The freeness property for locally nilpotent derivations of k[x,y,z]
aa r X i v : . [ m a t h . A C ] S e p THE FREENESS PROPERTYFOR LOCALLY NILPOTENT DERIVATIONS OF k[ x, y, z ] DANIEL DAIGLE
Abstract.
We prove Freudenburg’s Freeness Conjecture:
Let B be the polynomialring in three variables over a field of characteristic zero, let D : B → B be a nonzerolocally nilpotent derivation, and let A = ker( D ) . Then B is a free A -module, andthere exists a basis ( e i ) i ∈ N of B such that deg D ( e i ) = i for all i ∈ N . Introduction
Let B be an integral domain of characteristic zero. A derivation D : B → B is saidto be locally nilpotent if, for each b ∈ B , there exists n > D n ( b ) = 0. Wewrite LND( B ) for the set of locally nilpotent derivations D : B → B .If D ∈ LND( B ) \ { } , define a map deg D : B → N ∪ {−∞} by declaring thatdeg D (0) = −∞ and deg D ( b ) = max (cid:8) n ∈ N | D n ( b ) = 0 (cid:9) for all b ∈ B \ { } . Themap deg D is called the degree function associated to D ; it is well known that deg D hasthe following properties (see for instance [Dai, Prop. 4.8]): • deg D ( b ) = −∞ if and only if b = 0; • deg D ( b b ) = deg D ( b ) + deg D ( b ) for all b , b ∈ B ; • deg D ( b + b ) ≤ max (cid:0) deg D ( b ) , deg D ( b ) (cid:1) .A derivation D ∈ LND( B ) \ { } is said to have the freeness property if B is a freemodule over the ring A = ker( D ) = (cid:8) x ∈ B | D ( x ) = 0 (cid:9) , and there exists a basis( e i ) i ∈ N (of the A -module B ) satisfying deg D ( e i ) = i for all i ∈ N .Consider the polynomial ring B = k [ x , . . . , x n ], where k is a field of characteristiczero. It is easy to see that the statement( ∗ ) every D ∈ LND( B ) \ { } has the freeness property is true for n < n >
3. In section 9.4 of [Fre19], Gene Freudenburgconjectures that ( ∗ ) is true for n = 3. The aim of the present article is to prove thatFreudenburg’s conjecture is true: Main Theorem.
Let B be the polynomial ring in three variables over a field of char-acteristic zero, let D : B → B be a nonzero locally nilpotent derivation, and let A = ker( D ) . Then B is a free A -module, and there exists a basis ( e i ) i ∈ N of B such that deg D ( e i ) = i for all i ∈ N . Mathematics Subject Classification.
Primary: 13N15. Secondary: 14R10, 14R20.
Key words and phrases.
Locally nilpotent derivation, polynomial ring, affine space, free module.
Conventions.
The word “ring” means commutative, associative, unital ring. We use“ \ ” for set difference. If B is an algebra over a field k , the notation B = k [ n ] (where n ∈ N ) means that B is isomorphic as a k -algebra to a polynomial ring in n variablesover k . We adopt the convention that 0 ∈ N .1. Preliminaries
Recall the following (see for instance Tag 00NX in the Stacks Project):1.1.
Lemma.
For a finitely presented module M over a ring A , the following areequivalent: (a) M is flat; (b) M is projective; (c) for every maximal ideal m of A , M m is a free A m -module. Theorem ([Qui76], [Sus76]) . If A is a polynomial ring in finitely many variablesover a field, then every finitely generated projective A -module is free. Corollary.
Let A be a polynomial ring in finitely many variables over a field, let M be a finitely generated A -module, and let A → A ′ be a faithfully flat homomorphismof rings. If A ′ ⊗ A M is a flat A ′ -module, then M is a free A -module.Proof. Since A → A ′ is faithfully flat and A ′ ⊗ A M is a flat A ′ -module, M is a flat A -module by [Mat80, (4.E)]. Since M is finitely presented, M is a projective A -moduleby Lemma 1.1, so a free A -module by Thm 1.2. (cid:3) Let A be a subring of an integral domain B . One says that A is factorially closed in B if the conditions x, y ∈ B \ { } and xy ∈ A imply that x, y ∈ A . The following iswell known (see [Fre17, Pr. 1] or [Dai, Cor. 5.3]):1.4. Lemma. If B is an integral domain of characteristic zero and D ∈ LND( B ) then ker( D ) is factorially closed in B . The next fact is proved in [Miy85] under the assumption that k = C ; the generalcase follows by a straightforward application of Kambayashi’s theorem [Kam75].1.5. Theorem (Miyanishi) . If k is a field of characteristic zero and B = k [3] , then ker( D ) = k [2] for every D ∈ LND( B ) \ { } . Setup.
Let B be an integral domain of characteristic zero, D ∈ LND( B ) \{ } , and A = ker( D ). For each n ∈ N , define F n = (cid:8) b ∈ B | deg D ( b ) ≤ n (cid:9) and I n = A ∩ D n ( B );note that F n is an A -module, that I n is an ideal of A , and that A = F ⊆ F ⊆ F ⊆ · · · and A = I ⊇ I ⊇ I ⊇ · · · . We also define Gr D ( B ) = L n ∈ N F n / F n − , where F − = 0. It is well known that Gr D ( B )is a graded integral domain and that D induces a nonzero locally nilpotent derivationGr D ( D ) : Gr D ( B ) → Gr D ( B ) whose kernel is A = F / F − . HE FREENESS PROPERTY FOR LOCALLY NILPOTENT DERIVATIONS 3
Lemma.
With assumptions and notations as in , the following hold. (a)
For each n ∈ N , there is an isomorphism of A -modules F n / F n − ∼ = I n . (b) If A is Noetherian then, for each n ∈ N , F n is finitely generated as an A -module.Proof. (a) follows by observing that D n : F n → B is an A -linear map with kernel F n − and image I n . To prove (b), let n ≥ A is Noetherian) the ideal I n is finitely generated; then, by (a), F n / F n − is finitely generated as an A -module;considering the exact sequence of A -linear maps 0 → F n − → F n → F n / F n − →
0, wesee that if F n − is finitely generated then so is F n . So (b) follows by induction. (cid:3) Lemma.
Let the assumptions and notations be as in and consider the followingconditions: (a) D has the freeness property; (b) for each n ∈ N , F n / F n − is a free A -module; (c) for each n ∈ N , I n is a principal ideal of A ; (d) Gr D ( B ) is a free A -module.Then (a) ⇔ (b) ⇔ (c) ⇒ (d) . Moreover, we have (c) ⇔ (d) whenever A is a polynomialring in finitely many variables over a field.Proof. Since D = 0, there exists s ∈ B satisfying D ( s ) = 0 and D ( s ) = 0; thendeg D ( s ) = 1, so deg D ( s n ) = n for all n ∈ N , so F n / F n − = 0 and I n = 0 for all n ∈ N .Since (by Lemma 1.7) F n / F n − ∼ = I n for all n ∈ N , it is clear that (b) ⇔ (c).If (a) is true then there exists a basis ( e i ) i ∈ N of the A -module B such that deg D ( e i ) = i for all i ∈ N . Let i ∈ N . Let ¯ e i ∈ F i / F i − denote the canonical image of e i . Then { ¯ e i } generates the A -module F i / F i − . If a ∈ A \{ } , then deg D ( ae i ) = deg D ( a )+deg D ( e i ) = i , so a ¯ e i = 0, which shows that { ¯ e i } is a basis of the A -module F i / F i − . So (a) implies(b). It is also clear that (¯ e i ) i ∈ N is a basis of the A -module Gr D ( B ), so (a) implies (d).Suppose that (c) is true. For each i ∈ N , I i is a free A -module of rank 1 (since I i = 0), so F i / F i − ∼ = I i is free of rank 1. Choose e i ∈ F i such that { ¯ e i } is a basis of F i / F i − (where ¯ e i ∈ F i / F i − is the canonical image of e i ). Then it is easily verifiedthat ( e i ) i ∈ N is a basis of the A -module B , and clearly deg D ( e i ) = i for all i ∈ N . So (c)implies (a).Assume that A is a polynomial ring over a field. Suppose that (d) is true and let n ∈ N . By Lemma 1.7, F n / F n − is a finitely generated A -module; since F n / F n − isa direct summand of the free A -module Gr D ( B ), F n / F n − is projective, hence free byThm 1.2. So (d) implies (b). (cid:3) Proof of the main theorem
Let k be a field of characteristic zero, B = k [3] , D ∈ LND( B ) \ { } , and A = ker D .It has to be shown that D has the freeness property. We use the notations of 1.6. Note DANIEL DAIGLE that A = k [2] by Thm 1.5. Lemma 1.7 implies that, for each n ∈ N , the A -module F n is finitely generated. We begin by proving:2.1 . If k is algebraically closed then, for each n ∈ N , the A -module F n / F n − is free.Proof. Let n ∈ N . If n = 0 then the claim is trivial, so assume that n ≥
1. It suffices toshow that ( F n / F n − ) m is a free A m -module for every maximal ideal m of A . Indeed, ifthis is the case then Lemma 1.1 implies that F n / F n − is a projective A -module, hencea free A -module by Thm 1.2.Let m be a maximal ideal of A . Let us choose p ∈ m such that A/pA = k [1] (we have A = k [2] , and if we write A = k [ f, g ] then the fact that k is algebraically closed impliesthat m = ( f − λ, g − µ ) for some λ, µ ∈ k ; then p = f − λ has the desired property).Let π : B → B/pB be the canonical homomorphism of the quotient ring. Given b ∈ B ,write ¯ b = π ( b ); let ¯ A = π ( A ) and ¯ F m = π ( F m ) for all m .As A is factorially closed in B by Lemma 1.4, we obtain that p is irreducible in B andthat A ∩ pB = pA ; so B/pB is a domain and ¯ A = π ( A ) ∼ = A/ ( A ∩ pB ) = A/pA = k [1] is a PID.As F n is a finitely generated A -module by Lemma 1.7, ¯ F n is a finitely generated¯ A -module. Moreover, the ¯ A -module ¯ F n is torsion-free, since ¯ A and ¯ F n are included inthe integral domain B/pB . So ¯ F n is a free ¯ A -module of finite rank. Let us prove:(1) If ( v , . . . , v s ) is a family of elements of F n such that (¯ v , . . . , ¯ v s ) is a basisof the ¯ A -module ¯ F n , then ( v , . . . , v s ) is a basis of the A m -module ( F n ) m . Consider the A -submodule M of F n generated by v , . . . , v s .If ( v , . . . , v s ) is linearly dependent over A then there exist a , . . . , a s ∈ A not all zeroand such that P si =1 a i v i = 0; we may choose the a i so that a j / ∈ pA is true for at leastone j ; then a j / ∈ pB because A ∩ pB = pA , so ¯ a j = 0 and consequently P si =1 ¯ a i ¯ v i = 0,a contradiction. So ( v , . . . , v s ) is linearly independent over A and consequently M isa free A -module with basis ( v , . . . , v s ).The definition of M together with F n ∩ pB = p F n implies that F n = p F n + M. It follows that the A -module N = F n /M satisfies N = pN . Consequently, the A m -module N m satisfies N m = pN m . Since p ∈ m , p belongs to the maximal ideal of A m .Then Nakayama’s Lemma implies that N m = 0, so ( F n ) m = M m , so ( F n ) m is a free A m -module with basis ( v , . . . , v s ). This proves (1).Observe that ¯ F n is a free ¯ A -module of finite rank, and that ¯ F n − is a submodule of ¯ F n .By Thm 7.8 in Chap. III of [Lan93], there exists a basis ( w , . . . , w s ) of ¯ F n , an integer r such that 0 ≤ r ≤ s , and elements α , . . . , α r of ¯ A such that ( α w , . . . , α r w r ) is abasis of ¯ F n − over ¯ A . Choose v , . . . , v s ∈ F n such that ¯ v i = w i for all i = 1 , . . . , s , andchoose a , . . . , a r ∈ A such that ¯ a i = α i for all i = 1 , . . . , r . Applying (1) to both F n and F n − , we obtain that ( v , . . . , v s ) is a basis of ( F n ) m over A m and that ( a v , . . . , a r v r ) HE FREENESS PROPERTY FOR LOCALLY NILPOTENT DERIVATIONS 5 is a basis of ( F n − ) m over A m . For each i ∈ { , . . . , r } , the fact that a i v i ∈ ( F n − ) m implies that there exists a ′ i ∈ A \ m such that a ′ i a i v i ∈ F n − ; we have a ′ i , a i ∈ A \ { } ,so deg D ( a ′ i ) = 0 = deg D ( a i ), so n > deg D ( a ′ i a i v i ) = deg D ( a ′ i ) + deg D ( a i ) + deg D ( v i ) =deg D ( v i ), so v , . . . , v r ∈ F n − . It follows that ( v , . . . , v r ) is a basis of ( F n − ) m over A m . Consequently, ( F n / F n − ) m is a free A m -module. As explained at the beginning ofthe proof, we are done. (cid:3) Let us now prove the Main Theorem.Let ¯ k be an algebraic closure of k and let ¯ A = ¯ k ⊗ k A . Since A → ¯ A is a faithfully flathomomorphism of rings, ¯ A ⊗ A ( ) is an exact functor from the category of A -modulesto the category of ¯ A -modules. Let ¯ B = ¯ A ⊗ A B = ¯ k ⊗ k A ⊗ A B = ¯ k ⊗ k B = ¯ k [3] ,and note that applying ¯ A ⊗ A ( ) to the A -linear map D : B → B gives an ¯ A -linearmap ¯ D : ¯ B → ¯ B . As is well known, ¯ D is a locally nilpotent derivation of ¯ B and¯ D = 0. Define ¯ F n = ¯ A ⊗ A F n for all n ≥ −
1. Let n ∈ N ; applying ¯ A ⊗ A ( ) to theexact sequence of A -linear maps 0 → F n → B D n +1 −−−→ B gives the exact sequence of¯ A -linear maps 0 → ¯ F n → ¯ B ¯ D n +1 −−−→ ¯ B . Consequently, ¯ F n = (cid:8) x ∈ ¯ B | deg ¯ D ( x ) ≤ n (cid:9) for all n ≥
0. This together with 2.1 implies that ¯ F n / ¯ F n − is a free ¯ A -module (forevery n ∈ N ). We have ¯ A ⊗ A ( F n / F n − ) ∼ = ¯ F n / ¯ F n − by exactness of ¯ A ⊗ A ( ), so¯ A ⊗ A ( F n / F n − ) is a free (hence flat) ¯ A -module. So F n / F n − is a free A -module byCor. 1.3. Since this is true for every n ∈ N , Lemma 1.8 implies that D has the freenessproperty. The Theorem is proved. References [Dai] D. Daigle. Introduction to locally nilpotent derivations. Informal lecture notes prepared in2010, available at http://aix1.uottawa.ca/~ddaigle [Fre17] G. Freudenburg.
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