Topological noetherianity for cubic polynomials
Abstract.
Let be the space of complex cubic polynomials in infinitely many variables over the field . We show that this space is noetherian, meaning that any stable Zariski closed subset is cut out by finitely many orbits of equations. Our method relies on a careful analysis of an invariant of cubics we introduce called qrank. This result is motivated by recent work in representation stability, especially the theory of twisted commutative algebras. It is also connected to uniformity problems in commutative algebra in the vein of Stillman’s conjecture.
1. Introduction
Let be the space of degree polynomials in variables over an algebraically closed field of characteristic . Let be the inverse limit of the , equipped with the Zariski topology and its natural action (see §1.6). This paper is concerned with the following question:
Is the space noetherian with respect to the action? That is, can every Zariskiclosed stable subspace be defined by finitely many orbits of equations?
This question may seem somewhat esoteric, but it is motivated by recent work in the field of representation stability, in particular the theory of twisted commutative algebras; see §1.3. It is also connected to certain uniformity questions in commutative algebra in the spirit of (the now resolved) Stillman’s conjecture; see §1.2.
For the question is easy since one can explicitly determine the orbits on . For this is not possible, and the problem is much harder. The purpose of this paper is to settle the case:
Question 1 has an affirmative answer for .
In fact, we prove a quantitative result in finitely many variables that implies the theorem in the limit. This may be of independent interest; see §1.1 for details.
1.1. Overview of proof
The key concept in the proof, and the focus of most of this paper, is the following notion of rank for cubic forms.
Let with . We define the qrank^{1}^{1}1The q here is meant to indicate the presence of quadrics in the expression for . of , denoted , to be the minimal nonnegative integer for which there is an expression with and , or if no such exists (which can only happen if ).
For , the cubic
has qrank . This is proved in §4. In particular, infinite qrank is possible when .
The cubic has qrank 1, as follows from the identity
The cubic therefore has qrank at most , and we expect it is exactly .
The notion of qrank is similar to some other invariants in the literature:

Ananyan–Hochster [AH] define a homogeneous polynomial to have strength if it does not belong to an ideal generated by forms of strictly lower degree. For cubics, qrank is equal to strength plus one.
Let be the locus of forms with . This is the image of the map
The main theorem of [Eg] implies that the domain of the above map is noetherian, and so, by standard facts (see [Dr, §3]), its image is as well. It follows that any stable closed subset of of bounded qrank is cut out by finitely many orbits of equations. Theorem 1 then follows from the following result:
Any stable subset of containing forms of arbitrarily high qrank is Zariski dense.
To prove this theorem, one must show that if is a sequence in of unbounded qrank then for any there is a such that the orbitclosure of projects surjectively onto . We prove a quantitative version of this statement:
Let have qrank (in fact, suffices), and suppose . Then the orbit closure of surjects onto .
The proof of this theorem is really the heart of the paper. The idea is as follows. Suppose that has large qrank. We establish two key facts. First, after possibly degenerating (i.e., passing to a form in the orbitclosure) one can assume that the ’s and ’s are in separate sets of variables, while maintaining the assumption that has large qrank. This is useful when studying the orbit closoure, as it allows us to move the ’s and ’s independently. Second, we show that ’s have large rank in a very stong sense: namely, that within the linear span of the ’s there is a largedimensional subspace such that every nonzero element of it has large rank. The results of [Eg] then imply that the orbit closure of in surjects onto , and this yields the theorem.
1.2. Uniformity in commutative algebra
We now explain one source of motivation for Question 1. An ideal invariant is a rule that assigns to each homogeneous ideal in each standardgraded polynomial algebra (in finitely many variables) a quantity , such that only depends on the pair up to isomorphism. We say that is conestable if , that is, adjoining a new variable does not affect . The main theorem of [ESS] is (in part):
[[ESS]] The following are equivalent:

Let be a conestable ideal invariant that is upper semicontinuous in flat families, and let be a tuple of nonnegative integers. Then there exists an integer such that is either infinite or at most whenever is an ideal generated by elements of degrees . (Crucially, does not depend on .)

For every as above, the space
is noetherian.
Define an ideal invariant by taking to be the projective dimension of as an module. This is conestable and upper semicontinuous in flat families. The boundedness in Theorem 1.2(a) for this is exactly Stillman’s conjecture, proved in [AH].
Theorem 1.2 shows that Question 1 is intimately connected to uniformity questions in commutative algbera in the style of Stillman’s conjecture. The results of [ESS] are actually more precise: if part (b) holds for a single then part (a) holds for the corresponding . Thus, combined with Theorem 1, we obtain:
Let be a conestable ideal invariant that is upper semicontinuous in flat families. Then there exists an integer such that is either infinite or at most , whenever is generated by a single cubic form.
Given a positive integer there is an integer such that the following holds: if is a cubic hypersurface containing finitely many codimension linear subspaces then it contains at most such subspaces.
Given a positive integer there is an integer such that the following holds: if is a cubic hypersurface whose singular locus has codimension then its singular locus has degree at most .
It would be interesting if these results could be proved by means of classical algebraic geometry. It would also be interesting to determine the bound for some small values of .
1.3. Twisted commutative algebras
In this section we put . Our original motivation for considering Question 1 came from the theory of twisted commutative algebras. Recall that a twisted commutative algebra (tca) over the complex numbers is a commutative unital associative algebra equipped with a polynomial action of ; see [SS2] for background. The easiest examples of tca’s come by taking the symmetric algebra on a polynomial representation of : for example, or .
TCA’s have appeared in several applications in recent years, for instance:
A tca is noetherian if its module category is locally noetherian; explicitly, this means that any submodule of a finitely generated module is finitely generated. A major open question in the theory, first raised in [Sn], is:
Is every finitely generated tca noetherian?
So far, our knowledge on this question is extremely limited. For tca’s generated in degrees (or more generally, “bounded” tca’s), noetherianity was proved in [Sn]. (It was later reproved in the special case of modules in [CEF].) For the tca’s and , noetherianity was proved in [NSS]. No other cases are known. We remark that these known cases of noetherianity, limited though they are, have been crucial in applications.
Since noetherianity is such a difficult property to study, it is useful to consider a weaker notion. A tca is topologically noetherian if every radical ideal is the radical of a finitely generated ideal. The results of [Eg] show that tca’s generated in degrees are topologically noetherian. Topological noetherianity of the tca is equivalent to the noetherianity of the space appearing in Question 1. Thus Theorem 1 can be restated as:
The tca is topologically noetherian.
This is the first noetherianity result for an unbounded tca generated in degrees .
1.4. A result for tensors
Using similar methods, we can prove the following result:
The space is noetherian with respect to the action of the group , where denotes the completed tensor product.
We plan to write a short note containing the proof.
1.5. Outline of paper
In §2 we establish a number of basic facts about qrank. In §3 we use these facts to prove the main theorem. Finally, in §4, we compute the qrank of the cubic in Example 1.1. This example is not used in the proof of the main theorem, but we thought it worthwhile to include one nontrivial computation of our fundamental invariant.
1.6. Notation and terminology
Throughout we let be an algebraically closed field of characteristic . The symbols , , and will always denote vector spaces, perhaps infinite dimensional. We write for the space of degree polynomials on equipped with the Zariski topology. Precisely, we identify with the spectrum of the ring . When is infinite dimensional the elements of are certain infinite series and the functions on are polynomials in coefficients. Whenever we speak of the orbit of an element of , we mean its orbit.
Acknowledgements
We thank Bhargav Bhatt, Jan Draisma, Daniel Erman, Mircea Mustata, and Steven Sam for helpful discussions.
2. Basic properties of qrank
In this section, we establish a number of basic facts about qrank. Throughout will denote a vector space and a cubic in . Initially we allow to be infinite dimensional, but following Proposition 2 it will be finite dimensional (though this is often not necessary).
Our first result is immediate, but worthwhile to write out explicitly.
[Subadditivity] Suppose . Then
We defined qrank from an algebraic point of view (number of terms in a certain sum). We now give a geometric characterization of qrank that can, at times, be more useful.
We have if and only if there exists a linear subspace of of codimension at most such that .
Proof.
First suppose , and write . Then we can take . This clearly has the requisite properties.
Now suppose of codimension is given. Let be a basis for , and complete it to a basis of be adding vectors . Let be dual to . We can then write , where every term in uses one of the variables , and these variables do not appear in . Since by assumption and by its definition, we find . But only uses the variables , and these are coordinates on , so we must have . Thus every term of has one of the variables in it, and so we can write for appropriate , which shows . ∎
In the above proposition, means that the image of in is 0. It is equivalent to ask for all .
The next result shows that one does not lose too much qrank when passing to subspaces.
Suppose has codimension . Then for we have
Proof.
Our next result shows that if is infinite dimensional then the qrank of can be approximated by the qrank of for a large finite dimensional subspace of . This will be used at a key juncture to move from an infinite dimensional space down to a finite dimensional one.
Suppose (directed union). Then .
We first give two lemmas. In what follows, for a finite dimensional vector space we write for the Grassmannian of codimension subspaces of . For a point of , we write for the corresponding subspace of . By “variety” we mean a reduced scheme of finite type over .
Let be finite dimensional vector spaces, and let be a closed subvariety. Suppose that for every point of the space has codimension in . Then there is a unique map of varieties that on points is given by the formula .
Proof.
Let be the scheme of all linear maps , and let be the open subscheme of surjective linear maps. We identify with the quotient of by the group . The quotient map sends a surjection to its kernel. Let be the inverse image of . There is a natural map given by restricting. By assumption, every closed point of maps into under this map. Since is open, it follows that the map factors through a unique map of schemes . Since this map is equivariant, it descends to the desired map . If is a point of then it lifts to a point of , and the corresponding map has . The image of in is , which establishes the stated formula for our map. ∎
Let be an inverse system of nonempty proper varieties over . Then is nonempty.
Proof.
If then is a nonempty compact Hausdorff space, and the result follows from the wellknown (and easy) fact that an inverse limit of nonempty compact Hausdorff spaces is nonempty.
For a general field , we argure as follows. (We thank Bhargav Bhatt for this argument.) Let be the Zariski topological space underlying the scheme , and let be the inverse limit of the . Since each is a nonempty spectral space and the transition maps are spectral (being induced from a map of varieties), is also a nonempty spectral space [Stacks, Lemma 5.24.2, 5.24.5]. It therefore has some closed point . Let be the image of in .
We claim that is closed for all . Suppose not, and let be such that is not closed. Passing to a cofinal set in , we may as well assume is the unique minimal element. Let be the residue field of , and let be the direct limit of the . The point is then the image of a canonical map of schemes . Since is not closed, it admits some specialization, so we may choose a valuation ring in and a nonconstant map of schemes extending . Since is proper, the map extends uniquely to a map . By uniqueness, the ’s are compatible with the transition maps, and so we get an induced map extending the map . Since is induced from , it follows that is nonconstant. The image of the closed point in under is then a specialization of , contradicting the fact that is closed. This completes the claim that is closed.
Since is closed, it is the image of a unique map of schemes. By uniqueness, these maps are compatible, and so give an element of . ∎
Proof of Proposition 2.
First suppose that is finite dimensional for all . For we have by Proposition 2, and so either or stabilizes. If then by Proposition 2 and we are done. Thus suppose stabilizes. Replacing with a cofinal subset, we may as well assume is constant, say equal to , for all . We must show . Proposition 2 shows that , so it suffices to show .
Let be the closed subvariety consisting of all codimension subspaces such that . This is nonempty by Proposition 2 since has qrank . Suppose and is a point of , that is, is a codimension subspace of on which vanishes. Of course, then vanishes on , which has codimension at most in . Since has qrank exactly , it cannot vanish on a subspace of codimension less than (Proposition 2), and so must have codimension exactly . Thus by Lemma 2, intersecting with defines a map of varieties . This maps into , and so for we have a map . These maps clearly define an inverse system.
Appealing to Lemma 2 we see that is nonempty. Let be a point in this inverse limit, and put . Thus is a codimension subspace of on which vanishes, and for we have . It follows that is a codimension subspace of on which vanishes, which shows (Proposition 2).
We now treat the general case, where the may not be finite dimensional. Write with finite dimensional. Then , so
This completes the proof. ∎
For the remainder of this section we assume that is finite dimensional. If is dimensional then the qrank of any cubic in is obviously bounded above by . The next result gives an improved bound, and will be crucial in what follows.
Suppose . Then , where
Note that .
Proof.
Suppose that is a cubic. Eventually, we want to show that if has large qrank then its orbit under is large. For studying the orbit, it would be convenient if the ’s and the ’s were in separate sets of variables, as then they could be moved independently under the group. This motivates the following definition:
We say that a cubic is separable^{2}^{2}2This notion of separable is unrelated to the notion of separability of univariate polynomials. We do not expect this to cause confusion. if there is a direct sum decomposition and an expression with and .
Now, if we have a cubic of high qrank we cannot conclude, simply based on its high qrank, that it is separable. Fortunately, the following result shows that if we are willing to degenerate a bit (which is fine for our ultimate applications), then we can make it separable, while retaining high qrank.
Suppose that has qrank . Then the orbitclosure of contains a separable cubic satisfying .
Proof.
Let be a basis for . After possibly making a linear change of variables, we can write . Write , where is homogeneous of degree in the variables . Since has degree 3 in the variables , it can contain no other variables, and can thus be regarded as an element of . Therefore, by Proposition 2, we have . After possibly making a linear change of variables in , we can write for some . Let (resp. ) be the result of setting in (resp. ), for . We have by Proposition 2. Of course, , so . By subadditivity (Proposition 2), at least one of or has qrank .
We have where is a quadratic form in the variables with . Thus , and , is separable. We have where is a linear form in the variables with . Thus , and , is separable.
To complete the proof, it suffices to show that and belong to the orbitclosure of , as we can then take or . It is clear that is in the orbitclosure of , so it suffices to show that and are in the orbitclosure of . Consider the element of defined by
Then and . Thus . A similar construction shows that is in the orbitclosure of . ∎
Suppose that is a cubic of high qrank. One would like to be able to conclude that the then have high ranks as well. We now prove two results along this line. For a linear subspace , we let be the maximum of the ranks of elements of , and we let be the minimum of the ranks of the nonzero elements of (or 0 if ).
Suppose has qrank , and let be the span of the . Then for every subspace of we have
Proof.
We may as well assume that and are linearly independent. Thus . Let be a subspace of dimension . After making a linear change of variables in the ’s and ’s, we may as well assume that is the span of . Let . We must show that . Let have rank . Choose a basis of so that . If some for had a term of the form with then some linear combination of and would have rank , a contradiction. Thus every term of , for , has a variable of index , and so we can write where . But now
where . This shows , which completes the proof. ∎
In our eventual application, it is actually that is more important than . Fortuantely, the above result on automatically gives a result for , thanks to the following general proposition.
Let be a linear subspace and let be a positive integer. Suppose that
holds for all linear subspaces . Let and be positive integers satisfying
(1) 
Then there exists a dimensional linear subspace with .
Let be quadratic forms of rank . Suppose there is a linear combination of the ’s that has rank at least . Then there is a linear combination of the ’s satisfying .
Proof.
Let be a linear combination of the ’s with and minimal. Since , it follows that . Thus if then would have rank , contradicting the minimality of . Therefore . ∎
Proof of Proposition 2.
Let be a basis for so that is lexicographically minimal. In particular, this implies that . If then lexicographic minimality ensures that any nontrivial linear combination of has rank at least , and so we can take to be the span of these forms. Thus suppose that . In what follows, we put . Note that . In fact, , and so .
For , consider the following statement:

There exist linearly independent such that: (i) is a linear combination of ; (ii) ; and (iii) the span of has minrank at least .
We will prove by induction on . Of course, implies the proposition.
First consider the case . The statement asserts that there exists a nonzero linear combination of such that . Since the span of has codimension in , our assumption guarantees that some linear combination of these forms has rank at least . Since each form has rank , Lemma 2 ensures we can find with .
We now prove assuming . Let be the tuple given by . The span of has codimension in , and so our assumption guarantees that some linear combination has rank at least . By Lemma 2, we can ensure that this has rank at most . Thus (i) and (ii) in are established.
We now show that any nontrivial linear combination has rank at least , which will show that the ’s are linearly independent and establish (iii) in . If then the rank is at least by the assumption on . Thus assume . We have
Since , we thus see that has rank at least , which completes the proof. ∎
Proposition 2 is not specific to ranks of quadratic forms: it applies to any subadditive invariant on a vector space.
Suppose has qrank , let be the span of the ’s, and let and be positive integers such that (1) holds. Then there exists a dimensional linear subspace with .
3. Proof of Theorem 1
We now prove the main theorems of the paper. We require the following result (see [Eg, Proposition 3.3] and its proof):
Let be a point in , with finite dimensional. Write as , and let be the span of the . Let be a dimensional subspace of . Suppose that are linearly independent and that . Then the orbitclosure of surjects onto .
We begin by proving an analog of the above theorem for :
Suppose is finite dimensional. Let have qrank and let be a dimensional subspace of with
Then the orbitclosure of surjects onto .
Proof.
Applying Proposition 2, let be a separable cubic in the orbitclosure of satisfying . Write where and and and the ’s and ’s are linearly independent. Let be the span of the ’s. Put and . Note that
By Corollary 2 we can therefore find a dimensional subspace of with . Making a linear change of variables, we can assume is the span of . Let . This is in the orbitclosure of (and thus ) since it is obtained by setting for . It is crucial here that the ’s and ’s are in different sets of variables, so that setting some ’s to 0 does not change the ’s. By Theorem 3, the orbit closure of in surjects onto . Now let . Since we can write with and . Pick such that in the image of . Then is the image of , which completes the proof. ∎
[Theorem 1.1] Suppose that has qrank and let be a subspace of of dimension with Then the orbitclosure of surjects onto .
Proof.
By definition of , we have (for an integer ) if and only if . Thus the condition in the theorem is equivalent to , where is the left side of the inequality in the theorem. This expression is equal to plus lower order terms, and is therefore less than for ; in fact, is sufficient. Thus for it is enough that ; since , it is enough that . Thus for , the orbit closure of surjects onto . But it obviously then surjects onto smaller subspaces as well, so we only need to assume . ∎
[Theorem 1.1] Let be infinite dimensional. Suppose is Zariski closed, stable, and contains elements of arbitrarily high qrank. Then .
Proof.
It suffices to show that surjects onto for all finite dimensional . Thus let of dimension be given. Let be sufficiently large so that the inequality in Theorem 3 is satisfied and let have qrank at least . By Proposition 2, there exists a finite dimensional subspace of containing such that has qrank at least . Theorem 3 implies that the orbitclosure of surjects onto . Since surjects onto the orbit closure of , the result follows. ∎
It was explained in the introduction how this implies Theorem 1, so the proof is now complete.
4. A computation of qrank
Fix a positive integer , and consider the cubic
in the polynomial ring introduced in Example 1.1. We now show:
The above cubic has qrank .
It is clear that . To prove equality, it suffices by Proposition 2 to show that if is a codimension subspace of . This is exactly the content of the following proposition:
Let be a vector space of dimension and let be a collection of elements that span . Then is nonzero.
Proof.
Arrange the given elements in a matrix as follows:
Note that we are free to permute the rows and apply permutations within a row without changing the value of , e.g., we can switch the values of and , or switch with , without changing . We now proceed to find a basis for among the elements in the matrix according to the following threephase procedure.
Phase 1. Find a nonzero element of the matrix, and move it (using the permutations mentioned above) to the position. Now in rows find an element that is not in the span of (if one exists) and move it to the position. Now in rows