † , b . Then ab ∈ R ⇐⇒ a ′ b ∈ R ⇐⇒ val( b ) ≥ val( a † ) , since ab − a ′ b = cb ∈ R . If b ∈ Q , then cb ∈ Q and ab ∈ Q ⇐⇒ a ′ b ∈ Q ⇐⇒ val( b ) > val( a † ) . Definition 5.12.
For a ∈ O , let val ∂ ( a ) denoteval ∂ ( a ) = ( + ∞ if a ∈ Q − val( a † ) if a has a neutralizer a † Lemma 5.13. val ∂ ( a ) is well-defined: for any a ∈ O , exactly one of the following holds1. a ∈ Q a has a neutralizer a † ,and in case (2) the valuation val( a † ) is independent of the choice of a neutralizer a † . roof. If a / ∈ Q , then a neutralizer a † exists by Lemma 5.6. Conversely, if a neutralizer a † exists, then a † ∈ Q and aa † ∈ R \ Q . As Q is a subring, a / ∈ Q .Now suppose a † and b are two neutralizers of a . If a is wild, then Lemma 5.11 applies, so ab ∈ R \ Q = ⇒ val( b ) = val( a † ) . If a is tame, we can write c res( a ) = x + yε c res( a † ) = z + 0 ε c res( b ) = w + 0 ε for some x, y, z, w ∈ k , using the fact that a † , b ∈ Q . Then c res( aa † ) = xz + yzε c res( ab ) = xw + ywε, The fact that aa † , ab ∈ R \ Q implies that yz and yw are non-zero. Therefore z, w arenon-zero and res( a † ) and res( b ) are non-zero, implying val( a † ) = 0 = val( b ). Lemma 5.14.
For a ∈ O , • val ∂ ( a ) ≥ ⇐⇒ a ∈ R • val ∂ ( a ) > ⇐⇒ a ∈ Q .Proof. First suppose a ∈ Q . Then val ∂ ( a ) = + ∞ by definition. Next suppose that a ∈ R \ Q .Then 1 is a neutralizer of a , so val ∂ ( a ) = − val(1) = 0.Lastly, suppose a ∈ O \ R . Let a † be a neutralizer. By Lemma 5.7, c res( a † ) = 0, implyingres( a † ) = 0, val( a † ) >
0, and val ∂ ( a ) < Lemma 5.15. If a ∈ O and b ∈ Q , then val ∂ ( ab ) = val ∂ ( a ) + val( b ) unless the right hand side is positive, in which case val ∂ ( ab ) = + ∞ . Proof.
First suppose a ∈ Q . Then val ∂ ( a ) = + ∞ , and the conclusion says that ab ∈ Q ,which is true ( Q is a ring).Next, suppose that a ∈ R \ Q . Then val ∂ ( a ) = 0. As, a, ab ∈ R and b ∈ Q , we may write c res( a ) = x + yε c res( b ) = z + 0 ε c res( ab ) = xz + yzε. x, y, z ∈ k , with y = 0. Thenval( b ) > ⇒ z = res( b ) = 0 = ⇒ yz = 0 = ⇒ ab ∈ Q = ⇒ val ∂ ( ab ) = + ∞ val( b ) = 0 = ⇒ z = res( b ) = 0 = ⇒ yz = 0 = ⇒ ab ∈ R \ Q = ⇒ val ∂ ( ab ) = 0 . Lastly, suppose a ∈ O \ R , so a is wild. Take a neutralizer a † . We break into cases accordingto the sign of val ∂ ( a ) + val( b ). • If val ∂ ( a ) + val( b ) >
0, then val( b ) > val( a † ), so ab ∈ Q by Lemma 5.11. Thusval ∂ ( ab ) = + ∞ . • If val ∂ ( a ) + val( b ) = 0, then val( b ) = val( a † ), so ab ∈ R \ Q , by Lemma 5.11. Thusval ∂ ( ab ) = 0. • If val ∂ ( a ) + val( b ) <
0, then val( b ) < val( a † ), and ab / ∈ R by Lemma 5.11. On the otherhand, a, b ∈ O , so ab ∈ O \ R , and ab is wild (Lemma 4.8). Take a neutralizer ( ab ) † ,and let c = b ( ab ) † . Then b, ( ab ) † ∈ Q , so c ∈ Q . Also ac = ( ab )( ab ) † ∈ R \ Q , and so c is a neutralizer of a . Thenval ∂ ( ab ) = − val(( ab ) † ) = − val( c ) + val( b ) = val ∂ ( a ) + val( b ) . The next lemma says that for any γ , the set { x ∈ O : val ∂ ( x ) ≥ γ } is a subring of O . Lemma 5.16.
For any a, b ∈ O , let γ = min(val ∂ ( a ) , val ∂ ( b )) . Then val ∂ ( a + b ) ≥ γ val ∂ ( ab ) ≥ γ. Proof. If γ > γ = + ∞ ), this holds because Q is a ring. If γ = 0, this holds because R is a ring. So we may assume 0 > γ = val ∂ ( a ) ≤ val ∂ ( b ) , swapping a and b if necessary. The fact that val ∂ ( a ) < a is wild. Take a neutralizer a † of a . Then a † ∈ Q , so by Lemma 5.15,val ∂ ( a † b ) ≥ val( a † ) + val ∂ ( b ) = val ∂ ( b ) − val ∂ ( a ) ≥ , so a † b ∈ R by Lemma 5.14. Then a † ( a + b ) = aa † + a † b ∈ R + R = R,
31o val ∂ ( a † ( a + b )) ≥ . By Lemma 5.15, it follows that val ∂ ( a + b ) + val( a † ) ≥ , or equivalently, that val ∂ ( a + b ) ≥ − val( a † ) = val ∂ ( a ) = γ. Also, Lemma 5.11 shows that( a † b ∈ R and val( a † b ) ≥ val( a † )) = ⇒ aa † b ∈ R, as a is wild. Thus a † ( ab ) ∈ R , and val ∂ ( a † ( ab )) ≥
0. As in the case of a + b , this implies thatval ∂ ( ab ) + val( a † ) ≥ , or equivalently, that val ∂ ( ab ) ≥ γ .Later (Corollary 5.26), we will get an improved rule for val ∂ ( ab ), but for now we contentourselves with the following cases: Lemma 5.17. If a ∈ O and val( a ) + val ∂ ( a ) > , then a ∈ R , i.e., val ∂ ( a ) ≥ .Proof. We may assume a / ∈ R , so a is wild. Take a neutralizer a † . Then0 < val( a ) + val ∂ ( a ) = val( a ) − val( a † ) , and so val( a ) > val( a † ). Therefore a † /a is not integral over R . By Lemma 4.4, the inverse a/a † satisfies a monic quadratic polynomial equation over R . Therefore a = baa † + c ( a † ) for some b, c ∈ R . As aa † , b, c, a † are all in R , this implies a ∈ R . Lemma 5.18.
Let γ be a positive element of the valuation group, and a, b be elements of O .Suppose val( a ) > γ val( b ) > γ val ∂ ( a ) > − γ val ∂ ( b ) > − γ. Then ab ∈ R , i.e., val ∂ ( ab ) ≥ .Proof. Note that val( a ) + val ∂ ( a ) > b ) + val ∂ ( b ) >
0, so a , b ∈ R by Lemma 5.17.By Lemma 5.16, val ∂ ( a + b ) > − γ , and so similarly ( a + b ) ∈ R . Since R is an algebra overa field k of characteristic = 2, ab = ( a + b ) − a − b ∈ R. .4 Density Lemma 5.19.
For every non-zero a ∈ K , there is non-zero b ∈ Q such that val( b ) ≥ val( a ) .Proof. We may assume a ∈ O (otherwise take b = 1). By Corollary 3.9, there are non-zero x, y ∈ R such that x = ay . By surjectivity of c res : R → k [ ε ], there is z ∈ R , necessarilynon-zero, such that c res( z ) = ε . Set b = xz = ayz . Thenval( b ) ≥ val( a )because y, z ∈ R ⊆ O . And b = 0, because y, z = 0. Lastly, c res( b ) = c res( x ) c res( z ) = c res( x ) · ε = 0 , and so b ∈ Q . Lemma 5.20.
For every γ in the value group, there exists a with val( a ) > γ and val ∂ ( a ) < − γ .Proof. Increasing γ , we may assume 0 < γ = val( b ) for some b ∈ Q , by Lemma 5.19. Since ς is not weakly multi-valuation type (Definition 5.27 in [10]), the ball of valuative radius 2 γ cannot be contained in R . Therefore there is c ∈ K with val( c ) > γ and c / ∈ R . Let a = c/b .Then val( a ) = val( c ) − val( b ) > γ − γ = γ > . Therefore a ∈ O and val ∂ ( a ) is meaningful. If val ∂ ( a ) ≥ − γ , then0 ≤ val ∂ ( a ) + γ = val ∂ ( a ) + val( b ) ≤ val ∂ ( ab ) = val ∂ ( c ) , by Lemma 5.15. Then c ∈ R by Lemma 5.14, a contradiction. Lemma 5.21. If a, b ∈ m and val ∂ ( a ) < val ∂ ( b ) , then b = p + qa for some p, q ∈ Q .Proof. Applying Lemma 4.1 to the set { , a, b } we obtain one of three cases: • pa + qb for some p, q ∈ Q . This cannot happen, as pa + qb ∈ O m + O m = m . • a = p + qb for some p, q ∈ Q . By Lemma 5.16,val ∂ ( a ) = val ∂ ( p + qb ) ≥ min(val ∂ ( p ) , val ∂ ( q ) , val ∂ ( b )) = val ∂ ( b ) , as val ∂ ( p ) = val ∂ ( q ) = + ∞ . This contradicts the assumption.33 b = p + qa for some p, q ∈ Q . Proposition 5.22.
The set Q is dense in O , with respect to the valuation topology.Proof. We first show that the closure of Q contains m . Let b be some element of m . Let γ bea given positive element of the valuation group. We will find c ∈ Q such that val( c − b ) > γ .If b ∈ Q , we can take c = b . Otherwise, val ∂ ( b ) ≤
0. By Lemma 5.20, there is a such thatval( a ) > γ > ∂ ( a ) < val ∂ ( b ) ≤
0. In particular, a ∈ m . By Lemma 5.21, there are c, q ∈ Q such that b = c + qa. Then val( b − c ) = val( q ) + val( a ) ≥ val( a ) > γ , because q ∈ Q ⊆ O .Next let b be any element of O . Take d ∈ R such that c res( d ) = res( b ) + 0 ε. Then d ∈ Q and b − d ∈ m . So we can approximate b − d arbitrarily closely by elements of Q . Equivalently, b is in the closure of d + Q = Q . ∂ Let D be the Q -module D := O /Q and let ∂ : O ։ D be the natural Q -linear map. Proposition 5.23.
The Q -module structure on D extends to an O -module structure asfollows: for a, b ∈ O , a · ∂b := ∂ ( a ′ · b ) , where a ′ ∈ Q and val( a ′ − a ) + val ∂ ( b ) > . In particular, the choice of a ′ doesn’t matter.Proof. We first check that a · ∂b is well defined. We can find an a ′ ∈ Q such that val( a ′ − a ) + val ∂ ( b ) > a ′′ is another such choice, then a ′′ − a ′ ∈ Q andval( a ′′ − a ′ ) + val ∂ ( b ) >
0. By Lemma 5.15, it follows thatval ∂ (( a ′′ − a ′ ) b ) = + ∞ , and ( a ′′ − a ′ ) b ∈ Q by Lemma 5.14. Thus a ′′ b − a ′ b ∈ Q and ∂ ( a ′′ b ) = ∂ ( a ′ b ). So the actionof O on D is well-defined. Furthermore, the action of O on D extends the action of Q . (If a ∈ Q , we can take a ′ = a .)Next we check the module axioms. For the associative law( a · a ) ∂b ? = a ( a ∂b ) , take a ′ , a ′ ∈ Q such that val( a ′ a ′ − a a ) > − val ∂ ( b )val( a ′ − a ) > − val ∂ ( b )val( a ′ − a ) > − val ∂ ( b ) . Q and the fact that multiplication is continuous. As a ′ ∈ Q ,val ∂ ( a ′ b ) ≥ val( a ′ ) + val ∂ ( b ) ≥ val ∂ ( b ) , by Lemma 5.15, and so val( a ′ − a ) > − val ∂ ( b ) ≥ − val ∂ ( a ′ b ) . Thus ( a a ) ∂b = ∂ ( a ′ a ′ b ) = a ∂ ( a ′ b ) = a · ( a ∂b ) . The other three module axioms( a + a ) ∂b = a ∂b + a ∂ba ( ∂b + ∂b ) = ( a∂b ) + ( a∂b )1 ∂b = b are proven similarly: one replaces the a ’s with very close elements of Q . Proposition 5.24.
The map ∂ : O → D is a Q -linear derivation: • ∂q = 0 for q ∈ Q . • ∂ ( ab ) = a∂b + b∂a for a, b ∈ O .Proof. The map is Q -linear with kernel Q by construction. Note that I = 0, as R is a domainand R/I ∼ = k [ ε ] is not. Take non-zero u ∈ I . Then u ∈ Q and val( u ) >
0. Take γ a positiveelement of the value group such thatmin(val ∂ ( a ) , val ∂ ( b )) > − γ val( u ) < γ. By Proposition 5.22, we can find a Q , b Q ∈ Q and a ′ , b ′ ∈ O such that a = a Q + a ′ b = b Q + b ′ val( a ′ ) > γ val( b ′ ) > γ. By Lemmas 5.16 and 5.14,val ∂ ( a ′ ) ≥ min(val ∂ ( a ) , val ∂ ( − a Q )) = min(val ∂ ( a ) , + ∞ ) = val ∂ ( a ) > − γ. For the second distributive law, one must choose a ′ ∈ Q such thatval( a ′ − a ) + min { val ∂ ( b ) , val ∂ ( b ) , val ∂ ( b + b ) } > .
35f val ∂ ( a ′ /u ) ≤ − γ , thenval ∂ ( a ′ ) = val ∂ ( a ′ /u ) + val( u ) < − γ + γ = − γ < ∂ ( a ′ /u ) > − γ val ∂ ( b ′ /u ) > − γ, where the second line follows by a similar argument. Also,val( a ′ /u ) = val( a ′ ) − val( u ) > γ − γ = 2 γ val( b ′ /u ) = val( b ′ ) − val( u ) > γ − γ = 2 γ. Thus, by Lemma 5.18, ( a ′ /u )( b ′ /u ) ∈ R . Then a ′ b ′ = ( a ′ /u )( b ′ /u )( u ) ∈ R · u ⊆ I ⊆ Q, so we see that ∂ ( a ′ b ′ ) = 0 . Also, val(0 − a ′ ) + val ∂ ( b ′ ) = val( a ′ ) + val ∂ ( b ) > γ − γ > , so a ′ ∂b ′ = ∂ (0 · b ′ ) = 0. Similarly, b ′ ∂a ′ = 0. So ∂ ( a ′ b ′ ) = 0 = a ′ ∂b ′ + b ′ ∂a ′ . The other three equations ∂ ( a Q b ′ ) = a Q ∂b ′ + b ′ ∂a Q ∂ ( a ′ b Q ) = a ′ ∂b Q + b Q ∂a ′ ∂ ( a Q b Q ) = a Q ∂b Q + b Q ∂a Q hold by Q -linearity and the fact that ∂ vanishes on Q . Adding these four equations, weobtain the desired Leibniz rule ∂ ( ab ) = a∂b + b∂a. Note that Q = { a ∈ O : ∂a = 0 } . Let Γ be the value group of O . We defineval : D → Γ ≤ ∪ { + ∞} by the equation val( ∂a ) = val ∂ ( a ) . By Lemmas 5.14 and 5.16, this is well-defined, and satisfies the identitiesval( a + b ) ≥ min(val( a ) , val( b ))val( a ) = + ∞ ⇐⇒ a = 0for a, b ∈ D . 36 emma 5.25. For a ∈ O and b ∈ D , val( ab ) = val( a ) + val( b ) , unless the right hand side is positive, in which case val( ab ) = + ∞ . Proof.
Write b as ∂c for some c ∈ O . Take a ′ ∈ Q such that val( a − a ′ ) + val( ∂c ) > a − a ′ ) > val( a ). Then by definition, a∂c = ∂ ( a ′ c ). Also val( a ) = val( a ′ ). By Lemma 5.15,val( a∂c ) = val( ∂ ( a ′ c )) = val ∂ ( a ′ c ) = val( a ′ ) + val ∂ ( c ) = val( a ) + val( ∂c ) , unless the right hand side is positive, in which case val( ab ) = + ∞ .Using this and the fact that ∂ is a derivation, we get an improved version of the multi-plication statement in Lemma 5.16. Corollary 5.26. If a, b ∈ O , then val ∂ ( ab ) ≥ min(val( a ) + val ∂ ( b ) , val ∂ ( a ) + val( b )) . The O -module D of differentials shares many properties with K/ m . Lemma 5.27.
For any γ ≤ in the value group, there is b ∈ D such that val( b ) = γ .Proof. By Lemma 5.25, it suffices to show that { val( b ) : b ∈ D } has no lower bound, which follows by Lemma 5.20. Lemma 5.28. If a, b ∈ D and val( a ) < val( b ) , then b ∈ O · a .Proof. We may assume b = 0, in which case val( a ) < val( b ) ≤
0. By Proposition 5.22, O = m + Q . Therefore, we may write a = ∂a ′ and b = ∂b ′ for some a ′ , b ′ ∈ m . ByLemma 5.21, we can write b ′ = p + qa ′ for some p, q ∈ Q . Then ∂p = ∂q = 0, so b = ∂b ′ = q∂a ′ = qa. Of course, we can replace val( a ) < val( b ) with a non-strict inequality: Proposition 5.29. If a, b ∈ D and val( a ) ≤ val( b ) , then b ∈ O · a . roof. We may assume b, a = 0. By Lemma 5.27, there is z ∈ D with val( z ) < val( a ). Then a = αz and b = βz for some α, β ∈ O . By Lemma 5.25val( α ) = val( a ) − val( z )val( β ) = val( b ) − val( z ) . Then val( α ) ≤ val( β ), and so β ∈ O α as O is a valuation ring. Therefore b = βz = γαz = γa ∈ O a for some γ ∈ O . Proposition 5.30. D is divisible as an O -module: for any b ∈ D and non-zero a ∈ O , thereis x ∈ D such that ax = b .Proof. We may assume b = 0. By Lemma 5.27, there is c ∈ D such that val( c ) ≤ val( b ) − val( a ). Then val( ac ) ≤ val( b ), and so b ∈ O · ac ⊆ a · D. Proposition 5.31.
Let D be the O -submodule of x ∈ D such that val( x ) ≥ . • D is the image of R under ∂ . • Viewing k as the O -module O / m , there is a unique O -module isomorphism res : D → k such that c res( x ) = res( x ) + res ( ∂x ) ε for any x ∈ R .Proof. The first point is clear from Lemma 5.14:val( ∂x ) ≥ ⇐⇒ val ∂ ( x ) ≥ ⇐⇒ x ∈ R. Take some w ∈ R such that c res( w ) = 0 + ε . Then w / ∈ Q , soval( ∂w ) = val ∂ ( w ) = 0by Lemma 5.14. By Proposition 5.29, ∂w generates D as an O -module. Also, Ann O ( ∂w ) is m by Lemma 5.25. Thus there is an isomorphismres : D → O / m y∂w res( y ) . Now let x ∈ R be given. Then c res( x ) = s + tε s, t ∈ k . We already know that s = res( k ), and we must show thatres ( ∂x ) = t. Take y ∈ R with c res( y ) = t + 0 ε . Then y ∈ Q . Also, c res( x − wy ) = c res( x ) − c res( w ) c res( y ) = s + tε − εt = s, so x − wy ∈ Q . Then ∂x = ∂ ( wy ) = y∂w, and so res ( ∂x ) = res ( y∂w ) = res( y ) = t. This proves the formula c res( x ) = res( x ) + res ( ∂x ) ε. Finally, this formula uniquely determines res , because ∂ : R → D is onto. Proposition 5.32. If K is perfect, then char( K ) = 0 .Proof. Suppose for the sake of contradiction that char( K ) = p >
2. By construction, thederivation ∂ : O → D is onto. Also, D cannot vanish, since we have constructed a submodule D isomorphic to k . Therefore ∂a = 0 for some a ∈ O . By perfection of K , we can write a = b p . Then ∂a = ∂ ( b p ) = pb p − ∂b = 0 , a contradiction. Recall that a topology on a structure is definable if it admits a uniformly definable basis ofopen sets.
Theorem 6.1.
Let ( K, + , · , , , . . . ) be a field of characteristic 0, possibly with extra struc-ture. Suppose K has dp-rank 2 and is unstable.1. K does not admit two independent definable valuation rings.2. K admits a definable non-trivial V-topology.3. The canonical topology on K is definable. (See § We prove these statements in § roposition 6.2. Under the assumptions of Theorem 6.1, K admits a unique definablenon-trivial V-topology.Proof. Existence follows from Theorem 6.1.2. For uniqueness, suppose K admits two inde-pendent definable V-topologies. We may replace K with an ℵ -saturated elementary exten-sion. Then the two V-topologies are induced by externally definable valuation rings O , O ,by Proposition 3.5 in [4]. Replacing K with its Shelah expansion K Sh , we obtain two inde-pendent valuation rings. The Shelah expansion continues to have dp-rank 2—this is a simpleexercise using quantifier elimination in the Shelah expansion ([12], Proposition 3.23). Proposition 6.3.
Let ( K, O , O , . . . ) be a field with two definable valuation rings O , O ,and possibly additional structure. If dp-rk( K ) ≤ , then O and O are comparable.Proof. Suppose O , O are incomparable. The join O · O is itself a valuation ring. Let K ′ be the residue field of O · O . Then O and O induce two independent valuation rings O ′ and O ′ on K ′ . Indeed, there is an isomorphism between • The poset of valuation rings on K ′ . • The poset of valuation rings on K that are contained in O · O .Thus O ′ and O ′ are incomparable, and O ′ · O ′ must be the maximal valuation ring on K ′ ,which is K ′ itself. Thus O ′ and O ′ are incomparable and independent.Replacing ( K, O , O ) with ( K ′ , O ′ , O ′ ), we may assume that O and O are independent.Then we get a contradiction: • If dp-rk( K ) ≤
1, use Lemma 9.4.14 in [7]. • If char( K ) >
0, use Lemma 2.6 in [8]. • If dp-rk( K ) = 2 and char( K ) = 0, then Theorem 6.1 applies. Proposition 6.4.
Suppose parts 1 and 2 of Theorem 6.1 hold for all ranks. In other words,suppose the following hold: • No dp-finite field of characteristic 0 admits two independent definable valuation rings. • Every unstable dp-finite field of characteristic 0 admits a (non-trivial) definable V-topology.Then the Shelah conjecture holds for dp-finite fields: every dp-finite field is either finite,algebraically closed, real closed, or henselian. Therefore, the conjectured classification of([3], Theorem 3.11) holds.Proof.
As in Proposition 6.3, we conclude that any two definable valuation rings O , O ona dp-finite field are comparable. As in the proof of Theorem 2.8 in [8], this implies thehenselianity conjecture for dp-finite fields: every definable valuation ring on a dp-finite fieldis henselian.Because the Shelah expansion of a dp-finite structure is dp-finite, it follows that anyexternally definable valuation ring must also be henselian.40 laim . If K is dp-finite of characteristic 0, then one of the following holds: • K is algebraically closed. • K is real closed. • K is finite • K admits a definable non-trivial valuation ring. Proof.
We may replace K with a sufficiently saturated elementary extension. If K is stable,then K is finite or algebraically closed, by Proposition 7.2 in [5]. Otherwise, K admitsa non-trivial definable V-topology, by assumption. By Proposition 3.5 in [4], K admits anon-trivial externally definable valuation ring O . This valuation ring must be henselian.By Theorem 5.2 in [6], either K admits a definable valuation ring or K is real closed oralgebraically closed. (cid:3) Claim
This in turn implies the Shelah conjecture for dp-finite fields of characteristic 0. The caseof positive characteristic is Corollary 11.4 [8].The classification in ([3], Theorem 3.11) is proven conditional on the Shelah conjecture.(The proof is for strongly dependent fields, but can be restricted to the smaller class ofdp-finite fields.)
We review the setup from ([10], Part II). Let K be an unstable monster field, possibly withextra structure, with dp-rk( K ) ≤ magic subfield k (cid:22) K , i.e., a small model with the following property (Definition8.3 in [9]):For every k -linear subspace G ≤ ( K , +), if G is type-definable (over any smallset), then G = G .Magic subfields exist by ([8], Corollary 8.7).Let Λ denote the lattice of type-definable k -linear subspaces of K . Recall from ([8],Definition 9.13) that a strict n -cube in Λ is an injection P ow ( n ) ֒ → Λ S G S that preserves the unbounded lattice operations: G S ∪ S = G S + G S G S ∩ S = G S ∩ G S . If K is separably closed, then K is algebraically closed, because dp-finite fields are perfect.
41e call G ∅ the base of the cube; the base need not be 0.The reduced rank of Λ is the maximum r such that a strict r -cube exists (Definition 9.17in [8]). By Proposition 10.1.7 in [8], the reduced rank is 1 or 2.If r is the reduced rank, a pedestal in Λ is a group G ∈ Λ that is the base of a strict r -cube (Definition 8.4 in [9] ). Since r is small, we can describe what this means explicitly: • If r = 1, then an r -cube is a chain of length two, and a pedestal is any G ∈ Λ otherthan K itself. • If r = 2, then an r -cube is { G ∩ H, G, H, G + H } for two incomparable G, H ∈ Λ. Therefore, a pedestal is a group of the form G ∩ H where G, H are incomparable elements of Λ.
Fact 6.6 (Proposition 10.4.1 in [8]) . Non-zero pedestals exist.
In Theorem 9.3 of [10], we associated an r -inflator to any non-zero pedestal H . Fact 6.7.
Let H be a non-zero pedestal with associated r -inflator ς .1. ς is malleable.2. The fundamental ring R H of ς is given as R H = { x ∈ K : xH ⊆ H } .
3. If H is type-definable over a small model K containing k , then the infinitesimals J K are contained in the fundamental ideal I H .4. If H is type-definable over a small model K containing k , then R H · J K ⊆ J K , and so J K is a sub-ideal of the fundamental ideal I H .5. If ς ′ is obtained by mutating ς along a line K · ( a , . . . , a n ) , then ς ′ is the r -inflatorassociated to the group H ′ = ( a − H ) ∩ · · · ∩ ( a − n H ) In particular, H ′ is itself a non-zero pedestal. This follows from ([10], Theorem 9.3, Remark 9.5, Proposition 10.15), ([8], Proposi-tion 10.15.5, Lemma 10.20), and Lemma 6.9 below.
Remark . By construction ([8], Theorem 4.20.4, Definition 6.3), the family of basic neigh-borhoods is uniformly ind-definable across all models. In other words, there is a set offormulas { ψ i ( x ; ~z i ) } i ∈ I such that for any model K , the collection of basic neighborhoods on K is exactly { ψ i ( K ; ~c ) : i ∈ I, ~c ∈ K | ~z i | } . Pedestals were called “special groups” in §
10 of [8] emma 6.9. Let G be a non-zero pedestal, type-definable over a small model K (cid:22) K , with K extending k . Let R be the stabilizer ring of G : R = { x ∈ K : xG ⊆ G } . Let J K be the group of K -infinitesimals. Then J K is an ideal in R . The following proof was sketched in Remark 10.19 of [8].
Proof.
First of all, J K ⊆ R by Proposition 10.15.(2,5) in [8]. It remains to show that R · J K ⊆ J K . Take a non-zero element j ∈ G . Take a small model K ′ (cid:22) K with K ′ ⊇ K ∪ { j } . As G is type-definable over the larger model K ′ , we see that J K ′ · G ⊆ J K ′ by ([8], Proposition 10.4.3). Now for any ε ∈ J K ′ and a ∈ R , we have ε · a · j ∈ J K ′ · R · G ⊆ J K ′ · G ⊆ J K ′ , implying that ε · a ∈ j − J K ′ . As J K ′ is invariant under scaling by elements of ( K ′ ) × ([8],Remark 6.9.3), we see that j − J K ′ = J K ′ , and ε · a ∈ J K ′ . As a ∈ R and ε ∈ J K ′ were arbitrary, R · J K ′ ⊆ J K ′ . (7) Claim . If S ⊆ R is type-definable over K , and U is a K -definable basic neighborhood,then there is a K -definable basic neighborhood V such that S · V ⊆ U. Proof.
Since K ′ ⊇ K , the neighborhood U is K ′ -definable and contains J K ′ . Therefore, S · J K ′ ⊆ R · J K ′ ⊆ J K ′ ⊆ U, by (7). By compactness, there is a K -definable set S ′ ⊇ S , and a K ′ -definable basic neigh-borhood V ′ ⊇ J K ′ such that S ′ · V ′ ⊆ U. We can write V ′ as ψ i ( K ; ~b ) for one of the formulas ψ i in Remark 6.8. Since S ′ and U are K -definable, we can find ~c from K such that S ′ · ψ i ( K ; ~c ) ⊆ U. Take V = ψ i ( K ; ~c ). Then S · V ⊆ S ′ · V ⊆ U. (cid:3) Claim
Now by compactness, it follows that for any subset S ⊆ R that is type-definable over K ,we have S · J K ⊆ J K . As the ring R is K -invariant, it is a union of such subsets S , and therefore R · J K ⊆ J K , as desired. 43 .2 The valuation-type case In [9], we considered the case where the canonical topology is a V-topology. We say that K is valuation type if this holds. We showed in this case that • The canonical topology is a definable V-topology ([9], Lemma 7.1). • Any two definable valuation rings are dependent ([9], Lemma 9.5).Thus, the three parts of Theorem 6.1 are automatic in this case.
Fact 6.11 (Theorem 8.11 in [9]) . If K is a small submodel and if J K contains a non-zeroideal of some multi-valuation ring on K , then the canonical topology on K is a V-topology. This has several consequences:
Corollary 6.12.
Let G be a non-zero pedestal with stabilizer R and associated r -inflator ς .1. If R contains a non-zero ideal of a multi-valuation ring, then K is valuation type.2. If ς is weakly multi-valuation type, then K is valuation type.3. If r = 1 , then K is valuation type.4. If some mutation of ς is weakly multi-valuation type, then K is valuation type.Proof.
1. Lemma 6.9—the point is that if R ′ is a multi-valuation ring, and a R ′ ⊆ R, then a a R ′ ⊆ a R ⊆ J K for any non-zero a ∈ J K .2. R is the fundamental ring of ς , and “weakly multi-valuation type” means that thefundamental ring contains a non-zero multi-valuation ideal ([10], Definition 5.27).3. 1-inflators are multi-valuation type (Proposition 5.19 in [10]).4. If ς ′ is obtained from ς by mutation, then ς ′ is the r -inflator associated to some othernon-zero pedestal G ′ (Fact 6.7.5). In particular, if ς ′ is weakly of multi-valuation type,then G ′ shows that K is valuation type. Theorem 6.13. If K is not valuation type (and characteristic 0 and unstable), then thereis a small model K and a 2-inflator ς satisfying the Strong Assumptions of § J K are an ideal in the fundamental ring R of ς .Proof. By Fact 6.6, non-zero pedestals exist. Let G be some non-zero pedestal and ς be theassociated inflator. Then G satisfies the Weak Assumptions: • K has characteristic 0 by assumption. 44 ς is malleable by Fact 6.7.1. • r = 2 by Corollary 6.12.3. • No mutation of ς is weakly multi-valuation type, by Corollary 6.12.4.By Corollary 2.3, there is a mutation ς ′ of ς such that ς ′ is isotypic. Then ς ′ inherits theother properties from ς (see Remark 2.1), and therefore ς ′ satisfies the Strong Assumptions.By Fact 6.7.5, ς ′ is the 2-inflator coming from some other pedestal G ′ . Let K be a smallmodel containing k , and type-defining G ′ . By Fact 6.7.4, J K is an ideal in the fundamentalring of ς ′ .In the remainder of §
6, we therefore assume1. K is a monster model of an unstable field of dp-rank 2 and characteristic 0.2. k is a magic subfield.3. ς is a k -linear 2-inflator on K satisfying the Strong Assumptions of § R and I are the fundamental ring and ideal of ς , and D and ∂ are as in § J is the group of K -infinitesimals over some small model K (cid:22) K containing k . Inparticular, • J is type-definable • J is contained in every K -definable basic neighborhood. • J is non-zero ([8], Remark 6.9.1).6. J is an ideal in R , contained in the fundamental ideal I . Recall that the valuation ring O is the integral closure of R . Proposition 6.14.
Let O ′ be a valuation ring on K , independent from O . Then O ′ J .Proof. Assume for the sake of contradiction that O ′ ⊇ J . Take nonzero e ∈ J . Then R · e ⊆ J , so R ⊆ e − · J ⊆ e − O ′ . Let val ′ be the valuation from O ′ and let γ = val ′ ( e − ). Then x ∈ R = ⇒ val ′ ( x ) ≥ val ′ ( e − ) = γ. We claim that for all a ∈ K , x ∈ m = ⇒ val ′ ( x ) ≥ min( γ, γ/ x ∈ m . Then x − isn’t integral over R , so by Lemma 4.4, x + bx + c = 0for some b, c ∈ R . By Newton polygons,val ′ ( x ) ≥ min (cid:18) val ′ ( b ) , val ′ c (cid:19) ≥ min( γ, γ/ . On the other hand, O is independent from O ′ , so by the approximation theorem, there is x ∈ K with val( x ) > ′ ( x ) < min( γ, γ/ Corollary 6.15. If O , O are two 0-definable valuation rings on K , then O and O arenot independent.Proof. The definable set O i has full dp-rank for i = 1 ,
2. It follows that O i − O i is a0-definable basic neighborhood, and so J ⊆ O i − O i = O i for i = 1 ,
2. By Proposition 6.14, both O and O induce the same topology as the valuationring O —the integral closure of R . Let val : K → Γ be the valuation associated to O . We will show that the associated valuationtopology is definable. Lemma 6.16.
There is a type-definable set B ⊆ K and some γ ∈ Γ such that val( x ) > γ = ⇒ x ∈ B = ⇒ val( x ) ≥ . for x ∈ K .Proof. Let B be the set B = { x ∈ K | ∃ y, z ∈ J : x = yx + z } . Then B is type-definable. If x ∈ B , then x = yx + z for some y, z ∈ J ⊆ R , and so x lies in the integral closure O of R .Now take non-zero c ∈ J , and let γ = val( c ). Note c ∈ R ⊆ O , so γ ≥
0. Supposeval( x ) > γ . Then val( x/c ) >
0, so x/c ∈ m and c/x / ∈ O . By Lemma 4.4, either x/c or c/x satisfies a monic polynomial equation of degree 2 over R . As c/x is not in the integralclosure O of R , we see that x/c satisfies the equation:( x/c ) = ( x/c ) y + z for some y , z ∈ R . Then x = ( cy ) x + c z , and cy , c z ∈ J . Thus x ∈ B . 46ay that two subsets X, Y ⊆ K are “co-embeddable” if there exist a, b ∈ K × such that a · X ⊆ Yb · Y ⊆ X. This is an equivalence relation.
Remark . Suppose X and Y are co-embeddable, X is type-definable, and Y is ∨ -definable. Then there is a definable set Z co-embeddable with X and Y . Indeed, afterrescaling, we may assume X ⊆ Y. Then we may find a definable set Z interpolating X and Y , by compactness: X ⊆ Z ⊆ Y. Lemma 6.18.
There is a definable set B that is co-embeddable with O .Proof. By Lemma 6.16, there is a type-definable set B and γ ∈ Γ such thatval( x ) > γ = ⇒ x ∈ B = ⇒ val( x ) ≥ . Therefore B is co-embeddable with O . Let B be the ∨ -definable set B = { } ∪ { y ∈ K × : y − / ∈ B } . Note that val( y ) > ⇒ y ∈ B = ⇒ val( y ) ≥ − γ. Thus B is co-embeddable with O . By Remark 6.17, there is a definable set in the co-embeddability class of B , B , and O .Recall from Lemma 2.1(d) in [11], that a set S in a topological field K is bounded if andonly if for every open neighborhood U ∋
0, there is non-zero a ∈ K × such that a · S ⊆ U. Theorem 6.19.
The V-topology induced by O is definable.Proof. Take a definable set B that is co-embeddable with O . Then B is a bounded neighbor-hood of 0, with respect to the V-topology induced by O . Therefore, the following definablefamily is a neighborhood basis of 0, by Lemma 2.1(e) in [11]: { aB : a ∈ K × } . This proves definability, by Lemma 6.20 below.
Lemma 6.20.
Let ( K, + , · , . . . ) be a field, possibly with extra structure. Let τ be a fieldtopology on K . Then τ is definable if and only if there is a definable neighborhood basis of0. roof. If { U a } a ∈ Y is a definable basis of opens, then { U a : a ∈ Y and 0 ∈ U a } is a definable neighborhood basis of 0. Conversely, suppose { N a } a ∈ Y is a definable neighbor-hood basis of 0. Let N int a = { x ∈ K | ∃ b ∈ Y : x + N b ⊆ N a } . Then { N int a } a ∈ Y is a definable basis of open neighborhoods around 0, and { b + N int a : b ∈ K, a ∈ Y } . is a definable basis of open sets. Lemma 6.21.
There is nonnegative γ ∈ Γ , and a type-definable set S , such that for x ∈ K with val( x ) > γ , we have val( ∂x ) < − γ = ⇒ x ∈ S = ⇒ val( ∂x ) ≤ . Proof.
Take some non-zero c ∈ J . Take γ = val( c ). Then γ ≥
0, as J ⊆ R ⊆ O . Take e ∈ R such that c res( e ) = s + tε , with t = 0. Then e ∈ R \ Q . Let B be the openball of valuative radius γ . By Proposition 5.22, O = B + Q . Therefore there is e ∈ B with e − e ∈ Q . Then e ∈ R \ Q , and val( e ) > γ = val( c ).Let S be the type-definable set of x ∈ K such that ∃ y, z ∈ J : e = xy + z. Suppose val( x ) > γ and val( ∂x ) < − γ . Apply Lemma 4.1 to the set { xc, e, c } . There arethree cases: • c is generated by xc and e . This cannot happen, since R ⊆ O , since val( c ) < val( e )(by choice of e ), and since val( c ) < val( xc ) as val( x ) > γ ≥ • xc is generated by e and c . As e, c ∈ R , this would imply xc ∈ R . But c ∈ J ⊆ I ⊆ Q ,so by Lemma 5.15, val( ∂ ( xc )) = val( c ) + val( ∂x ) < val( c ) − γ = 0 . By Lemma 5.14, xc / ∈ R , a contradiction. • e is generated by c and xc . Then e = xcy + cz , for some y , z ∈ R . If y = cy and z = cz , then y, z ∈ J (as J ⊳ R ), and e = xy + z .So x ∈ S . 48onversely, suppose x ∈ S and val( x ) > γ . Then there are y, z ∈ J ⊆ I ⊆ Q such that e = xy + z. Now Q is a subring, and y, z ∈ Q , e / ∈ Q . Therefore x / ∈ Q . On the other hand, val( x ) >γ ≥
0, so x ∈ O . Therefore x ∈ O \ Q , which implies val( ∂x ) ≤ Lemma 6.22.
Some ∨ -definable set is co-embeddable with R .Proof. Take γ and S as in Lemma 6.21. By Theorem 6.19, we can find γ ′ and a definableset B such that val( x ) > γ ′ = ⇒ x ∈ B = ⇒ val( x ) > γ. Claim . If val( x ) > γ ′ and val( ∂x ) >
0, then x ∈ B \ S . Proof.
Because val( x ) > γ ′ , we have x ∈ B and val( x ) > γ , and so Lemma 6.21 applies.Then val( ∂x ) > x / ∈ S , by the contrapositive to Lemma 6.21. (cid:3) Claim
Claim . If x ∈ B \ S , then val( x ) > γ and val( ∂x ) ≥ − γ . Proof.
The fact that x is in B implies that val( x ) > γ , and thus that Lemma 6.21 applies.By the contrapositive to Lemma 6.21, x / ∈ S implies val( ∂x ) ≥ γ . (cid:3) Claim
By Lemma 5.19, there is b ∈ Q such that val( b ) > γ ′ . Then bR ⊆ B \ S. (8)Indeed, if x ∈ R , then val( bx ) = val( b ) + val( x ) > γ ′ + 0val( ∂ ( bx )) ≥ val( b ) + val( ∂x ) > γ ′ + 0 ≥ , and Claim 6.23 applies. Also, b · ( B \ S ) ⊆ R. (9)Indeed, if x ∈ B \ S , then val( x ) > γ val( ∂x ) ≥ − γ by Claim 6.24. But then bx ∈ R :val( bx ) = val( b ) + val( x ) > γ ′ + γ ≥ ∂ ( bx )) ≥ val( b ) + val( ∂x ) > γ ′ − γ ≥ . By (8)-(9), the ∨ -definable set B \ S is co-embeddable with R . Lemma 6.25.
The set J is bounded with respect to the canonical topology on K : for anybasic neighborhood U , there is a ∈ K × such that aJ ⊆ U . roof. Recall that J is the set J K of K -infinitesimals. Let K ′ be a small model containing K and defining U . Then U contains the group J K ′ of K ′ -infinitesimals. By Corollary 8.9 in[9], there is non-zero a such that a · J K ⊆ J K ′ ⊆ U. Recall from [11], §
2, that a ring topology is locally bounded if there is a bounded neigh-borhood of 0. If R is a proper subring of a field K = Frac( R ), then R induces a locallybounded ring topology on K , as in [11], Example 1.2 and Theorem 2.2(a). Theorem 6.26.
The canonical topology on K is locally bounded, definable, and induced by R .Proof. Note that J and R are co-embeddable, as cR ⊆ J ⊆ R for any non-zero c ∈ J . By Lemma 6.22, some ∨ -definable set U is co-embeddable with R and J . As J itself is type-definable, we can take U to be definable by Remark 6.17. Rescaling U , we may assume J ⊆ U . By compactness, there is a K -definable basic neighborhood V such that J ⊆ V ⊆ U, as J is the directed intersection of such neighborhoods. Therefore U is a neighborhood of0. Also, U is bounded, because it is co-embeddable with the bounded set J . Therefore thecanonical topology is locally bounded. By Lemma 2.1(e) in [11], the family { aU : a ∈ K × } is a neighborhood basis of 0. Then the canonical topology is definable by Lemma 6.20. Thefamily { aR : a ∈ K × } is also a neighborhood basis of 0, because U and R are co-embeddable.Once the canonical topology is definable on the monster, it is uniformly definable on allmodels: Theorem 6.27.
1. There is a formula ϕ ( x ; ~y ) such that for every small model K (cid:22) K , the family of sets { ϕ ( K ; ~b ) : ~b ∈ K | ~y | } is a neighborhood basis of 0 for the canonical topology on K .2. If K, K ′ are two small submodels, then K and K ′ with their canonical topologies are“locally equivalent” in the sense of [11]. roof. Let { ψ i ( x ; ~z i ) } i ∈ I be as in Remark 6.8, so that { ψ i ( K ; ~c ) : i ∈ I, ~c ∈ K | ~z i | } is the set of basic neighborhoods on any K ≡ K .On K , Theorem 6.26 gives a ( K -)definable neighborhood basis N . By saturation, theremust be a finite subset I ⊆ I such that every set in N has the form ψ i ( K ; ~c ) for some i ∈ I .The fact that N is a neighborhood basis implies that ∀ j ∈ I ∀ ~c ∃ i ∈ I ∃ ~e : ψ i ( K ; ~e ) ⊆ ψ j ( K ; ~c ) . This is a small conjunction of first-order sentences, so it holds in submodels K (cid:22) K . Thenfor any small model K , the family { ψ i ( K ; ~c ) : i ∈ I , ~c ∈ K | ~z i | } is a neighborhood basis of 0. Because I is finite, this can be written as { ϕ ( K ; ~b ) : ~b ∈ K | ~y | } for some formula ϕ ( x ; ~y ).This proves the first point. The second point is immediate, because local sentences canbe evaluated on a neighborhood basis ([11], Theorem 1.1(a)). In Theorem 6.13, we can weaken the assumption char( K ) = 0 to char( K ) = 2; the sameproof works. But the positive characteristic case then leads to a contradiction: Proposition 6.28.
Let K be a monster model of an unstable field with dp-rk( K ) ≤ . If K is not of valuation type, then char( K ) is 0 or 2.Proof. Suppose char( K ) >
2. As in the proof of Theorem 6.13, there would be a 2-inflatorsatisfying the Strong Assumptions of § K ) = 0 by Proposition 5.32.Therefore Theorem 6.29. If K is a field with dp-rk( K ) ≤ and char( K ) > , then either K is stable,or K is valuation type. Perhaps this can be proven in characteristic 2 as well.51
Reduced rank and generators
Let R be a noncommutative ring, and M be an R -module. Say that M has property W n ifthe following holds: for any a , a , . . . , a n ∈ M , there is some 0 ≤ i ≤ n such that a i ∈ R · a + · · · + R · a i − + R · a i +1 + · · · + R · a n . In other words, any submodule of M generated by a set S of size n + 1 is generated by an n -element subset of S . Remark . This property appeared in Lemma 4.1, which said that K has property W asa Q -module or R -module. Lemma 7.2.
1. If M , . . . , M n are non-zero, then M ⊕ · · · ⊕ M n does not have property W n .2. If M has property W n and N ≤ M , then N has property W n .3. If M has property W n and N ≤ M , then M/N has property W n .Proof.
1. Take a i a non-zero element of M i , viewed as an element of the direct sum. Then { a , a , . . . , a n } violates property W n .2. Clear.3. Given a i ∈ M/N , lift them to ˜ a i ∈ M , apply property W n in M to obtain i and r , . . . , r n ∈ R such that˜ a i = r ˜ a + · · · + r i − ˜ a i + r i +1 ˜ a i +1 + · · · + r n ˜ a n , and then project back to M/N . Proposition 7.3. M has property W n if and only if the reduced rank of Sub R ( M ) is at most n .Proof. If the reduced rank of Sub R ( M ) is greater than n , then there is a strict ( n + 1)-cubein M . This corresponds to submodules M − ≤ M + ≤ M and an isomorphism M + /M − ∼ = N ⊕ · · · ⊕ N n where the N i are non-zero R -modules. By Lemma 7.2, the right hand side does not satisfy W n , and therefore neither do M + or M .Conversely, suppose W n fails, witnessed by a , . . . , a n ∈ M . Let N i = M · a i . Then N + · · · + N n > N + · · · + N i − + N i +1 + · · · + N n , for any 0 ≤ i ≤ n . By Proposition 6.3.2 in [10], Sub R ( M ) has reduced rank greater than n . 52 Diffeovaluation data In §
8, all fields will have characteristic 0, and all rings will be Q -algebras. K/ m ’s Let K be a valued field with valuation ring O , maximal ideal m , and residue field k = O / m . Definition 8.1. A mock K/ m is a divisible O -module D extending k satisfying the followingproperty: for any x, y ∈ D , x ∈ O · y or y ∈ O · x. Note that K/ m is naturally a mock K/ m .There is a theory T whose models are pairs ( K, D ), where K is a valued field and k ֒ → D is a mock K/ m . Proposition 8.2.
Let ( K, D ) be a model of T . If ( K, D ) is countable or ℵ -resplendent,then D is isomorphic (as an extension of k ) to K/ m .Proof. The resplendent case follows from the countable case. Assume
K, D are countable.Then the value group Γ has countable cofinality. Take a sequence a , a , . . . in K such that a = 1 and the sequenceval( a ) , val( a ) , . . . is descending with no lower bound. Then O = O · a ⊆ O · a ⊆ · · · and the union of this chain is K .By divisibility, we can find a sequence b , b , . . . in D such that • b is the image of 1 under the embedding k ֒ → D . • b i − = ( a i − /a i ) b i , for all i ≥ i , the b i are all non-zero. Define f i : O · a i → D by f i ( x ) = ( x/a i ) b i . If x ∈ O · a i , then f i ( x ) = ( x/a i ) b i = ( x/a i )( a i /a i +1 ) b i +1 = ( x/a i +1 ) b i +1 = f i +1 ( x ) . Therefore the f i glue together to yield a morphism f : K → D. Moreover, f ( a i ) = f i ( a i ) = ( a i /a i ) b i = b i for all i .53 laim . For any x ∈ K , f ( x ) = 0 ⇐⇒ val( x ) > . Proof.
First suppose val( x ) ≥
0. Then x ∈ O = O · a = dom( f ), and so f ( x ) = f ( x ) = ( x/a ) b = xb . By choice of b , the annihilator Ann O ( b ) is exactly m , and so f ( x ) = 0 ⇐⇒ xb = 0 ⇐⇒ x ∈ m ⇐⇒ val( x ) > . Next suppose val( x ) ≤
0. Then 1 /x ∈ O , and so f (1) = f ((1 /x ) x ) = (1 /x ) f ( x ) , because f is O -linear. By the first case,val(1) = 0 = ⇒ f (1) = 0 = ⇒ f ( x ) = 0 . (cid:3) Claim
Therefore ker( f ) = m , and f induces an embedding K/ m ֒ → D. Restricted to k = O / m , this embeding is( x + m ) f ( x ) = f ( x ) = ( x/a ) b = xb . By choice of b , this is the given embedding of k into D .It remains to show that f is onto. Suppose not. Take an element y of D that is not inthe image of f . Then y / ∈ O · b i , for any i , since O · b i is the image of f i . By definition of mock K/ m , it follows that b i ∈ O · y for all i . In particular, there are c i ∈ O such that b i = c i y . Take i large enough thatval( a i c ) <
0. Then (1 /a i ) y = 1 a i c c y = 1 a i c b = 0 , because Ann( b ) = m . But then b = f i ( a ) = f i (1) = (1 /a i ) b i = (1 /a i ) c i y = c i (1 /a i ) y = c i , contradicting the choice of b . Corollary 8.4.
Let D be a mock K/ m . Then there is a map val : D → Γ ∪ { + ∞} with thefollowing properties: . val( x ) ≤ or val( x ) = + ∞ for all x ∈ D .2. val( x ) = + ∞ if and only if x = 0 .3. val( x ) ≥ if and only if x is in the image of k ֒ → D .4. For any a ∈ O and x ∈ D , val( ax ) = ( val( a ) + val( x ) if val( a ) + val( x ) ≤ ∞ if val( a ) + val( x ) > .
5. For any x, y ∈ D , val( x + y ) ≥ min(val( x ) , val( y )) .
6. If val( x ) ≤ val( y ) , then y ∈ O · x .7. For any γ ∈ Γ , there is x ∈ D such that val( x ) ≤ γ .Proof. Let D ∗ be the image of k \ { } in D . We claim that • For every non-zero x in D , there is an a ∈ O such that ax ∈ D ∗ . • If ax ∈ D ∗ and bx ∈ D ∗ , then val( a ) = val( b ). • If we define val : D → Γ ∪ { + ∞} asval( x ) = ( + ∞ if x = 0 − val( a ) if ax ∈ D ∗ , then val satisfies the listed conditions.These three claims can be expressed by a first-order sentence, so we may pass to a resplendentelementary extension. Then we may assume D is K/ m , in which case the three claims arestraightforward. Definition 8.5. A diffeovalued field is a structure ( K, O , D, ∂ ) where • ( K, O ) is a valued field (of equicharacteristic 0). • k ֒ → D is a mock K/ m . • ∂ : O → D is a derivation.The theory of diffeovalued fields is first-order.55 efinition 8.6. A normalization of a diffeovalued field is a choice of an isomorphism D ∼ = K/ m (respecting the embedding k ֒ → D ).Every sufficiently resplendent diffeovalued field admits a normalization, by Proposi-tion 8.2. Definition 8.7. A normalized diffeovalued field is a diffeovalued field with a choice of anormalization.Equivalently, a normalized diffeovalued field is a valued field ( K, O ) with a derivation ∂ : O → K/ m . Definition 8.8.
Let K be a normalized diffeovalued field. A lifting is a derivation δ : K → K such that ∂x = ( δx ) + m for x ∈ O . Proposition 8.9.
Let K be a sufficiently resplendent diffeovalued field. Suppose the valuegroup Γ is p -divisible for at least one prime. Then K admits a normalization and a lifting.Proof. The normalization comes from Proposition 8.2, and the lifting comes from Corol-lary A.21 in the appendix.Some assumption on the value group is necessary: Proposition A.22 in the appendix givesan example of a normalized diffeovalued field which cannot be lifted, even after passing toan elementary extension.
Definition 8.10. A lifted diffeovalued field is a normalized diffeovalued field with a choiceof a lifting.Equivalently, a lifted diffeovalued field is a field with a derivation and a valuation. Lemma 8.11.
Let K be a diffeovalued field. If x ∈ O × , then ∂ ( x − ) = − x − ∂x .Proof. This follows as usual from0 = ∂ ( xx − ) = x − ∂x + x∂ ( x − ) . Definition 8.12.
A diffeovalued field is dense if for every x ∈ D , the fiber { y ∈ O : ∂y = x } is dense in O , with respect to the valuation topology.The theory of dense diffeovalued fields is first-order.56 emark . Denseness implies that the value group Γ is non-trivial.
Proof.
If the valuation is trivial, then O = K and the valuation topology is discrete. Thenevery fiber { y ∈ K : ∂y = x } is dense in O , hence equal to O . This is absurd unless D is a singleton. But D contains asubmodule isomorphic to k ∼ = K . Fix a dense diffeovalued field K . Define R = { x ∈ K : val( x ) ≥ ∂x ) ≥ } Q = { x ∈ K : val( x ) ≥ ∂x ) > } I = { x ∈ K : val( x ) > ∂x ) > } . Note that in the definition of
Q, I ,val( ∂x ) > ⇐⇒ ∂x = 0 , because D is a mock K/ m . In fact, Q is merely the kernel of ∂ : O → D . Lemma 8.14. R, Q are proper subrings of K .2. I is a proper ideal in R and in Q .3. Frac( Q ) = Frac( R ) = K .4. Q is a local ring with maximal ideal I .5. I = 0 .Proof.
1. Easy. Properness holds because
R, Q ⊆ O , and O is a proper subring by Remark 8.13.2. Easy. Properness holds because 1 / ∈ I .3. As Q ⊆ R , it suffices to show Frac( Q ) = K . Given a ∈ K , we must show a ∈ Frac( Q ).Replacing a with a − , we may assume a ∈ O . If a ∈ Q , we are done. Otherwise,val( ∂a ) ≤
0. By density, there is b such thatval( b ) > − val( ∂a ) b = 0 ∂b = 0 . b ∈ Q , and so ab ∈ O . Also, ∂ ( ab ) = a∂b + b∂a = b∂a = 0 . because ∂b = 0, and val( b ) is high enough for b to annihilate ∂a . Thus ab ∈ Q and a = ( ab ) /b ∈ Frac( Q ).4. Since I is a proper ideal in Q , it suffices to show x ∈ Q \ I = ⇒ x − ∈ Q. Suppose x ∈ Q \ I , so that val( x ) = 0 and ∂x = 0. Then x − ∈ O , and ∂ ( x − ) = − x − ∂x = 0 , by Lemma 8.11. So x − ∈ Q .5. By (4), Q/I is a field. If I = 0, then Q is a field, and Q = K by (3). This contradicts(1). Proposition 8.15.
Let K be a dense diffeovalued field. There is a locally bounded fieldtopology on K characterized by the fact that either of the following are a neighborhood basisof 0: { aR : a ∈ K × }{ aQ : a ∈ K × } . The topology is locally bounded, non-discrete, and Hausdorff.Proof.
The ring Q induces a locally bounded, non-discrete, Hausdorff field topology because Q is a proper local subring of K = Frac( Q ). For example, see Theorem 2.2(b) in [11]. If a is a non-zero element of I , then R · a ⊆ R · I = I ⊆ Q, so R and Q induce the same topology. Definition 8.16.
The diffeovaluation topology is the topology induced by Q or R as inProposition 8.15. Proposition 8.17.
The diffeovaluation topology is not a V-topology.Proof.
In a V-topology, the following local sentence holds, where
U, V range over neighbor-hoods of 0: ∀ U ∃ V ∀ x, y : (( xy ∈ V ) → ( x ∈ U or y ∈ U )) . However, this fails for U = R . Indeed, suppose V = aR is such that xy ∈ aR = ⇒ ( x ∈ R ∨ y ∈ R ) . aR , we may assume a ∈ R . By density, there is x such thatval( x ) > val( a )val( ∂x ) < . Let y = a/x . Then xy = a ∈ aR. On the other hand, x / ∈ R by choice of val( ∂x ), and y / ∈ O ⊇ R , by choice of val( x ). Definition 8.18. A DV-topology is a field topology that is locally equivalent to a densediffeovaluation topology.
For lifted diffeovalued fields, we can characterize density and the diffeovaluation topologymore naturally.
Proposition 8.19.
Let ( K, δ, val) be a lifted diffeovalued field. Then K is dense if and onlyif for every a, b ∈ K and every γ ∈ Γ , there is x ∈ K such that val( x − a ) > γ val( δx − b ) > γ Proof.
Unwinding the definition, density says that we can solve equations of the formval( x − a ) > γ val( δx − b ) > a ∈ O . So the listed conditions certainly imply density. Conversely, suppose densityholds. Claim . For any γ ∈ Γ and b ∈ K , there is x such thatval( x ) > γ val( δx − b ) > γ. Proof.
Take some non-zero a such that val( a ) > γ .By density, there is y such that val( y ) > max( γ − val( δa ) , δy − b/a ) > . Let x = ya . Then val( x ) = val( ya ) = val( y ) + val( a ) > γ val( aδy − b ) > val( a ) > γ. yδa ) = val( y ) + val( δa ) > γ. So we see that val( δx − b ) = val( yδa + ( aδy − b )) > γ. (cid:3) Claim
Now given any a, b, γ , we can find ε such thatval( ε ) > γ val( δε + δa − b ) > γ. Set x = a + ε . Then val( x − a ) = val( ε ) > γ val( δx − b ) = val( δε + δa − b ) > γ. Proposition 8.21. If ( K, δ, val) is a dense, lifted diffeovalued field, then sets of the form { x ∈ K : val( x − a ) > γ and val( δx − a ) > γ } for γ , γ ∈ Γ , a , a ∈ K form a basis of opens in the diffeovaluation topology.Proof. Let B γ = { x ∈ K : val( x ) > γ and val( δx ) > γ } . It suffices to show that the B γ form a neighborhood basis of 0. First of all, one sees by a straightforward calculation that B γ is an R -submodule of K . The strong form of density in Proposition 8.19 can be used toshow that B γ is strictly bigger than { } ; for example take a, b very small relative to γ , andthen take x with x − a and δx − b very small relative to a, b, γ . Thus B γ is a neighborhoodof 0 in the diffeovaluation topology. Conversely, given any a ∈ K × , we claim that B γ ⊆ a − · R for sufficiently large γ . Indeed, if γ is sufficiently large relative to a , and x ∈ B γ , thenval( ax ) = val( a ) + val( x ) ≥ val( a ) + γ ≥ δ ( ax )) = val( aδx + xδa ) ≥ γ + min(val( a ) , val( δa )) ≥ . Fix a dense diffeovalued field K . Let D be the image of the embedding k ֒ → D , and letres ′ : D → k be the inverse of this embedding. Let Q, R, I be as in § a ∈ R ,then ∂a ∈ D , and so res ′ ( ∂a ) makes sense. Lemma 8.22.
1. If a ∈ O and b ∈ R , then res ′ ( a∂b ) = res( a ) res ′ ( ∂b ) . . The quotient Q/I is isomorphic (as a ring) to k via the map res( − ) . Therefore, wecan regard k -modules as Q -modules.3. R and I are Q -submodules of K .4. For x ∈ R , let c res( x ) = (res( x ) , res ′ ( ∂x )) ∈ k . Then c res induces an isomorphism of Q -modules from R/I to k . In particular, R/I is a semisimple Q -module of length 2.Proof.
1. This holds because res ′ is an O -linear map from D to k , and the O -modulestructure on k comes from res : O → k .2. The ring homomorphism res : Q → k is onto, by density. Indeed, given any x ∈ k , wecan find y ∈ O such that res( x ) = y∂x = 0 . Then x ∈ Q and res( x ) = y . The kernel of res : Q → k is I , by definition of Q and I .3. R is a Q -module because R is a superring of Q . I is a Q -module because I is an idealin R .4. The map c res : R → k is obviously Z -linear. It is surjective by density. The kernel is I , by definition of R and I . For Q -linearity, suppose x ∈ Q and y ∈ R . Thenres( xy ) = res( x ) res( y ) . Also, x ∈ Q implies ∂x = 0, and sores ′ ( ∂ ( xy )) = res ′ ( x∂y + y∂x ) = res ′ ( x∂y ) = res( x ) res ′ ( ∂y )by part (1). Thus c res( xy ) = res( x ) · (res( y ) , res ′ ( ∂y )) = res( x ) c res( y ) . Lemma 8.23.
For any a, b, c ∈ K , the Q -submodule generated by { a, b, c } is generated by atwo-element subset of { a, b, c } .Proof. Without loss of generality, val( c ) ≤ val( a ) and val( c ) ≤ val( b ). Rescaling, we mayassume c = 1. Then a, b ∈ O .Without loss of generality, val( ∂a ) ≤ val( ∂b ). Now, if val( ∂b ) >
0, then b ∈ Q ·
1, and weare done. So we may assume val( ∂a ) ≤ val( ∂b ) ≤ . Take x ∈ O such that ∂b = x ∂a . By density, there is some x such that ∂x = 0 andval( x − x ) > − val( ∂a ) ≥ . x − x ) ≥
0, so x − x ∈ O . As x ∈ O , we see x ∈ O . Then x ∈ Q .Now val( x − x ) is large enough that x − x annihilates ∂a , so x∂a = x ∂a = ∂b. Then ∂ ( xa ) = x∂a + a∂x = ∂b + 0 . Let y = b − xa . Then ∂y = ∂b − ∂ ( xa ) = 0 . Also, b, x, a ∈ O , and so y ∈ O . Thus y ∈ Q . Then b = xa + y ∈ Q · a + Q · . Theorem 8.24.
Let K be a dense diffeovalued field. There is a malleable Q -linear 2-inflator Dir K ( K ) → Dir k ( k )Sub K ( K n ) → Sub k ( k n ) V
7→ { ( c res( x ) , . . . , c res( x n )) : ~x ∈ V ∩ R n } , where c res( x ) = (res( x ) , res ′ ( ∂x )) .Proof. By Proposition 7.3 and Lemma 8.23, the reduced rank of Sub Q ( K ) is at most 2. As R/I is a semisimple Q -module of length 2, we see that Sub Q ( K ) has reduced rank exactlytwo. Let • C the category of Q -modules. • F : K Vect → Q Mod the forgetful functor. • G : Q Mod → Q Vect the forgetful functor.Then Assumptions 8.1 and 8.11 of [10] hold. Applying Propositions 8.9 and 8.12 in [10], weobtain a malleable 2-inflator Dir K ( K ) → Dir Q ( R/I )Sub K ( K n ) → Sub Q (( R/I ) n ) V ( V ∩ R n + I n ) /I n . Now ( V ∩ R n + I n ) /I n can be described as the image of V ∩ R n under the projection R n ։ ( R/I ) n . Under the isomorphism Dir Q ( R/I ) ∼ = Dir Q ( k ) ∼ = Dir k ( k ), this is exactly { ( c res( x ) , . . . , c res( x n )) : ~x ∈ V ∩ R n } . Definition 8.25. A diffeovaluation inflator on K is a 2-inflator on K arising from a densediffeovaluation on K via Theorem 8.24. 62 .7 Characterization of diffeovaluation inflators We can summarize § Theorem 8.26.
Let ς be an isotypic, malleable 2-inflator on a field K of characteristic 0.If no mutation of ς is weakly multi-valuation type, then ς is a diffeovaluation inflator.Proof. By Corollary 4.5, we have a valuation ring O . By Propositions 5.23 and 5.24, we havean O -module D and a derivation ∂ : O → D . By Propositions 5.29, 5.30, and 5.31 (withProposition 4.9), D is a mock K/ m , and the sets R, I, Q of § R = { x ∈ O : val( x ) ≥ } I = { x ∈ m : ∂x = 0 } Q = { x ∈ O : ∂x = 0 } . Thus we have a diffeovaluation, and the sets
R, I, Q agree with the ones defined in § ∂ : O → D is surjective, by its construction in § Q is dense in O . Every other fiber of ∂ is a translate of Q , by surjectivity. Therefore everyfiber is dense. So the diffeovaluation data is dense.Finally, by Propositions 3.8 and 5.31, ς has the same form as the diffeovaluation inflatorconstructed in Theorem 8.24.Under the Weak Assumptions of § Corollary 8.27.
Let ς be a malleable 2-inflator on a field K of characteristic 0. Then somemutation ς ′ of ς is either weakly multi-valuation type, or a diffeovaluation inflator.Proof. If no mutation of ς is weakly multi-valuation type, then ς satisfies the Weak Assump-tions of §
2. By Corollary 2.3, there is some mutation ς ′ which is isotypic. By Remark 2.1, ς ′ satisfies the Strong Assumptions of § ς ′ is a diffeovaluation inflator by Theo-rem 8.26.We can use Theorem 8.26 to characterize diffeovaluation inflators. We first need somelemmas. Lemma 8.28.
Let K be a field of characteristic 0 and O , . . . , O n be some valuation rings on K . Let b be an element of K . Then there is non-zero q ∈ Q such that / ( b − q ) ∈ O ∩· · ·∩O n .Proof. Let m i be the maximal ideal of O i . We need non-zero q such that b − q / ∈ m i for1 ≤ i ≤ n . For each i , let B i = { q ∈ Q : b − q ∈ m i } . We must show that B ∪ · · · ∪ B n ∪ { } fails to cover all of Q . There are three possibilitiesfor each B i : • If b / ∈ Q + m i , then B i is empty. • Otherwise, if O i has residue characteristic 0, then B i is a singleton.63 Otherwise, if O i has residue characteristic p >
0, then B i is a p -adic ball in Q (of radius1 /p ).We can take q = 1 /n !, for n ≫ Lemma 8.29.
Let K be a dense diffeovalued field. As usual, let R = { x ∈ O : val( ∂x ) ≥ } . Let S be a multi-valuation ring on K . If aS ⊆ R , then a = 0 .Proof. Note that a = a · ∈ a · S ⊆ R , so a ∈ R ⊆ O .Suppose a = 0. By density, we can find b ∈ K such thatval( b ) > ∂b ) < min(val( ∂a ) , − val( a )Then b ∈ m . By Lemma 8.28, there is non-zero q ∈ Q such that 1 / ( b − q ) ∈ S . Then ab − q ∈ aS ⊆ R. Now b − q ∈ O × because O is equicharacteristic 0 and q = 0. By Lemma 8.11, ∂ (cid:18) b − q (cid:19) = − ∂b ( b − q ) , and val (cid:18) − a∂b ( b − q ) (cid:19) = val( a ) + val( ∂b ) , because the right hand side is less than 0. Then ∂ (cid:18) ab − q (cid:19) = ∂ab − q + − a∂b ( b − q ) . But val (cid:18) ∂ab − q (cid:19) = val( ∂a )val (cid:18) − a∂b ( b − q ) (cid:19) = val( a ) + val( ∂b ) < val( ∂a ) . So val (cid:18) ∂ (cid:18) ab − q (cid:19)(cid:19) = val( a ) + val( ∂b ) < . So a/ ( b − q ) / ∈ R , a contradiction. 64 roposition 8.30. Let K be a dense diffeovalued field, and let ς : Dir K ( K ) → Dir k ( k ) bethe induced 2-inflator. Then no mutation of ς is weakly multi-valuation type.Proof. Recall that ς is the 2-inflator induced by the pedestal I in Sub Q ( K ). Suppose ς ′ isthe mutation of ς along the line K · ( a , a , . . . , a m ). By Proposition 10.15 in [10], ς ′ is the2-inflator induced by the pedestal I ′ = a − I ∩ · · · ∩ a − m I. Note that I ′ is non-zero, because it is open in the diffeovaluation topology on K .By Proposition 8.10 in [10], the fundamental ring R ′ of ς ′ is the “stabilizer” R ′ = { x ∈ K : xI ′ ⊆ I ′ } . Suppose for the sake of contradiction that ς ′ is weakly multi-valuation type. Then there isa multivaluation ring S on K such that R ′ contains a non-zero S -module. So there is somenon-zero a ∈ K such that a · S ⊆ R ′ . Let b be a non-zero element in I ′ . Choose i so that a i = 0. Then a i · b · a · S ⊆ a i · b · R ′ ⊆ a i · I ′ ⊆ I ⊆ R. By Lemma 8.29, a i ba = 0, which is absurd. Theorem 8.31.
Let ς be a 2-inflator on a field K of characteristic 0. Then ς is a diffeoval-uation inflator if and only if the following conditions hold: • ς is malleable. • ς is isotypic. • No mutation of ς is weakly multi-valuation type.Proof. If the listed properties hold, then ς is a diffeovaluation inflator by Theorem 8.26.Conversely, suppose ς : Dir K ( K ) → Dir k ( k ) is a diffeovaluation inflator. Then ς is plainlyisotypic, and malleable by Theorem 8.24. The final property holds by Proposition 8.30. Let ( K , + , · , . . . ) be a field, possibly with extra structure. Assume • K is sufficiently resplendent • dp-rk( K ) ≤ K ) = 0. • K is unstable, and the canonical topology is not a V-topology.65 heorem 9.1. There is a valuation val : K → Γ and a derivation δ : K → K such that forevery a, b ∈ K and γ ∈ Γ , the set { x ∈ K : val( x − a ) > γ and val( δx − b ) > γ } is non-empty, and these sets form a basis for the canonical topology on K .Proof. By resplendence and the uniform definability of the canonical topology (Theorem 6.27),we may replace K with an elementarily equivalent field K . By Propositions 8.19 and 8.21, itsuffices to produce a dense, lifted diffeovaluation structure on K such that the diffeovaluationtopology agrees with the canonical topology.Take a magic subfield k (cid:22) K . By Corollary 4.6 in [1], there is a prime p such that theembedding k × / ( k × ) p ֒ → K × / ( K × ) p is an isomorphism, and so K × = k × · ( K × ) p . Thus K × /k × is p -divisible.By Theorem 6.13, there is a k -linear 2-inflator ς on K satisfying the Strong Assumptionsof § R induces the canonical topology on K .By Theorem 8.26 (and its proof), ς is the 2-inflator induced by some diffeovaluation data( O , D, ∂ ), and R = { x ∈ O : val( ∂x ) ≥ } . Thus, the canonical topology agrees with the diffeovaluation topology.Note that R is a k -algebra, and therefore k ⊆ R ⊆ O . So the value group K × / O × is aquotient of K × /k × , and is p -divisible.Let K + be the expansion of K by the diffeovaluation data. By Theorem 6.27, there is asentence σ holding in K + , expressing that • the diffeovaluation is dense • the value group is p -divisible • the diffeovaluation topology agrees with the canonical topology (of the reduct).Let K be a sufficiently resplendent elementary extension of K + . Then σ holds in K , and K admits a lifting, by Proposition 8.9.Recall from Definition 8.18 that a DV-topology is a field topology that is “locally equivalent”in the sense of [11] to a diffeovaluation topology on a dense diffeovalued field.
Corollary 9.2. If K is a field of dp-rank 2 and characteristic 0, then one of the followingholds: • K is stable. • The canonical topology on K is a V-topology. • The canonical topology on K is a DV-topology.Proof. Theorem 9.1 and Theorem 6.27.2. 66
As outlined in §
10 of [9], it would be very helpful if the Valuation Conjecture 1.2 were true.Unfortunately, algebraically closed dense diffeovalued fields turn out to be a counterexample,as hinted by Theorem 9.1 and Corollary 9.2.
Theorem 10.1.
Let
ADVF be the theory of algebraically closed, dense diffeovalued fields ofresidue characteristic 0. Then
ADVF is consistent, complete, unstable, has dp-rank 2, andis not valuation type.
This will take some work to prove. In order to get a cleaner quantifier elimination result,it helps to work in a slightly different theory expanding ADVF.
Definition 10.2.
Let ( K, O , m ) be a valued field of residue characteristic 0. • Let M be an O -module. An M -valued log derivation on K is a group homomorphism ∂ log : K × → ( M, +)such that ( x + y ) ∂ log( x + y ) = x∂ log x + y∂ log y for x, y ∈ O . • A truncated log derivation on K is a log derivation taking values in K/ m .If ∂ log : K × → ( M, +) is a log derivation, and we define ∂x = x · ∂ log x for x ∈ O , then ∂ : O → M is a derivation. Definition 10.3.
LDVF is the theory of ( K, val , ∂ log), where • ( K, val) | = ACVF , • ∂ log is a truncated log derivation on K . • Every fiber of ∂ log is dense in K , with respect to the valuation topology. Remark . The notion of “log derivation” used here is probably related to a standard con-struction of logarithmic differentials in log geometry. Specifically, an M -valued log derivati-ion on O is probably the same thing as an O -linear morphism Ω O / Q (log Γ > ) → M , wherethe module of log differentials Ω O / Q (log Γ > ) is as defined in § Recall that if
L/K is an extension of fields of characteristic 0, if V is an L -vector space, andif ∂ : K → V is a derivation, then we can extend ∂ to a derivation ∂ ′ : L → V . In the casewhere L = K ( t ) (a pure transcendental extension), we can arrange for ∂ ′ t to equal any valuewe want in V . Lemma 10.5.
There is an algebraically closed dense, lifted diffeovalued field. In other words,there is an algebraically closed field K of characteristic 0, a non-trivial valuation val : K → Γ with residue characteristic 0, and a derivation ∂ : K → K such that for any a, b ∈ K and γ ∈ Γ , there is x ∈ K such that val( x − a ) ≥ γ val( ∂x − b ) ≥ γ. (Compare with Proposition 8.19.)Proof. Choose some extension of the t -adic valuation on Q ( t ) to Q ( t ) alg . Let K be the com-pletion of Q ( t ) alg . Then K is an algebraically closed field with a complete rank 1 valuationof residue characteristic 0. Moreover, the valuation topology on K is metrizable, separable,and complete. As ( K, +) is a non-discrete topological group, it has no isolated points. As K is a perfect Polish space, it is uncountable.Let { U i × V i } i ∈ N be a countable basis of opens in K × K . Let K (cid:22) K be a countableelementary substructure which defines the U i and V i , and is dense in K . Recursively choose t i , s i ∈ K such that • t i ∈ U i and s i ∈ V i • t i is transcendental over K ( t , t , . . . , t i − ).This is possible because K ( t , t , . . . , t i − ) is countable, so at least one transcendental t ′ ∈ K exists. Replacing t ′ with its inverse, we may arrange for t ′ ∈ O . Then, K -definability of U i ensures that there are a, b ∈ K × such that a · O + b ⊆ U i . Take t i = at ′ + b .Take the trivial derivation K → K and extend it successively to K ( t ), K ( t , t ),. . . , arranging for ∂t i = s i . This determines a derivation K ( t , t , . . . ) → K , which we canthen extend to a derivation K → K . The collection of t i witnesses the required densitystatement. Lemma 10.6.
Let ( K, val , δ ) be an algebraically closed, dense, lifted diffeovalued field. Let ∂ log : K × → K/ m be the composition K × → K ։ K/ m , where the first map is the usual log derivation x ( δx ) /x , and the second map is the quotientmap x x + m . Then ∂ log is a truncated log derivation, and ( K, val , ∂ log) is a model ofLDVF. roof. If we set ∂x = x · ∂ log x for x ∈ O , then ∂ : O → K/ m is exactly the composition K δ → K ։ K/ m . Thus ∂ is a derivation, and ∂ log is a truncated log derivation.By choice of ( K, val), it is a model of ACVF , . Finally, we verify the density axiom.Given b ∈ K , we must show that the fiber (cid:26) x ∈ K : δxx ∈ b + m (cid:27) is dense in K , or equivalently, dense in K × . Fix a ∈ K × and γ in the value group. Bycontinuity of division, there is γ ′ such that for any x, y ∈ K ,(val( x − a ) > γ ′ and val( y − ab ) > γ ′ ) = ⇒ val (cid:16) yx − b (cid:17) > . By choice of ( K, val , δ ), there is x such thatval( x − a ) > max( γ ′ , γ )val( δx − ab ) > γ ′ . Then val( x − a ) > γ val (cid:18) δxx − b (cid:19) > . Thus x is within γ of a , and ∂ log x = b + m . So ( K, val , ∂ log) is a model of LDVF.As an immediate corollary, Proposition 10.7.
The theory LDVF is consistent.
Lemma 10.8.
Let K be a model of ACVF , . Let S be a subset of K and S ′ be a subset of K/ m . Let D ⊆ K be definable over S ∪ S ′ . Then one of the following holds: • D has interior. • D is finite, and every element is field-theoretically algebraic over S .Proof. Replacing K with an elementary extension, and we may assume that K is a monstermodel. We may assume S, S ′ are finite. For A ⊆ K , let A alg denote the field-theoreticalgebraic closure, i.e., the algebraic closure of the subfield generated by A .69he swiss cheese decomposition ensures that D has interior unless D is finite. So we mayassume D is finite. Then D ⊆ acl( S ∪ S ′ ). Let { x , . . . , x n } enumerate the elements of S ′ .Enlarging S ′ , we may assume x = 0. Let π : K → K/ m be the quotient map. The fibersof π are infinite, and thus uncountable, by saturation of the monster. Therefore we can find y i , y ′ i ∈ π − ( x i ) such that y i / ∈ ( Sy y · · · y i − ) alg y ′ i / ∈ ( Sy y · · · y n y ′ y ′ · · · y ′ i − ) alg . Then the sequence y , y , . . . , y ′ , y ′ , . . . , y ′ n is a sequence of independent transcendentals over S alg .We arranged for y , y ′ ∈ π − ( x ) = π − (0) = m . Therefore, y and y ′ have non-trivialvaluation. Then M := ( Sy y · · · y n ) alg (cid:22) KM ′ := ( Sy ′ y ′ · y ′ n ) alg (cid:22) K, by model completeness of ACVF. Note that x i ∈ dcl eq ( y i ), and so S ′ ⊆ dcl eq ( M ). Then D is M -definable, and so D ⊆ M because D is finite. Similarly, D ⊆ M ′ .On the other hand, we arranged for the following to hold in the ACF reduct: y y · · · y n | ⌣ S y ′ y ′ · · · y ′ n . Therefore M ∩ M ′ = S alg , and so D ⊆ S alg . Lemma 10.9.
Let K be a model of ACVF. Let P ( x ) = a n x n + · · · + a x + a be a poly-nomial such that min i ≤ n (val( a i )) = 0 . Then the number of roots of P in O , counted withmultiplicities, is equal to the largest i such that val( a i ) = 0 .Proof. This is a basic statement about Newton polygons. Let r , . . . , r n be the roots of P ( x ),counted with multiplicity. Reordering, we may assume r , . . . , r m ∈ O , and r m +1 , . . . , r n / ∈ O .Then P ( x ) = cQ ( x ) Q ( x ) = m Y i =1 ( x − r i ) · n Y i = m +1 (1 − x/r i ) . for some c ∈ K × . Then Q ( x ) ∈ O [ x ], and its reduction modulo m is m Y i =1 ( x − res( r i )) , a nonzero polynomial in k [ x ], of degree m . If we write Q ( x ) = b n x n + · · · + b x + b , thenmin i ≤ n (val( b i )) = 0, and so val( c ) = 0. Thenmax { i ≤ n : val( a i ) = 0 } = max { i ≤ n : val( b i ) = 0 } = m. , : Lemma 10.10.
Let K be a model of ACVF , . Let B be a ball. Let P ( x ) be a polynomialin K [ x ] . If P has two distinct zeros in B , then P ′ ( x ) has a zero in B .Proof. We may assume P is non-zero; otherwise the result it trivial. Let r , r be twozeros in B . Shrinking B to the smallest ball containing r , r , we may assume B is aclosed ball. Shifting everything by an affine transformation, we may assume B = O . Let P ( x ) = a n x n + · · · + a x + a . Multiplying P by a constant from K × , we may assumemin i ≤ n val( a i ) = 0. The polynomial P ( x ) has at least two roots in O , so by Lemma 10.9,there is some m ≥ a m ) = 0. Note P ′ ( x ) = na n x n − + · · · + 2 a x + a . Also, val( ia i ) = val( a i ) for i ≥
1, because of residue characteristic 0. Therefore, val( ia i ) ≥
0, and val( ma m ) = 0. So the coefficient of x m − has valuation 0 for some m ≥
2. ByLemma 10.9, P ′ ( x ) has at least 2 - 1 roots in O . Lemma 10.11.
Let ( K, O , m ) be a valued field, let M be an O -module, and let ∂ log : K × → M be a log derivation. If x, y ∈ K satisfy val( x − y ) ≤ max(val( x ) , val( y )) , (10) then x/ ( x − y ) , y/ ( x − y ) ∈ O , and ∂ log( x − y ) = xx − y · ∂ log( x ) − yx − y · ∂ log( y ) . Proof.
First note that val( x − y ) ≤ min(val( x ) , val( y )) . This is automatic if val( x ) = val( y ), and equivalent to (10) otherwise.Recall the derivation ∂ : O → M given by ∂x = x · ∂ log x . Then0 = ∂ (1) = ∂ (cid:18) x − yx − y (cid:19) = ∂ (cid:18) xx − y (cid:19) − ∂ (cid:18) yx − y (cid:19) = xx − y ∂ log (cid:18) xx − y (cid:19) − yx − y ∂ log (cid:18) yx − y (cid:19) = xx − y [ ∂ log( x ) − ∂ log( x − y )] − yx − y [ ∂ log( y ) − ∂ log( x − y )]= (cid:18) xx − y · ∂ log( x ) − yx − y · ∂ log( y ) (cid:19) − (cid:18) xx − y − yx − y (cid:19) ∂ log( x − y )= (cid:18) xx − y · ∂ log( x ) − yx − y · ∂ log( y ) (cid:19) − ∂ log( x − y ) . roposition 10.12. Let ( K, O ) be an algebraically closed field with a log derivation ∂ log : K → M , for some O -module M . Suppose that ∂ log vanishes on some subfield F ⊆ K , and K is algebraic over F (so that K = F alg ). Then ∂ log vanishes on K .Proof. Increasing F , we may assume F is maximal among subfields on which ∂ log vanishes.Suppose for the sake of contradiction that F ( K . Take minimal n > F hasa finite extension of degree n . If P ( x ) ∈ F [ x ] has degree ≤ n , then one of the followinghappens: • P ( x ) factors into linear polynomials • P ( x ) is irreducible of degree n . Claim . If a ∈ K and [ F ( a ) : F ] = n , then • F ( a ) × is generated by F × and the elements a − b with b ∈ F . • There is b ∈ F such that ∂ log( a − b ) = 0. Proof.
Every element of F ( a ) is of the form P ( a ) for some polynomial P ( x ) ∈ F [ x ] of degreeless than n . Then P splits into linear factors, so P ( a ) = c ( a − b )( a − b ) · · · ( a − b n )for some c, b , b , . . . , b n ∈ F . This proves the first point. If ∂ log( a − b ) = 0 for all b ∈ F ,then ∂ log must vanish on F ( a ) × , contradicting the choice of F . (cid:3) Claim
Take an arbitrary extension
L/F of degree n , and break into cases: • If val( L ) is strictly larger than val( F ), take γ ∈ val( L ) \ val( F ). The inequality | val( L ) / val( F ) | ≤ [ L : F ]implies that mγ ∈ val( F ) for some m ≤ n . Take c ∈ F with val( c ) = mγ . Thepolynomial x m − c has no roots in F , so m = n and x n − c is irreducible. Take a ∈ K such that a n = c . Note that ∂ log( a ) = (1 /n ) ∂ log( c ) = 0 , because c ∈ F and the residue characteristic is 0.By Claim 10.13, there is b ∈ F ( a ) such that ∂ log( a − b ) = 0. Now val( b ) = γ = val( a ),by choice of γ , and so val( a − b ) = min(val( a ) , val( b )) . Also ∂ log( a ) = ∂ log( b ) = 0. By Lemma 10.11, ∂ log( a − b ) = 0, a contradiction.72 If res( L ) is strictly larger than res( F ), take α ∈ res( L ) \ res( F ). The inequality[res( L ) : res( F )] ≤ [ L : F ]implies that [res( F )( α ) : res( F )] ≤ n . Let x m + β m − x m − + · · · + β x + β be the monic irreducible polynomial of α over res( F ). Because of residue characteristic0, this polynomial is separable, and so mα m − + ( m − β m − α m − + · · · + 2 β α + β = 0 . (11)Take b i ∈ F with res b i = β i , and let P ( x ) be the polynomial x m + b m − x m − + · · · + b x + b ∈ F [ x ] . Then P ( x ) is irreducible, and so m = [res( F )( α ) : res( F )] = n . Let a ∈ K be the rootof P ( x ) with res( a ) = α . Then a n + b n − a n − + · · · + b a + b = 0 . Applying the derivation ∂ : O → M , which vanishes on the b i , we obtain( na n − + ( n − b n − a n − + · · · + 2 b a + b ) ∂a = 0 . The expression inside the parentheses has nonzero residue, by (11), and so it is anelement of O × . Therefore ∂a = 0. Now res( a ) = α / ∈ res( F ), so res( a ) = 0 and a isinvertible as well. Therefore ∂ log a = ( ∂a ) /a = 0.By Claim 10.13, there is some b ∈ F such that ∂ log( a − b ) = 0. Thenval( a − b ) ≤ max(val( a ) , val( b )) . (Otherwise, res( a ) = res( b ) ∈ F , contradicting the choice of a and α .) By Lemma 10.11, ∂ log( a − b ) = 0, a contradiction. • Lastly, suppose that
L/F is an immediate extension. By maximality of F , there is a ∈ L with ∂ log( a ) = 0. Let C be the collection of balls containing a , with center andradius from F . Let I be the intersection T C . As usual, I ∩ F = ∅ . (Suppose b ∈ I ∩ F .Then rv ( a − b ) = rv ( b ′ − b ) for some b ′ ∈ F , because the extension is immediate. Theball centered around b ′ of radius val( a − b ′ ) does not contain b , contradicting the choiceof b .)Let P ( x ) be the minimal polynomial of a over F . Then P ( x ) has degree n . Let a , . . . , a n be the roots of P ( x ), with a = a . Note that P ′ ( x ) has degree n −
1, andtherefore splits over F . So no root of P ′ ( x ) is in I . By Rolle’s Theorem (Lemma 10.10), a is the unique root of P ( x ) in I . 73herefore I has empty intersection with the finite set { , a , . . . , a n } . We can find b ∈ F such thatval( a − b ) > max(val(0 − b ) , val( a − b ) , val( a − b ) , . . . , val( a n − b )) . Take c ∈ F with val( a − b ) = val( c ), and let e i = ( a i − b ) /c . Then val( e ) = 0, andval( e i ) < i >
1. The e i are the roots of the irreducible polynomial Q ( x ) = P ( cx + b ) = s n x n + s n − x n − + · · · + s x + s ∈ F [ x ] . By Newton polygons, val( s ) = val( s ) < val( s i ) for i >
1. Then we can apply ∂ tothe equation ( s n /s ) e n + · · · + ( s /s ) e + e + ( s /s ) = 0 , and obtain ( n ( s n /s ) e n − + · · · + 2( s /s ) e + 1) ∂e = 0 , because the coefficients s n /s lie in F , where ∂ vanishes. But the expression in paren-theses has valuation 0, because e ∈ O and s i /s ∈ m for 2 ≤ i ≤ n . Therefore ∂e = 0. As val( e ) = 0, we have e ∈ O × as well, and then ∂ log e = ( ∂e ) /e = 0.Then ∂ log( a − b ) = ∂ log e + ∂ log c = 0, as c ∈ F . Finally,val( a − b ) > val( b ) = val( a ) , and so ∂ log( a − b ) and ∂ log( b ) determine ∂ log( a ), by Lemma 10.11. Thus ∂ log( a ) = 0,contradicting the choice of a .Proposition 10.12 is probably a consequence of Lemma 6.5.12 and Claim 6.5.14 in [2],but I am not entirely certain. Let L be the language for ACVF , with two sorts, K and K/ m , and the following functionsand relations: • The field operations on K , including the constants 0, 1, and division . • The O -module structure on K/ m , i.e., the group structure (including 0 and negation)and the multiplication map O × K/ m → K/ m , understood as a partial function on K × K/ m . • All ∅ -definable relations on K and K/ m .Then ACVF , has quantifier elimination in L , because we Morleyized. If K | = ACVF , ,an L -substructure of K consists of a pair ( F, D ), where74 F is a subfield of K . • D is an O F -submodule of K/ m .Note that D need not contain the image of F under K ։ K/ m , as we did not include thismap as one of the functions in the signature.Let L be the language for LDVF obtained by expanding L with a function symbol forthe map ∂ log : K × → K/ m . If K is a model of LDVF, then an L -substructure is a pair( F, D ), where • F is a subfield of K • D is an O F -submodule of K/ m • D contains ∂ log x for x ∈ F . Lemma 10.14.
Let K be a model of LDVF . Let K ′ be a | K | + -saturated model of LDVF .Let ( F, D ) be a proper L -substructure of K . Let f : ( F, D ) ֒ → K ′ be an L -embedding(an isomorphism onto a substructure of K ′ ). Then f can be extended to an L -embedding f ′ : ( F ′ , D ′ ) ֒ → K ′ for some strictly larger L -substructure ( F ′ , D ′ ) .Proof. First suppose
D < K/ m . Note that ( F, K/ m ) is an L -substructure of K . By quan-tifier elimination of ACVF , in the language L , we can extend f to an L -embedding f ′ : ( F, K/ m ) ֒ → K ′ . Then f ′ is already an L -embedding, because f ′ ( ∂ log( x )) = f ( ∂ log( x )) = ∂ log( f ( x )) = ∂ log( f ′ ( x ))for any x ∈ F . (The first equation holds because x ∈ F = ⇒ ∂ log( x ) ∈ D , and f ′ extends f on D .)So we may assume D = K/ m , and F < K . Claim . If F ′ is a subfield of K containing F , and f ′ : ( F ′ , K/ m ) ֒ → K ′ is an L -embedding extending f , then • f ′ induces a map from the valuation ring of F ′ to the valuation ring of K ′ , and so wecan regard K ′ / m ′ as a module over the valuation ring of F ′ . • If ∆ : F ′ → K ′ / m ′ is defined by∆( x ) = f ′ ( ∂ log( x )) − ∂ log( f ′ ( x )) , then ∆ is a log derivation F ′ → K/ m . • ∆ vanishes on F . • If ∆ vanishes on F ′ , then f ′ is an L -embedding.75 roof. The first point is clear, since f ′ is a partial elementary map in the ACVF reduct. Thesecond point is a direct calculation:∆( xy ) = f ′ ( ∂ log( xy )) − ∂ log( f ′ ( xy ))= f ′ ( ∂ log( x ) + ∂ log( y )) − ∂ log( f ′ ( x ) f ′ ( y ))= f ′ ( ∂ log( x )) + f ′ ( ∂ log( y )) − ∂ log( f ′ ( x )) − ∂ log( f ′ ( y ))= ∆( x ) + ∆( y )( x + y )∆( x + y ) = f ′ ( x + y )∆( x + y )= f ′ ( x + y ) f ′ ( ∂ log( x + y )) − f ′ ( x + y ) ∂ log( f ′ ( x + y ))= f ′ (( x + y ) ∂ log( x + y )) − ( f ′ ( x ) + f ′ ( y )) ∂ log( f ′ ( x ) + f ′ ( y ))= f ′ ( x∂ log x + y∂ log y ) − f ′ ( x ) ∂ log( f ′ ( x )) − f ′ ( y ) ∂ log( f ′ ( y ))= f ′ ( x ) f ′ ( ∂ log x ) + f ′ ( y ) f ′ ( ∂ log y ) − f ′ ( x ) ∂ log( f ′ ( x )) − f ′ ( y ) ∂ log( f ′ ( y ))= f ′ ( x )∆( x ) + f ′ ( y )∆( y ) . The third point expresses that f is an L -embedding. The fourth point is clear. (cid:3) Claim
Next suppose F = F alg . By quantifier elimination of ACVF , , we can extend f to an L -embedding f ′ : ( F alg , K/ m ) ֒ → K ′ . Let ∆ : F alg → K ′ / m ′ be as in Claim 10.15. Then ∆vanishes on F , and therefore on F alg , by Proposition 10.12. Therefore f ′ is an L -embedding.Finally, suppose that F = F alg . Take a transcendental a ∈ K \ F . Let ~s be an infinitetuple enumerating F , and ~t be an infinite tuple enumerating K/ m . Let Σ( x ; ~y ; ~z ) be thecomplete L -type of ( a ; ~s ; ~t ). Note that a ′ ∈ K ′ satisfies Σ( x ; f ( ~s ); f ( ~t )), if and only if thereis an L -embedding f ′ : ( F ( a ) , K/ m ) ֒ → K ′ extending f and sending a a ′ .For any b ∈ F , let ψ b ( x ) be the type in K ′ asserting that ∂ log( x − f ( b )) = f ( ∂ log( a − b )) . (The right hand side makes sense, because we arranged D = K/ m .) Claim . For any b ∈ F , the type Σ( x ; f ( ~s ); f ( ~t )) ∪ { ψ b ( x ) } is realized in K ′ . Proof.
By saturation, it suffices to show finite satisfiability. Suppose ϕ ( x ; ~y ; ~z ) is an L -formula satisfied by ( a ; ~s ; ~t ). We must find a ′ ∈ K ′ satisfying ϕ ( a ′ ; f ( ~s ); f ( ~t )) ∧ ψ b ( a ′ )The definable set ϕ ( K ; ~s ; ~t )) has interior, by Lemma 10.8 and transcendence of a over ~s . As f is a partial elementary map in the ACVF reduct, the definable set ϕ ( K ′ ; f ( ~s ); f ( ~t )) hasinterior as well. By the density axiom of LDVF, there is x ∈ K ′ such that x + f ( b ) ∈ ϕ ( K ′ ; f ( ~s ); f ( ~t )) ∂ log( x ) = f ( ∂ log( a − b )) . Take a ′ = x + f ( b ). (cid:3) Claim laim . The type Σ( x ; f ( ~s ); f ( ~t )) ∪ { ψ b ( x ) : b ∈ F } is realized in K ′ . Proof.
By saturation, it suffices to show finite satisfiability. Let b , . . . , b n be elements of F .We claim that the type Σ( x ; f ( ~s ); f ( ~t )) ∪ { ψ b ( x ) , . . . , ψ b n ( x ) } is realized in K ′ . Without loss of generality,val( a − b ) ≥ val( a − b ) ≥ · · · ≥ val( a − b n ) . By Claim 10.16 there is a ′ realizing Σ( x ; f ( ~s ); f ( ~t )) ∪ { ψ b ( x ) } . Let f ′ : ( F ( a ) , K/ m ) → K ′ be the L -embedding extending f and sending a to a ′ . Note that ∂ log( f ′ ( a − b )) = ∂ log( f ′ ( a ) − f ′ ( b )) = ∂ log( a ′ − f ( b )) ∗ = f ( ∂ log( a − b )) = f ′ ( ∂ log( a − b )) . The starred equation holds because of ψ b ( a ′ ). By Claim 10.15, there is a log derivation F ( a ) → K ′ / m ′ given by ∆( x ) := f ′ ( ∂ log( x )) − ∂ log( f ′ ( x )) . Then ∆( a − b ) = 0. Also, ∆( b i − b ) = 0 for any i , because b i − b ∈ F . Moreover,val(( a − b ) − ( b i − b )) = val( a − b i ) ≤ val( a − b ) , and so ∆( a − b i ) = 0, by Lemma 10.11. Then for each i , ∂ log( a ′ − f ( b i )) = ∂ log( f ′ ( a ) − f ′ ( b i )) = ∂ log( f ′ ( a − b i )) = f ′ ( ∂ log( a − b i )) = f ( ∂ log( a − b i )) . Therefore ψ b i ( a ′ ) holds.Using Claim 10.17, take a ′ ∈ K ′ realizingΣ( x ; f ( ~s ); f ( ~t )) ∪ { ψ b ( x ) : b ∈ F } . Let f ′ : ( F ( a ) , K/ m ) ֒ → K ′ be the L -embedding extending f and mapping a to a ′ . Let∆( x ) = f ′ ( ∂ log( x )) − ∂ log( f ′ ( x )) as in Claim 10.15. The statement ψ b ( a ′ ) implies that ∂ log( a ′ − f ( b )) = f ( ∂ log( a − b )) . Therefore ∂ log( f ′ ( a − b )) = ∂ log( f ′ ( a ) − f ′ ( b )) = ∂ log( a ′ − f ( b )) = f ( ∂ log( a − b )) = f ′ ( ∂ log( a − b )) . So ∆( a − b ) = 0 for any b ∈ F . As F is algebraically closed and a is transcendental, themultiplicative group F ( a ) × is generated by F × ∪ { a − b : b ∈ F } . Then ∆ must vanish on F ( a ), because it vanishes on the generators. By Claim 10.15, f ′ isan L -embedding. 77 heorem 10.18. LDVF has quantifier elimination in the language L .Proof. This follows from Lemma 10.14, by well-known model-theoretic techniques.
Corollary 10.19.
LDVF is complete.Proof.
LDVF is consistent by Proposition 10.7. Take two models M and M , viewed as L -structures. Note that ( Q ,
0) is an L -substructure of M i , for each i . The identity map( Q , → ( Q ,
0) is an isomorphism of L -structures: • For the L -structure, this holds because ACVF , is complete. • For the map ∂ log, this holds because ∂ log is trivial on both copies of Q .By quantifier elimination, M ≡ M . Corollary 10.20.
Let
ADVF be the theory of algebraically closed dense diffeovalued fields ( K, ∂, D ) .1. If ( K, ∂ log) is a model of
LDVF , then ( K, ∂, K/ m ) is a model of ADVF .2. Up to elementary equivalence, every model of
ADVF arises in this way.3.
ADVF is complete.Proof.
It suffices to prove Point 2. Let (
K, ∂, val) be a model of ADVF. Passing to aresplendent elementary extension, we may assume that a lifting exists, by Proposition 8.9.So we obtain a lifted diffeovalued field (
K, δ, val). Define ∂ log : K × → K/ m ∂ log( x ) = δxx + m . Then (
K, ∂ log , val) | = LDVF, by Lemma 10.6. The derivation ∂ : O → K/ m determined by ∂ log is the original derivation ∂ . Let T be the theory of dense, lifted diffeovalued fields. By Proposition 8.19, a model of T is afield K with a non-trivial valuation (of residue characteristic 0) and a derivation δ : K → K such that for any a, b ∈ K and γ in the value group, there is x ∈ K such thatval( x − a ) ≥ γ val( δx − b ) ≥ γ. The theory T probably has no nice properties, other than being consistent.78 emark . Lemma 10.6 says that if (
K, δ, val) | = T , and we define ∂ log( x ) := δxx + m , then ( K, ∂ log , val) | = LDVF. As LDVF is complete, it follows that every sufficiently re-splendent model of LDVF can be expanded to a model of T .Fix some one-sorted language for T . Say that ϕ ( ~x ) is a ACVF-formula if ϕ ( ~x ) is definedin the ACVF-reduct, and similarly for LDVF-formulas . Let tp
ACVF ( ~a/B ) be the set of allACVF-formulas with parameters in B , satisfied by ~a . Define tp LDVF ( ~a/B ) similarly.If M | = T and ~a is a tuple in M , define δ ( ~a ) coordinatewise. Lemma 10.22.
Let M , M be two models of T . Let ~a be a tuple in M and ~b be a tuple of thesame length in M . Suppose that tp ACV F ( ~aδ ( ~a ) / ∅ ) = tp ACV F ( ~bδ ( ~b ) / ∅ ) . Then tp LDV F ( ~a/ ∅ ) =tp LDV F ( ~b/ ∅ ) .Proof. After replacing M and M with elementary extensions, there is an isomorphism ofthe ACVF-reducts f : ( M , val) → ( M , val)such that f ( ~a ) = ~b and f ( δ ( ~a )) = δ ( ~b ). With L as in the previous section, this induces anisomorphism of L -structures f ′ : ( M , M / m M ) → ( M , M / m M ) . Restricting the first sort, we obtain an L -isomorphism f ′′ : ( Q ( ~a ) , M / m M ) → ( Q ( ~b ) , M / m M ) , because f ( ~a ) = ~b . We claim that f ′′ preserves ∂ log. If c is in Q [ ~a ], then c = P ( a , . . . , a n )for some P ( x , . . . , x n ) ∈ Q [ x , . . . , x n ]. Then δ ( c ) = n X i =1 ∂P∂x i ( ~a ) · δ ( a i ) . Since f sends δ ( a i ) to δ ( b i ), f ( δ ( c )) = n X i =1 ∂P∂x i ( ~b ) · δ ( b i ) = δ ( P ( ~b )) = δ ( f ( c )) . Then f ( δ ( c )) = δ ( f ( c )), implying that f ′ ( ∂ log( c )) = ∂ log( f ′ ( c )). More generally, if c ∈ Q ( ~a ), then c = c /c for c i ∈ Q [ ~a ], and f ′ ( ∂ log( c )) = f ′ ( ∂ log( c ) − ∂ log( c )) = f ′ ( ∂ log( c )) − f ′ ( ∂ log( c ))= ∂ log( f ′ ( c )) − ∂ log( f ′ ( c )) = ∂ log( f ′ ( c ) /f ′ ( c ))= ∂ log( f ′ ( c /c )) = ∂ log( f ′ ( c )) . Thus f ′ ( ∂ log( c )) = ∂ log( f ′ ( c )) for c ∈ Q ( ~a ), and f ′′ is an L -isomorphism. By quantifierelimination of LDVF in the language L , it follows that tp LDVF ( ~a/ ∅ ) = tp LDVF ( ~b/ ∅ ).79 emma 10.23. For every LDVF-formula ϕ ( ~x ) , there is an ACVF-formula ψ ( ~x ; ~y ) such that T ⊢ ϕ ( ~x ) ⇐⇒ ψ ( ~x ; δ ( ~x )) . Proof.
A standard compactness argument.
Theorem 10.24. If ( K, ∂ log , val) | = LDVF , then dp-rk( K ) ≤ .Proof. Suppose there is an ict-pattern of depth 3: ϕ ( x ; ~b , ) , ϕ ( x ; ~b , ) , . . .ϕ ( x ; ~b , ) , ϕ ( x ; ~b , ) , . . .ϕ ( x ; ~b , ) , ϕ ( x ; ~b , ) , . . . Replacing K with an elementary extension, we may assume that K can be expanded to amodel of T , by Remark 10.21. Applying Lemma 10.23, we obtain an ict-pattern of depth 3, ϕ ′ ( x, y ; ~c , ) , ϕ ′ ( x, y ; ~c , ) , . . .ϕ ′ ( x, y ; ~c , ) , ϕ ′ ( x, y ; ~c , ) , . . .ϕ ′ ( x, y ; ~c , ) , ϕ ′ ( x, y ; ~c , ) , . . . made of ACVF-formulas. But in ACVF, the set K has dp-rank less than 3. Lemma 10.25. If R is an interpretable integral domain in some structure, and K =Frac( R ) , then dp-rk( K ) = dp-rk( R ) .Proof. Work in a monster model. The inequality dp-rk( R ) ≤ dp-rk( K ) is clear. Conversely,suppose there is an ict-pattern of depth κ in K . Then there are formulas { ϕ α ( x ; y α ) } α<κ ,coefficients { b α,i } α<κ, i<ω , and witnesses { a η } η : κ → ω in K , such that ϕ α ( a η , b α,i ) ⇐⇒ η ( α ) = i. For any x , . . . , x n ∈ K , we can find a non-zero common denominator s ∈ R such that { x s, . . . , x n s } ⊆ R. By saturation, we can find some non-zero s ∈ R such that sa η lies in R for every η . Let a ′ η = sa η , and let ψ α ( x, y, z ) be the formula ψ α ( x, y, z ) ≡ ϕ α ( x/y, z ) . Then there is an ict-pattern of depth κ in R : ψ α ( a ′ η , s, b α,i ) ⇐⇒ η ( α ) = i. heorem 10.26. If ( K, ∂ log , val) | = LDVF , then K is unstable, not of valuation type, andhas dp-rank exactly 2.Proof. Consider the reduct (
K, ∂, val), a dense diffeovalued field by Corollary 10.20. Let R be the ring R = { x ∈ O : ∂x ∈ O / m } , as in § { aR : a ∈ K × } form a neighborhood basis for adefinable non-trivial Hausdorff topology. So certainly K is unstable. Also Frac( R ) = K , sodp-rk( R ) = dp-rk( K ) by Lemma 10.25. Therefore R − R = R is a basic neighborhood in thecanonical topology. So the canonical topology is finer than the diffeovaluation topology. Ifthe canonical topology is a V-topology, so is every coarsening, by Theorem 3.2 in [11]. Butthe diffeovaluation topology is not a V-topology, by Proposition 8.17. So ( K, ∂ log , val) doesnot have valuation type. Then dp-rk( K ) = 1, because dp-minimal fields have valuation type(Theorem 9.3.28 in [7]). Therefore, K has dp-rank 2. Remark . The same argument applies to intermediate reducts between the full modelof LDVF, and the reduct ( K, + , · , R ).The structure ( K, + , · , O , R ) is similar to Example 7.1 in [4]: both are NIP valued fieldsin which some infinite definable set has empty interior. (In our case, the set is R .) Unlike[4], our example is not a pure valued field, but an expansion.
11 Concluding remarks
The example of §
10 derails some promising strategies to attack the Shelah conjecture andhenselianity conjecture. For example, it disproves our “valuation conjecture” (Conjec-ture 1.2), which would have implied the Shelah conjecture ([9], Theorem 9.9).Consider the even simpler conjecture:
Conjecture 11.1. If ( K, + , · , O , . . . ) is a dp-finite valued field, and S ⊆ K is a definable setof full dp-rank ( dp-rk( S ) = dp-rk( K ) ), then S has non-empty interior. Conjecture 11.1 would imply the Henselianity conjecture for dp-finite fields, by Theo-rem 7.5 in [4]. To the best of my knowledge, all the known results on the henselianityconjecture use this strategy.However, the theory LDVF of §
10 is a counterexample to Conjecture 11.1. Indeed, thedefinable set R has full rank, but empty interior with respect to O .Conjecture 11.1 probably holds for pure valued fields ( K, + , · , O ), but this is probablyimpossible to prove without first classifying dp-finite valued fields using some other strategy.(The purity assumption is hard to use in proofs.)It seems we need a new strategy to attack the dp-finite Shelah and henselianity conjec-tures. Perhaps the analysis of § n ≥
1, there should be a class of “field topologiesof type W n ,” cut out by a local sentence (in the sense of [11]). The canonical topology on81n unstable dp-finite field K should be a definable W n -topology for some n ≤ dp-rk( K ).For n = 1, a W -topology should be the same thing as a V-topology. For n = 2, a W -topology should either be a DV-topology in the sense of § n = 3, there should be four types: • A topology generated by three independent V-topologies. • A topology generated by a V-topology and an independent DV-topology. • A topology that is like a DV-topology, but with basic opens { x ∈ K : val( x − a ) ≥ γ, val( δ x − b ) ≥ γ, val( δ x − c ) ≥ γ } . for two derivations δ , δ : K → K . • A topology that is like a DV-topology, but involving second derivatives, with basicopens { x ∈ K : val( x − a ) ≥ γ, val( δx − b ) ≥ γ, val( δ x − c ) ≥ γ } . Now suppose K is a field of dp-rank 3. Using results from [9], it should be possible to provethat the squaring map f ( x ) = x is an open map from K × to K × . This should exclude thefirst two cases. In the latter two cases, it should be possible to show that the W -topologyhas a unique V-topology coarsening. This would imply that K admits a unique definableV-topology. Generalizations of these arguments should work for n > A Appendix: Resplendent lifting
In the appendix, we assume that all rings are Q -algebras, all fields extend Q , and all valuedfields have residue characteristic 0. A.1 Extending derivations
Say that an ordered abelian group Γ is Z -less if it satisfies the following equivalent conditions: • For every a > + is the minimal convex subgroup containing a and ∆ − isthe maximal convex subgroup avoiding a , then ∆ + / ∆ − = Z . • For every a > b ∈ Γ such that(1 / a < b < (2 / a, i.e., a < b < a . 82 For every a > p < q in Q , there is b ∈ Γ such that pa < b < qa.
For example, if Γ is p -divisible for some prime p , then Γ is Z -less. Remark
A.1 . Let Γ ′ > Γ be an extension of ordered abelian groups. Suppose Γ ′ / Γ is torsion(i.e., Γ ′ ≤ Γ ⊗ Z Q ). If Γ is Z -less, then Γ ′ is Z -less. Lemma A.2.
Let
L/K be an algebraic extension of valued fields. Suppose that the valuegroup of K is Z -less. Let a be a nonzero element of O L with positive valuation. Suppose that a n ∈ K . Then there are b, c ∈ O L such that • a = bc n • bc n − is in K . • c n is in K .Proof. By Z -lessness, there is γ ∈ Γ K such that n − n val( a ) < γ < val( a ) . Let e ∈ K have val( e ) = γ . Let b = e n a − n and c = ae − . Thenval( b ) = n · γ − ( n −
1) val( a ) > c ) = val( ae − ) = val( a ) − γ > bc n = e n a − n a n e − n = abc n − = e n a − n a n − e − n = e ∈ K.c n = a n e − n ∈ K. Proposition A.3.
Let ( K, O ) be an algebraically closed field with a derivation ∂ : O → M for some O -module M . Suppose that ∂ vanishes on some subfield F ⊆ K , and K is algebraicover F (so that K = F alg ). Suppose val( F ) is Z -less. Then ∂ vanishes on K . The proof is nearly identical to the proof of Proposition 10.12.
Proof.
Increasing F , we may assume F is maximal among subfields on which ∂ vanishes.Suppose for the sake of contradiction that F ( K . Take minimal n > F hasa finite extension of degree n . If P ( x ) ∈ F [ x ] has degree ≤ n , then one of the followinghappens: • P ( x ) factors into linear polynomials • P ( x ) is irreducible of degree n .Take an arbitrary extension L/F of degree n , and break into cases:83 If val( L ) is strictly larger than val( F ), take γ ∈ val( L ) \ val( F ). The inequality | val( L ) / val( F ) | ≤ [ L : F ]implies that mγ ∈ val( F ) for some m ≤ n . Take c ∈ F with val( c ) = mγ . Thepolynomial x m − c has no roots in F , so m = n and x n − c is irreducible. Take a ∈ K such that a n = c .By maximality of F , there is some x ∈ O F ( a ) such that ∂x = 0. We can write x = y + y a + · · · + y n − a n − for some y i ∈ F . Note that iγ / ∈ val( F ) for i < n , and so the non-zero terms y i a i havepairwise distinct valuations. Therefore0 ≤ val( x ) = min i (val( y i a i )) . So every y i a i is in O , and ∂x = n − X i =0 ∂ ( y i a i ) = n − X i =1 ∂ ( y i a i ) . For each i >
0, we have val( y i a i ) = 0. Then Lemma A.2 gives b, c ∈ O F ( a ) and e ∈ O F such that e = bc n − y i a i = bc n = ecc n ∈ F. Therefore ∂ ( y i a i ) = ∂ ( ec ) = e∂ ( c ) = bc n − ∂ ( c ) = bn ∂ ( c n ) = 0 , as e, c n ∈ O F , and 1 /n ∈ O K (by the assumption of residue characteristic 0). So ∂ ( x ) = 0, a contradiction. • If res( L ) is strictly larger than res( F ), take α ∈ res( L ) \ res( F ). The inequality[res( L ) : res( F )] ≤ [ L : F ]implies that [res( F )( α ) : res( F )] ≤ n . Let x m + β m − x m − + · · · + β x + β be the monic irreducible polynomial of α over res( F ). Because of residue characteristic0, this polynomial is separable, and so mα m − + ( m − β m − α m − + · · · + 2 β α + β = 0 . (12)84ake b i ∈ F with res b i = β i , and let P ( x ) be the polynomial x m + b m − x m − + · · · + b x + b ∈ F [ x ] . Then P ( x ) is irreducible, and so m = [res( F )( α ) : res( F )] = n . Let a ∈ K be the rootof P ( x ) with res( a ) = α . Then a n + b n − a n − + · · · + b a + b = 0 . Applying the derivation ∂ : O → M , which vanishes on the b i , we obtain( na n − + ( n − b n − a n − + · · · + 2 b a + b ) ∂a = 0 . The expression inside the parentheses has nonzero residue, by (12), and so it is anelement of O × . Therefore ∂a = 0.Now if x = y + y a + · · · + y n − a n − is any element of F ( a ), then val( x ) = min i val( y i ). To see this, one reduces to the casewhere min i (val( y i )) = 0; thenres( y ) + res( y ) α + · · · + res( y n − ) α n − = 0 , by linear independence of { , α, . . . , α n − } over res( K ).Consequently, O F ( a ) = O F [ a ]. As ∂ vanishes on O F and a , it vanishes on O F ( a ) ,contradicting the maximality of F . • Lastly, suppose that
L/F is an immediate extension. By maximality of F , there is a ∈ O L with ∂ ( a ) = 0. Let C be the collection of balls containing a , with center andradius from F . Let I be the intersection T C . As usual, I ∩ F = ∅ .Let P ( x ) be the minimal polynomial of a over F . Then P ( x ) has degree n . Let a , . . . , a n be the roots of P ( x ), with a = a . Note that P ′ ( x ) has degree n −
1, andtherefore splits over F . So no root of P ′ ( x ) is in I . By Rolle’s Theorem (Lemma 10.10), a is the unique root of P ( x ) in I .Therefore I has empty intersection with the finite set { , a , . . . , a n } . We can find b ∈ F such thatval( a − b ) > max(val(0 − b ) , val( a − b ) , val( a − b ) , . . . , val( a n − b )) . Take c ∈ F with val( a − b ) = val( c ), and let e i = ( a i − b ) /c . Then val( e ) = 0, andval( e i ) < i >
1. The e i are the roots of the irreducible polynomial Q ( x ) = P ( cx + b ) = s n x n + s n − x n − + · · · + s x + s ∈ F [ x ] .
85y Newton polygons, val( s ) = val( s ) < val( s i ) for i >
1. Then we can apply ∂ tothe equation ( s n /s ) e n + · · · + ( s /s ) e + e + ( s /s ) = 0 , and obtain ( n ( s n /s ) e n − + · · · + 2( s /s ) e + 1) ∂e = 0 , because the coefficients s n /s lie in F , where ∂ vanishes. But the expression in paren-theses has valuation 0, because e ∈ O and s i /s ∈ m for 2 ≤ i ≤ n . Therefore ∂e = 0.Meanwhile, val( a − b ) > val( b ) implies that0 ≤ val( a ) = min(val( a − b ) , val( b )) = min(val( c ) , val( b )) , and so b, c ∈ O F . But a = b + e c , and so ∂ ( a ) = c∂e = 0, contradicting the choice of a . A.2 Review of K¨ahler differentials If A → B is a morphism of (commutative unital) rings, then Ω B/A denotes the module ofK¨ahler differentials. This is the B -module generated by terms db for b ∈ B , subject to therelations d ( b + b ) = db + db d ( b b ) = b db + b db da = 0 if a ∈ A. If M is a B -module, there is an isomorphismHom B (Ω B/A , M ) ∼ = Der A ( B, M )natural in M , where Der A ( B, M ) denotes the set of A -linear derivations B → M .The following facts about K¨ahler differentials are well-known: Fact A.4. If A → B → C is a morphism of rings, then Ω B/A ⊗ B C → Ω C/A → Ω C/B → is exact. Fact A.5. If A → B is a morphism of rings and S ⊆ B is a multiplicative subset, then S − Ω B/A ∼ = Ω S − B/A . Fact A.6. If L/K is an extension of (characteristic 0) fields and { t i } i ∈ I is a transcendencebasis (possibly infinite), then { dt i } i ∈ I is an L -basis of Ω L/K . emark A.7 . If O is a valuation ring and M is an O -module, the following are equivalent:1. M is flat.2. M is torsionless.3. Every finitely-generated submodule of M is free.4. M is a direct limit of free modules.5. The natural map M → M ⊗ O K is an injection.We will use two flatness results from ([2], Corollary 6.5.21 and Theorem 6.5.15). Fact A.8. If O is a valuation ring with residue characteristic 0, then Ω O / Q is flat as an O -module. Fact A.9.
Let O ′ / O be an extension of valuation rings. Suppose Frac( O ) | = ACF . Then Ω O ′ / O is flat as an O -module. A.3 Flatness and extensions
Lemma A.10.
Let O be a valuation ring and → A → B → C → be a short exactsequence of O -modules. • If B is flat, then A is flat. • If A and C are flat, then B is flat.Proof. By Remark A.7, an O -module is flat if and only if it is torsionless.For the first point: submodules of torsionless modules are torsionless.For the second point, suppose that A and C are torsionless. For any nonzero r ∈ O ,there is a diagram 0 / / A / / (cid:15) (cid:15) B / / (cid:15) (cid:15) C / / (cid:15) (cid:15) / / A / / B / / C / / r . Because A and C are torsionless, the outer vertical maps are injective. By the snake lemma, the inner verticalmap is injective. As r is arbitrary, C is torsionless. Lemma A.11.
Let O ′ / O be an extension of valuation rings. Let M be an O -module. Then M is flat (as an O -module) if and only if M ⊗ O O ′ is flat (as an O ′ -module). roof. If M is flat, then M is a direct limit of free O -modules, and so M ⊗ O O ′ is a directlimit of free O ′ -modules. Conversely, suppose that M is not flat. Then there is an injection O /I ֒ → M for some non-zero proper ideal I in O . As O ′ is torsionless over O , it is flat as an O -module.Therefore the functor − ⊗ O O ′ is exact, and the map( O /I ) ⊗ O O ′ ֒ → M ⊗ O O ′ is injective. But ( O /I ) ⊗ O O ′ ∼ = O ′ /I O ′ . The ideal I O ′ is non-trivial, because it containsthe non-trivial elements of I . And I O ′ is a proper ideal, because it is generated by elementsof positive valuation. Therefore O ′ /I O ′ is not torsionless, and neither is the larger module M ⊗ O O ′ . Lemma A.12.
Let K ⊆ K ⊆ K be a chain of three fields (of characteristic 0). Then themap Ω K /K ⊗ K K → Ω K /K is injective.Proof. Let B be a transcendence basis of K /K , and B ′ be a transcendence basis of K /K extending B . By Fact A.6, the set { dt : t ∈ B } is a K -linear basis of Ω K /K ⊗ K K , andthe set { dt : t ∈ B ′ } is a K -linear basis of Ω K /K . The map in question is induced by theinclusion B ֒ → B ′ , and is therefore injective. Lemma A.13.
Let O ′ / O be an extension of valuation rings (with residue characteristic 0).Then the map Ω O / Q ⊗ O O ′ → Ω O ′ / Q is injective.Proof. This follows from the commuting diagramΩ O / Q ⊗ O O ′ (cid:31) (cid:127) / / (cid:15) (cid:15) (Ω O / Q ⊗ O O ′ ) ⊗ O ′ K ′ (cid:15) (cid:15) (Ω O / Q ⊗ O K ) ⊗ K K ′ (cid:15) (cid:15) Ω K/ Q ⊗ K K ′ (cid:127) _ (cid:15) (cid:15) Ω O ′ / Q (cid:31) (cid:127) / / Ω O ′ / Q ⊗ O ′ K ′ Ω O ′ / Q ⊗ O ′ K ′ Ω K ′ / Q , where the left horizontal arrows are injective by flatness (Fact A.8), the right horizon-tal arrows are isomorphisms by Fact A.5, and the rightmost vertical map is injective byLemma A.12. Lemma A.14.
Let O ⊆ O ⊆ O be a chain of two valuation ring extensions . Then themap Ω O / O ⊗ O O → Ω O / O is injective. Meaning that the inclusions are local homomorphisms. roof. There is a commutative diagram0 / / Ω O / Q ⊗ O O / / Ω O / Q ⊗ O O / / (cid:127) _ (cid:15) (cid:15) Ω O / O ⊗ O O (cid:15) (cid:15) / / / / Ω O / Q ⊗ O O / / Ω O / Q / / Ω O / O / / → Ω O / Q ⊗ O O → Ω O / Q → Ω O / O → − ⊗ O O yields the exactness of the toprow of (13). The middle vertical map of (13) is injective by Lemma A.13. The snake lemmathen implies that the right vertical map is injective. Definition A.15.
Let O ′ / O be an extension of valuation rings. Then O ′ / O is pseudosmooth if Ω O ′ / O is flat (as an O ′ -module). Proposition A.16.
Let O ⊆ O ⊆ O be a chain of two valuation ring extensions.1. If O / O and O / O are pseudosmooth, then O / O is pseudosmooth.2. If O / O is pseudosmooth, then O / O is pseudosmooth.Proof. The sequence 0 → Ω O / O ⊗ O O → Ω O / O → Ω O / O → O / O is pseudosmooth ⇐⇒ Ω O / O is flat ⇐⇒ Ω O / O ⊗ O O is flat O / O is pseudosmooth ⇐⇒ Ω O / O is flat O / O is pseudosmooth ⇐⇒ Ω O / O is flat,using Lemma A.11 in the first line. The desired statements follow from Lemma A.10. Proposition A.17.
Let O ′ / O be an extension of valued fields of residue characteristic 0.Suppose the value group of O is Z -less. Then Ω O ′ / O is flat as an O ′ -module.Proof. We must show that O ′ / O is pseudosmooth. By Proposition A.16.2, we may replace O ′ with a larger valued field. Since valuations can be extended along any field extension, wemay assume that Frac( O ′ ) contains the algebraic closure of Frac( O ). Let O ′′ be the inducedvaluation ring on Frac( O ) alg . Then O ⊆ O ′′ ⊆ O ′ . Now O ′ / O ′′ is pseudosmooth by Fact A.9,as Frac( O ′′ ) is algebraically closed. By Proposition A.16.1, it remains to show that O ′′ / O ispseudosmooth. In fact, Ω O ′′ / O vanishes, by Proposition A.3.89 .4 Resplendent lifting in the Z -less case Lemma A.18.
Let R be a ring, and A _(cid:127) f ′ (cid:15) (cid:15) g ′ / / M f (cid:15) (cid:15) (cid:15) (cid:15) A ⊕ B g / / N be a diagram of R -modules, with f surjective, and f ′ the inclusion of the first factor. If B is free, then there is a diagonal map h : A ⊕ B → M making the diagram commute: A _(cid:127) f ′ (cid:15) (cid:15) g ′ / / M f (cid:15) (cid:15) (cid:15) (cid:15) A ⊕ B h ; ; ✈✈✈✈✈✈✈✈✈ g / / N. The proof is well-known but included for completeness.
Proof.
Consider the morphism B → N given by x g (0 , x ). Because B is free, there is h : B → M lifting this, so that g (0 , x ) = f ( h ( x ))for x ∈ B . Define h : A ⊕ B → M by the formula h ( x, y ) = g ′ ( x ) + h ( y ). Then h ( f ′ ( x )) = h ( x,
0) = g ′ ( x ) f ( h ( x, y )) = f ( g ′ ( x )) + f ( h ( y )) = g ( f ′ ( x )) + g (0 , y ) = g ( x,
0) + g (0 , y ) = g ( x, y ) . Lemma A.19.
Let R be a valuation ring. Let A _(cid:127) f ′ (cid:15) (cid:15) g ′ / / M f (cid:15) (cid:15) (cid:15) (cid:15) B g / / N be a diagram of R -modules, with f ′ : A ֒ → B injective and f : M ։ N surjective. Supposethe following hold: • coker( f ′ ) is flat. • Let ( M, N ) be the two-sorted structure with the R -module structure on M and N , andthe surjection f : M ։ N . Then ( M, N ) is ( | A | + | B | + | R | ) + -saturated.Then there is a morphism h : B → M making the diagram commute A _(cid:127) f ′ (cid:15) (cid:15) g ′ / / M f (cid:15) (cid:15) (cid:15) (cid:15) B h > > ⑥⑥⑥⑥⑥⑥⑥⑥ g / / N. roof. Without loss of generality, f ′ : A ֒ → B is an inclusion. Then B/A is flat.Let ~x = h x b i b ∈ B be a tuple of variables indexed by B . For every submodule C ⊆ B containing A , let Σ C ( ~x ) be the ∗ -type in ( M, N ) asserting the following: • If b ∈ C , then x b ∈ M and f ( x b ) = g ( b ). • If b ∈ A , then x b = g ′ ( x b ). • If r ∈ R and b ∈ C , then rx b = x rb . • If b, b ′ ∈ C , then x b + b ′ = x b + x b ′ .Then Σ C ( ~x ) is realized in ( M, N ) if and only if there is a morphism h C : C → M such thatthe diagram commutes A g ′ / / ⊆ (cid:15) (cid:15) M f (cid:15) (cid:15) C h C > > ⑥⑥⑥⑥ g | C / / N (14)It suffices to realize Σ B ( ~x ). This type is a directed union of the types { Σ C ( ~x ) : C/A is finitely generated } . By saturation, it suffices to realize the types in this family. Suppose
C/A is finitely generated.Then
C/A injects into
B/A , so
C/A is free by Remark A.7. The sequence0 → A → C → C/A → Z -lessness is a conjunction of first-order axioms, so it is preserved in elementaryequivalence of ordered abelian groups. Theorem A.20.
Let ( K, O ) be a valued field with residue characteristic 0 and Z -less valuegroup. Let f : M ։ N be a surjective morphism of O -modules. Let ∂ : O → N be aderivation. Consider the three-sorted structure ( O , M, N ) with the ring structure on O , themodule structures on M, N , the epimorphism f , and the derivation ∂ . • If the structure ( O , M, N ) is sufficiently saturated and resplendent, then there is aderivation δ : O → M making the diagram commute: M f (cid:15) (cid:15) (cid:15) (cid:15) O δ > > ⑥⑥⑥⑥⑥⑥⑥⑥ ∂ / / N • In general, such a lifting exists after passing to an elementary extension. roof. The two statements are clearly equivalent, by definition of resplendence and exis-tence of resplendent elementary extensions. We prove the second statement. Consider anelementary chain( O , M, N ) = ( O , M , N ) (cid:22) ( O , M , N ) (cid:22) ( O , M , N ) (cid:22) · · · where each structure is saturated over the previous structure. Let f i , ∂ i be the structuremaps in ( O i , M i , N i ). We will recursively build a sequence of derivations δ i : O i → M i +1 such that f i +1 ( δ i ( x )) = ∂ i ( x ) δ i +1 ( x ) = δ i ( x )for i ≥ x ∈ O i . If this can be done successfully, then the union of the δ i ’s is the desiredlifting of ∂ on the structure S i ( O i , M i , N i ), an elementary extension of ( O , M, N ), and weare done.At step i = 0, we must find an O -linear map Ω O / Q → M making the diagram commute M f (cid:15) (cid:15) ✤✤✤ Ω O / Q ❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧ ∂ / / N ⊆ / / N . (15)At step i >
1, we must find an O i -linear map Ω O i / Q → M i +1 making the diagram commute( O i ⊗ O i − Ω O i − / Q ) (cid:15) (cid:15) δ i − / / M i ⊆ / / M i +1 f i +1 (cid:15) (cid:15) Ω O i / Q ✐✐✐✐✐✐✐✐✐✐✐ ∂ i / / N i ⊆ / / N i +1 . (16)The dashed map exists in both cases by Lemma A.19. For (15), Fact A.8 shows that Ω O / Q is flat. For (16), the map O i ⊗ O i − Ω O i − / Q → Ω O i / Q is injective by Lemma A.13, the cokernel is Ω O i / O i − by Fact A.4, and Ω O i / O i − is flat byProposition A.17. Corollary A.21.
Let ( K, ∂ ) be a sufficiently resplendent normalized diffeovalued field. Ifthe value group of K is Z -less, or p -divisible for some p , then ( K, ∂ ) admits a lifting.Proof. If the value group is p -divisible, then it is Z -less. Theorem A.20 allows us to lift thegiven derivation ∂ : O → K/ m to a derivation δ : O → K . This corresponds to an O -linearmap Ω O / Q → K , which in turn yields a K -linear mapΩ K/ Q ∼ = Ω O / Q ⊗ O K → K by Fact A.5. Thus δ : O → K extends to a derivation δ : K → K .92 .5 An unliftable example The assumption that Γ is Z -less is necessary in Corollary A.21. Proposition A.22.
There is a valued field ( K, O , m ) of residue characteristic 0, and aderivation ∂ : O → K/ m , such that in any elementary extension ( K ∗ , O ∗ , m ∗ , ∂ ) (cid:23) ( K, O , m , ∂ ) ,there is no derivation δ : O ∗ → K ∗ making the diagram commute: K ∗ (cid:15) (cid:15) O ∗ δ ; ; ✈✈✈✈✈✈✈✈✈ ∂ / / K ∗ / m ∗ One can even take ( K, O ) to be dp-minimal as a pure valued field.Proof. Let Z + Z ω be the free abelian group on two generators 1 , ω , ordered so that ω > n · n ∈ Z . In other words, Z + Z ω is the lexicographic product Z × Z , with generators ω := (1 ,
0) and 1 := (0 , L be the Hahn field Q alg (( t Z + Z ω )), and let K be the relative algebraic closure of Q ( t, t ω ) in L . Let O L , O K denote the valuation rings on L and K , and m L , m K denotetheir maximal ideals. The valued field ( K, O K ) is henselian with residue characteristic 0,algebraically closed residue field, and dp-minimal value group, so ( K, O K ) is a dp-minimalvalued field.Let val ′ be the coarsening of val by the convex subgroup Z ≤ Z + Z ω , and let p ⊳ O L bethe associated maximal ideal. If val( x ) = i + jω , then val ′ ( x ) = j . Moreover, for any x , x ∈ p ⇐⇒ val ′ ( x ) > ⇐⇒ val( x ) > Z . Note that for x ∈ L , val ′ ( x ) ≥ ⇒ x ∈ m L + Q alg [ t − ] , (17)because one can split x = P i,j a i,j t i + jω as x = X i + jω ≤ a i,j t i + jω + X i + jω> a i,j t i + jω . The assumption val ′ ( x ) ≥ i + jω with j = 0. Thesupport is well-ordered, so the first sum is finite, and belongs to Q alg [ t − ]. The other sum isin m L , proving (17).Choose u = 1 + a t + a t + · · · ∈ t Q alg [[ t ]] ⊆ Q alg (( t )) = Q alg (( t Z )) ⊆ Q alg (( t Z + Z ω )) = L such that u v (mod p ) for all v ∈ K . Such a u exists because 1 + t Q alg [[ t ]] is uncountable, K is countable, and the elements of Q alg (( t )) are pairwise distinct modulo p . (The valuationval ′ restricts to the trivial valuation on Q alg (( t )).)93onsider the derivation ∂ : L → L X i,j a i,j t i + jω X i,j a i,j jt i +( j − ω . Note that for x ∈ L , x ∈ O L = ⇒ val ′ ( ∂ x ) ≥ ∂ : L → L be the derivation ∂ x := u∂ x . Let ∂ be the composition O K ֒ → L ∂ → L → L/ m L . We claim that ∂ factors through the inclusion K/ m K ֒ → L/ m L . Indeed, x ∈ O K = ⇒ x ∈ O L = ⇒ val ′ ( ∂ x ) ≥ ⇐⇒ val ′ ( u∂ x ) ≥ , because val( u ) = 0. By (17),val ′ ( u∂ x ) ≥ ⇒ u∂ x ∈ m L + Q alg [ t − ]= ⇒ ∂x ∈ ( Q alg [ t − ] + m L ) / m L ⊆ ( K + m L ) / m L ∼ = K/ ( K ∩ m K ) = K/ m K . So ∂ is a well-defined derivation from O K to K/ m K .We claim that the following first-order statement σ holds in the structure ( K, O K , ∂ ):There is an a ∈ O K such that for every a ′ ∈ K , there are b, c ∈ O K such that a = bc and for every b ′ , c ′ ∈ K , the following identities do not all hold: b ′ ≡ ∂b (mod m K ) c ′ ≡ ∂c (mod m K ) a ′ = bc ′ + cb ′ . Before proving this, note that this would complete the proof: • The statement σ is first-order, so it remains true in any elementary extension of( K, O , ∂ ). • If the lifting δ : O → K exists, the statement σ is false, because an adversary canchoose a ′ = δab ′ = δbc ′ = δc. and the three equations would hold. 94e now prove σ . For our opening move, we choose a = t ω ∈ O K . The opponent chooses a ′ ∈ K . Note ∂ a = ∂ t ω = 1 ∂ a = u∂ a = u. By choice of u , we know that a ′ − u / ∈ p , so val( a ′ − u ) < n for some n ∈ Z .For our next move, we take b = t n and c = t ω − n . The condition a = bc holds, so wehaven’t lost the game yet. The opponent chooses b ′ , c ′ ∈ K . Suppose that all three identitieshold: b ′ ≡ ∂b (mod m K ) c ′ ≡ ∂c (mod m K ) a ′ = bc ′ + cb ′ . Then ∂ b ≡ b ′ (mod m L ) ∂ b ≡ c ′ (mod m L ) . Now b, c are divisible by t n , so c∂ b ≡ cb ′ (mod t n m L ) b∂ c ≡ bc ′ (mod t n m L )Adding the two equations, and using the identities a ′ = bc ′ + cb ′ ∂ ( bc ) = b∂ c + c∂ b, we obtain ∂ ( bc ) ≡ a ′ (mod t n m L )On the other hand, ∂ ( bc ) = ∂ ( a ) = u, so u ≡ a ′ (mod t n m L ). Then val( u − a ′ ) > n , contradicting the choice of n . So it is impossiblefor all three identities to hold, and we have won the game. This proves the sentence σ andcompletes the proof. Acknowledgments.
The author would like to thank Meng Chen, Hagen Knaf, and Franz-Viktor Kuhlmann for some helpful information on K¨ahler differentials.
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