An equivariant Hilbert basis theorem
Abstract.
We prove a version of the Hilbert basis theorem in the setting of equivariant algebraic geometry: given a group acting on a finite type morphism of schemes , if is topologically noetherian, then so is .
2010 Mathematics Subject Classification:
14L301. Introduction
A topological space equipped with an action of a group is noetherian if every descending chain of stable closed subsets of stabilizes. If is a scheme equipped with an action of , one says is topologically noetherian if the topological space is noetherian. The notion of noetherianity has received much attention in recent years due to its connection to representation stability. We recall a few examples:

Cohen [cohen1, cohen2] proved that the scheme (or even , for any ) is topologically noetherian^{1}^{1}1In fact, Cohen’s result is stronger and applies at the level of rings., and used this to prove certain results in universal algebra. This result was rediscovered some decades later by Aschenbrenner, Hillar, and Sullivant [aschenbrenner, hillar], with applications to combinatorial algebra and algebraic statistics.

Draisma–Eggermont [draismaeggermont] considered the scheme of matrices, and showed that is topologically noetherian for any , with . This result was crucial to their study the equations of socalled Plücker varieties. See [draismakuttler] for related results and applications.

Following work of Eggermont [eggermont] and Derksen–Eggermont–Snowden [DES], Draisma [draisma] proved that if is any polynomial representation of then is topologically noetherian. This result has been applied by the present authors [ess] to prove a general finiteness result in commutative algebra.
Unfortunately, there are few general tools for proving that a space is equivariantly noetherian. The purpose of this paper is to establish one such tool: we view our main theorem as a version of the classical Hilbert basis theorem in the setting of equivariant noetherianity.
Recall that the classical Hilbert basis theorem states that if is a noetherian ring then the polynomial ring is again noetherian. This can be recast in the language of schemes as follows: if is a noetherian scheme and is a finite type map of schemes then is noetherian. Our main theorem is the following equivariant version of this statement.
Theorem 1.1.
Let be a equivariant finite type map of schemes. Suppose is topologically noetherian. Then is topologically noetherian.
We emphasize that there are no finiteness assumptions in the above theorem except for those stated. Indeed, the theorem is most interesting for infinite dimensional schemes like those mentioned above. We note the following useful corollary:
Corollary 1.2.
Let be a scheme equipped with an action of a group and a commuting algebraic action of a finitetype algebraic group . Suppose that is topologically noetherian. Then it is also topologically noetherian.
Example 1.3.
Let be a scheme that is topologically noetherian. Then any finite rank equivariant vector bundle over is also noetherian. For example, if is one of , , or , and is a finite length algebraic representation of (in the sense of [infrank]) then is noetherian by [eggsnow] (extending the main result of [draisma]), and so any finite sum of finite tensor products of the rank tautological bundle and its dual is also noetherian. ∎
1.1. Overview of proof
Let notation be as in Theorem 1.1. The idea of the proof is as follows. First, we can proceed by noetherian induction on , that is, we can assume that for every stable closed subset of , the space is topologically noetherian. This allows us to freely pass to open subsets of . Second, finite type schemes over an arbitrary base behave similarly (at least in the ways that we care about) to finite type schemes over a field, assuming we are always allowed to replace the base with an open subset. Combining these two observations, we can in effect pretend that is the spectrum of a field, and then the result is obvious. The technical details of the actual proof are more complicated, but this is at least the intuition.
1.2. Application
We mention one (now defunct) application of our theorem. In [ess], we used Draisma’s theorem mentioned above to prove a vast generalization of Stillman’s conjecture: we showed that any invariant of ideals satisfying certain natural conditions is “degreewise bounded” (Stillman’s conjecture being the case where the invariant is projective dimension). A preliminary version of [ess], written prior to Draisma’s theorem, proved that Draisma’s theorem was in fact equivalent to our generalization of Stillman’s conjecture. Our proof that “generalized Stillman” implies Draisma’s theorem required Corollary 1.2, which is what propelled us to prove Theorem 1.1 in the first place.
Acknowledgments
We thank Bhargav Bhatt for helpful conversations.
2. Preliminaries from topology
Proposition 2.1.
Let be an open map of topological spaces and let be a dense open subset of . Then is a dense open subset of .
Proof.
Suppose that is not dense in , so that its closure is a proper closed subset of . Let be the complement of . This is an open subset of that does not meet . It follows that the open set is disjoint from the dense open set , which is a contradiction. ∎
Proposition 2.2.
Let be a topological space, let be a proper closed set, and let be a dense open subset of . Then is a proper closed subset of .
Proof.
Suppose . The closure of is then . On the other hand, the closure of is contained in . This is a contradiction, so . ∎
Proposition 2.3.
Let be a topological space, and let be subspaces whose union is . Suppose that each is noetherian. Then is noetherian.
Proof.
Let be a descending chain of closed sets in . Then is a descending chain in , and thus stabilizes by noetherianity. Since is the union of the finitely many stable chains , it too is stable. ∎
Remark 2.4.
Let be a topological space on which acts. We can then consider the quotient space . The stable open (or closed, or irreducible) subsets of correspond to the open (or closed, or irreducible) subsets of . Thus we can translate properties of to usual properties of . For instance, is noetherian if and only if is noetherian. In this way, the above results can be applied in the noetherian setting. ∎
3. Dimension vectors
Let be the set of finite sequences , of variable length, where and each is a nonnegative integer. By convention, for of length , we put for . We order lexicographically, so that if , or if and , and so on. The unique minimal element of is the sequence of length 0; it has for all . The following result allows for induction on elements of :
Proposition 3.1.
For each , let be a boolean value. Suppose that is true for all implies is true. Then is true for all .
Proof.
Let be the set of for which is false. Since is a wellorder, if were nonempty then there would be a minimal element . But then is true for all , so would be true as well, a contradiction. So is empty. ∎
Let be a finite type scheme over a separably closed field , and let be the irreducible components of , ordered by dimension (with largest). We define the dimension vector of , denoted , to be the sequence , where . We regard it as an element of . It is invariant under passing to a larger separably closed extension. For a finite type morphism of schemes and , we let be , where is a separably closed point at . We say that is constant if is independent of , and then write for the common value.
4. Preliminaries from algebraic geometry
Proposition 4.1.
Let be an affine scheme, and write (directed union) where each is finitely generated as a algebra. Put .

Let be a morphism of finite presentation. Then there exists and a morphism of finite type such that .

Let and be schemes of finite type over , and let be a morphism of schemes over . Then there exists and a morphism such that is the base change of to .

Let be a finite type morphism such that is flat. Then there exists some in such that is flat.

Let be a scheme of finite presentation over and let be a closed subscheme of that is also of finite presentation over . Then there exists , a finite type scheme over , and a closed subscheme of such that is the base change of .
Proof.
Parts (a) and (b) are parts of [stacks, Tag 01ZM]. Part (c) is a special case of [stacks, Tag 05LY]. For part (d), we argue as follows. Let be given as in (d), and descend to over . The closed immersion is of finite presentation [stacks, Tags 01TY, 02FV], and so is an module of finite presentation. By [stacks, Tag 01ZR], after possibly replacing with a larger index, there is a finite presentation quasicoherent sheaf on and a morphism whose base change to is the surjection . By [stacks, Tag 01ZL, parts (2), (3)], after replacing with a larger index, we can find an inverse to the map where is the kernel of , and hence the original map defines a closed subscheme of whose base change to is . ∎
Proposition 4.2.
Let be a finite type morphism, with reduced. Then there is a dense open subset of such that is flat of finite presentation.
Proof.
This follows from a general version of generic flatness [stacks, Tag 052B]. ∎
Proposition 4.3.
Let be a finite type morphism of noetherian schemes. Then there are open sets of with dense union such that is constant.
Proof.
This proof follows [stacks, Tag 055A] closely (this proposition is really just a refinement of loc. cit.). Since is noetherian, we can replace it with an open dense subscheme in which no two irreducible components intersect. Thus is the disjoint union of its irreducible components, so we may just assume is irreducible. By [stacks, Tag 0551], after replacing with a nonempty open subset, we can find a surjective finite étale morphism with irreducible such that all irreducible components of are geometrically irreducible, where is the generic point of and . Since is open, we may as well replace with . We may further assume is integral, as is insensitive to nilpotents.
Let be the irreducible components of . These are all geometrically irreducible by our reductions. Let be the closure of in . After replacing with a nonempty open subset, we can assume is the union of the [stacks, Tag 054Y]. After shrinking again, we can assume that is geometrically irreducible for all [stacks, Tag 0559]. After shrinking yet again, we can assume that each fiber of has at least irreducible components [stacks, Tag 0554]. Since , it follows that for every , the fiber has exactly irreducible components, namely the , and they are each geometrically irreducible. Finally, by [stacks, Tag 05F6], we can find an open subset of such that the fibers of has constant dimension for each . ∎
Proposition 4.4.
Let be a finite type morphism of nonempty reduced schemes. Then there is a nonempty open subset such that is flat of finite presentation and constant.
Proof.
By Proposition 4.2, after replacing with a dense open subscheme, we can assume is flat of finite presentation. Replacing with some affine open, we can assume is affine. By Proposition 4.1, we can find a finite type morphism with noetherian such that is the base change of along a morphism . We may as well replace with the schemetheoretic image of , which is just the reduced subscheme structure on [stacks, Tag 056B]. In particular, has dense image. By Proposition 4.3, there is a nonempty open subset of such that is constant. Since is dense in it must meet . Therefore is a nonempty open subset of such that is constant (and still flat of finite presentation). ∎
Proposition 4.5.
Let be a flat morphism of finite presentation, let be an open dense subset of , and let be a proper closed subset of . Then is a proper closed subset of .
Proof.
Proposition 4.6.
Let be a flat finite type morphism of noetherian schemes. Assume there is a dense subset of such that is constant for . Then there is an open dense subset of such that is constant.
Proof.
Applying Proposition 4.3, there are open subsets of such that is dense and is constant. Since is dense, it meets each , and so is independent of . It follows that is constant. ∎
Proposition 4.7.
Let be a finite type morphism of reduced noetherian schemes that is flat and constant. Let be a proper closed subscheme of . Then there exists a nonempty open subset of such that is constant and .
Proof.
By Proposition 4.3, there are open subsets of such that is constant and is dense in . By Proposition 4.5, is a proper closed subset of . Thus is a proper subset of for some . Put for this . Thus is a proper closed subset of and is constant. It remains to show that . Since both are constant, we can verify this over a generic point of .
So assume that is a reduced finite type scheme over a field and is a closed subscheme. Let be the irreducible components of which have largest possible dimension . Suppose one of them is not an irreducible component of . Then with repeated times, but has instances of , so . In the other case, all of the are irreducible components of . Then both and begin with instances of , and we replace and with the union and of their components not equal to one of . In particular, , and by induction on dimension, we have which implies . ∎
Proposition 4.8.
Let be a flat morphism of finite presentation between reduced schemes that is constant. Let be a proper closed subscheme of . Then there is a nonempty open subset of such that is flat of finite presentation and constant with .
Proof.
Since is a closed subscheme of and is finite type over , it follows that is finite type over . Thus, by Proposition 4.2, we can find an open dense subset of such that is flat of finite presentation. By Proposition 4.5, is a proper closed subset of . Thus we may as well replace with , and just assume that is flat and of finite presentation over .
Replace with an affine open so that is still a proper closed subscheme of . By Proposition 4.1(d), there is a noetherian scheme , a finite type morphism , a closed subscheme of , and a morphism such that is the pullback of . By Proposition 4.1(c), we can assume that is flat. We may as well replace with the schemetheoretic image of , which is just with the reduced subscheme structure, so we can assume that is dense in . Since is constant for and is dense, it follows from Proposition 4.6 that there is a dense open subset of such that is constant. By Proposition 4.5, is a proper closed subset of . By Proposition 4.7, there is a nonempty open subset of such that is constant with . Since is dense in , it meets , and so is a nonempty open subset of . Clearly, is constant with . ∎
Corollary 4.9.
Let be a equivariant flat morphism of finite presentation between reduced schemes that is constant. Let be a proper closed equivariant subscheme of . Then there is a nonempty equivariant open subset of such that is flat of finite presentation and constant with .
Proof.
By Proposition 4.8, there exists an open set , not necessarily equivariant, with the desired property. Now take . ∎
5. Proof of main results
Consider the following statement, for .
Statement . Let be a equivariant map of schemes that is flat of finite presentation and constant with . Suppose that is topologically noetherian, and that for every proper stable closed subset of the scheme is topologically noetherian. Then is topologically noetherian.
Lemma 5.1.
Statement is true for all .
Proof.
We proceed by induction on (Proposition 3.1). Thus let and as in Statement be given, and assume holds for all . Since the statement is topological, we may assume and are reduced. It suffices to show that every proper stable closed subset of is topologically noetherian. Thus let such a be given. By Corollary 4.9 there exists a nonempty stable open subset of such that is flat of finite presentation and constant with . Thus by , we have that is noetherian. Since is topologically noetherian, by the assumptions of , the space is also topologically noetherian. It follows that is topologically noetherian (Proposition 2.3), which completes the proof. ∎
Proof of Theorem 1.1.
Let be the given equivariant map of schemes. Since the statement is topological, we can assume and are reduced. We proceed by noetherian induction on : that is, we assume that for every proper closed subset of the space is topologically noetherian. By Proposition 4.6 and Corollary 4.9 there is a nonempty stable open subset of such that is flat of finite presentation and constant. Put . By , it follows that is topologically noetherian. By the inductive hypothesis, is topologically noetherian. Thus is topologically noetherian (Proposition 2.3). ∎
Proof of Corollary 1.2.
Suppose that is topologically noetherian, where is an arbitrary group and is a finite type algebraic group acting algebraically on . Consider the action map . Let act on by , and let act on by . Then is equivariant. Since is finite type, so is . The theorem therefore implies that is topologically noetherian. If is a descending chain of stable closed subsets then is a descending chain of stable closed subsets, and thus stabilizes. Thus stabilizes, and so is topologically noetherian. ∎