Big polynomial rings and Stillman's conjecture
aa r X i v : . [ m a t h . A C ] J un BIG POLYNOMIAL RINGS AND STILLMAN’S CONJECTURE
DANIEL ERMAN, STEVEN V SAM, AND ANDREW SNOWDEN Introduction
The purpose of this paper is to prove that certain limits of polynomial rings are themselvespolynomial rings, and show how this observation can be used to deduce some interestingresults in commutative algebra. In particular, we give two new proofs of Stillman’s conjecture[PS09, Problem 3.14]. The first is similar to that of Ananyan–Hochster [AH16], though morestreamlined; in particular, it establishes the existence of small subalgebras. The second proofis completely different, and relies on a recent noetherianity result of Draisma [Dra17].1.1.
Polynomiality results.
For a commutative ring A , let A JJ x , x , . . . KK be the inverselimit of the standard-graded polynomial rings A [ x , . . . , x n ] in the category of graded rings.A degree d element of this ring is a (possibly infinite) formal A -linear combination of degree d monomials in the variables { x i } i ≥ . Fix a field k , and let R = k JJ x , x , . . . KK . Our firstpolynomiality theorem is: Theorem 1.1.
Assume k is perfect. Then R is (isomorphic to) a polynomial ring. The set of variables in the polynomial ring is uncountable; hence the phrase “big poly-nomial rings” in the title of the paper. We deduce Theorem 1.1 from the following generalcriterion. For a graded ring R , we write R + for the ideal of positive degree elements. Theorem 1.2.
Let R be a graded ring with R = k a perfect field. Assume: • Characteristic zero: R has enough derivations (Definition 2.1), that is, for everynon-zero x ∈ R + there is a derivation ∂ of negative degree such that ∂ ( x ) = 0 . • Positive characteristic: R has enough Hasse derivations (see Definition 2.10).Then R is a polynomial ring. Precisely, for any set of positive degree homogeneous elements { f i } i ∈ I whose images in R + /R form a k -basis, the k -algebra homomorphism k [ X i ] i ∈ I → R taking X i to f i is an isomorphism. The proof of Theorem 1.2 is elementary: essentially, if one had an algebraic relation amongsome of the f i , then one could apply an appropriate (Hasse) derivation to get a lower degreerelation, and eventually reach a contradiction. To prove Theorem 1.1, we simply observe that(Hasse) derivatives with respect to the variables x i extend continuously to R and furnish itwith enough (Hasse) derivations.The inverse limit R is one way to make sense of a limit of finite polynomial rings. Adifferent way is through the use of ultrapowers, or, more generally, ultraproducts (see § Date : June 9, 2018.2010
Mathematics Subject Classification. In this paper, all graded rings are supported in non-negative degrees. for background). Let S be the graded ultrapower of the standard-graded polynomial ring k [ x , x , . . . ]. We also prove: Theorem 1.3.
Assume k is perfect. Then S is a polynomial ring. This also follows quickly from Theorem 1.2. The perfectness hypotheses in this sectioncan be relaxed: for instance, Theorems 1.1 and 1.3 hold if [ k : k p ] is finite, see Remarks 2.12and 5.4. In fact, some time after completing this paper, we succeeded in removing thesehypotheses entirely; see [ESSb].1.2. Connection to the work of Ananyan–Hochster.
We recall (and slightly extend)the notion of strength from [AH16]:
Definition 1.4.
Let R be a graded ring with R = k a field, and let f be a homogeneouselement of R . The strength of f is the minimal integer k ≥ − f = P k +1 i =1 g i h i with g i and h i homogeneous elements of R of positive degree, or ∞ if no such decomposition exists. The collective strength of a set of homogeneous elements { f i } i ∈ I of R is the minimal strength of a non-trivial homogeneous k -linear combination. (cid:3) Example 1.5. (a) In k [ x , . . . , x n ], non-zero elements of degree 1 have infinite strength,while non-zero elements of degree > < n .(b) In R , there are a wealth of interesting elements of infinite strength, such as P i ≥ x di (if d is invertible in k ).(c) In any graded ring R , the ideal R is exactly the ideal of finite strength elements. (cid:3) Many of the results of Ananyan–Hochster are instances of the following general principle:elements in a polynomial ring of sufficiently large collective strength behave approximatelylike independent variables. Theorem 1.1 shows that this approximation becomes exact in thelimiting ring R . Indeed, suppose { f i } i ∈ I are elements of R + that form a basis modulo R .Thus no linear combination of the f i belongs to R , i.e., has finite strength (Example 1.5(c)),and so { f i } has infinite collective strength. The Ananyan–Hochster principle thus suggeststhat the { f i } should be independent variables, and this is exactly the content of Theorem 1.1.1.3. Stillman’s conjecture via ultraproducts.
While ultraproducts may be less familiarto some readers than inverse limits, Theorem 1.3 leads to our most efficient proof of Stillman’sconjecture. As in [AH16] (see § Theorem 1.6.
Fix integers d , . . . , d r . Then there exists an integer N with the followingproperty. If k is a perfect field, and f , . . . , f r ∈ k [ x , . . . , x n ] are polynomials of degrees d , . . . , d r with collective strength at least N , then f , . . . , f r is a regular sequence. Ananyan–Hochster prove this theorem (in the case where k is algebraically closed) viaa multi-tiered induction, where elements of increasingly high strength obtain an array ofincreasingly nice properties. Our proof using Theorem 1.3 is more direct. Here is the idea.Suppose that f ,i , . . . , f r,i ∈ k [ x , x , . . . ], for i ∈ N , are polynomials of the given degrees withcollective strength tending to infinity. It suffices to show that f ,i , . . . , f r,i eventually forms aregular sequence. For each j , the sequence f j, • defines an element f j in the ultraproduct ring S . It is easy to see that f , . . . , f r has infinite collective strength (Proposition 4.6). Thus, byTheorem 1.3, f , . . . , f r are independent variables in S , and hence form a regular sequence. IG POLYNOMIAL RINGS AND STILLMAN’S CONJECTURE 3
We then apply a result (Corollary 4.10) comparing codimension in S to codimension in k [ x , x , . . . ] to conclude that f ,i , . . . , f r,i is eventually a regular sequence.As in [AH16], we show that the bound in Theorem 1.6 (and Stillman’s conjecture as well)is independent of the field k . To do so, we prove a generalization of Theorem 1.3 (see § S is replaced with an ultraproduct of polynomial rings with variable coefficient fields.1.4. Stillman’s conjecture via inverse limits.
Returning to the inverse limit, Theo-rem 1.1 enables a proof of Stillman’s conjecture that follows the very general rubric inalgebraic geometry of proving a result generically, spreading out to an open set, and theninductively treating proper closed subsets. The basic idea in characteristic zero is as fol-lows. Suppose that A is a characteristic 0 domain with fraction field K , and M is a finitelypresented A JJ x , x , . . . KK -module. Then K ⊗ A M is a finitely presented module over thering K ⊗ A A JJ x , x , . . . KK . While K ⊗ A A JJ x , x , . . . KK is not isomorphic to K JJ x , x , . . . KK ,Theorem 1.2 shows it is also an abstract polynomial ring.It then follows from simple homological properties of infinite polynomial rings that K ⊗ A M has a finite length resolution by finite free modules. A flatness argument produces an opendense subset U of Spec( A ) such that M y has the same Betti table as K ⊗ A M for all y ∈ U .We can then restrict our attention to Spec( A ) \ U , and apply the same argument. This showsthat there is some (perhaps infinite) stratification of Spec( A ) such that on each stratum thefibers of M have the same Betti table.We apply this as follows. Fix positive integers d , . . . , d r , and let A be the symmetricalgebra on the vector space Sym d ( k ∞ ) ⊕ · · · ⊕ Sym d r ( k ∞ ). Then Spec( A ) is the space offorms f , . . . , f r ∈ k JJ x , x , . . . KK of degrees d , . . . , d r . We let M be the universal module A JJ x , x , . . . KK / ( f , . . . , f r ). The stratification constructed in the previous paragraph can bemade compatible with the GL ∞ action on Spec( A ). A recent theorem of Draisma [Dra17]asserts that Spec( A ) is GL ∞ -noetherian, and hence this stratification is finite. We concludethat there are only finitely many resolution types for ideals generated by f , . . . , f r of thegiven degrees. This, in particular, implies Stillman’s conjecture in characteristic zero.The same idea works in positive characteristic, but when K fails to be perfect, we need tobootstrap from the perfect case to produce the open subset with constant Betti numbers.1.5. Connections to other work.
The Milnor–Moore theorem [MM65], and generaliza-tions [Sj¨o80], establish that certain commutative graded rings are polynomial rings via prop-erties of a comultiplication. While this, and its extensions to non-commutative rings, can beapplied to examples in commutative algebra, it is of a fairly distinct nature from the criteriain the present paper.Theorem 1.1 is an example of the meta-principle that inverse limits of free objects tend tobe free themselves. See [Ser97, § I.4.2, Corollary 4] for an example of this principle with pro- p -groups. Alexandru Chirvasitu informed us that he can prove a non-commutative versionof Theorem 1.1 where polynomial rings are replaced by non-commutative polynomial rings.The use of ultraproducts in commutative algebra was famously employed in [vdDS84] toestablish a variety of bounds (with the number of variables fixed). See [Sch10] for morediscussion and examples.The Gr¨obner theory of the inverse limit ring k JJ x , x , . . . KK was studied by Snellmanin [Sne98b, Sne98]. Shortly after a draft of this article was posted, [DLL18] applied The-orem 1.1 to obtain finiteness results for grevlex Gr¨obner bases over R , and then used this DANIEL ERMAN, STEVEN V SAM, AND ANDREW SNOWDEN to answer some questions raised by Snellman and to give a generic initial ideal proof ofStillman’s Conjecture.The use of GL ∞ -noetherianity of spaces to prove the existence of uniform bounds inalgebraic geometry has been used in several papers. See [Dra14] for a survey.1.6. Outline. In §
2, we establish our polynomiality criteria (summarized in Theorem 1.2).In §
3, we prove some easy results concerning dimension theory in polynomial rings with aninfinite number of variables. In §
4, we prove that the ultraproduct ring is a polynomial ring(Theorem 1.3), and use this to deduce our first proof of Stillman’s conjecture. Finally, in §
5, we prove that the inverse limit ring is a polynomial ring (Theorem 1.1), and use this todeduce our second proof of Stillman’s conjecture.
Acknowledgements.
We thank Craig Huneke and Gregory G. Smith for useful conversa-tions. We also thank Alexandru Chirvasitu for informing us about his work on the non-commutative analogue of Theorem 1.1 and the reference in [Ser97].2.
Criteria for polynomiality
Let R be a graded ring with R = k a field. We say that R is a polynomial ring if thereare elements { x i } i ∈ I of R , each homogeneous of positive degree, such that the natural map k [ X i ] → R sending X i to x i is an isomorphism. The x i ’s need not have degree 1, and theset I need not be finite. The purpose of this section is to characterize polynomial rings viaderivations.2.1. Characteristic 0.
We first treat the case where k has characteristic 0, for which thefollowing definition and theorem constitute our criterion for polynomiality. We say that aderivation ∂ of a graded ring R is homogeneous of degree d if deg ∂ ( x ) = deg( x ) + d forall homogeneous x ∈ R . Definition 2.1.
Let R be a graded ring with R = k a field. We say that R has enoughderivations if for every non-zero homogeneous element x of positive degree there is a ho-mogeneous derivation ∂ of negative degree such that ∂ ( x ) = 0. (cid:3) Theorem 2.2.
Let R be a graded k -algebra with R = k a field of characteristic . Then R is a polynomial ring if and only if R has enough derivations.Proof. In this proof, “derivation” will mean “homogeneous derivation of negative degree.”It is clear that a polynomial ring has enough derivations. We prove the converse.Let E be a set of homogeneous elements of R + that gives a basis of R + /R . By Nakayama’slemma, E generates R as a k -algebra, so it suffices to show that E is algebraically independent.Let E ≤ d (resp. E d ) be the set of elements in E of degree ≤ d (resp. d ). We prove that E ≤ d isalgebraically independent for all d by induction on d . Suppose that we have shown E ≤ d − isalgebraically independent. To prove that E ≤ d is algebraically independent, it suffices to provethe following statement: if E ≤ d − ⊂ E ⊂ E ≤ d is algebraically independent and x ∈ E d \ E ,then E ′ = E ∪ { x } is algebraically independent. Indeed, this statement implies that all setsof the form E ≤ d − ∪ E ′′ with E ′′ a finite subset of E d are algebraically independent, whichimplies that E ≤ d is algebraically independent.Thus let E , E ′ , and x as above be given. Let A ⊂ R be the k -subalgebra generated by E .To prove that E ′ is algebraically independent, it suffices to show that if 0 = P ni =0 a i x i with IG POLYNOMIAL RINGS AND STILLMAN’S CONJECTURE 5 a i ∈ A then a i = 0 for all i . Before proceeding, we note that if ∂ is any derivation of R then ∂ ( E ≤ d ) ⊂ A since ∂ decreases degrees, and so ∂ ( A ) ⊂ A and ∂ ( x ) ∈ A .Suppose that 0 = P ni =0 a i x i with a i ∈ A and a n = 0. Of all such relations, choose oneof minimal degree (i.e., with deg( a n x n ) minimal). Suppose that a n has positive degree. Byassumption, there exists a derivation ∂ such that ∂ ( a n ) = 0. Applying ∂ to our given relationyields 0 = ∂ ( a n ) x n + P n − i =0 b i x i where the b i are elements of A . This is a contradiction, since ∂ ( a n ) has smaller degree than a n . Thus deg( a n ) = 0, and so we may assume a n = 1.Since E is linearly independent modulo R , we see that x / ∈ A , and so n ≥ nx + a n − is non-zero. It follows that there exists a derivation ∂ such that ∂ ( nx + a n − ) = 0. Applying ∂ to our original relation gives 0 = ∂ ( nx + a n − ) x n − + P n − i =0 b i x i for some b i ∈ A . This is asmaller degree relation, which is a contradiction. We thus see that no relation 0 = P ni =0 a i x i exists with a n non-zero, which completes the proof. (cid:3) Positive characteristic.
Theorem 2.2 obviously fails in characteristic p : since p thpowers are killed by every derivation, no reduced ring has enough derivations. The mostobvious adjustment would be to ask that if x is a homogeneous element of R that is not a p th power then there is a derivation ∂ such that ∂ ( x ) = 0. The following two examples showthat this condition is insufficient to conclude that R is a polynomial ring. Example 2.3.
Let R = k [ x ] / ( x p ) where k is perfect of characteristic p and x has degree 1.Then ddx is a well-defined derivation on R , and thus R has enough derivations. (cid:3) Example 2.4.
Let R = k [ x, y, yx p ] where k is perfect of characteristic p , x has degree 1,and y has degree p + 1. Then ddx and x p ddy are well-defined derivations on R , and everyhomogeneous element of R that is not a p th power is not annihilated by one of them. (cid:3) To extend our criterion to the positive characteristic case, we employ the following exten-sion of the notion of a derivation (see [Gol03, pp. 27–29] for additional discussion).
Definition 2.5.
Let R be a k -algebra. A Hasse derivation on R is a sequence ∂ • = ( ∂ n ) n ≥ where each ∂ n is a k -linear endomorphism of R such that ∂ is the identity and ∂ n ( xy ) = X i + j = n ∂ i ( x ) ∂ j ( y )holds for all x, y ∈ R . If R is graded then we say ∂ • is homogeneous of degree d if ∂ n ( x )has degree deg( x ) + nd for all homogeneous x ∈ R . (cid:3) Remark 2.6.
Giving a Hasse derivation on R is equivalent to giving a ring homomorphism ϕ : R → R J t K such that the constant term of ϕ ( x ) is x . If ∂ • is a Hasse derivation, then theassociated ring homomorphism is defined by ϕ ( x ) = P i ≥ ∂ i ( x ) t i . (cid:3) Example 2.7.
Suppose R = k [ x ], with k any field. Define ∂ n ( x k ) = (cid:0) kn (cid:1) x k − n . (Notethat ∂ n = n ! d n dx n if n ! is invertible in k .) Then ∂ • is a Hasse derivation, called the Hassederivative . If R is graded with x of degree d then ∂ • is homogeneous of degree − d . Thehomomorphism ϕ : R → R J t K associated to the Hasse derivative is given by x x + t . (cid:3) Remark 2.8.
Curiously, Hasse derivatives also play a key role in Draisma’s [Dra17], wherethey are closely related to his directional derivatives. (cid:3)
DANIEL ERMAN, STEVEN V SAM, AND ANDREW SNOWDEN
Lemma 2.9.
Let R be a k -algebra, where k is a field of characteristic p , and let ∂ • be aHasse derivation on R . Let q be a power of p . Then for x ∈ R and n ∈ N we have ∂ n ( x q ) = ( ( ∂ n/q x ) q if q | n if q ∤ n . Proof.
We have ∂ n ( x q ) = X ( i ,...,i q ) i + ··· + i q = n ∂ i ( x ) · · · ∂ i q ( x ) . If i , . . . , i q are not all equal then the orbit of ( i , . . . , i q ) under the symmetric group S q hascardinality divisible by p . All elements of this orbit contribute equally to the sum, and thusthey all cancel. We thus see that the only surviving term occurs when n is a multiple of q and i = · · · = i q = n/q ; this term is ( ∂ n/q x ) q . (cid:3) The following definition and theorem constitute our criterion for polynomiality in positivecharacteristic.
Definition 2.10.
Let R be a graded ring with R = k a field of characteristic p >
0. We saythat R has enough Hasse derivations if the following condition holds: if x is a positivedegree homogeneous element of R such that x k R p (the k -span of the set R p ) then thereexists a homogeneous Hasse derivation ∂ • of R of negative degree such that ∂ ( x ) = 0. (cid:3) Theorem 2.11.
Let R be a graded ring with R = k a perfect field of characteristic p > .Then R is a polynomial ring if and only if it has enough Hasse derivations.Proof. In this proof, “Hasse derivation” will mean “homogeneous Hasse derivation of negativedegree.” We note that since k is perfect, k R p = R p . If R is a polynomial ring then it hasenough Hasse derivations; one can see this using Hasse derivatives (Example 2.7). We nowprove the converse.We first show that R is reduced. Suppose not, and let x ∈ R be a non-zero homogeneousnilpotent element of minimal degree. Note that x R p , for if x = y p then y would be alower degree nilpotent element. Let r be such that x p r = 0 and let ∂ • be a Hasse derivationsuch that ∂ ( x ) = 0. Then 0 = ∂ p r ( x p r ) = ( ∂ x ) p r (Lemma 2.9), and so ∂ ( x ) is nilpotent,contradicting the minimality of x . Thus R is reduced.Let E be a set of homogeneous elements of R + that forms a basis for R + /R . It sufficesto prove that E is algebraically independent. For E ⊂ E , consider the following statement: A E : Given distinct elements x , . . . , x r ∈ E and a polynomial F ∈ k [ X , . . . , X r ] suchthat F ( x , . . . , x r ) ∈ R p , we have F ∈ k [ X , . . . , X r ] p .Observe that if A E holds then E is algebraically independent. Indeed, suppose that F ( x , . . . , x r ) =0 is a minimal degree algebraic relation among distinct elements of E . Since 0 ∈ R p , we seethat F ( X , . . . , X r ) = G ( X , . . . , X r ) p for some G by A E , and so G ( x , . . . , x r ) p = 0. Since R is reduced, it follows that G ( x , . . . , x r ) = 0, contradicting the minimality of F . Thus toprove the theorem it suffices to prove A E .We prove that A E holds for all E by induction on E in the following manner. Let E ≤ d (resp. E d ) be the set of elements of E of degree ≤ d (resp. d ). Suppose that E ≤ d − ⊂ E ⊂ E ≤ d and let E ′ = E ∪ { x } for some x ∈ E d \ E . Assuming A E , we prove A E ′ . This will establish A E for all E by the same logic used in the proof of Theorem 2.2. IG POLYNOMIAL RINGS AND STILLMAN’S CONJECTURE 7
Fix E , E ′ , and x as above. Let A be the k -subalgebra of R generated by E . We claimthat A E ′ can be reduced to the following statement, for all n and m : B n,m : If P ni =0 a i x i ∈ R p with a i ∈ A and deg( a n ) ≤ m then a i ∈ R p and ia i = 0 for all i .Indeed, suppose B n,m holds for all n and m , and suppose F ( x , . . . , x r ) ∈ R p for distinctelements x , . . . , x r ∈ E ′ . We may as well suppose x r = x and x , . . . , x r − ∈ E . Write F ( X , . . . , X r ) = P ni =0 G i ( X , . . . , X r − ) X ir for polynomials G i . By B n,m , we see that G i ( x , . . . , x r − ) ∈ R p for all i and G i ( x , . . . , x r − ) = 0 if p ∤ i . By A E , it follows that G i ( X , . . . , X r − ) = G ′ i ( X , . . . , X r − ) p for some polynomial G ′ i and that G i ( X , . . . , X r ) = 0if p ∤ i . We thus find F ( X , . . . , X r ) = (cid:18) X ≤ i ≤ np | i G ′ i/p ( X , . . . , X r − ) X i/pr (cid:19) p , which establishes A E ′ .We now prove B n,m by induction on n and m . Clearly, B ,m holds for all m . We notethat if B n,m holds and P ni =0 a i x i = 0 with deg( a n ) ≤ m then a i = 0 for all i ; the proof is thesame as the proof given above that A E implies algebraic independence of E . We also notethat if ∂ • is any Hasse derivation then ∂ n ( E ≤ d ) ⊂ A for all n >
0, and so ∂ n ( A ) ⊂ A and ∂ n ( x ) ∈ A .We now prove B ,m for all m by induction on m . First suppose m = 0, and suppose that ax + b ∈ R p with a ∈ k and b ∈ A . We thus see that ax + b = 0 in R + /R . Since E is linearlyindependent in R + /R , it follows that a = 0, and so B , holds. Now suppose B ,m − holds,and let us prove B ,m . Thus suppose that ax + b = y p for some y ∈ R p with a, b ∈ A and deg( a ) = m . If ∂ • is any Hasse derivation of R then ∂ ( a ) x + ( a∂ ( x ) + ∂ ( b )) = 0(Lemma 2.9). Since deg( ∂ ( a )) < m , we see that ∂ ( a ) = 0 by B ,m − . Since this holds forall ∂ • , we find a ∈ R p . Suppose a = 0, and let q be the maximal power of p such that a ∈ R q (this exists since deg( a ) > a = c q , and note c R p . Let ∂ • be a Hasse derivationsuch that ∂ ( c ) = 0; note then that ∂ q ( a ) = ( ∂ c ) q = 0 (Lemma 2.9). Again by Lemma 2.9,we have ∂ q ( a ) x + ( a∂ q ( x ) + ∂ q ( b )) = ∂ q ( y p ) = ( ∂ q/p y ) p ∈ R p By B ,m − , we have ∂ q ( a ) = 0, a contradiction. Thus a = 0 and B ,m holds.We now prove B n,m for n ≥
2, assuming B n − , • and B n,m − . Thus suppose that P ni =0 a i x i ∈ R p with a i ∈ A and deg( a n ) ≤ m . Let ∂ • be a Hasse derivation of R . Applying ∂ , we find0 = ∂ ( a n ) x n + ( na n ∂ ( x ) + ∂ ( a n − )) x n − + · · · , where the remaining terms have degree ≤ n − x . By B n,m − , all the above coefficientsvanish. Thus ∂ ( a n ) = 0 for all ∂ • , and so a n ∈ R p . We now see that the coefficient of x n − is ∂ ( na n x + a n − ). Since this vanishes for all ∂ • , we find na n x + a n − ∈ R p , and so na n = 0by B , • . In particular, p | n if a n = 0, so a n x n ∈ R p , and hence P n − i =0 a i x i ∈ R p . Thus by B n − , • we have a i ∈ R p and ia i = 0 for all 0 ≤ i ≤ n −
1. This proves B n,m . (cid:3) Remark 2.12.
The perfectness hypothesis in Theorem 2.11 can be omitted. Indeed, letting K be the perfection of k , the theorem shows that K ⊗ k R is a polynomial ring, which impliesthat R is a polynomial ring. (cid:3) DANIEL ERMAN, STEVEN V SAM, AND ANDREW SNOWDEN Dimension theory in polynomial rings
Fix a field k . For a ring A and a (possibly infinite) set U , we let A [ U ] be the polynomialalgebra over A in variables U . We aim to prove a number of basic results on codimensionin rings of the form A [ U ] where A is a finitely generated k -algebra. All of these results arestandard when U is finite. We do not impose any gradings in this section.For a prime ideal p in a commutative ring R , the codimension (or height ) of p is themaximum integer c for which there exists a chain of prime ideals p ( · · · ( p c = p , or ∞ ifsuch chains exist with c arbitrarily large. The codimension of an arbitrary non-unital ideal I of R is the minimum of the codimensions of primes containing I , or ∞ if I is not containedin any prime of finite codimension. This will be denoted codim R ( I ). We start with a basicfact that we will cite often. Proposition 3.1.
Let A ⊂ B be a flat integral extension of rings. For any ideal I ⊂ B , wehave codim B ( I ) = codim A ( A ∩ I ) .Proof. We first prove the statement assuming that I = p is prime. Suppose that codim B ( p ) ≥ c . Let p ( p ( · · · ( p c = p be a chain of distinct prime ideals. Let q i = A ∩ p i . Byincomparability [AK13, Theorem 14.3(2)], the q i are distinct and thus codim A ( q c ) ≥ c .In particular, if codim B ( p ) = ∞ , this shows that codim A ( A ∩ p ) = ∞ . Now supposethat codim B ( p ) is finite and equal to c . If there were some longer chain of primes leadingup to q c , then by going down for flat extensions [AK13, Theorem 14.11], we would havecodim B ( p c ) > c , which is a contradiction. Thus codim A ( q c ) = c which finishes the specialcase when I is prime.Now consider the general case. Given a prime p containing I , we have just shown thatcodim A ( A ∩ p ) = codim B ( p ). On the other hand, given a prime q containing A ∩ I , using[AK13, Theorem 14.3(4)], there is a prime p ⊃ I such that A ∩ p = q . In particular, wededuce that codim B ( I ) = codim A ( A ∩ I ). (cid:3) Proposition 3.2.
Let A be a finitely generated k -algebra and let p be a prime ideal of A [ U ] of finite codimension. Then p is finitely generated.Proof. Let c be the codimension of p . We prove the statement by induction on c . Firstsuppose that c = 0. If p = 0, then we are done. Otherwise, choose a nonzero element g ∈ p .Let V ⊂ U be a finite subset such that g belongs to A [ V ]. Let p ′ = A [ U ]( A [ V ] ∩ p ). Then wehave p ′ ⊆ p . Since p is prime, so is A [ V ] ∩ p , and hence so is p ′ since A [ U ] is obtained from A [ V ] by adjoining variables. In particular, we have p ′ = p , and so p is finitely generated.Now suppose c >
0. Choose a prime ideal q ⊂ p of codimension c −
1. By induction, weknow that q is finitely generated; let f , . . . , f r be generators. Let g ∈ p \ q and let V ⊂ U be a finite subset such that the f i ’s and g belong to A [ V ]. Let p ′ = A [ U ]( A [ V ] ∩ p ). Notethat q = A [ U ]( A [ V ] ∩ q ). We thus have q ⊂ p ′ ⊂ p . Since p is prime, so is its contractionto A [ V ], and so is the extension of this back to A [ U ], since A [ U ] is obtained from A [ V ] byadjoining variables. Thus p ′ is either q or p ; however, it is not q since it contains g . Thus p = p ′ , which shows that p is finitely generated. (cid:3) Proposition 3.3.
Let A be a finitely generated k -algebra and let V ⊂ U be sets. If I is a finitely generated ideal of A [ V ] , and J is its extension to A [ U ] , then codim A [ V ] ( I ) =codim A [ U ] ( J ) .Proof. We note that the result is classical if U is finite (as can be seen, for example, usingHilbert polynomials). We will use this twice in the proof of the general case. IG POLYNOMIAL RINGS AND STILLMAN’S CONJECTURE 9
First suppose that V is finite and I = p is prime. Note then that J = q is prime as well.If p ( · · · ( p c = p is a chain of primes in A [ V ], then letting q i be the extension of p i ,we get a chain of primes q ( · · · ( q c = q in A [ U ], and so codim A [ U ] ( q ) ≥ c , which showscodim A [ U ] ( q ) ≥ codim A [ V ] ( p ).Next suppose that q ( · · · ( q c = q is a chain of primes in A [ U ]. For each 0 < i ≤ c ,pick f i ∈ q i \ q i − . Let V ′ be a finite subset of U containing V and such that each f i belongsto A [ V ′ ]. Then q • ∩ A [ V ′ ] is a strict chain of primes ideals in A [ V ′ ], and so we see thatcodim A [ V ′ ] ( p A [ V ′ ]) ≥ c (note that the contraction of q to A [ V ′ ] is equal to the extension of p to A [ V ′ ]). However, codim A [ V ′ ] ( p A [ V ′ ]) = codim A [ V ] ( p ) by classical theory. We thus see thatcodim A [ V ] ( p ) ≥ c , and so codim A [ V ] ( p ) ≥ codim A [ U ] ( q ). In particular, we have equality andthis case has been proven.Next, suppose still that V is finite, but let I be an arbitrary ideal. If p is a codimension c prime of A [ V ] containing I then p A [ U ] is a codimension c prime of A [ U ], by the previousparagraph, containing J . We thus see that codim A [ U ] ( J ) ≤ codim A [ V ] ( I ). Next, supposethat q is a codimension c prime of A [ U ] containing J . By Proposition 3.2, there is a finitesubset V ′ of U (which we can assume contains V ) such that q is the extension of an ideal(necessarily prime) q ′ of A [ V ′ ]. By the previous paragraph, q ′ has codimension c in A [ V ′ ].Since q ′ clearly contains the extension of I to A [ V ′ ], we see that codim A [ V ′ ] ( IA [ V ′ ]) ≤ c . Butcodim A [ V ′ ] ( IA [ V ′ ]) = codim A [ V ] ( I ) by classical theory, and so codim A [ V ] ( I ) ≤ c . We thus seethat codim A [ V ] ( I ) ≤ codim A [ U ] ( J ).Finally, we treat the case where V is arbitrary. Since I is finitely generated, there is afinite subset V of V such that I is the extension of an ideal I of A [ V ]. Thuscodim A [ V ] ( I ) = codim A [ V ] ( I ) = codim A [ U ] ( J ) , by two applications of the case where V is finite. (cid:3) Corollary 3.4.
Let A be a finitely generated k -algebra. Every finitely generated ideal of A [ U ] has finite codimension.Proof. Let J be a finitely generated ideal of A [ U ]. Then J is the extension of an ideal I ofsome A [ V ] with V ⊂ U finite. Since codim A [ V ] ( I ) ≤ dim A [ V ] < ∞ , Proposition 3.3 impliesthat codim A [ U ] ( J ) is finite as well. (cid:3) Corollary 3.5.
Let U be a set, and let f , . . . , f r ∈ k [ U ] . Then f , . . . , f r form a regularsequence if and only if the ideal ( f , . . . , f r ) has codimension r .Proof. Let V be a finite subset of U such that f , . . . , f r ∈ k [ V ]. Let I (resp. J ) be the idealof k [ V ] (resp. k [ U ]) generated by the f i . The f i form a regular sequence in k [ V ] (or in k [ U ])if and only the Koszul complex on the f i is exact; however, since k [ U ] ⊆ k [ V ] is faithfullyflat, the Koszul complex on the f i is exact over k [ U ] if and only if it exact over k [ V ]. (cid:3) Corollary 3.6.
Let A be a finitely generated k -algebra, let U be a set, and let J be a finitelygenerated ideal of A [ U ] containing a nonzerodivisor f . Let J be the image of J in A [ U ] / ( f ) .Then codim A [ U ] / ( f ) ( J ) = codim A [ U ] ( J ) − .Proof. Let V be a sufficiently large finite subset of U such that A [ V ] contains f and somefinite generating set of J ; thus J is the extension of some ideal I of A [ V ]. Let B = A [ V ],which is a finitely generated k -algebra, and note that A [ U ] = B [ U ′ ], where U ′ = U \ V . Let I be the image of I in B = B/ ( f ). Since J is the extension of I to B [ U ′ ] = A [ U ] / ( f ), weobtain codim B ( I ) = codim A [ U ] / ( f ) ( J ) and codim B ( I ) = codim A [ U ] ( J ) by two applications of Proposition 3.3. Finally since codim B ( I ) = codim B ( I ) − (cid:3) Proposition 3.7.
Let A be a finitely generated k -algebra and let U be a set. Let J be afinitely generated ideal of A [ U ][ y ] . Then A [ U ] ∩ J is also a finitely generated ideal.Proof. Let V be a finite subset of U such that J is the extension of an ideal I of A [ V ][ y ].Then A [ V ] ∩ I is finitely generated since A [ V ] is noetherian. One easily sees that A [ U ] ∩ J is the extension of A [ V ] ∩ I , which proves the result. (cid:3) Corollary 3.8.
Let A be a finitely generated k -algebra, let U be a set, let R = A [ U ] , and let S = R [ y ] . Let I be a finitely generated ideal of S . Suppose that I contains a positive degreemonic polynomial, that is, an element of the form y n + P n − i =0 a i y i with a i ∈ R and n > .Then codim R ( R ∩ I ) = codim S ( I ) − .Proof. Let f ∈ I be a monic polynomial. Let I be the image of I in S/ ( f ). Then R → S/ ( f )is a finite flat extension of rings and R ∩ I is the contraction of I along this map. We thus seethat codim R ( R ∩ I ) = codim S/ ( f ) ( I ) by Proposition 3.1. But codim S/ ( f ) ( I ) = codim S ( I ) − (cid:3) Stillman’s conjecture via the ultraproduct ring
Background on ultraproducts.
For more details and references on ultraproducts,see [Sch10, § I be an infinite set. We fix a non-principal ultrafilter F on I , which isa collection of subsets of I satisfying the following properties:(a) F contains no finite sets,(b) if A ∈ F and B ∈ F , then A ∩ B ∈ F ,(c) if A ∈ F and A ⊆ B , then B ∈ F ,(d) for all A ⊆ I , either A ∈ F or I \ A ∈ F (but not both).We think of the sets in F as neighborhoods of some hypothetical (and non-existent) point ∗ of I , and refer to them as such. We say that some condition holds near ∗ if it holds in someneighborhood of ∗ .Given a family of sets { X i } i ∈ I , their ultraproduct is the quotient of the usual product Q i ∈ I X i in which two sequences ( x i ) and ( y i ) are identified if the equality x i = y i holds near ∗ . If x is an element of the ultraproduct, we will write x i for the i th coordinate of x , keepingin mind that this is only well-defined in sufficiently small neighborhoods of ∗ ; in other words,we can think of x as a germ of a function around ∗ .Suppose that each X i is a graded abelian group. We define the graded ultraproduct of the X i ’s to be the subgroup of the usual ultraproduct consisting of elements x such thatdeg( x i ) is bounded near ∗ . The graded ultraproduct is a graded abelian group; in fact, it isthe ultraproduct of the X i ’s in the category of graded abelian groups. The degree d pieceof the graded ultraproduct is the usual ultraproduct of the degree d pieces of the X i ’s. Weapply this construction in particular to the case where the X i ’s are graded rings; the gradedultraproduct is then again a graded ring. Example 4.1. If K is the ultraproduct of { k i } i ∈ I , then the graded ultraproduct of k i [ x , . . . , x n ](with standard grading) is K [ x , . . . , x n ] (also with standard grading). (cid:3) In this subsection, we develop a few basic properties of graded ultraproduct rings. Webegin with a simple observation on adjoining variables to ultraproducts.
IG POLYNOMIAL RINGS AND STILLMAN’S CONJECTURE 11
Proposition 4.2.
Let { R i } i ∈ I be a family of graded rings with graded ultraproduct S . Let y be a variable of degree , and let e S be the graded ultraproduct of the rings R i [ y ] . Then thenatural map S [ y ] → e S is an isomorphism.Proof. Suppose that f = P dk =0 a k y k is an element of S [ y ], and let g be its image in e S . Then g i = P dk =0 a k,i y k . If g = 0 then, passing to some neighborhood of ∗ , we can assume g i = 0for all i , which implies that a k,i = 0 for all i and k , which implies that a k = 0 for all k , whichshows that f = 0. Thus the map is injective.Next, suppose that g is an element of e S of degree d . Then we can write g i = P dk =0 a k,i y k for each i , where the a k,i ’s are elements of R i . Let a k be the element of S defined by thesequence ( a k,i ). Then g is the image of f = P dk =0 a k y k , and so the map is surjective. (cid:3) We next prove a simple result on base change:
Proposition 4.3.
Let k ′ / k be a finite field extension. Let { R i } i ∈ I be a family of graded k -algebras with graded ultraproduct S . Let R ′ i = k ′ ⊗ k R i , and let S ′ be the graded ultraproductof { R ′ i } i ∈ I . Then the natural map k ′ ⊗ k S → S ′ is an isomorphism.Proof. Let ǫ , . . . , ǫ d be a basis for k ′ over k . We note that R ′ i is free as an R i -module withbasis ǫ ⊗ , . . . , ǫ d ⊗
1; similarly for k ′ ⊗ k S over S . We claim that the ǫ ’s are also a basis for S ′ over S . Given f = ( f i ) ∈ S ′ , we can decompose f i uniquely as ǫ ⊗ f i, + ǫ ⊗ f i, + · · · + ǫ d ⊗ f i,d where f i,j ∈ R i for all i, j . We define g j = ( f i,j ) ∈ S for 1 ≤ j ≤ d , and we have the uniquedecomposition f = g ǫ + · · · + g d ǫ d in S ′ . (cid:3) We now examine how ideals in an ultraproduct relate to ideals in the original rings.Given a family of graded rings { R i } i ∈ I and a family of ideals { I i } i ∈ I , we say that the I i are uniformly finitely generated if there exists an integer n such that I i is generated by atmost n elements for all i ∈ I . Proposition 4.4.
Let { R i } i ∈ I be a family of graded rings with graded ultraproduct S . (a) Suppose that { I i } is a uniformly finitely generated family of homogeneous ideals. Thentheir graded ultraproduct I is a finitely generated ideal of S . (b) Suppose that { I i } and { J i } are two uniformly finitely generated families of homoge-neous ideals whose graded ultraproducts are equal. Then I i = J i for all i in someneighborhood. (c) Suppose that I is a finitely generated homogeneous ideal of S . Then there exists auniformly finitely generated family of homogeneous ideals { I i } with ultraproduct I .Proof. (a) Suppose that each I i is generated by ≤ n elements; pick generators f ,i , . . . , f n,i ofeach I i . Let f , . . . , f n be the elements of S defined by these sequences. We claim that I isgenerated by f , . . . , f n . Indeed, suppose that g is a homogeneous element of I ; thus, passingto a small enough neighborhood, we see that each g i is an element of I i , and can thus bewritten as P nk =1 a k,i f k,i for some homogeneous elements a k,i ∈ R i . Let a k be the element of S defined by the sequence a k,i . Then g = P nk =1 a k f k , proving the claim. (Note that for k fixed,each a k,i is homogeneous of some degree, but that the degree may depend on i . However, thedegree is bounded by the degree of g , and so in any small enough neighborhood, the degreeof a k,i will be independent of i .)(b) Suppose that I i and J i are each generated by at most n elements for all i , and pickgenerators f ,i , . . . , f n,i and g ,i , . . . , g n,i . Let f , . . . , f n and g , . . . , g n be the elements of S these sequences define. By (a), the f k ’s and g k ’s generate the same ideal of S . Thus we havean expression g k = P nj =1 a j f j for some a j ∈ S , and so g k,i = P nj =1 a j,i f j,i holds for all i insome neighboorhood of ∗ , and so g k,i belongs to the ideal I i for all such i . Since there areonly finitely many f ’s and g ’s, we can pass to some common neighboorhood of ∗ so that g k,i ∈ I i and f k,i ∈ J i for all i and k , and so I i = J i .(c) Let I be generated by f , . . . , f n . Let f k be represented by some sequence ( f k,i ), andlet I i be the ideal of R i generated by f ,i , . . . , f n,i . Then the argument in (a) shows that I isthe ultraproduct of the I i ’s. (cid:3) Due to this proposition, we can unambiguously speak of the germ of a finitely generatedhomogeneous ideal I of S . We denote these ideals by I i , keeping in mind that they are onlywell-defined for i sufficiently close to ∗ . We next show that this construction interacts wellwith contraction. Proposition 4.5.
Let { R i } be a family of graded rings with graded ultraproduct S , and let { R ′ i } be a family of graded subrings of { R i } with graded ultraproduct S ′ . Let I be a finitelygenerated homogeneous ideal of S , and suppose that S ′ ∩ I is a finitely generated ideal of S ′ .Then ( S ′ ∩ I ) i = S ′ i ∩ I i in a neighborhood of ∗ .Proof. Let g , . . . , g m be generators for S ′ ∩ I . Then g k,i belongs to S ′ i ∩ I i (in some neigh-borhood of ∗ ), and so ( S ′ ∩ I ) i is contained in S ′ i ∩ I i (in some neighborhood of ∗ ), since theformer is generated by g ,i , . . . , g m,i . We now claim that the inclusion ( S ′ ∩ I ) i ⊂ S ′ i ∩ I i isan equality in some neighborhood of ∗ . Assume not. Then we can find a sequence ( h i ) suchthat h i ∈ S ′ i ∩ I i for all i , but in any neighborhood of ∗ there exists i such that h i ( S ′ ∩ I ) i .Let h ∈ S be the element defined by ( h i ). Then h ∈ S ′ , since h i ∈ R ′ i for all i , and h ∈ I ,since h i ∈ I i for all i . Thus h ∈ S ′ ∩ I , and so h = P mk =1 a k g k for some a k ∈ S ′ . But then h i = P mk =1 a k,i g k,i holds in some neighborhood of ∗ , which shows that h i ∈ ( S ′ ∩ I ) i in someneighborhood of ∗ , a contradiction. (cid:3) We close this subsection with a result on strength in ultraproducts:
Proposition 4.6.
Let { R i } be a family of graded rings with graded ultraproduct S . Supposethat the degree piece of R i is a field k i , so that the degree piece of S is the ultraproduct K of these fields. Let f , . . . , f r ∈ S . Suppose that the collective strength of f ,i , . . . , f r,i tendsto infinity as i → ∗ . Then f , . . . , f r has infinite collective strength.Proof. Suppose we have a relation P rj =1 a j f j = P sk =1 g k h k where a i ∈ K are not all zero and g k and h k are elements of positive degree. Represent everything by sequences: a j = ( a j,i ), g j = ( g j,i ), and h j = ( h j,i ). Then, by definition of the ultraproduct, we have P rj =1 a j,i f j,i = P sk =1 g k,i h k,i for all i sufficiently close to ∗ , and moreover, not all the a j,i vanish. But thisshows that f ,i , . . . , f r,i has collective strength < s in this neighborhood of ∗ . (cid:3) The main theorems on ultraproduct rings.
Let { k i } i ∈ I be a family of perfect fieldswith ultraproduct K . The field K is also perfect, as if K has characteristic p >
0, then k i isperfect of characteristic of p for all i sufficiently close to ∗ , and so one can take p th roots in K . Let R i = k i [ x , x , . . . ] with standard grading, and let S be the graded ultraproduct ofthe family { R i } i ∈ I . Theorem 4.7.
The ring S is a polynomial ring. IG POLYNOMIAL RINGS AND STILLMAN’S CONJECTURE 13
Proof.
We use the criteria of §
2. First suppose that K has characteristic 0, and let us provethat S has enough derivations (Definition 2.1). Let f ∈ S be a non-zero homogeneouselement of degree d >
0. Passing to a neighborhood of ∗ , we can assume that each k i hascharacteristic 0 or characteristic p with p > d , and that f i = 0. For each i , let a ( i ) be anindex such that x a ( i ) appears in some monomial in f i , and let ∂ i be the derivation ddx a ( i ) of R i . The derivations ( ∂ i ) define a derivation ∂ on S . Since ∂ i ( f i ) = 0 near ∗ , we see that ∂ ( f ) = 0, and so S has enough derivations. Thus S is a polynomial ring (Theorem 2.2).Now suppose that K has characteristic p , and let us prove that S has enough Hassederivations (Definition 2.10). Let f ∈ S be a homogeneous element of positive degree thatis not a p th power. Passing to a neighborhood of ∗ , we can assume that each f i is not a p thpower. For each i , let a ( i ) be an index such that x a ( i ) appears in some monomial in f i withexponent not divisible by p , and let ∂ i be the Hasse derivative on R i with respect to x a ( i ) (Example 2.7). The Hasse derivations ∂ i on the R i induce a Hasse derivation ∂ on S . Since ∂ i ( f i ) = 0 near ∗ , we see that ∂ ( f ) = 0, and so S has enough Hasse derivations. Thus S is apolynomial ring (Theorem 2.11). (cid:3) Theorem 4.8. If I ⊂ S is a finitely generated ideal, then codim S ( I ) = codim R i ( I i ) for all i sufficiently close to ∗ .Proof. Let c = codim S ( I ), which is finite by Corollary 3.4. If c = 0 then I = 0, and so I i = 0for all i sufficiently close to ∗ , and so the formula holds. We now proceed by induction on c .Suppose the result holds for c −
1, and let I be an ideal of S of codimension c >
0. Let f ∈ I be a non-zero homogeneous element. We would like for k i > deg f to hold sufficientlyclose to ∗ . Suppose this is not the case. Then, since the size of the k i is bounded near ∗ ,there must exist a single q such that k i = F q for all i sufficiently close to ∗ . It follows that K = F q . Indeed, an element of K is a sequence ( x i ) i ∈ I , and as each x i can only take finitelymany values the sequence must be constant in a neighborhood of ∗ . Let e > q e > deg f , let k ′ i = F q e , let R ′ i = k ′ i [ x , x , . . . ], and let S ′ be the ultraproduct ofthe R ′ i . By Proposition 4.3, the natural map F q e ⊗ F q S → S ′ is an isomorphism. Write I ′ and I ′ i for the extension of the ideals I and I i to S ′ and R ′ i , respectively. We have thatcodim R i ( I i ) = codim R ′ i ( I ′ i ) and codim S ′ ( I ′ ) = codim S ( I ). It thus suffices to prove thatcodim S ′ ( I ′ ) = codim R ′ i ( I ′ i ) for all i sufficiently close to ∗ . Relabeling, we have reduced to thecase where k i > deg f holds in a neighborhood of ∗ .For each i , let γ i be an automorphism of R i such that γ i ( f i ) is monic in x , at least for i sufficiently close to ∗ (see Lemma 4.9). The family { γ i } induces an automorphism γ of S .Since codimension is invariant under automorphisms, we may replace I with γ ( I ), and so wecan assume that f i is monic in x for all i sufficiently close to ∗ .Let R ′ i = k i [ x , . . . ] and let S ′ be the ultraproduct of { R ′ i } . We have R i = R ′ i [ x ] for each i , and so S ∼ = S ′ [ x ] by Proposition 4.2. Under this identification, f corresponds to a monicpolynomial in S ′ [ x ]. Let I ′ be the contraction of I to S ′ , which is finitely generated byProposition 3.7. We note that I ′ i is the contraction of I i to R ′ i , for all i sufficiently close to ∗ ,by Proposition 4.5. Corollary 3.8 implies that codim S ′ ( I ′ ) = codim S ( I ) − c −
1. Thus,by the inductive hypothesis, we have codim R ′ i ( I ′ i ) = c − i sufficiently close to ∗ . ByCorollary 3.8 again, codim R i ( I i ) = codim R ′ i ( I ′ i ) + 1 = c . The result follows. (cid:3) Lemma 4.9.
Let k be a field, let R = k [ x , x , . . . ] , and let f ∈ R be a non-zero homogeneouselement. If k > deg f then there exists an automorphism γ of R (as a graded k -algebra)such that γ ( f ) is monic in x . Proof.
We may assume that f lies in k [ x , . . . , x n ] for some n . Let d = deg f . We consideran automorphism of the form γ ( x i ) = x i for i = 1 or i > n and γ ( x i ) = x i − a i x for2 ≤ i ≤ n , where a i ∈ k . The coefficient of x d in γ ( f ) can be viewed as an inhomogeneouspolynomial g ( a , . . . , a n ), with deg( g ) ≤ d . Thus, as long as k > d , we can find some choiceof a , . . . , a n where g ( a , . . . , a n ) = 0, (cid:3) Corollary 4.10.
Let f , . . . , f r be homogeneous elements of S . Then f , . . . , f r form a regularsequence in S if and only if f ,i , . . . , f r,i form a regular sequence in R i for all i sufficientlyclose to ∗ .Proof. This follows from Theorem 4.8 and Corollary 3.5. (cid:3)
Stillman’s conjecture.Theorem 4.11.
Given positive integers d , . . . , d r there exists an integer N = N ( d , . . . , d r ) with the following property. If f , . . . , f r are homogeneous elements of k [ x , . . . , x n ] , for anyperfect field k and any n , of degrees d , . . . , d r and collective strength at least N then f , . . . , f r form a regular sequence.Proof. Suppose that the theorem is false. Then for each j ∈ N , we can find f j, , . . . , f j,r in k j [ x , x , . . . ], with k j perfect, which fails to be a regular sequence and where the collectivestrength goes to ∞ as j → ∞ . Choose some function n : I → N where n ( i ) → ∞ as i → ∗ .For each i ∈ I , we let f i, , . . . , f i,r be any of the collections in our sequence of collectivestrength at least n ( i ). We let f = ( f i, ) , . . . , f r = ( f i,r ) be the corresponding collection in S .By Proposition 4.6, f , . . . , f r has infinite collective strength. However, by Corollary 4.10, f , . . . , f r fail to be a regular sequence. This contradicts Theorem 1.1. (cid:3) For completeness, we now illustrate how Theorem 4.11 implies the existence of smallsubalgebras and Stillman’s conjecture. This implication is essentially the same as in [AH16].
Theorem 4.12.
Given positive integers d , . . . , d r there exists an integer s = s ( d , . . . , d r ) with the following property. If f , . . . , f r are homogeneous elements of k [ x , . . . , x n ] , for anyperfect field k and any n , with deg( f i ) = d i , then: (a) There exists a regular sequence g , . . . , g s in k [ x , . . . , x n ] , where each g i is homo-geneous of degree at most max( d , . . . , d r ) , such that f , . . . , f r are contained in thesubalgebra k [ g , . . . , g s ] . (b) The ideal ( f , . . . , f r ) has projective dimension at most s .Proof. (a) To each sequence d = ( d , . . . , d r ) we attach a monomial y ( d ) = y b y b · · · where b j is the number of times j appears in d . If there is an ideal ( f , . . . , f r ) of type d thatfails to be a regular sequence, then by Theorem 4.11 there is some N , depending only on d ,such that some k -linear homogeneous combination of the f i has strength ≤ N . Without lossof generality, we may replace one of our elements with this linear combination, and call it f i . Taking f i = P Nj =1 a j g j , and replacing f i by the g j , we get an ideal of type d ′ and where y ( d ) < y ( d ′ ) in the revlex order, and where the difference in total degree is at most N −
1. Inparticular, given y ( d ) there are only a finite number of possible monomials y ( d ′ ) that couldarise in this way. The descendants of y ( d ) thus form a tree with finitely many branches ateach node and with no infinite chains, and there are thus only finitely many descendants of y ( d ). Letting s be the max total degree of a descendant of y ( d ), it follows that f , . . . , f r can be embedded in a subalgebra k [ g , . . . , g s ] where the g i form a regular sequence. IG POLYNOMIAL RINGS AND STILLMAN’S CONJECTURE 15 (b) Choose g , . . . , g s as in (a). Since g , . . . , g s form a regular sequence, the extension k [ g , . . . , g s ] ⊆ k [ x , . . . , x n ] is flat. Thus, if G is the minimal free resolution of ( f , . . . , f s )over k [ g , . . . , g s ], then the extension of G to k [ x , . . . , x n ] is the minimal free resolution ofthis ideal over k [ x , . . . , x n ]. In particular, the projective dimension of ( f , . . . , f s ) is ≤ s . (cid:3) The inverse limit ring
Inverse limit polynomial ring.
Recall that A JJ x , x , . . . KK denotes the inverse limitof the standard-graded polynomial rings A [ x , . . . , x n ] in the category of graded rings. Welet K denote a ring containing A , and we write α n : K ⊗ A A JJ x , x , . . . KK → K [ x , . . . , x n ] forthe natural surjection. We set R = K ⊗ A A JJ x , x , . . . KK .The following hypothesis will be used repeatedly. Hypothesis 5.1. A is an integral domain with fraction field K . If the characteristic p of K is positive, we assume also that the Frobenius map on A is surjective (so that K isperfect). (cid:3) Remark 5.2. If A is normal and its fraction field K is perfect, then because a /p satisfiesthe integral equation x p − a , it lies in A . Thus, we can often arrange to satisfy Hypothesis 5.1by replacing A with its integral closure in an algebraic closure of K . (cid:3) The following is an analogue of Theorem 4.8 and Corollary 4.10. It implies Theorem 1.1.
Theorem 5.3.
Suppose Hypothesis 5.1 holds. Then R is a polynomial ring.Proof. We use the criteria of §
2. If p = 0, then the partial derivatives ddx i show that R hasenough derivations.Now suppose that p >
0. We claim that the Hasse derivatives corresponding to ddx i (Example 2.7) provide R with enough Hasse derivations. Let f ∈ R be such that ddx i f = 0for all i . This implies that f ∈ K ⊗ A A JJ x p , x p , . . . KK . In particular, we can write f = g/a where a ∈ A and g ∈ A JJ x p , x p , . . . KK . Since the Frobenius map is surjective on A and K , both g and a are p th powers, which implies that f is also a p th power. (cid:3) Remark 5.4.
The perfectness hypothesis in Theorem 5.3 can be relaxed. For example,suppose k is a field of characteristic p such that k is a finite extension of the subfield k p ,and let R = k JJ x , x , . . . KK . Then k R p consists exactly of all (possibly infinite) k -linearcombinations of p th powers of monomials; this uses the hypothesis on k . Thus if f k R p then some Hasse derivative will not kill f , and so R has enough derivations, and is thus apolynomial ring by Remark 2.12. (cid:3) Theorem 5.5.
Suppose Hypothesis 5.1 holds. Let f , . . . , f s ∈ R and let I = ( f , . . . , f s ) . (a) For any n ≫ , we have that codim R ( I ) = codim K [ x ,...,x n ] ( α n ( I )) . (b) The sequence f , . . . , f s forms a regular sequence if and only if α n ( f ) , . . . , α n ( f s ) forms a regular sequence for all n ≫ . (c) If α n ( f ) , . . . , α n ( f s ) forms a regular sequence for some n , then α m ( f ) , . . . , α m ( f s ) forms a regular sequence for all m ≥ n .Proof. (a) We prove this by induction on c = codim R ( I ), which is finite by Corollary 3.4. If c = 0 then I = 0 and the statement is immediate. Now let c > f ∈ I nonzero. Let n large enough so that α n ( f ) = 0. We would like for K to be greater than deg f . Supposethat this is not the case. This implies that A is finite, and since a finite integral domain is a field, we have A = K and R = K JJ x , x , . . . KK . Let K ′ be a finite extension of K and write R ′ = K ′ JJ x , x , . . . KK . By an argument similar to the proof of Proposition 4.3, we see thatthe natural map K ′ ⊗ K R → R ′ is an isomorphism. If I ′ is the extension of I to R ′ , then wehave that codim R ( I ) = codim R ′ ( I ′ ) and codim K [ x ,...,x n ] ( α n ( I )) = codim K ′ [ x ,...,x n ] ( α n ( I ′ )). Itthus suffices to verify the claim after replacing K by K ′ .Relabeling if necessary, we may therefore assume that K is larger than the degree of f .Lemma 4.9 implies that there is a graded K -algebra automorphism γ of K [ x , . . . , x n ] suchthat γα n ( f ) is monic over K [ x , . . . , x n ]. If γ ′ is the automorphism of R which acts by γ on x , . . . , x n and which acts trivially on the other x i , then α n ( γ ′ f ) = γα n f . We may thusassume that f is monic over K ⊗ A A JJ x , x , . . . KK . The rest of the proof is essentially identicalto the proof of Theorem 4.8.(b) This is an immediate consequence of (a) and Corollary 3.5.(c) By Corollary 3.5, we have codim α n ( I ) = s and it suffices to show that codim α n +1 ( I ) = s . Since K [ x , . . . , x n +1 ] / ( α n +1 ( I ) + ( x n +1 )) is isomorphic to K [ x , . . . , x n ] /α n ( I ), the prin-cipal ideal theorem implies that codim α n +1 ( I ) is either s or s + 1. However, α n +1 ( I ) isgenerated by s elements, so its codimension is at most s . Thus codim α n +1 ( I ) = s . (cid:3) Definition 5.6.
Fix a ring A , a field k , and a point y ∈ Spec( A )( k ). For f ∈ A JJ x , x , . . . KK ,we let f y denote the image of f in k JJ x , x , . . . KK . Similarly, for an A JJ x , x , . . . KK -module M ,we let M y = k JJ x , x , . . . KK ⊗ A JJ x ,x ,... KK M . We note that k ⊗ A A JJ x , x , . . . KK is not generallyisomorphic to k JJ x , x , . . . KK .If instead f ∈ A [ x , . . . , x n ], then we let f y denote its image in k [ x , . . . , x n ]. If M is an A [ x , . . . , x n ]-module, then we let M y = k [ x , . . . , x n ] ⊗ A [ x ,...,x n ] M . (cid:3) Corollary 5.7.
Suppose Hypothesis 5.1 holds. Let f , . . . , f s ∈ A JJ x , x , . . . KK be elementswhose images in R form a regular sequence. There exists a dense open set U ⊆ Spec( A ) such that for any algebraically closed field k and any y ∈ U ( k ) , the elements f ,y , . . . , f s,y form a regular sequence in k JJ x , x , . . . KK .Proof. By Theorem 5.5, there is some n so that α n ( f ) , . . . , α n ( f s ) ∈ K [ x , . . . , x n ] formsa regular sequence. Let g i = α n ( f i ), considered as an element of A [ x , . . . , x n ]. Let Q = A [ x , . . . , x n ] / ( g , . . . , g s ) and let π : Spec( Q ) → Spec( A ). Since the generic fiber of π hasdimension n − s , it follows that the locus U ⊆ Spec( A ) of points whose fibers have dimension n − s is dense and Zariski open by semicontinuity of fiber dimension [Stacks, 05F6].Let k be an algebraically closed field and let y ∈ U ( k ). Since dim( Q ⊗ A k ) = n − s ,it follows that g ,y , . . . , g s,y forms a regular sequence. But g i,y equals α n ( f i,y ), and thus byTheorem 5.5(c) and (b), we have that f ,y , . . . , f s,y forms a regular sequence. (cid:3) Lemma 5.8. If k is a perfect field and f , . . . , f s ∈ k JJ x , x , . . . KK is a regular sequence, then i ′ : k [ f , . . . , f s ] → k JJ x , x , . . . KK is faithfully flat.Proof. By Theorem 5.3, we can write k JJ x , x , . . . KK ∼ = k [ V ]. There is a finite subset H ⊆ V such that f , . . . , f s ∈ k [ H ]. Since the f i form a regular sequence, we can extend this to amaximal regular sequence, f , . . . , f s , g , . . . , g r on k [ H ]. The map i ′ factors as k [ f , . . . , f s ] i −→ k [ f , . . . , f s , g , . . . , g r ] i −→ k [ H ] i −→ k JJ x , x , . . . KK . For each extension, the larger ring is free over the smaller ring. Both i and i are extensionsof polynomial rings. For i , freeness follows from [BH93, Proposition 2.2.11] (the statementthere is for a local ring, but the proof also works for a graded ring). (cid:3) IG POLYNOMIAL RINGS AND STILLMAN’S CONJECTURE 17
Constant Betti tables over an open subset.
For a graded ring R with R = k afield, we set β i,j ( M ) = dim k Tor Ri ( M, k ) j . The Betti table of M is the collection of all β i,j . Definition 5.9.
Let A be a commutative ring and let U ⊆ Spec( A ) be a locally closedsubset. Let M be a finitely presented, graded module over either A JJ x , x , . . . KK or over apolynomial ring over A . We say that M has a constant Betti table over U if for everyalgebraically closed field k and every y ∈ U ( k ), the Betti table of M y is the same. (Recallthat M y is defined in Definition 5.6). (cid:3) The following lemma, which is likely known to experts, shows that a finitely presentedmodule over a finite polynomial ring has a constant Betti table over an open subset.
Lemma 5.10.
Let A be an integral domain and let R = A [ y , . . . , y r ] be a graded polynomialring over A , with deg( y i ) ≥ for ≤ i ≤ r . If M ′ is a finitely presented, graded R -module,then M ′ has a constant Betti table over some dense, open subset U ⊆ Spec( A ) .Proof. Let K be the fraction field of A and let G ′ K = [0 → G ′ K,p ∂ p → · · · ∂ → G ′ K, ] be theminimal free resolution of K ⊗ A M ′ over K [ y , . . . , y r ]. Represent each ∂ i by a matrix ϕ i . The entries of each ϕ i have positive degree and, by multiplying by an element in A if needed, we may assume that the entries of each ϕ i also lie in R . These matrices canthen be used to define a bounded, graded complex G ′ of free R -modules. By construction, K ⊗ A coker( G ′ → G ′ ) is isomorphic to K ⊗ A M ′ . Since both M ′ and coker( G ′ → G ′ ) arefinitely presented, this isomorphism extends to an isomorphism over A g for some g ∈ A .Let N be the direct sum of the homology modules H i ( G ′ ) for 1 ≤ i ≤ p . Since N is a finitelygenerated module over the finitely presented extension A → A [ y , . . . , y r ], [Stacks, 051S]implies that there is some h ∈ A such that N h is free over A h . But taking homologycommutes with localization, so K ⊗ A h N h is isomorphic to the homology of the acycliccomplex K ⊗ A G ′ . Since K ⊗ A h N h = 0 and N h is free, this implies that N h = 0 and G ′ h isacyclic. By a similar argument, there exists k ∈ A such that the cokernel of G ′ is free over A k . In sum, if f = ghk , then G ′ f is a free resolution of M ′ f , and M ′ f is a flat A f -module.Let U = Spec( A f ). Let k be a field and y ∈ U ( k ). Since M ′ f is flat over A f , k ⊗ A f G ′ f is afree resolution of M ′ y . The resolution is minimal since each entry of ϕ i had positive degree,and this remains true under localization at f and specialization to k . The Betti table of M ′ y is thus determined by the free modules in G ′ f , and so it does not depend on y . (cid:3) Lemma 5.11.
Suppose Hypothesis 5.1 holds. Let M be a finitely presented, graded A JJ x , x , . . . KK -module. There exist: (a) elements f , . . . , f s ∈ A JJ x , x , . . . KK whose images in R form a regular sequence, and (b) an element g ∈ A and a finitely presented A g [ f , . . . , f s ] -module M ′ such that theextension of M ′ to A g ⊗ A A JJ x , x , . . . KK is isomorphic to A g ⊗ A M .Proof. Let U be a set of homogeneous elements of R + such that R = K [ U ]. Since any f ∈ R can be written as a fraction with numerator in A JJ x , x , . . . KK and denominator in A , we mayrescale each element of U so that it lies in A JJ x , x , . . . KK . Since R = K [ U ] is a polynomialring, for any element f ∈ R , there is a finite subset U ′ ⊆ U , and an element γ ∈ A such that f lies in A γ [ U ′ ]. The same holds for any finite collection of elements in R .Let Φ be a finite presentation matrix of M . By the above discussion, we can find distinctelements f , . . . , f s ∈ U and g ∈ A such that each entry of Φ lies in A g [ f , . . . , f s ]. Let Φ ′ bethe same matrix as Φ, but considered as a map of graded, free A g [ f , . . . , f s ]-modules and let M ′ be the cokernel of Φ ′ . By construction, the extension of M ′ to A g ⊗ A A JJ x , x , . . . KK is isomorphic to A g ⊗ A M . The elements f , . . . , f s ∈ R form a regular sequence as they are“variables” (elements of U ). (cid:3) Theorem 5.12.
Suppose Hypothesis 5.1 holds. If M is a finitely presented, graded A JJ x , x , . . . KK -module, then M has a constant Betti table over some dense open subset U ⊆ Spec( A ) .Proof. We apply Lemma 5.11, and let M ′ be the A g [ f , . . . , f s ]-module satisfying the con-clusion of that lemma. Applying Lemma 5.10 to M ′ , we can assume that M ′ has constantBetti table over a dense open subset U ⊆ Spec( A g ). By Corollary 5.7, we can find a denseopen subset U ⊆ Spec( A g ) where for all algebraically closed fields k and all y ∈ U ( k ), thesequence f ,y , . . . , f s,y forms a regular sequence. We let U = U ∩ U .Let k be an algebraically closed field and let y ∈ U ( k ). We have a commutative diagram A g [ f , . . . , f s ] / / (cid:15) (cid:15) k [ f ,y , . . . , f s,y ] i ′ (cid:15) (cid:15) A JJ x , x , . . . KK / / A g ⊗ A A JJ x , x , . . . KK / / k JJ x , x , . . . KK where the extension of the k [ f ,y , . . . , f s,y ]-module M ′ y by i ′ is M y . By Lemma 5.8, i ′ isfaithfully flat, and thus the Betti table of M ′ y is the same as the Betti table of M y . Since M ′ has a constant Betti table over U , the module M also has a constant Betti table over U . (cid:3) Corollary 5.13. (We do not assume Hypothesis 5.1.) Let A be an integral domain. If M is a finitely presented, graded A JJ x , x , . . . KK -module, then M has a constant Betti table oversome dense open subset U ⊆ Spec( A ) .Proof. Let A be the integral closure of A in an algebraic closure of K , and let K be thefraction field of A . Let M be the extension of M to A JJ x , x , . . . KK . Since A and K satisfyHypothesis 5.1 (see Remark 5.2), Theorem 5.12 implies that M has a constant Betti tableover some dense open subset U ⊆ Spec( A ).Since the integral morphism Spec( A ) → Spec( A ) is closed [Stacks, 01WM], the image of U in Spec( A ) contains a dense open set U . Let k be an algebraically closed field and let y ∈ U ( k ). By integrality, there is k -point y ′ lying over y , and by definition of U , y ′ ∈ U ( k ).The map y ′ → y induces an isomorphism of M y ′ and M y as k JJ x , x , . . . KK -modules, and theytherefore have the same Betti table. Thus M has a constant Betti table over U . (cid:3) Connection with GL-noetherianity and Stillman’s conjecture.
We now combineCorollary 5.13 with [Dra17] to prove Stillman’s conjecture.Throughout this section we fix a ground field k . Fix degrees d , . . . , d r . Let S be the set ofpairs ( α, i ) where 1 ≤ i ≤ r and α ranges over all exponent vectors of degree d i in the variables x , x , . . . . Let A = k [ c α,i | ( α, i ) ∈ S ]. For 1 ≤ i ≤ r , let e f i = P c α,i x α ∈ A JJ x , x , . . . KK be auniversal polynomial of degree d i . We let Q = A JJ x , x , . . . KK / ( e f , . . . , e f r ). If k ′ is a field over k , then there is a bijection between Spec( A )( k ′ ) and tuples f , . . . , f r ∈ k ′ JJ x , x , . . . KK wheredeg( f i ) = d i ; with notation from Definition 5.6, this bijection is given by y ∈ Spec( A )( k ′ ) ↔ e f ,y , . . . , e f r,y .There is a natural change of basis action by the group scheme GL ∞ on Spec( A ). The A JJ x , x , . . . KK -module Q is equivariant with respect to this action. IG POLYNOMIAL RINGS AND STILLMAN’S CONJECTURE 19
Theorem 5.14.
The space
Spec( A ) decomposes into a finite disjoint union of locally closedsubsets { U j } such that Q has a constant Betti table over U j for each j . In particular, there areonly finitely many distinct Betti tables among all ideals ( f , . . . , f r ) ⊆ k ′ [ x , . . . , x n ] generatedin degrees d , . . . , d r , for all n and all fields k ′ over k .Proof. Applying Corollary 5.13, we have that Q has a constant Betti table over a dense, opensubset U ′ ⊆ Spec( A ). Let U be the union of all GL ∞ translates of U ′ . Since Betti tables are GL ∞ -invariant, Q has a constant Betti table over U . By [Dra17, Theorem 1], Spec( A ) \ U consists of finitely many irreducible components, each of which is GL ∞ -invariant. Passingto a component, we can apply the same argument. Continuing in this way, we obtain thedesired stratification of Spec( A ), and it is finite by [Dra17, Theorem 1].For any field k ′ , the canonical map k ′ [ x , . . . , x n ] ⊗ k ′ k ′ JJ x n +1 , . . . KK → k ′ JJ x , x , . . . KK is anisomorphism. It follows that for f , . . . , f r ∈ k ′ [ x , . . . , x n ] the Betti table of k ′ [ x , . . . , x n ] / ( f , . . . , f r )is the same as that of k ′ JJ x , x , . . . KK / ( f , . . . , f r ), and this implies the final statement of thetheorem for algebraically closed fields k ′ . To get the statement for arbitrary k ′ , we let k ′ be an algebraic closure of k ′ and note that the extension k ′ [ x , . . . , x n ] → k ′ [ x , . . . , x n ] isfaithfully flat, and hence Betti tables are unchanged under this extension. (cid:3) Remark 5.15.
It would be interesting to extend [Dra17, Theorem 1] to spaces over Z , asthis would yield characteristic free bounds in the above result. (cid:3) Remark 5.16.
Theorem 5.14 slightly generalizes Stillman’s conjecture, as it also applies toideals ( f , . . . , f r ) in k JJ x , x , . . . KK that use an infinite number of variables. (cid:3) Remark 5.17.
The proof of Theorem 5.14 is much less self-contained than our ultraproductproof, however it is distinctly different in character: it does not rely on the notion of strength,but rather on a generalized noetherianity principle. This is pursued in more detail in [ESS]to obtain generalizations of Stillman’s conjecture. (cid:3)
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E-mail address : [email protected] URL : http://math.wisc.edu/~derman/ Department of Mathematics, University of Wisconsin, Madison, WI
E-mail address : [email protected] URL : http://math.wisc.edu/~svs/ Department of Mathematics, University of Michigan, Ann Arbor, MI
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