Exponential ideals and a Nullstellensatz
aa r X i v : . [ m a t h . A C ] A p r EXPONENTIAL IDEALS AND A NULLSTELLENSATZ
FRANC¸ OISE POINT ( † ) AND NATHALIE REGNAULT
Abstract.
We prove a version of a Nullstellensatz for partial exponential fields(
K, E ), even though the ring of exponential polynomials K [ X ] E is not a Hilbertring. We show that under certain natural conditions one can embed an ideal of K [ X ] E into an exponential ideal. In case the ideal consists of exponential polyno-mials with one iteration of the exponential function, we show that these conditionscan be met. We apply our results to the case of ordered exponential fields. MSC classification: primary: 03C60 (secondary: 12L12, 12D15).Key words: exponential ideals, Nullstellensatz, partial exponential fields.1.
Introduction
There is an extensive literature on
Nullstellens¨atze for expansions of fields (ordered,differential, p -valued) and several versions of abstract Nullstellens¨atze attempting toencompass those cases in a general framework (see for instance [14], [18]). When K is a (pure) algebraically closed field, Hilbert’s Nullstellensatz establishes the equiva-lence between the following two properties: a system of polynomial equations (withcoefficients in K ) has a common solution (in K ) and the ideal generated by thesepolynomials is nontrivial in the polynomial ring K [ x , . . . , x n ].In this note we will give a version of a Nullstellensatz for exponential fields, namelyfields ( K, E ) endowed with an exponential function E , where E is possibly partiallydefined. In [13], K. Manders investigated the notion of exponential ideals and pin-pointed several obstructions to develop an analog of the classical Nullstenllensatz.Let us readily note the following differences. In (ordinary) polynomial rings, in orderto prove a Nullstellensatz [6], one may proceed as follows. Any field K has an alge-braic closure and given a maximal ideal M of K [ X ], the field K [ X ] /M is algebraicover K . Using that K [ X ], with X := ( X , . . . , X n ), is a Hilbert ring, namely a ringwhere any prime ideal is an intersection of maximal ideals, one proves by inductionon n ≥
1, that K [ X ] /M is algebraic over K , for M a maximal ideal of K [ X ] [6,Corollary]. In contrast, the ring of exponential polynomials is not a Hilbert ring (seesubsection 2.4). However there is a notion of E -algebraic closure over a subfield F ofan exponential field F and that closure operation gives rise to a dimension functionon definable subsets [8]. An element a ∈ F belongs to the E -algebraic closure of F ,whenever one can find a tuple of elements ( a, a , . . . , a n ) with the property that it isa regular zero of some E -variety defined over F [12]. One shows that the E -algebraic Date : April 23, 2020.( † ) Research Director at the ”Fonds de la Recherche Scientifique (F.R.S.-F.N.R.S.)”. closure has the exchange property by relating it, as for the classical algebraic closure,to the vanishing of ( E )-derivations [8]. Let us now describe our main result.Let ( R, E ) be an exponential field, namely a field endowed with a morphism E from its additive group ( R, + ,
0) to its multiplicative group ( R ∗ , .,
1) with E (0) = 1.Using the construction of the free E -ring R [ X ] E of exponential polynomials over R on X as an increasing union of group rings R ℓ with R := R [ X ] [4], we show thefollowing. Theorem (Later Theorem 3.10) Let f , . . . , f m , g ∈ R [ X ] E . and let ℓ be minimalsuch that f , . . . , f m , g ∈ R ℓ . Suppose that there is a maximal ideal M of R ℓ con-taining f , . . . , f m and such that for any h ∈ R ℓ − ∩ M , E ( h ) − ∈ M ( † ). Assumethat g vanishes at each common zero of f , . . . , f m in any partial exponential fieldcontaining ( R, E ). Then g belongs to M .In case f , . . . , f m , g ∈ R , we get a stronger result which follows from the theoremabove and of the construction of exponential ideals (Proposition 3.5). Corollary (Later Corollary 3.11) Assume that f , . . . , f m , g ∈ R and let I be theideal of R generated by f , . . . , f m . Assume that g vanishes at each common zeroof f , . . . , f m in any partial exponential field containing ( R, E ). Then g belongs to J ( I ), the Jacobson radical of I .There are some known limitations to improve the above result. P. D’Aquino, A.Fornasiero, G. Terzo observed that in the field of complex numbers endowed withthe classical exponential function exp , the following fails [3]. Let c ∈ C \ exp (2 π Z ).Consider f ( X ) = exp ( X ) − c and g ( X ) = exp ( iX ) −
1. Let I be the ideal in C [ X ] E generated by f and g . Then even though I is a nontrivial ideal of C [ X ] E , f and g have no common zero in C (or in any pseudo-exponential field (as defined by B.Zilber) containing C ).The paper is organized as follows. In section 2, we recall the construction of freeexponential rings (and the complexity function they can be endowed with), exponen-tial ideals and the E -algebraic closure (ecl). We note that free exponential rings arenot Hilbert rings.In section 3, we prove our main result (Theorem 3.10) by constructing step by stepmaximal exponential ideals, containing a given proper ideal under condition ( † ). Wealso apply our techniques to the ordered case, considering real exponential ideals. Acknowledgments:
These results are part of the PhD thesis of Nathalie Regnault[16]. 2.
Preliminaries
Throughout, all our rings R will be commutative rings of characteristic 0 withidentity 1. Let N ∗ := N \ { } , R ∗ := R \ { } . We will work in the language of rings L rings := { + , ., − , , } augmented by a unary function E ; let L E := L rings ∪ { E } . Definition 2.1. [4] An E -ring ( R, E ) is a ring R equipped with a map E : ( R, + , → ( R ∗ , · ,
1) satisfying E (0) = 1 and ∀ x ∀ y ( E ( x + y ) = E ( x ) .E ( y )). XPONENTIAL IDEALS AND A NULLSTELLENSATZ 3 An E -domain is an E -ring with no zero-divisors; an E -field is an E -domain whichis a field.We will need to work also with partial E -rings, namely rings where the exponentialfunction is only partially defined and so the corresponding language contains a unarypredicate for the domain of the exponential function. Definition 2.2.
A partial E -ring is a two-sorted structure ( R, A ( R ) , + R , · , + A , E ),where ( A ( R ) , + A ,
0) is a monoid and E : ( A ( R ) , + A , → ( R, · ,
1) is a morphism ofmonoids. We identify A ( R ) with a subset of R . We get the corresponding notions ofpartial E -domains and partial E -fields.Let R be a partial E -ring, then E ( A ( R )) is a subset of the set of invertible elementsof R .The above definition of partial E -rings is basically the one of [8], except that in[8, Definition 2.2], one requires in addition that R is a Q -algebra, A ( R ) is a groupand also a Q -vector space and as such endowed with scalar multiplication · q for each q ∈ Q . Examples 2.1.
We recall below classical examples of exponential rings and fieldswhere the exponential function is possibly only partially defined.(1) Let R be the field of real numbers (respectively C the field of complexes)endowed with the exponential function exp ( x ) := P i ≥ x i i ! .(2) Let Z p be the ring of p -adic integers endowed with the exponential function x exp ( px ), p > Z be the ring of integers endowed with the exponential function x x only defined on the positive integers.(4) Let Q p be the field of p -adic numbers (respectively C p the completion ofthe algebraic closure of Q p ) endowed with the exponential function exp ( px )restricted to Z p (respectively to O p the valuation ring of C p ), p > K, E ) be a partial E -field. Consider the field of Laurent series K (( t )). Write r ∈ K [[ t ]] as r + r where r ∈ K and r ∈ t.K [[ t ]]. For r ∈ A ( K ), extend E on K [[ t ]] as follows: E ( r + r ) := E ( r ) .exp ( r ). By Neumann’s Lemma, the series exp ( r ) ∈ K (( t )) [5, chapter 8, section 5, Lemma]. So K (( t )) can be endowed with astructure of a partial E -field.More generally let G be an abelian totally ordered group and consider the Hahnfield K (( G )). Let M v denote the maximal ideal of K (( G )). Then K (( G )) canbe endowed with a structure of a partial E -field defining E on the elements r of thevaluation ring which can be decomposed as r + r with r ∈ A ( K ) and r ∈ M v . Then exp ( r ) ∈ K (( G )) by Neumann’s lemma cited above and E ( r + r ) := E ( r ) .exp ( r ),for r ∈ A ( K ).2.1. Free exponential rings.
For the convenience of the reader not familiar withthe construction of the ring of exponential polynomials, we recall its constructionbelow.
POINT AND REGNAULT
The construction of free E -rings is due to B. Dahn, on X := X , . . . , X n , denotedby Z [ X ] E , or over an E -ring ( R, E ), denoted by R [ X ] E ) (see for instance in [4], [12]).(When n = 1, we will use the variable X .)The elements of these rings are called E -polynomials in the indeterminates X .The ring R [ X ] E is constructed by stages as follows: let R − := R , R := R [ X ] and A the ideal generated by X in R [ X ]. We have R = R ⊕ A . For k ≥
0, let t A k be amultiplicative copy of the additive group A k .Then, set R k +1 := R k [ t A k ] and let A k +1 be the free R k − submodule generated by t a with a ∈ A k − { } . Then R k +1 = R k ⊕ A k +1 .By induction on k ≥
0, one shows the following isomorphism: R k +1 ∼ = R [ t A ⊕···⊕ A k ] , using the fact that R [ t A ⊕···⊕ A k ] ∼ = R [ t A ⊕···⊕ A k − ][ t A k ] [12, Lemma 2].We define the map E − : R − → R as the map E on R composed by the embeddingof R − into R . Then for k ≥
0, we define the map E k : R k → R k +1 as follows: E k ( r ′ + a ) = E k − ( r ′ ) .t a , where r ′ ∈ R k − and a ∈ A k . Since R k − and A k are indirect summand this is well-defined. Moreover for k ≤ ℓ , E ℓ extends E k on R k .Finally let R [ X ] E := S k ≥ R k and extend E on R [ X ] E (and by abuse of notationwe will still denote it by E ) by setting E ( f ) := E k ( f ) for f ∈ R k .Using the construction of R [ X ] E as an increasing union of group rings, one candefine on the elements of R [ X ] E an analogue of the degree function for ordinarypolynomials which measures the complexity of the elements; it will take its valuesin the class On of ordinals and was described for instance in [4, 1.9] for exponentialpolynomials in one variable. We would like it to be defined for exponential polyno-mials in more than one variable and so we slightly adapt the construction given in[4].For p ∈ R [ X ], let us denote by totdeg X ( p ) the total degree of p , namely themaximum of { P nj =1 i j : for each monomial X i . . . X i n n occurring (nontrivially) in p } .Then one defines a height function h (with values in N ) which detects at whichstage of the construction the (non-zero) element is introduced.Let p ( X ) ∈ R [ X ] E , then h ( p ( X )) = k , if p ∈ R k \ R k − , k > h ( p ( X )) = 0 if p ∈ R [ X ]. Using the freeness of the construction, one defines a function rkrk : R [ X ] E → N :If p = 0, set rk ( p ) := 0,if p ∈ R [ X ] \ { } , set rk ( p ) := totdeg X ( p ) + 1 andif p ∈ R k \ R k − , k >
0, let p = P di =1 r i .E ( a i ), where r i ∈ R k − , a i ∈ A k − \ { } .Set rk ( p ) := d .Finally, one defines the complexity function ordord : R [ X ] E → On as follows. For k ≥
0, write p ∈ R k as p = P ki =0 p i with p ∈ R , p i ∈ A i , 1 ≤ i ≤ k .Define ord ( p ) := P ki =0 ω i .rk ( p i ). XPONENTIAL IDEALS AND A NULLSTELLENSATZ 5
Remark 2.1.
Let p ∈ R k \ R k − , k ≥ p = P ki =1 p i with p i ∈ A i ,1 ≤ i ≤ k . Then there is q ∈ R [ X ] E such that ord ( E ( q ) .p ) < ord ( p ) (the proof isexactly the same as the one in [4, Lemma 1.10]).Note that if we had started with R a partial E -ring, then we can endow the ringof E -polynomials R [ X ] E with a structure of a partial E -ring, defining for f ∈ R k ,written as f + f , with f ∈ A ( R ) and f ∈ A ⊕ A ⊕ · · · ⊕ A k , we define E ( f )as E ( f ) .t f ∈ R k +1 , k ≥ −
1. Also, if we start with R a Q -algebra, then A ( R ) isendowed with a Q -vector space structure.Similarly one can define the construction of free E -rings on countably many in-determinates X ij , 1 ≤ i ≤ n , j ∈ ω in order to define differential E -polynomials.Suppose now that D is a derivative on R . We set D ( X ij ) := X i ( j +1) , we identify X i with X i . and we denote the ring of differential polynomials in X := X , . . . , X n by R { X } . We will denote the corresponding E -ring by R { X } E .2.2. Exponential ideals.Definition 2.3.
Let (
R, E ) be an E -ring (respectively a partial E -ring).An exponential ideal (or E -ideal) J of ( R, E ) is an ideal such that for any h ∈ J , E ( h ) − ∈ J . In case ( R, E ) is a partial E -ring, J is an exponential ideal if for all h ∈ J ∩ A ( R ), E ( h ) − ∈ J .An example of an E -ideal is the set of exponential polynomials in R [ X ] E whichvanish on a subset of R n .Let J be an exponential ideal of ( R, E ). Then the ring
R/J can be endowed withan exponential function E J : R/J → R/J sending g + J to E ( g ) + J , g ∈ R . This iswell-defined since for h, h ′ ∈ J , E ( h ) − E ( h ′ ) = ( E ( h − h ′ ) − .E ( h ′ ) ∈ J . Definition 2.4. An E -derivation D on ( R, E ) is a derivation on R which satisfiesin addition ∀ x D ( E ( x )) = E ( x ) .D ( x ) ( ⋆ ). If R is a partial E -ring, we make theconvention that the equation ( ⋆ ) is only satisfied for the elements x ∈ A ( R ).Assume that we have an E -derivation on R . A differential ideal I in R is an ideal I with the property that f ∈ I implies that D ( f ) ∈ I .Let ( R, E, D ) be an E -ring endowed with an E -derivation and let J be an expo-nential differential ideal of ( R, E, D ), then the ring
R/J is again an E -differentialring.For p ( X ) ∈ R { X } E , we extend D by setting D ( E ( p )) := D ( p ) .E ( p ). Then usingthe function ord which measures the complexity of p ( X ), one can show that R { X } E isagain an E -ring endowed with an E -derivation. Denote by ∂ X j p the partial derivativeof p with respect to the variable X j , 1 ≤ j ≤ n . Lemma 2.5. [15, Lemma 2.6]
Let p ∈ R [ X ] E . Then there exist p δ ∈ R [ X ] E suchthat D ( p ( X )) = P nj =1 D ( X j ) ∂ X j p + p δ and the elements p δ have the property that if D is trivial on R , then p δ = 0 . POINT AND REGNAULT
Notation 2.1.
Given f , . . . , f n ∈ R [ X ] E , ¯ f := ( f , . . . , f n ), we will denote by J ¯ f ( X ),the exponential polynomial:= (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂ X f · · · ∂ X n f ... . . . ... ∂ X f n · · · ∂ X n f n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Definition 2.6. [8, Definition 3.1] Let (
B, E ) ⊆ ( R, E ) be partial E-domains. A
Khovanskii system over B is a quantifier-free L E ( B )-formula of the form H ¯ f ( x , . . . , x n ) := n ^ i =1 f i ( x , . . . , x n ) = 0 ∧ J ¯ f ( x , . . . , x n ) = 0 , for some f , . . . , f n ∈ B [ X ] E . Let a ∈ R . Then a ∈ ecl R ( B ) if H ¯ f ( a , . . . , a n ) holds for some a , . . . , a n ∈ R with a = a , where H ¯ f is a Khovanskii system, f , . . . , f n ∈ B [ X ] E (assuming that a i ∈ A ( R ), 1 ≤ i ≤ n , if necessary for the f ′ i s to be defined).As recalled in the introduction, ecl is a well-behaved E -algebraic closure operator[12], [8, Lemma 3.3], satisfying the exchange property [17], [8, Theorem 1.1]. A.Wilkie used it in his proof of the model-completeness of ( R , exp ) the field of reals withthe exponential function. Then J. Kirby extracted ecl from this o-minimal setting andshowed that it coincides with another closure operator cl defined using E -derivations[8, Propositions 4.7, 7.1]. The fact that cl is a closure operator satisfying the exchangeproperty is relatively easy. So the operator cl on subsets of an E -field F induces apregeometry and we get a notion of dimension dim. In order to show that cl ⊆ ecl,one uses Ax’s result on the Schanuel property for differential fields of characteristic0 [8, Theorem 5.1]. Lemma 2.7. [15, Lemma 2.14]
Let ( F , E ) ⊆ ( F , E ) be two partial E -fields andsuppose we have an E -derivation D on F . Then we can extend D in a unique wayon ecl F ( F ) . (cid:3) Lemma 2.8.
Let ( F , E ) ⊆ ( F , E ) be two partial E -fields and suppose we have an E -derivation D on F . Assume that c ∈ F is such that there is an E -polynomial p ( X ) ∈ F [ X ] E such that p ( c ) = 0 ( X a single variable). Then we can extend D toan E -derivation on c in a unique way. Proof: This statement follows from Kirby’s results on ecl [8, Proposition 4.7] andcan be proven as the preceding lemma. However, we will give an alternative proofhere that only uses the complexity function ord on F [ X ] E . Recall that in section2.1, we have defined the ring of exponential polynomials F [ X ] E by induction setting R = F [ X ] = F ⊕ A and R i := R i − [ t A i − ] = R i − ⊕ A i , i > ord ( p ) is minimal such that p ( c ) = 0. Write p as: p = p + P ki =1 p i , with p ∈ F [ X ] and p i ∈ A i , i > p is non trivial. Suppose otherwise that p = 0, thenby Lemma 1.10 in [4] (see also Remark 2.1), there exists q ∈ F [ X ] E such that ord ( E ( q ) .p ) < ord ( p ). Since if p ( c ) = 0, then E ( q ( c )) .p ( c ) = 0, we get a contradiction.We may assume further that p is a monic polynomial. XPONENTIAL IDEALS AND A NULLSTELLENSATZ 7
By Lemma 2.5, D ( p ( u )) = D ( u ) ∂p ( u ) + p δ ( u ) and p δ have the property that if D is trivial on R , then p δ = 0. Since p = 0, ord ( ∂p ) < ord ( p ) since X a single variable.So, by minimality of p , ∂p ( u ) = 0. If now we assume D to be trivial on F , p δ = 0and so D ( u ) = 0. (cid:3) Let P be a prime E -ideal of F [ X ] E and consider the partial E -domain F [ X ] E / P .Let F be the fraction field of F [ X ] E / P . It is a partial E -field. Lemma 2.9.
Let P be a prime E -ideal of F [ X ] E and let F be the fraction field of F [ X ] E / P . Then F is included in ecl F ( F ) . Proof: Let u := X + P ∈ F [ X ] E / P . Then let p ( X ) ∈ P of minimal order, so wehave p ( u ) = 0 and w.l.o.g. we may assume that p is monic. Let D be an E -derivationon F and consider the partial E -domain extension F [ X ] E / P of F . Then by Lemmaabove D ( u ) is completely determined and if D is trivial on F , then D ( u ) = 0. So, u ∈ cl F ( F ) and by [8, Proposition 7.1], u ∈ ecl F ( F ). Since, ecl F ( F ) is a partial E -subfield containing u [8, Lemma 3.3], it contains F . (cid:3) Remark 2.2.
The above lemma could also have been proven using [12, Theorem 15and Corollary]. In case the length of X is bigger than 1, this theorem of Macintyrealso implies that dim( F [ X ] E / P ) < n . Set a i := X i + P and suppose by the way ofcontradiction that a , . . . , a n are ecl-independent. Then P is an E -ideal closed underall partial E -derivations and so it is equal to F [ X ] E , a contradiction.2.3. Group rings and augmentation ideals.
Now let S be any ring of character-istic 0 (not necessarily a polynomial ring) and let G be a torsion-free abelian group.Recall that the torsion-free rank of G is the dimension of the Q -vector space G ⊗ Q . Definition 2.10.
We consider the group ring S := S [ G ] and we define a map φ a from S → S : P i r i .g i → P i r i , with g i ∈ G , r i ∈ S . The kernel of the map φ a iscalled the augmentation ideal of S .Recall that the augmentation ideal is generated by elements of the form g − g ∈ G (write P i r i .g i as P i r i . ( g i −
1) + P i r i ).2.4. Hilbert rings.
The Jacobson radical J ( R ) of a ring R is by definition theintersection of all maximal ideals of R . It is a definable subset of R , namely J ( R ) = { u ∈ R : ∀ z ∃ y (1 + u.z ) .y = 1 } . Given an ideal I of R , we denote by J ( I )the intersection of all maximal ideals that contain I , so this is equal to { u ∈ R : ∀ z ∃ y (1 + u.z ) .y − ∈ I } . We denote by rad ( I ) the intersection of all prime ideals of R that contain I and one shows that rad ( I ) = { u ∈ R : ∃ n ∈ N u n ∈ I } .A Hilbert ring (also called Jacobson ring) is a ring R where any prime ideal is theintersection of maximal ideals. Therefore for any ideal I , rad ( I ) = J ( I ).The terminology of Hilbert ring comes from the fact that in a Hilbert ring, HilbertNullstellensatz holds, namely if a polynomial vanishes at every zero of an ideal I ,then f ∈ rad ( I ).An easy observation is that if R is a Hilbert ring, then for any ideal I of R , then R/I is also a Hilbert ring [6, Theorem 1].
POINT AND REGNAULT
Another property of a Hilbert ring is that R is Hilbert if and only if every maximalideal of the polynomial ring R [ X ] contracts to a maximal ideal of R [6, Theorem 5].One can also define the Rabinowitsch ’s spectrum
RSpec ( A ) of a ring A , namelythe set of prime ideals of A of the form A ∩ M , where M is a maximal ideal of A [ Y ].Then one can show that rad ( I ) is equal to T P ∈ RSpec ( A ) ,I ⊆ P P [7, Proposition 1.11].Following a result of Krull for polynomial rings, one can show [9, Proposition 1],that if F is a field and G a group of torsion-free rank α ≥ ω , then the group ring F [ G ] is a Hilbert ring if and only if | F | > α . When F is not a field, a necessary andsufficient condition was obtained by Krempa and Okninski [9, Theorem 4]. Let usstate below part of their result. Theorem 2.11. [9, Theorem 4]
Let G be a group of infinite torsion-free rank α andlet A be a ring. If A [ G ] is a Hilbert ring, then all homomorphic images of A havecardinality exceeding α . Corollary 2.12.
The ring R [ X ] E is not a Hilbert ring. Proof: Let R be an E -ring and consider the group rings R ℓ , ℓ ≥
1. Recall that R ℓ ∼ = R [ t A ⊕···⊕ A ℓ ] (see section 2.1) and R [ X ] E ∼ = R [ t L ℓ ≥ A ℓ ]. Let α ℓ be the torsion-free rank of the (multiplicative) group t A ⊕···⊕ A ℓ . Then all the homomorphic imagesof R do not have cardinality > α ℓ and so the first condition of Theorem 4 in [9] failsand so R ℓ is not a Hilbert ring. A similar reasoning applies for R [ X ] E . (cid:3) Exponential ideals and a Nullstellensatz
Embedding an E -ideal into a maximal ideal that is an E -ideal.Notation 3.1. Let B be a ring of characteristic 0, let G be a torsion-free abeliangroup and let B be the group ring B [ G ]. Let I be an ideal of B . Then composethe augmentation map φ a : B → B with the map sending B to B /I . Denote thecomposition of these two maps: φ aI and denote by I the kernel of φ aI in B . Lemma 3.1.
Let B = B [ G ] be the group ring B [ G ] . Let I be an ideal of B andlet φ a , φ aI and I as in Notation 3.1. Then, (1) I ∩ B = I , (2) if I prime, then I prime, (3) if I maximal, then I maximal.Proof. (1) Let r ∈ B such that φ aI ( r ) = 0. Then φ a ( r ) = r + I = I hence r ∈ I .(2) Assume that I is a prime ideal. Let f.h ∈ I . Then φ a ( f ) .φ a ( h ) ∈ I , thereforeone of them belongs to I (as I is prime), hence φ aI ( f ) = 0 or φ aI ( h ) = 0.(3) Assume I a maximal ideal. Let us show that B /I is a field. Let f = P i r i g i ∈ B \ I . Since f / ∈ I , φ a ( f ) + I is invertible in B /I (since B /I is a XPONENTIAL IDEALS AND A NULLSTELLENSATZ 9 field), hence there is s ∈ B such that ( P i r i + I ) . ( s + I ) = 1 + I . Therefore,( f + I ) . ( s + I ) = ( X i r i ( g i −
1) + X i r i + I ) . ( s + I )= ( X i r i + I ) . ( s + I ) = 1 + I . (cid:3) Now we will place ourselves in the group rings R n , n ≥
1, defined in 2 .
1, assumingthat (
R, E ) is an E -field and keeping the same notations as in subsection 2 .
1. (Inparticular all ideals of R n are, as additive groups, Q -vector spaces.) Given an idealof R n , we want to find a natural condition under which we can extend it to anexponential ideal of R [ X ] E . We do it by steps, using the above lemma. In orderto extend a proper ideal I n of R n to a proper ideal of R n +1 , we will modify theaugmentation ideal map “along I n ”. We will require on I n the following property:( † ) n u ∈ I n ∩ R n − ⇒ E ( u ) − ∈ I n . Notation 3.2.
Recall that R n = R n − ⊕ A n and denote by π A n (respectively by π R n − ) the projection on A n (respectively on R n − ). Recall also that I n ∩ R n − is adivisible abelian subgroup of I n . Therefore I n − := I n ∩ R n − has a direct summand˜ I n ⊆ I n in I n , namely I n = ˜ I n ⊕ I n − Note that π A n is injective on ˜ I n : let u, v ∈ ˜ I n , and write u = u + u , v = v + v ,with u , v ∈ R n − and u , v ∈ A n . Suppose u = v , then u − v = u − v ∈ R n − ∩ I n ∩ ˜ I n = I n − ∩ ˜ I n = { } , consequently u = v .Let A n = π A n ( ˜ I n ) ⊕ ˜ A n . In the statement of the following lemma, we will use Notation 3.2.
Lemma 3.2.
Let I n be a proper ideal of R n and let u ∈ R n [ t A n ] , then u can berewritten in a unique way as X i r i E ( u i ) where r i ∈ R n , u i ∈ ˜ I n ⊕ ˜ A n , and for i = j , u i = u j . In other words, the group ring R n [ t A n ] is isomorphic to the group ring R n [ E ( ˜ I n ⊕ ˜ A n )] .Proof. Let u = P i r i .t a i ∈ R n [ t A n ], where r i ∈ R n and a i ∈ A n . Decompose a i as a i + a i with a i ∈ π A n ( ˜ I n ) and a i ∈ ˜ A n . Since π A n is injective on ˜ I n , there exists a unique f i ∈ ˜ I n such that a i = π A n ( f i ) and so f i = π R n − ( f i ) + a i . Set f i := π R n − ( f i ).We have E ( f i ) = E ( f i ) .t a i (since E ( a i ) as been defined as t a i (see section 2.1)),and t a i = t a i .t a i = E ( − f i ) .E ( f i ) .t a i . Observe that both E ( − f i ) ∈ R n , E ( f i ) ∈ R n and E ( f i ) ∈ R n [ t A n ]. Moreover, since a i ∈ A n , t a i = E ( a i ). So we may re-write u as X i ( r i E ( − f i )) E ( f i ) t a i = X i ( r i E ( − f i )) E ( f i + a i )with r i E ( − f i ) ∈ R n and f i + a i ∈ ˜ I n ⊕ ˜ A n . Such expression is unique since theprojection π A n on ˜ I n is injective: for f = g ∈ ˜ I n , π A n ( f ) = π A n ( g ) ∈ A n . So if u i = u j ∈ ˜ I n ⊕ ˜ A n , then π A n ( u i ) = π A n ( u j ). (cid:3) Proposition 3.3.
Let n ∈ N and I n be a proper ideal of R n with the property ( † ) n .Then, I n embeds in a (proper) ideal I n +1 of R n +1 such that E ( f ) − ∈ I n +1 for any f ∈ I n , ( † ) n +1 and I n +1 ∩ R n = I n ( †† ) n +1 . Moreover if I n is prime (respectively maximal), then I n +1 is prime (respectively max-imal). Proof: Let I n be a proper ideal of R n , n ∈ N (in particular I n ∩ R − = { } ). By thepreceding lemma, any u ∈ R n +1 can be rewritten in a unique way as P ℓi =1 r i E ( u i ),where r i ∈ R n , u i ∈ ˜ I n ⊕ ˜ A n , and the u i ’s are distinct. So the map φ sending P ℓi =1 r i E ( u i ) to P ℓi =1 r i ∈ R n is well-defined and it is a ring morphism from R n [ E ( ˜ I n ⊕ ˜ A n )] to R n .The kernel ker ( φ ) contains { E ( f ) − f ∈ ˜ I n } . Let φ I n be the map sending P ℓi =1 r i E ( u i ) to P ℓi =1 r i + I n ∈ R n /I n .Define I n +1 as ker ( φ I n ). By Lemma 3.1, I n +1 is an ideal of R n +1 with the propertythat ker ( φ I n ) ∩ R n = I n ( †† ) n +1 .It remains to show that I n +1 contains E ( f ) − f ∈ I n ( † ) n +1 . Let f ∈ I n and write f as f + f with f ∈ I n − and f ∈ ˜ I n .Then E ( f ) − E ( f ) − .E ( f ) + ( E ( f ) − E ( f ) − ∈ ker ( φ ) and by assumption ( † ) n , E ( f ) − ∈ I n for f ∈ I n − . So, E ( f ) − ∈ ker ( φ I n ).Applying Lemma 3.1 with S = R n and G = E ( ˜ I n ⊕ ˜ A n ), if I n is prime (respectivelymaximal), then I n +1 is prime (respectively maximal). (cid:3) Corollary 3.4.
Let I n be a proper ideal of R n , n ≥ , with the property ( † ) n .Then, I n embeds in a (proper) E -ideal I E of R [ X ] E such that I E ∩ R n = I n . More-over if I n is prime (respectively maximal), then I E is prime (respectively maximal).Proof. Proposition 3.3 allows to construct a proper (respectively prime, maximal)ideal I n +1 of R n +1 containing E ( I n ) − †† n +1 ). Therefore we mayreiterate the construction. Then let I E := S n ≥ n I n . It is an E -ideal by construction,and it is proper because for all n ≥ n , I E ∩ R n = I n . If I n is prime, then each I n is prime for n ≥ n (by Proposition 3.3) and so I E is prime as a union of a chain ofprime ideals. If I is maximal, then for n ≥ n , each I n is maximal by Proposition3.3 and I E is maximal as a chain of maximal ideals. (cid:3) XPONENTIAL IDEALS AND A NULLSTELLENSATZ 11
A natural question is when an ideal I ⊆ R n satisfies the condition ( † ) n . Assumethat ( R, E ) has an extension (
S, E ) where there is a tuple of elements ¯ α with theproperty that for any f ∈ I , f ( ¯ α ) = 0. Then consider I ¯ α := { g ∈ R n : g ( ¯ α ) = 0 } .By definition I ⊆ I ¯ α , I ¯ α is a (prime) ideal and for any f ∈ I ∩ R n − , E ( f ) − ∈ I ¯ α .In the following proposition, we will examine the condition ( † ) that we put on anideal I of R in order to embed it in an exponential ideal. (This corresponds to thecase of exponential polynomials with only one iteration of E .) In this particular case,we can use the fact that R is a Noetherian ring. Proposition 3.5.
Let I be a proper ideal of R . Then we can embed I in an ideal J of R such that for any f ∈ J ∩ R , E ( f ) − ∈ J . Therefore we can embed I in a(proper) E -ideal of R [ X ] E . Proof: Let I be a proper ideal of R (in particular I ∩ R − = { } ) and set I := R ∩ I . Using Notation 3.2, we get that ˜ I = I . Recall that ˜ A is a direct summandof π A ( I ) in A . By Lemma 3.2, R ∼ = R [ E ( I ⊕ ˜ A )].Set R ′ := R [ E ( ˜ A )], so we get R ∼ = R ′ [ E ( I )]. Consider the ideal I ′ := I ∩ R ′ .(Note that I = I ′ ∩ R , since R ′ ∩ R = R .) Let u ∈ R and write it as P j r j .E ( u j ),where r j ∈ R ′ , u j ∈ I , with the u j ’s distinct. The map φ + sending u to P ℓj =1 r j ∈ R ′ is well-defined. Define J as ker ( φ + I ′ ), then J is an ideal of R containing E ( f ) − f ∈ I with the property that J ∩ R ′ = I ′ by Lemma 3.1. (It implies that J ∩ R = J ∩ R ′ ∩ R = I ′ ∩ R = I .Let u ∈ h J, I i ∩ R ′ ; then u = P i u i a i with u i ∈ J, a i ∈ I . Since u ∈ R ′ , we havethat φ + ( u ) = u . So u = P i φ + ( u i ) .φ + ( a i ). But φ + ( u i ) ∈ I ′ and so since φ + ( a i ) ∈ R ′ and I ′ is an ideal, we get that u ∈ I ′ . In particular h J, I i is proper.So we repeat the same procedure replacing I by h J, I i∩ R . Since R is noetherian,the process will stop. So we get a proper ideal ˜ J containing I with the property thatfor any f ∈ ˜ J ∩ R , E ( f ) − ∈ ˜ J .
So we may apply Corollary 3.4 and embed ˜ J inan exponential ideal of R [ X ] E . (cid:3) Rabinowitsch’s trick.
Recall that Rabinowitsch’s trick corresponds to the in-troduction of an extra variable, allowing one to deduce the algebraic strong Nullstel-lensatz from the weak one. Given f ( ¯ X ) , . . . , f m ( ¯ X ) ∈ R [ X ] and another polynomial g ( ¯ X ) ∈ R [ X ] vanishing on all common zeroes of f , . . . , f m , introducing the newvariable Y , one gets: f , . . . , f m , − Y g do not have any common zeroes.By the weak Nullstellensatz, this implies that the ideal generated by these poly-nomials is not proper. So one expresses (by an equality) that 1 belongs to the idealgenerated by f , . . . , f m and 1 − Y.g . Then one substitutes g − to Y in the equality,and clears denominators. This entails that some power of g belongs to the idealgenerated by f , . . . , f m in R [ X ].To mimick Rabinowitsch’s trick in R [ X ] E , where ( R, E ) is as previously an E -field, we proceed as follows introducing a ”non- E ” variable, extending R [ X ] E to R [ X ] E ⊗ R R [ Y ]. This partial E -ring is isomorphic to the following chain of partial E -rings. Recall that R − := R and R := R [ X ]. Denote by S := R [ X , Y ] = R [ Y ] ∼ = R ⊗ R R [ Y ]. Let S := S [ t A ] ∼ = R ⊗ R R [ Y ], and by induction on n ≥
2, let S n = S n − [ t A n − ] ∼ = R n ⊗ R R [ Y ]. Let S := S n ≥ S n . Then S ∼ = R [ X ] E ⊗ R R [ Y ].Now we consider an ideal J of R [ X ] E ⊗ R R [ Y ] and we want to extend it into an E -ideal, namely an ideal containing E ( f ) −
1, for f ∈ J ∩ R [ X ] E . Proposition 3.6.
Let J n be a proper ideal of S n , let I n := J n ∩ R n . Suppose that I n satisfies ( † ) n as an ideal of R n . Then J n embeds into a proper ideal J n +1 of S n +1 such that I n +1 := J n +1 ∩ R n +1 satisfies ( † ) n +1 and ( †† ) n +1 . Moreover if J n is prime(respectively maximal), then J n +1 is prime (respectively maximal). Proof: By Lemma 3.2, R n [ t A n ] ∼ = R n [ E ( ˜ I n ⊕ ˜ A n )] (see Notation 3.2). Since S n +1 = S n [ t A n ] ∼ = R n [ t A n ] ⊗ R R [ Y ], we get that S n +1 ∼ = S n [ E ( ˜ I n ⊕ ˜ A n )]. Rewrite s ∈ S n +1 as P ℓi =1 s i .E ( u i ) : s i ∈ S n , u i ∈ ˜ I n ⊕ ˜ A n , with u i distinct. Let φ be the map sending P ℓi =1 s i .E ( u i ) to P ℓi =1 s i ∈ S n and let φ ↾ R n +1 be the restriction of the map φ to R n +1 . Let φ J n be the map sending P ℓi =1 s i .E ( u i ) to P ℓi =1 s i + J n ∈ S n /J n . Define J n +1 as the kernel of φ J n ; by Lemma 3.1, it is an ideal of S n +1 and ker ( φ J n ) ∩ S n = J n . Furthermore ker ( φ J n ) contains ker (( φ ↾ R n +1 ) I n ) and so by Proposition 3.3, itcontains E ( f ) − f ∈ I n .Applying Lemma 3.1 with B = S n and G = E ( ˜ I n ⊕ ˜ A n ), if J n is prime (respectivelymaximal), then J n +1 is prime (respectively maximal). (cid:3) Corollary 3.7.
Let J be a proper ideal of S n with n ≥ chosen minimal such.Assume that J ∩ R n satisfies the property ( † ) n (as an ideal of R n ). Then J embedsinto a proper exponential ideal J E of S [ X ] E with J E ∩ S n = J . Moreover whenever J is prime (respectively maximal) in S n , then J E is prime (respectively maximal) in S [ X ] E . We apply Proposition 3.6. (cid:3)
Main results.Corollary 3.8. ”Weak Nullstellensatz”
Let ( R, E ) be an E -field and f , . . . , f m ∈ R [ ¯ X ] E . Let n ∈ N be chosen minimal such that f , . . . , f m ∈ R n . Assume the ideal I n generated by f , . . . , f m is proper and that there is a maximal ideal M n of R n containing I n with the property ( † ) n (as an ideal of R n ). Then f , . . . , f m have acommon zero in an E -field extending ( R, E ) .Proof. By Proposition 3.6, M n embeds into an E -ideal M E of R [ ¯ X ] E which is amaximal ideal. The quotient R [ ¯ X ] E /M E is an E -field in which ( X + M E , . . . , X n + M E ) is a common zero of f , . . . , f m . (cid:3) Definition 3.9.
Let I be an ideal of R ℓ , ℓ ≥
1. We define J E ( I ) as the intersectionof all maximal ideals M of R ℓ containing I with the property( † ) ℓ . If there are nosuch maximal ideals, then we set J E ( I ) = R ℓ . Theorem 3.10.
Let f , . . . , f m , g ∈ R [ X ] E and let ℓ be minimal such that f , . . . , f m , g ∈ R ℓ . Let I be the ideal of R ℓ generated by f , . . . , f m . Suppose that J E ( I ) is a properideal. Assume that g vanishes at each common zero of f , . . . , f m in any partialexponential field containing ( R, E ) . Then g belongs to J E ( I ) . XPONENTIAL IDEALS AND A NULLSTELLENSATZ 13
Proof: Consider the ring S and the element 1 − Y.g , then 1 − Y.g ∈ S ℓ . Since J E ( I )is proper, there is a maximal ideal M of R ℓ containing I with property ( † ) ℓ . Let J bean ideal of S ℓ containing M and 1 − Y.g . Assume that J is proper, so J ∩ R ℓ = M andit embeds into a maximal ideal of S ℓ . Then, we embed it in an exponential maximalideal ˜ M E of S , by Proposition 3.6. Since ˜ M E ∩ R [ X ] E is a prime ideal, the quotient R [ X ] E / ˜ M E ∩ R [ X ] E is an E -domain containing ( R, E ) which embeds in the partial E -field S / ˜ M E where f , . . . , f m , − gY have a common zero, a contradiction.Therefore we have1 = X i t i ( X, Y ) .h i ( X ) + (1 − Y.g ) .r ( X, Y ) , with h i ( X ) ∈ M , t i ( X, Y ) , r ( X, Y ) ∈ S ℓ . Note that the elements of S ℓ , ℓ ≥
1, areof the form S [ t A ⊕ ... ⊕ A ℓ − ] ∼ = R ℓ ⊗ R R [ Y ]. So we may substitute g − to Y and finda sufficiently big power g d , d >
0, of g such that by multiplying each t i ( X, g − ), r ( X, Y ), we obtain again an element in S ℓ . Since t i ( X, g − ) .g d ∈ R ℓ , we get g d = X i ( t i ( X, g − ) .g d ) .h i ( X ) ∈ R ℓ . Therefore g d ∈ M and since M is maximal, g ∈ M . Since we can do that for anymaximal ideal M containing I with property ( † ) ℓ , g ∈ J E ( I ). (cid:3) Remark 3.1.
The problem with replacing in the theorem above, maximal ideals M with prime ideals P is that J ∩ R ℓ might be bigger than P and so the condition ( † ) ℓ might not hold anymore.Using Proposition 3.5, we may deduce a stronger result in the case when the E -polynomials have only one iteration of the exponential function (namely the case ℓ = 1). Corollary 3.11.
Let f , . . . , f m , g ∈ R and let I be the ideal of R generated by f , . . . , f m . Assume that g vanishes at each common zero of f , . . . , f m in any partialexponential field containing ( R, E ) . Then g belongs to J ( I ) , the Jacobson radical of I . Proof: We apply Theorem 3.10 and we may replace J E ( I ) by the ordinary Jacobsonradical of I since by Proposition 3.5, every maximal ideal of R satisfies ( † ) . (cid:3) Corollary 3.12.
Let f , . . . , f n ∈ R k and assume that k ∈ N is minimal such.Assume that the ideal I := ( f , . . . , f n ) of R k generated by f , . . . , f n is disjoint from { J m ¯ f : m ∈ N ∗ } . Let P be an ideal of R k containing I and maximal for the propertyof being disjoint from { J m ¯ f : m ∈ N ∗ } . Suppose P satisfies ( † ) k . Then f , . . . , f n havea (non singular) zero ¯ α in a partial E -field ( F, E ) extending ( R, E ) with J ¯ f ( ¯ α ) = 0 .Moreover ¯ α ∈ ecl F ( R ) . Proof: It is well-known that P is a prime ideal of R k . Since we assumed that itsatisfies the hypothesis that for any f ∈ P ∩ R k − , E ( f ) − ∈ P , we may applyCorollary 3.4 and extend P to a prime E -ideal ˜ P of R [ X ] E with ˜ P ∩ R k = P . Then the element ¯ α := ( X + ˜ P , . . . , X n + ˜ P ) satisfies the formula H ¯ f in the E -domain R [ X ] E / ˜ P . Let F be the fraction field of R [ X ] E / ˜ P . Then we have that ¯ α belongs toecl F ( R ) (see Definition 2.6). (cid:3) Real Nullstellensatz.
In this section we adapt our previous results to thecontext of ordered E -fields. First let us quickly recall some terminology and basicfacts. Definition 3.13. [11, p.279]
Let R be a ring. Then R is formally real if one can puta total order < on R such that for a, b, c ∈ R , a < b ⇒ a + c < b + c and < a, b ⇒ < ab. An ordered E -ring (respectively an ordered E -field) is an E -ring (respectively E -field)which is an ordered ring (respectively ordered field).An ideal I of R is said to be real if for any u , . . . , u n ∈ R such that P ni =1 u i ∈ I ,then u i ∈ I for all i = 1 , . . . , n .Denote by Σ the set of sums of squares in R and let I be an ideal of R . Then thereal radical rad < ( I ) of I is defined as follows rad < ( I ) := { f ∈ R : f m + s ∈ I for some m ∈ N and s ∈ Σ } . An ideal I of R is radical if whenever a n ∈ I , then a ∈ I , for a ∈ R and n ∈ N ∗ . Proposition 3.14. [11, Theorem 17.11]
A ring R is formally real iff − / ∈ Σ . Theorem 3.15. [1, Lemmas 4.1.5, 4.1.6 and Proposition 4.1.7]
Let R be a ring, and I an ideal of R . • Suppose I is prime. Then I is real iff the fraction field of R/I is formallyreal. • The real radical rad < ( I ) of I is the smallest real ideal containing I . • If I is real, then I is radical. Lemma 3.16. [10, Lemma 1.2 & Remark 1.3]
Let R be a formally real ring. Let I ⊆ R be an ideal, maximal with respect to the property that I is disjoint from .Then I is a prime real ideal. Definition 3.17.
Let (
R, E ) be an E -field. A real exponential ideal I of ( R, E ) is areal ideal which is also an exponential ideal.
Examples 3.1.
Let R be an ordered E -domain, then the set of exponential polyno-mials in R [ X ] E which vanish on a subset of R n is a real exponential ideal. Lemma 3.18.
Let B be a ring of characteristic , G a torsion-free abelian groupand B be the group ring B [ G ] . Let I be a real ideal of B and let φ a , φ aI as inNotation 3.1. The kernel I in B of the map φ aI is a real ideal of B . Proof: Suppose that P ni =1 u i ∈ I , so P ni =1 φ aI ( u i ) = 0. Therefore, P ni =1 φ a ( u i ) ∈ I . Since I is real, it implies that φ a ( u i ) ∈ I for all 1 ≤ i ≤ n , equivalently that φ aI ( u i ) = 0 for all 1 ≤ i ≤ n . (cid:3) XPONENTIAL IDEALS AND A NULLSTELLENSATZ 15
Lemma 3.19.
Let n ∈ N , I n ⊆ R n be a prime real ideal with the property ( † ) n . Then I n embeds in a prime real ideal I n +1 of R n +1 with properties ( † ) n +1 and ( †† ) n +1 .Proof. By Lemma 3.18 and Proposition 3.3. (cid:3)
Corollary 3.20.
Let ( R, E, < ) be an ordered E -field and f , . . . , f m ∈ R [ ¯ X ] E and let k ∈ N be chosen minimal such that f , · · · , f m ∈ R k . Assume the ideal I generatedby f , · · · , f m is disjoint from the set . Let P be an ideal of R k maximal forthe property of containing I and being disjoint from . Suppose that P satisfies ( † ) k Then f , · · · , f m have a common zero in a partial ordered E -domain extending ( R, E ) . Proof: It is well-known that such ideal P is a prime real ideal (see Lemma 3.16) Itremains to apply Lemma 3.19 and Corollary 3.4, checking that a union of a chain ofreal ideals is a real ideal. So we embed P into P E , an exponential prime, real ideal.The quotient R [ X ] E / P E is an orderable E -domain, where f , · · · , f m have a commonzero. (cid:3) Remark 3.2.
Note that when the exponential function satisfies the growth condition: E ( f ) ≥ f for any f ∈ R [ X ] E / P E , then by results of Dahn and Wolter [2,Theorem 24], one can embed this partial E -domain in a real-closed E -field where theexponential function is surjective on the positive elements. References [1] Bochnak, J. & Coste, M. & Roy, M.F.
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Department of Mathematics (De Vinci), UMons, 20, place du Parc 7000 Mons,Belgium
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