Classifying local Artinian Gorenstein algebras

The classification of local Artinian Gorenstein algebras is equivalent to the study of orbits of a certain non-reductive group action on a polynomial ring. We give an explicit formula for the orbits and their tangent spaces. We apply our technique to analyse when an algebra is isomorphic to its associated graded algebra. We classify algebras with Hilbert function (1, 3, 3, 3, 1), obtaining finitely many isomorphism types, and those with Hilbert function (1, 2, 2, 2, 1, 1, 1). We consider fields of arbitrary, large enough, characteristic.

Let k be a field.So far we do not impose any conditions such as k being algebraically closed or on its characteristic (see Section 3), however the reader may freely assume that k = C.
Let (A, m A , k) be a local Artinian Gorenstein k-algebra.As the notation suggests, throughout the paper we assume that k → A/m A is an isomorphism.This assumption is automatic if k is algebraically closed.It is standard and it is used to discard examples such as k = R ⊂ A = C.
An important numerical invariant of A is the Hilbert function H A (i) = dim k m i A /m i+1 A .The socle degree d is the maximal number such that H A (d) = 0.If H A (d − i) = H A (i) for all i, then the associated graded algebra gr A = i 0 m i A /m i+1 A is also Gorenstein, see [Iar94, Proposition 1.7].
We may now formulate our motivating problems: 1. What is the classification up to isomorphism of local Artinian Gorenstein algebras with Hilbert function H A ?
2. If H A is symmetric, what are the sufficient conditions for A ≃ gr A?
The first of these problems naturally leads to the second: suppose we are classifying A whose Hilbert function H A is symmetric.Since gr A is in this case Gorenstein, one natural way is to first determine possible gr A and then classify A having fixed gr A. Note that it is convenient to classify graded Gorenstein algebras by using ideas from projective geometry: Waring, border and smoothable ranks, and secant varieties, see [Ger96,IK99] for the overview of methods and [LO13,BB14] for the state of the art on secant varieties.See [MZ15,Sta96] and references therein for information about the possible Hilbert functions.See [Iar94,Appendix] for a list of possible Hilbert functions for small values of dim k A.
Our main goal is to refine, simplify and make explicit the existing theory.Having done this, we are gratified with classification results in several cases: 1. We reprove and extend the result of Elias and Rossi: Proposition 1.1 ([ER12, ER15], Corollary 3.14).Let k be a field of characteristic other than 2 or 3. A local Artin Gorenstein k-algebra with Hilbert function (1, n, n, 1) or Hilbert function 1, n, n+1 2 , n, 1 is isomorphic to its associated graded algebras.
The original statement is proven in [ER12,ER15] for fields of characteristic zero.In Example 3.15 we show that the statement is not true for fields of characteristic two or three.
In Proposition 1.1 the condition on the Hilbert function can be rephrased by saying that the algebra is compressed, see Section 3.2.The algebras appearing in Proposition 1.1 are precisely compressed algebras of socle degree three and four.A analogous statement for compressed algebras of higher socle degree is false.However, we show that compressed algebras of all socle degrees are near to the graded algebras, see Corollary 3.13.
3. An algebra is canonically graded if it isomorphic to its associated graded algebra.We investigate the set of (dual socle generators of) algebras which are canonically graded.In Proposition 3.19 we show that this set is irreducible, but in general neither open nor closed in the parameter space.This answers a question of Elias and Rossi [ER15, Remark 3.6].
We now explain our approach.In short, the classification problem may be reduced to representation theory, see [Ems78].Assume for a moment for simplicity of presentation that the characteristic of k is zero.Let n = H A (1).Then A may be seen as a quotient of a fixed power series ring S = k[[α 1 , . . ., α n ]].The ring S acts on a polynomial ring P = k[x 1 , . . ., x n ] by letting α i act as the partial derivative with respect to x i .It turns out that A = S/ Ann(f ) for an element f ∈ P , called the dual socle generator of A. Then the problem of classifying algebras reduces to classifying quotients of S, which in turn boils down to classifying f ∈ P .There is a certain group G acting on P such that S/ Ann(f ) ≃ S/ Ann(g) if and only if f and g lie in the same G-orbit.We summarise this (for fields of arbitrary characteristic) in Proposition 2.14.
Our contribution to the above is in providing explicit formulas for all objects involved and in applying Lie-theoretic ideas.In particular, we give formulas for the G-orbit and its tangent space, thus we are able to compute dimensions of orbits.We also find the unipotent radical of G and analyse its orbits, which are closed.
The explicit formulas make several known results, such as those by Elias and Rossi, corollaries of the presented theory.This is, in our opinion, the most important contribution of this paper.Accordingly, we have tried to keep the paper as elementary as possible.
The reader willing to get a quick idea of the contents of this paper should analyse Example 2.16, referring to previous results if necessary.Otherwise we recommend to read through the paper linearly.The paper is organized as follows.First, we recall and develop the abstract theory of inverse systems; in particular we give an explicit formula for the automorphism group action in Proposition 2.9.Then we explain the link between classifying algebras and elements of P in Proposition 2.14.We give some examples, in particular Example 2.16, and then discuss the basic Lie theory in Section 2.4.Finally, in Section 3 we present the applications mentioned above.Apart from the conjecture above, there are several natural questions that are in our opinion worth considering: 1. What are H such that there are finitely many isomorphism types of Artinian Gorenstein algebras with Hilbert function H? What is the classification in these cases?
2. An Artinian Gorenstein algebra is rigid if it cannot be deformed to a non-isomorphic algebra, see [Sch73].The G-orbit of a rigid algebra should be large.For k algebraically closed, are there rigid k-algebras other than k?
3. Given the Hilbert function, are there non-trivial bounds on the dimensions of G-orbits of algebras with this Hilbert function?
4. Can the above ideas of classification be generalised effectively to non-Gorenstein algebras?
2 Preliminaries and theoretical results 2.1 Power series ring S and its dual P In this section we introduce our main objects of study: the power series ring S and its action on the dual divided power (or polynomial) ring P .
By N 0 we denote the set of non-negative integers.Let k be an algebraically closed field of arbitrary characteristic.Let S be a power series ring over k of dimension dim S = n and let m be its maximal ideal.By ord (σ) we denote the order of a non-zero σ ∈ S i.e. the largest i such that σ ∈ m i .Then ord (σ) = 0 if and only if σ is invertible.Let S ∨ = Hom k (S, k) be the space of functionals on S. We denote the pairing between S and S ∨ by −, − : S × S ∨ → k.
Definition 2.1.The dual space P ⊂ S ∨ is the linear subspace of functionals eventually equal to zero: On P we have a structure of S-module via precomposition: for every σ ∈ S and f ∈ P the element σ f ∈ P is defined via the equation This action is called contraction.
Existence of contraction is a special case of the following construction, which is basic and foundational for our approach.Let L : S → S be a k-linear map.Assume that there exists an integer s such that L(m i ) ⊂ m i+s for all i.Then the dual map L ∨ : S ∨ → S ∨ restricts to L ∨ : P → P .Explicitly, L ∨ is given by the equation To obtain contraction with respect to σ we use L(τ ) = στ , the multiplication by σ.Later in this paper we will also consider maps L which are automorphisms or derivations.
To get a down to earth description of P , choose α 1 , . . ., α n ∈ S so that S = k[[α 1 , . . ., α n ]].Write α a to denote α a 1 1 . . .α an n .For every a ∈ N n 0 there is a unique element x [a] ∈ P dual to α a , given by Additionally, we define x i as the functional dual to α i , so that x i = x [(0,...0,1,0,...,0)] with one on i-th position.
Let us make a few remarks: 1.The functionals x [a] form a basis of P .
2. The contraction action is given by the formula Therefore our definition agrees with the standard, see [IK99, Definition 1.1, p. 4].
We say that x [a] has degree a i .We will freely speak about constant forms, linear forms, (divided) polynomials of bounded degree etc.
We endow P with a topology, which is the Zariski topology of an affine space.It will be used when speaking about general polynomials and closed orbits, but for most of the time it is not important.Now we will give a ring structure on P .It will be used crucially in Proposition 2.9.For multi Definition 2.2.We define multiplication on P by In this way P is a divided power ring.
The multiplicative structure on P can be defined in a coordinate-free manned using a natural comultiplication on S. We refer to [Eis95, §A2.4] for details in much greater generality.
Example 2.3.Suppose that k is of characteristic 3. Then P is not isomorphic to a polynomial ring.Indeed, 1 is not in the subring generated by x 1 , . . ., x n .
Note that linear forms from S act on P as derivatives.Therefore we can interpret S as lying inside the ring of differential operators on P .We will need the following related fact.
Lemma 2.4.Let σ ∈ S and denote by σ (i) its i-th partial derivative.For every f ∈ P we have Proof.Since the formula is linear in σ and f we may assume these are monomials.Let σ = α r i τ , where α i does not appear in τ .Then i g where g is a monomial in variables other than x i .Then 4) is valid in this case also.
Remark 2.5.Lemma 2.4 applied to σ = α i shows that α i (x i • f ) − x i • (α i f ) = f .This can be rephrased more abstractly by saying that α i and x i interpreted as linear operators on P generate a Weyl algebra.
Example 2.3 shows that P with its ring structure has certain properties distinguishing it from the polynomial ring, for example it contains nilpotent elements.Similar phenomena do not occur in degrees lower than the characteristic or in characteristic zero, as we show in Proposition 2.6 and Proposition 2.8 below.
Proposition 2.6.Let P d be the linear span of {x [a] | a d}.Then P d is an ideal of P , for all d.Let k be a field of characteristic p.The ring P/P p is isomorphic to the truncated polynomial ring.In fact is an isomorphism.
Proof.Since Ω maps a basis of P/I p to a basis of k[x 1 , . . ., x n ]/(x 1 , . . ., x n ) p , it is clearly well defined and bijective.The fact that Ω is a k-algebra homomorphism reduces to the equality Characteristic zero case.In this paragraph we assume that k is of characteristic zero.This case is technically easier, but there are two competing conventions: contraction and partial differentiation.These agree up to an isomorphism.The main aim of this section is clarify this isomorphism and provide a dictionary between divided power rings used in the paper and polynomial rings in characteristic zero.Contraction was already defined above, now we define the action of S via partial differentiation.
Definition 2.7.Let k[x 1 , . . ., x n ] be a polynomial ring.There is a (unique) action of S on k[x 1 , . . ., x n ] such that the element α i acts a ∂ ∂x i .For f ∈ k[x 1 , . . ., x n ] and σ ∈ S we denote this action as σ • f .The following Proposition 2.8 shows that in characteristic zero the ring P is in fact polynomial and the isomorphism identifies the S-module structure on P with that from Definition 2.7 above.
Proposition 2.8.Suppose that k is of characteristic zero.Let k[x 1 , . . ., x n ] be a polynomial ring with S-module structure as defined in 2.7.Let Ω : P → k[x 1 , . . ., x n ] be defined via Then Ω is an isomorphism of rings and an isomorphism of S-modules.
Proof.The map Ω is an isomorphism of k-algebras by the same argument as in Proposition 2.6.We leave the check that Ω is a S-module homomorphism to the reader.
Summarising, we get the following corresponding notions.

Automorphisms and derivations of the power series ring
Let as before S = k[[α 1 , . . ., α n ]] be a power series ring with maximal ideal m.This ring has a huge automorphism group: for every choice of elements σ 1 , . . ., σ n ∈ m whose images span m/m 2 there is a unique automorphism ϕ : S → S such that ϕ(α i ) = σ i .Note that ϕ preserves m and its powers.Therefore the dual map ϕ ∨ : S ∨ → S ∨ restricts to ϕ ∨ : P → P .The map ϕ ∨ is defined (using the pairing of Definition 2.1) via the condition Now we will describe this action explicitly.
Proof.We need to show that σ, ϕ ∨ (f ) = ϕ(σ), f for all σ ∈ S. Since f ∈ P , it is enough to check this for all σ ∈ k[α 1 , . . ., α n ].By linearity, we may assume that σ = α a .For every g ∈ P let ε(g) = 1, g ∈ k.We have Consider a term of this sum.Observe that for every Indeed it is enough to check the above equality for g = x [c] and both sides are zero unless c = a − b, thus it is enough to check the case g = x [a−b] , which is straightforward.Moreover note that if b a, then the right hand side is zero for all g, because ε is zero for all x [c] with non-zero c.We can use (6) and remove the restriction b a, Consider now a derivation D : S → S.Here and elsewhere a derivation is a k-linear map satisfying the Leibnitz rule.The derivation D gives rise to a dual map D ∨ : P → P .We wish to describe it explicitly.
Proposition 2.10.Let D : S → S be a derivation and Proof.The proof is similar, though easier, to the proof of Proposition 2.9.
We say that D lowers the degree.

Characteristic zero case.
Let k be a field of characteristic zero.By x a we denote the monomial x a 1 1 . . .x an n in the polynomial ring k[x 1 , . . ., x n ].Then, in the notation of Proposition 2.8, we have Clearly, an automorphism of S gives rise to an linear map k We may restate Proposition 2.9 and Proposition 2.10 as Corollary 2.12.Let ϕ : S → S be an automorphism.
Let D : S → S be a derivation and ] and consider a linear map ϕ : S → S given by ϕ(α 1 ) = α 1 and ϕ(α Let us check this equality using Proposition 2.12.We have We have which indeed agrees with our previous computation.When ϕ is not linear, ϕ ∨ is not an endomorphism of k[x 1 , x 2 ] and computing it directly from definition becomes harder.For example, if ϕ(α 1 ) = α 1 and ϕ(α 2 ) = α 2 + α 2 1 , then

Local Artinian Gorenstein algebras
Let (A, m A , k) be a local Artinian Gorenstein algebra.Recall that the socle degree of A is defined as Note that A may be presented as a quotient of a power series ring S over k if and only dim S H A (1). Therefore it is most convenient to analyse the case when dim S = H A (1), which we will usually do.
Recall that by Macaulay's inverse systems, for every local Gorenstein quotient A = S/I there is an f ∈ P such that I = Ann(f ), see [IK99,Iar94], [Eis95, Chapter 21] and [Mac94,Chapter IV].Such an f is called a dual socle generator of A. Given I, the choice of f is non-unique, but any two choices differ by a unit in S. Conversely, for f ∈ P we denote which is called the apolar algebra of f and Ann(f ) is called the apolar ideal of f .It is important that the apolar algebra of the top degree form of f is a quotient of gr A, see [Iar94, Chapter 1].We may compare A and f as follows: As mentioned in the introduction, our interest is in determining when two local Artinian Gorenstein quotients of the power series ring are isomorphic.The following Proposition 2.14 connects this problem with the results of the previous section.It is well-known, see e.g.[Ems78].Let S * denote the group of invertible elements of S and let be the group generated by Aut(S) and S * in the group of linear operators on S. As the notation suggests, the group G is a semidirect product of those groups: indeed ϕ , where ϕ is an automorphism, s ∈ S is invertible and µ s denotes multiplication by s.By Equation (2) we have an action of G on P .Here S * acts by contraction and Aut(S) acts as described in Proposition 2.9.
Proposition 2.14.Let A = S/I and B = S/J be two local Artinian Gorenstein algebras.Choose f, g ∈ P so that I = Ann (f ) and J = Ann (g).The following conditions are equivalent: 1.A and B are isomorphic, 2. there exists an automorphism ϕ : S → S such that ϕ(I) = J, 3. there exists an automorphism ϕ : S → S such that ϕ ∨ (f ) = σ g, for an invertible element σ ∈ S.
4. f and g lie in the same G-orbit of P .
Proof.Taking an isomorphism A ≃ B, one obtains ϕ ′ : S → B = S/I, which can be lifted to an automorphism of S by choosing lifts of linear forms.This proves 1. ⇐⇒ 2. 2. ⇐⇒ 3. Let ϕ be as in 2. Then Ann (ϕ ∨ (f )) = ϕ(Ann (f )) = ϕ(I) = J.Therefore the principal S-submodules of P generated by ϕ ∨ (f ) and g are equal, so that there is an invertible element σ ∈ S such that ϕ ∨ (f ) = σ g.The argument can be reversed.
Finally, 4. is just a reformulation of 3.
The invertible element in Point 3 cannot be discarded, see Example 2.16 and Example 3.27.The outcome of this theorem is that we are interested in the orbits of elements of P under the G action.
Example 2.15 (quadrics).Let f ∈ P 2 be a quadric of maximal rank.Then G•f is the set of (divided) polynomials of degree two, whose quadric part is of maximal rank.
Example 2.16 (compressed cubics).Assume that k has characteristic not equal to two.
We claim that there is an element ϕ ∈ G such that ϕ ∨ (f 3 ) = f .This implies that Apolar (f ) ≃ Apolar (f 3 ) = gr Apolar (f ).We say that the apolar algebra of f is canonically graded.
Let A = Apolar (f 3 ).Since H A (2) = n = H S (1), every linear form in P may be obtained as δ f for some operator δ ∈ S of order two, see Remark 3.11 below.We pick operators D 1 , . . ., D n so that Here we use the assumption on the characteristic.
Let ϕ : S → S be an automorphism defined via ϕ(α i ) = α i + D i .Since (D i D j ) f = 0 by degree reasons, the explicit formula in Proposition 2.9 takes the form The missing term f 0 is a constant, so that we may pick an order three operator σ ∈ S with The isomorphism Apolar (f ) ≃ Apolar (f 3 ) was first proved by Elias and Rossi, see [ER12, Thm 3.3].See Example 2.22 for a more conceptual proof for k = C and Section 3.2 for generalisations.In particular Corollary 3.14 extends this example to cubic and quartic forms.
Remark 2.17 (Graded algebras).In the setup of Proposition 2.14 one could specialize to homogeneous ideals I, J and homogeneous polynomials f, g ∈ P .Then Condition 1. is equivalent to the fact that f and g lie in the same GL(m/m 2 )-orbit.The proof of Proposition 2.14 easily restricts to this case, see [Ger96].Now we turn our attention to derivations.The notation and motivation come from Lie theory, see section on characteristic zero below.Let aut denote the space of derivations of S preserving m, i.e. derivations such that D(m) ⊆ m.Let S ⊂ Hom k (S, S) be given by sending σ ∈ S to multiplication by σ.Let where the sum is taken in the space of linear maps from S to S. Then g acts on P as defined in Equation (2).The space g is naturally the tangent space to the group G. Similarly, aut • f is naturally the tangent space of the orbit Aut(S) • f and gf is the tangent space to the orbit G•f for every f ∈ P .Sometimes it is more convenient to work with S than with P .For each subspace W ⊆ P we may consider the orthogonal space W ⊥ ⊆ S. Below we describe the linear space (gf ) ⊥ .For σ ∈ S by σ (i) we denote the i-th partial derivative of σ.We use the convention that deg(0) < 0.
Proposition 2.18 (tangent space description).Let f ∈ P .Then Suppose further that f ∈ P is homogeneous of degree d.Then (gf ) ⊥ is spanned by homogeneous operators and Proof.Let D ∈ aut and D i := D(α i ).By Proposition 2.10 we have For any δ ∈ m we may choose D so that D i = δ and all other D j are zero.This proves the description of aut • f .Now gf = Sf + span Now let σ ∈ S be an operator such that σ, gf = 0.This is equivalent to σ (gf ) = 0, which simplifies to σ f = 0 and (σm) (x i f ) = 0 for all i.We have σ ( f , thus we get equivalent conditions: and the claim follows.Finally, if f is homogeneous of degree d and σ ∈ S is homogeneous of degree at most d then σ (i) f has no constant term and so deg(σ Remark 2.19.Let f ∈ P be homogeneous of degree d.Let i d and K i := (gf ) ⊥ i .Proposition 2.18 gives a useful connection of K i with the conormal sequence.Namely, let I = Ann (f ) and B = Apolar (f ) = S/I.We have (I 2 ) i ⊆ K i and the quotient space fits into the conormal sequence of S → B, making that sequence exact: This is expected from the point of view of deformation theory.Recall that by [Har10, Theorem 5.1] the deformations of B over k[ε]/ε 2 are in one-to-one correspondence with elements of a k-linear space T 1 (B/k, B).On the other hand, this space fits [Har10, Prop 3.10] into the sequence Since B is Gorenstein, Hom B (−, B) is exact and we have T 1 (B/k, B) i ≃ Hom(K/I 2 , B) i for all i 0. The restriction i 0 appears because Hom(K/I 2 , B) is the tangent space to deformations of B inside S, whereas T 1 (B/k, B) parameterises all deformations.Now we introduce a certain subgroup G + of G.It plays an important part in characteristic zero, because G + -orbits are closed in P .This group is also very useful in applications, because it preserves the top degree form, which allows induction on the degree type arguments.
Each automorphism of S induces a linear map on its cotangent space: we have a restriction Aut(S) → GL(m/m 2 ).Let us denote by Aut + (S) the group of automorphisms which act as identity on the tangent space: We have the following sequence of groups: We define Note that we have the following exact sequence: Correspondingly, let aut denote the space of derivations preserving m, i.e. derivations such that D(m) ⊆ m.Let aut + denote the space of derivations such that D(m) ⊆ m 2 .Denoting by gl m/m 2 the space of linear endomorphisms of m/m 2 , we have we following sequence of linear spaces: We define Following the proof of Proposition 2.18 we get the following proposition.
Proposition 2.20.Let f ∈ P .Then If f is homogeneous of degree d then g + f is spanned by homogeneous polynomials and

Characteristic zero
For simplicity we restrict to k = C in this section, to freely speak about Lie groups and algebras.
As a general reference we suggest [FH91].One technical problem is that the group Aut(S) is infinite-dimensional.In this section we implicitly replace S with S/m D and Aut(S) with the group Aut(S/m D ), where D ≫ 0 is larger than the degree of any element of P considered.Then Aut(S) and G become Lie groups.
Note that the group G + ⊆ GL(S) is unipotent: the map ϕ − id is nilpotent for every ϕ ∈ G + .Therefore, we get the following theorem.
Theorem 2.21.For every f ∈ P the orbit G + •f is closed in P , in both Euclidean and Zariski topologies.
Classically, the Lie algebra corresponding to Aut(S) is aut.Moreover by the exact sequences (8) and (10), the Lie algebra of Aut + (S) is aut + .Finally, the Lie algebra of G is aut + S = g and the Lie algebra of G + is aut + + m = g + .In particular We now give another proof of Elias-Rossi theorem on canonically graded algebras, in the following Example 2.22.
We claim that there is an element ϕ of G such that ϕ ∨ (f 3 ) = f .This proves that Apolar (f ) ≃ Apolar (f 3 ) = gr Apolar (f ).We say that the apolar algebra of f is canonically graded.
In fact, we claim that already G + •f 3 is the whole space: From the explicit formula in Proposition 2.9 we see that G + •f 3 ⊆ f 3 + P 2 .Then it is a Zariski closed subset.To prove equality it enough to prove that these spaces have the same dimension, in other words that g + f 3 = P 2 .

Applications
In this part of the paper we give several new classification results using the above theory.The two most important ones are the t-compressed algebras in Section 3.2, and algebras having Hilbert function (1, 3, 3, 3, 1) in Example 3.27.We begin and end with some examples computable by hand, which are nevertheless useful, see in particular the reference in Example 3.1.
We make the following assumption Assumption.The field k is algebraically closed.It has characteristic zero or greater than the degree of all considered elements of P .
For example, when we analyse cubics, we assume that k has characteristic different from 2 or 3. When analysing degree d polynomials, as in Example 3.1, we assume that the characteristic is at least d + 1 or 0. The fact that k is algebraically closed is assumed because of various geometrical arguments involved, such as in Proposition 3.3, and because of assumptions required for the references: Example 3.27 refers to [LO13] and [CJN15] which both work over an algebraically closed field.It seems plausible that the classification is feasible over non-algebraically closed fields, however then there will be more isomorphism types: already the classifications of homogeneous quadrics over C and R differ.In contrast, the assumption that k has large enough characteristic is hardly avoidable since key results are false without it, see Example 3.15 and compare with Example 2.16.This assumption is frequently used to guarantee that a non-constant element of S has a non-zero derivative, which is needed to effectively apply Proposition 2.18, see Proposition 3.12.
Under Assumption, we may always think of P as a polynomial ring, see Proposition 2.6.Before we begin, we make the following observation.As shown by Equation (9), the group G is build from G + , GL(m/m 2 ) and k * .We will be mostly concerned with G + , in particular we will analyse its orbits.This often suffices to prove important results as in the cubic case above (Example 2.16) or below for t-compressed algebras.However sometimes the action of G + is not sufficient and the difference between this group and G becomes the substance of the result.This is the case for algebras having Hilbert function (1, 3, 3, 3, 1).The group G + is also convenient because it allows degree-by-degree analysis which is our main technical tool, see Proposition 3.3.
When classifying Artinian Gorenstein quotients of S with Hilbert function H we will usually assume that dim S = H(1).This does not reduce generality and simplifies the argument a bit.

Basic examples and tools
In this section we analyse the G-orbits of "easy" homogeneous polynomials: those with with small Waring and border Waring rank.
Example 3.1 (Rank one).The simplest example seems to be the case f = ℓ d for some linear form ℓ. We may assume ℓ = x 1 and apply Proposition 2.20 to see that or, in an invariant form, g + ℓ d = span ℓ r • m | m ∈ P 1 , r + 1 < d .Using Proposition 2.9 one can even compute the orbit itself.For example, when d = 4, the orbit is equal to This As above, the apolar ideal of f is monomial, equal to (α d , β 2 ).Using Proposition 2.20 we easily get that g + f ⊥ <d = (β 3 ) <d , so g + f is spanned by monomials x a y b , where b 2 and a + b < d.Note that in contrast with Example 3.1, the equality (g + f ) <d does not hold.This manifests the fact that Apolar (f ) is a deformation of Apolar x d + y d .
Before we analyse the rank two polynomials i.e. x d + y d , we need a few more observations.Let for every g ∈ P the symbol tdf(g) denote the top degree form of g, so that for example Proposition 3.3.Let f ∈ P .Then the top degree form of every element of G + •f is equal to the top degree form of f .Moreover, If f is homogeneous, then both sides of (11) are equal to the set of homogeneous elements of g + f .
Proof.Consider the S-action on P d .This action descents to an S/m d+1 action.Further in the proof we implicitly replace S by S/m d+1 , thus also replacing Aut(S) and G by appropriate truncations.Let ϕ ∈ G + .Since (id −ϕ) m i ⊆ m i+1 for all i, we have (id −ϕ) d+1 = 0.By our global assumption on the characteristic D := log(ϕ) is well-defined and ϕ = exp(D).We get an injective map exp : g + → G + with left inverse log.Since exp is algebraic we see by dimension count that its image is open in G + .Since log is Zariski-continuous, we get that log(G + ) ⊆ g + , then exp : g + → G + is an isomorphism.Therefore By Remark 2.11 the derivation D ∈ g + lowers the degree, we see that tdf(ϕ . This proves (11).Finally, if f is homogeneous then g + f is equal to the span tdf(h) | h ∈ g + f by Proposition 2.20, and the last claim follows.
For an elementary proof, at least for the subgroup Aut + (S), see [Mat87, Proposition 1.2].
The following almost tautological Corollary 3.4 enables one to prove that a given apolar algebra is canonically graded inductively, by lowering the degree of the remainder.
Corollary 3.4.Let F and f be polynomials.Suppose that the leading form of F − f lies in g + F .Then there is an element ϕ ∈ G + such that deg(ϕ Proof.Let G be the leading form of f − F and e be its degree.By Proposition 3.3 we may find ϕ ∈ Aut + (S) such that tdf(ϕ ∨ (F ) − F ) = −G, so that ϕ ∨ (F ) ≡ F − G mod P e−1 .By the same proposition we have deg(ϕ Example 3.5 (Rank two).Let P = k dp [x, y], S = k[[α, β]] and F = x d +y d for some d 2. Then H Apolar(F ) = (1, 2, 2, 2, . . ., 2, 1).We claim that the orbit G + •F consists precisely of polynomials f having leading form F and such that H Apolar(f ) = H Apolar(F ) .
Let us first compute (g + F ) ⊥ .Since F is homogeneous, Proposition 2.20 shows that Since Ann (F ) = (αβ, α d − β d ), we see that Now we proceed to the description of G + •F .It is clear that every f ∈ G + •F satisfies the conditions given.Conversely, suppose that f ∈ P satisfies these conditions: the leading form of f is F and H Apolar(f ) = H Apolar(F ) .If d = 2, then the claim follows from Example 2.15, so we may assume d 3.By applying α d−2 and β d−2 to f , we see that x 2 and y 2 are leading forms of partials of f .Since H Apolar(f ) (2) = 2, these are the only leading forms of partials of degree two.
Let G be the leading form of f −F .Suppose that G contains a monomial x a y b , where a, b 2. Then α a−1 β b−1 f = xy + l, where l is linear, then we get a contradiction with the conclusion of the previous paragraph.
Since G contains no monomials of the form x a y b with a, b 2, we see that G is annihilated by (g + F ) ⊥ , so it lies in g + F .By Corollary 3.4 we may find u ∈ G + such that deg(uf Thus, replacing f by uf we lower the degree of f − F .Repeating, we arrive to the point where f − F = 0, so that f = F .
The analysis made in Example 3.5 may be generalized to obtain the following Proposition 3.6.
Proposition 3.6.Let f ∈ P be a polynomial with leading form F .Let I = Ann (F ).Fix an integer t 0 and assume that 1. dim k Apolar (f ) = dim k Apolar (F ).

we have
Then there is an element F + g ∈ G + • f such that deg(g) < t.Equivalently, Apolar (f ) ≃ Apolar (F + g) for some polynomial g of degree less than t.
Proof.We apply induction with respect to deg(f Since Apolar (F ) is a quotient of gr Apolar (f ), the first condition implies that Apolar (F ) ≃ gr Apolar (f ) , Thus for every element i ∈ I = Ann (F ) we may find σ ∈ S such that (i + σ) f = 0 and ord (σ) > ord (i).
Let G be the leading form of f − F and t r d − 1 be its degree.We will now prove that G ∈ g + F .By assumption, it is enough to show that I 2 r annihilates G.The ideal I is homogeneous so it is enough to show that for any elements i, j ∈ I such that deg(ij) = r we have and the claims follow., n, 1).Then the apolar algebra of f is isomorphic to an apolar algebra of F + g, where g is a polynomial of degree at most three.
Indeed, let I = Ann (F ).Then the generators of I <d are α i α j for i = j, thus the ideals I <d and (I 2 ) <d are monomial.Also the ideal (g + F ) ⊥ is monomial by its description from Proposition 2.20.The only monomials of degree at least 4 which do not lie in I 2 are of the form α d−1 i α j and these do not lie in (g + F ) ⊥ .Therefore I 2 i = (g + F ) ⊥ i for all 4 i d − 1.The assumptions of Proposition 3.6 are satisfied with t = 4 and the claim follows.
It is worth noting that for every g of degree at most three the apolar algebra of F + g has Hilbert function (1, n, n, . . ., n, 1), so the above considerations give a full classification of polynomials in P with this Hilbert function and leading form F .
Since (g + F ) ⊥ 3 = span α i α j α l | i < j < l , we may also assume that f = F + i<j<l λ ijl x i x j x l for some λ ijl ∈ k, see also Example 3.27.

Compressed algebras and generalisations
In this section we use Proposition 2.18 to obtain a generalisation of wonderful results of Elias and Rossi on being canonically graded (see Corollary 3.14) and answer their question stated in [ER15, Remark 3.6].
Example 2.16 is concerned with a degree three polynomial f such that the Hilbert function of Apolar (f ) is maximal i.e. equal to (1, n, n, 1) for n = H S (1).Below we generalise the results obtained in this example to polynomials of arbitrary degree.
Recall that a local Artinian Gorenstein algebra A of socle degree Here we introduce a slightly more general notation.
Definition 3.8 (t-compressed).Let A = S/I be a local Artinian Gorenstein algebra of socle degree d.Let t 1.Then A is called t-compressed if the following conditions are satisfied: Example 3.9.Let n = 2. Then H A = (1, 2, 2, 1, 1) is not t-compressed, for any t.The function Note that it is always true that H A (d − 1) H A (1) H S (1), thus both conditions above assert that the Hilbert function is maximal possible.Therefore they are open in P d .
Remark 3.10.The maximal value of t, for which t-compressed algebras exists, is t = ⌊d/2⌋.Every compressed algebra is t-compressed for t = ⌊d/2⌋ but not vice versa.If A is graded, then The following technical Remark 3.11 will be useful later.Up to some extent, it explains the importance of the second condition in the definition of t-compressed algebras.
Remark 3.11.Let A = Apolar (f ) be a t-compressed algebra with maximal ideal m A .We have The definition of t-compressed algebras explains itself in the following Proposition 3.12.Proposition 3.12.Let f ∈ P be a polynomial of degree d 3 and A be its apolar algebra.Suppose that A is t-compressed.Then the G + -orbit of f contains f + P t+1 .In particular Proof.First we show that P t+1 ⊆ g + f , i.e. that no non-zero operator of order at most t + 1 lies in (g + f ) ⊥ .Pick such an operator.By Proposition 2.20 it is not constant.Let σ ′ be any of its non-zero partial derivatives.Proposition 2.20 asserts that deg(σ ′ f ) 1. Let ℓ := σ ′ f .By Remark 3.11 every linear polynomial is contained in m d−1 f .Thus we may choose a δ ∈ m d−1 such that δ f = ℓ.Then (σ ′ − δ) f = 0. Since d 3, we have d − 1 > ⌊d/2⌋ t, so that σ − δ is an operator of order at most t annihilating f .This contradicts the fact that H A (i) = H S (i) for all i t.Therefore P t+1 ⊆ g + f .Second, pick a polynomial g ∈ f + P t+1 .We prove that g ∈ G + • f by induction on deg(g − f ).The top degree form of g − f lies in g + f .Using Corollary 3.4 we find ϕ ∈ G + such that deg(ϕ For completeness, we state the following consequence of the previous result.
Corollary 3.13.Let f ∈ P be a polynomial of degree d 3 and A be its apolar algebra.Suppose that A is compressed.Then A ≃ Apolar f ⌊d/2⌋+2 .Proof.The algebra A is ⌊d/2⌋-compressed and the claim follows from Proposition 3.12.
As a corollary we reobtain the result of Elias and Rossi, see [ER15, Thm 3.1].
Corollary 3.14.Suppose that A is a compressed Artinian Gorenstein local k-algebra of socle degree d 4. Then A is canonically graded i.e. isomorphic to its associated graded algebra gr A.
Proof.The case d 2 is easy and left to the reader.We assume d 3, so that 3 d 4.
Fix n = H A (1) and choose f ∈ P = k dp [x 1 , . . ., x n ] such that A ≃ Apolar (f ).This is possible by the existence of standard form, see [Iar94, Thm 5.3AB] or [IM15].Let f d be the top degree part of f .Since ⌊d/2⌋ + 2 = d, Corollary 3.13 implies that f d ∈ G + •f .Therefore the apolar algebras of f and f d are isomorphic.The algebra Apolar (f d ) is a quotient of gr Apolar (f ).Since dim k gr Apolar (f ) = dim k Apolar (f ) = dim k Apolar (f d ) it follows that Apolar (f ) ≃ Apolar (f d ) ≃ gr Apolar (f ) , which was to be proved.
The above Corollary 3.14 holds under the assumptions that k is algebraically closed and of characteristic not equal to 2 or 3.The assumption that k is algebraically closed is unnecessary as proven for cubics in Example 2.16, the cases of quartics is similar.Surprisingly, the assumption on the characteristic is necessary.
Example 3.15 (compressed cubics in characteristic two).Let k be a field of characteristic two.Let f 3 ∈ P 3 be a cubic form such that H Apolar(f 3 ) = (1, n, n, 1) and α 2 1 f 3 = 0. Then there is a degree three polynomial f with leading form f 3 , whose apolar algebra is compressed but not canonically graded.
Indeed, take σ = α 2 1 .Then all derivatives of σ are zero because the characteristic is two.By Proposition 2.18 the element σ lies in (gf 3 ) ⊥ .Thus gf 3 does not contain P 2 and so G•f 3 does not contain f 3 + P 2 .Taking any f ∈ f 3 + P 2 outside the orbit yields the desired polynomial.In fact, one can explicitly check that i is an example.A similar example shows that over a field of characteristic three there are compressed quartics which are not canonically graded.
Compressed algebras of socle degree d 5 in two variables.As noted in [ER15, Example 3.4], the claim of Corollary 3.14 is false for d = 5.Below we explain this from the point of view of our theory.First, we give an example of a compressed algebra of socle degree five, which is not canonically graded.

Example 3.16 (H
2 are all partials of F we have H A (2) = 3 and the algebra A is compressed.It is crucial to note that Indeed, the only nontrivial derivative of this element is 4α 3 2 ∈ Ann (F ), so (12) follows from Proposition 2.18.Therefore G•F is strictly contained in F + P 4 .Pick any element f ∈ F + P 4 not lying in G•F .The associated graded of Apolar (f ) is equal to Apolar (F ) but Apolar (f ) is not isomorphic to Apolar (F ), thus Apolar (f ) is not canonically graded.In fact one may pick On the positive side, Corollary 3.13 implies that every compressed algebra of socle degree 5 is isomorphic to Apolar (f ) for some f = f 5 + f 4 .Thus we may always remove the cubic part of f .But even more is true: a general compressed algebra of socle degree five is canonically graded, as shown in Example 3.17 below.
Example 3.17 (H A = (1, 2, 3, 3, 2, 1), general).Let f ∈ P = k dp [x 1 , x 2 ] be a general polynomial of degree five with respect to the natural affine space structure on k dp [x 1 , x 2 ] 5 .Then Apolar (f ) and Apolar (f 5 ) are compressed by [Iar84].The ideal Ann (f 5 ) is a complete intersection of a cubic and a quartic.Since we assumed that f is general, we may also assume that the cubic generator c of Ann (f 5 ) is not a power of a linear form.
We claim that Apolar (f ) is canonically graded.Equivalently, we claim that We will show that (g + f 5 ) ⊥ 4 = 0.This space is spanned by homogeneous elements.Suppose that this space contains a non-zero homogeneous σ.Proposition 2.20 implies that σ is nonconstant and that all partial derivatives of σ annihilate f 5 .These derivatives have degree at most three, thus they are multiples of the cubic generator c of Ann (f 5 ).This implies that all partial derivatives of σ are proportional, so that σ is a power of a linear form.Then also c is a power of a linear form, which contradicts earlier assumption.We conclude that no non-zero σ ∈ (g + f 5 ) ⊥ 4 exists.Now, the equation (13) follows from Corollary 3.4.
Example 3.18 (general polynomials in two variables of large degree).Let F ∈ P = k dp [x 1 , x 2 ] be a homogeneous form of degree d 9 and assume that no linear form annihilates F .The ideal Ann (F ) = (q 1 , q 2 ) is a complete intersection.Let We claim that the apolar algebra of a general (in the sense explained in Example 3.17) polynomial f ∈ P with leading form F is not canonically graded.Indeed, we shall prove that G•F ∩ (F + P <d ) is strictly contained in F + P <d which is the same as to show that K := (gF ) ⊥ <d is non-zero.
The sequence (7) from Remark 2.19 becomes where the rightmost map has degree −1.Note that By assumption d − 2 > 6 = dim (A ⊕ A) d−2 , so that (K/I 2 ) d−1 = 0 and our claim follows.In view of Proposition 3.19 it is natural to ask for which degrees and numbers of variables the answer to the question above is positive; the apolar algebra of a general polynomial in P d is a canonically graded algebra.Example 3.17 shows that this happens for n = 2, d = 5.However, the list of cases where the answer may be positive is short, as we shall see now.
Let us now analyse gF .By Proposition 2.18 the equations of gF are given by σ ∈ S satisfying σ F = 0 and σ (i) F = 0 for all i = 1, . . ., n.
Consider σ ∈ S homogeneous of degree d − 1.The space Ann (F ) d−2 is of codimension Thus for fixed i the condition σ (i) F = 0 amounts to at most n+1 2 linear conditions on the coefficients of σ.Summing over i = 1, . . ., n we get n so that there exists a non-zero σ satisfying (15).Since σ = (d − 1) −1 α i σ (i) also σ F = 0. Thus σ ∈ (gF ) ⊥ is a non-zero element.Therefore gF does not contain the whole P d−1 , which contradicts (14).
We have some computational evidence that in all cases listed in Proposition 3.20 a general polynomial indeed gives a canonically graded algebra.Of course for d 4 it is the result of Elias and Rossi reproved in Corollary 3.14.We put forward the following conjecture.Here n = dim S is the number of variables.
We have some supporting evidence in all cases d = 5, n 6 and d = 6, n = 2. Namely, for a pseudo-randomly chosen form F a machine computation of (gF ) ⊥ via Proposition 2.18 reveals that gF ⊃ P d−1 .
Thus arguing by semicontinuity we get that the map G × P d → P d is dominating and the conjecture follows.However we find the use of computer here far from satisfactory, since it dims the reasons for this result.Hopefully someone would invent a computer-free proof.
Improvements using symmetric decomposition.The condition H A (d − 1) = n in definition of t-compressed algebras may be slightly weakened, which is sometimes useful.The price one pays is a more technical assumption.We present the result below.An example of use of Proposition 3.22 is given in Example 3.29.For information on the symmetric decomposition ∆ r of the Hilbert function, see [Iar94], [CJN15] or [BJMR14, Section 2].In short, the Hilbert function H of an apolar algebra of socle degree d admits a canonical decomposition Proposition 3.22.Let f ∈ P be a polynomial of degree d 3 and A its apolar algebra.Let ∆ • be the symmetric decomposition of H A .Suppose that 1. H A (r) = H S (r) = r+n−1 r for all 0 r t.

Moreover, the assumption H
Thus, for every ℓ ∈ P 1 we have a δ of order greater than t such that δ f = ℓ.Now we repeat the proof of Proposition 3.12 with one difference: instead of referring to Remark 3.11 we use Equation ( 16) to obtain, for a given linear form ℓ, an element δ ∈ S of order greater than t and such that δ f = ℓ.

Further examples
Below we present some more involved examples, which employ the tools developed in the previous section.
By multiplying variables by suitable constants, we may assume c = 0 or c = 1 and obtain three possibilities: Note that the three types appearing above are pairwise non-isomorphic.Indeed, an isomorphism may only occur between Apolar x 3 y and Apolar x 3 y + y 3 .Suppose that such exists, so that there is ϕ ∈ G such that ϕ ∨ (x 3 y) = x 3 y + y 3 .Write ϕ = gu, for g linear and u ∈ G + .Then g preserves x 3 y and a direct check shows that g is diagonal.Then u ∨ (x 3 y) = x 3 y + cy 3 for some non-zero c ∈ k and y 3 = c −1 tdf(u(x 3 y) − x 3 y) ∈ g + (x 3 y), by Proposition 3.3.This is a contradiction with Example 3.2.This example was analysed, among others, in [Cas10], where Casnati classifies all Artinian Gorenstein algebras of length at most 9. Example 3.27 (Hilbert function (1, 3, 3, 3, 1)).Consider now a polynomial f ∈ P = k dp [x, y, z] whose Hilbert function is (1, 3, 3, 3, 1).Let F denote the leading form of f .By [LO13] or [CJN15, Prop 4.9] the form F is linearly equivalent to one of the following: Since Apolar (f ) is 1-compressed, we have Apolar (f ) ≃ Apolar (f 3 ); we may assume that the quadratic part is zero.In fact by the explicit description of top degree form in Proposition 3.3 we see that Recall that G/G + is the product of the group of linear transformations and k * acting by multiplication.
The case F 1 .Example 3.7 shows that (g + F ) ⊥ 3 is spanned by αβγ.Therefore we may assume f = F 1 + c • xyz for some c ∈ k.By multiplying variables by suitable constants and then multiplying whole f by a constant, we may assume c = 0 or c = 1.As before, we get two non-isomorphic algebras.Summarising, we got two isomorphism types: f 1,0 = x 4 + y 4 + z 4 , f 1,1 = x 4 + y 4 + z 4 + xyz.
Note that f 1,0 is canonically graded, whereas f 1,1 is a complete intersection.

Conclusion:
There are 11 isomorphism types of algebras with Hilbert function (1, 3, 3, 3, 1).We computed the tangent spaces to the corresponding orbits in characteristic zero, using a computer implementation of the description in Proposition 2.18.The dimensions of the orbits are as follows: The closure of the orbit of f 1,1 = x 4 + y 4 + z 4 + xyz is contained in GL 3 (x 4 + y 4 + z 4 ) + P 3 , which is irreducible of dimension 29.Since the orbit itself has dimension 29 it follows that it is dense inside.Hence the orbit closure contains GL 3 (x 4 + y 4 + z 4 ) + P 3 .Moreover, the set GL 3 (x 4 + y 4 + z 4 ) is dense inside the set σ 3 of forms F whose apolar algebra has Hilbert function (1, 3, 3, 3, 1).Thus the orbit of f 1,1 is dense inside the set of polynomials with Hilbert function (1, 3, 3, 3, 1).Therefore, the latter set is irreducible and of dimension 29.
It would be interesting to see which specializations between different isomorphism types are possible.There are some obstructions.For example, the GL 3 -orbit of x 3 y + x 2 z 2 has smaller dimension than the GL 3 -orbit of x 3 y + z 4 .Thus x 3 y + x 2 z 2 + y 3 + yz 2 does not specialize to x 3 y + z 4 even though its G-orbit has higher dimension.
Example 3.28 (Hilbert function (1, 3, 4, 3, 1)).Consider now a polynomial f ∈ k dp [x, y, z] whose apolar algebra has Hilbert function H = (1, 3, 4, 3, 1).Let F be the leading form of f .Since H is symmetric, the apolar algebra of F also has Hilbert function H.In particular it is annihilated by 2-dimensional space of quadrics.Denote this space by Q. Suppose that the quadrics in Q do not share a common factor, then they are a complete intersection.By looking at the Betti numbers, we see that Ann (F ) itself is a complete intersection of Q and a cubic.From this point of view, the classification of F using our ideas seems ineffective compared to classifying ideals directly.Therefore we will not attempt a full classification.Instead we show that there are infinitely many isomorphism types and discuss non-canonically graded algebras.
Let V = V (Q) ⊂ P 2 be the zero set of Q inside the projective space with coordinates α, β, γ.
for some non-zero numbers λ i .Then the Hilbert function of Apolar (f ) is (1, 3, 4, 3, 1) and V (Q) is a set of four points, no three of them collinear.Suppose that F λ and F λ ′ are in the same G-orbit.By Remark 2.17 there is an element of GL(m/m 2 ) mapping F λ to F λ ′ .Such element stabilizes V (Q), which is a set of four points, no three of them collinear.But the only elements of GL(m/m 2 ) stabilizing such set of four points are the scalar matrices.Therefore we conclude that the set of isomorphism classes of F λ is the set of quadruples λ • up to homothety.This set is in bijection with (k * ) 4 /k * ≃ (k * ) 3 , thus infinite.It fact the set of F λ with λ 4 = 1 is a threefold in the moduli space of finite algebras with a fixed basis, see [Poo08b] for construction of this space.
Non-canonically graded algebras.We now classify forms F such that all polynomials f with leading form F lie in G•F .As in Example 3.27 we see that Thus we investigate (g + F ) ⊥ 3 using Proposition 2.20.Let us suppose it is non-zero and pick a non-zero element σ ∈ (g + F ) ⊥ 3 .Then σ (i) ∈ Ann (F ) 2 = Q for all i.Since Q is two-dimensional

Conjecture 1. 2 (
Conjecture 3.21).Assume that k is a field of characteristic not equal to 2, 3, 5. Say that an algebra A has type (n, d) if H A (1) = n and the socle degree of A is d.Then a general algebra of type (n, d) is canonically graded if and only if (n, d) belongs to the following list: (a) d 4 and n arbitrary, (b) d = 5 and n 6, (c) d = 6 and n = 2.(d) d arbitrary and n = 1.

Example 3. 7 (
Rank n).Let f ∈ P = k dp [x 1 , . . ., x n ] be a polynomial of degree d 4 with leading form F = x d 1 + . . .+ x d n and Hilbert function H Apolar(f ) = (1, n, n, n, . . . Proposition 3.20.Assume that k is a field of characteristic zero.Suppose that n = dim S and d are such that the apolar algebra of a general polynomial in P d is canonically graded.Then (n, d) belongs to the following list: 1. d 4 and n arbitrary, 2. d = 5 and n 6, 3. d = 6 and n = 2, 4. d arbitrary and n = 1.Proof.Fix n, d outside the above list and suppose that G • P d is dense in P d .Since we are over a field of characteristic zero, the tangent map G × P d → P d at a general point (g, F ) is surjective.Then by G-action, this map is surjective also at (1, F ).Its image is gF + P d .Thus gF + P d = P d .

Conjecture 3. 21 .
Assume that k is a field of characteristic not equal to 2, 3, 5.The pairs (n, d) such that the apolar algebra of a general polynomial in P d is canonically graded are precisely the pairs listed below: 1. d 4 and n arbitrary, 2. d = 5 and n 6, 3. d = 6 and n = 2. 4. d arbitrary and n = 1.
orbit dimensionG•(x 4 + y 4 + z 4 + xyz) 29 G•(x 4 + y 4 + z 4 ) 28 G• x 3 y + z 4 + y 3 + y 2 z 28 G• x 3 y + z 4 + y 3 27 G• x 3 y + z 4 + y 2 z 27 G• x 3 y + z 4 26 orbit dimension G• x 3 y + x 2 z 2 + y 3 + yz 2 27 G• x 3 y + x 2 z 2 + y 3 26 G• x 3 y + x 2 z 2 + y 2 z 26 G• x 3 y + x 2 z 2 + yz 2 25 G• x 3 y + x 2 z 2 24 The set of canonically graded algebras is constructible but not necessarily open nor closed.Again consider P d as an affine space with Zariski topology.Suppose that there exists a polynomial of degree d whose apolar algebra is not canonically graded.Elias and Rossi asked in [ER15, Remark 3.6] whether in this case also the apolar algebra of a general polynomial of this degree is not canonically graded.The following proposition answers this question negatively.Proposition 3.19.Fix the degree d and the number of variables n = dim S. Consider the set G ⊂ P d of dual socle generators of canonically graded algebras of socle degree d.This set is irreducible and constructible, but in general it is neither open nor closed.Proof.The set G = G•P d is the image of G×P d , thus irreducible and constructible.Example 3.16 and Example 3.17 together show that for n = 2, d = 5 this set is not closed.Example 3.18 shows that for n = 2, d 9 this set is not open.