Classifying local Artinian Gorenstein algebras

Abstract

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.

1 Introduction

The problem of classifying Artinian Gorenstein algebras up to isomorphism is hard and long studied, a comprehensive reference is [29]. The results usually rest on additional assumptions: small length [5, 7–9, 37] or being a codimension two complete intersection [3, 15, 18]. See also [11, 12, 20, 26] for other approaches.

Let k be a field. So far we do not impose any conditions such as k being algebraically closed or on its characteristic (see Sect. 3), however the reader may freely assume that $k = \mathbb {C}$.

Let be a local Artinian Gorenstein k-algebra. As the notation suggests, throughout the paper we assume that $k\rightarrow A/\mathfrak {m}_A$ is an isomorphism. This assumption is automatic if k is algebraically closed. It is standard and it is used to discard examples such as $k = \mathbb {R} \subset A = \mathbb {C}$.

An important numerical invariant of A is the Hilbert function . The socle degree d is the maximal number such that $H_A(d)\ne 0$. If for all i, then the associated graded algebra is also Gorenstein, see [28, 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 ?

The first of these problems naturally leads to the second: suppose we are classifying A whose Hilbert function $H_A$ is symmetric. Since is in this case Gorenstein, one natural way is to first determine possible and then classify A having fixed . 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 [22, 29] for the overview of methods and [4, 32] for the state of the art on secant varieties. See [36, 40] and references therein for information about the possible Hilbert functions. See [28, 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

([16, 17], 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 $\left( 1, n, \left( {\begin{array}{c}n+1\\ 2\end{array}}\right) , n, 1\right)$ is isomorphic to its associated graded algebras.

The original statement is proven in [16, 17] 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 Sect. 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.

• 2. We prove that general algebras with Hilbert function (1, 2, 3, 3, 2, 1) and general algebras with Hilbert function $(1, 2, 2, 2, \ldots , 2, 1)$ are isomorphic to their associated graded algebras (see Example 3.17 or Example 3.25 for a precise meaning of the word general).

• 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 [17, Remark 3.6].

We investigate density of the set canonically graded algebras and in Proposition 3.20 we prove the only if part of the following conjecture.

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) $d\le 4$ and n arbitrary,

• (b) $d = 5$ and $n\le 6$,

• (c) $d = 6$ and $n = 2$.

• (d) d arbitrary and $n = 1$.

• 4. We classify algebras with Hilbert function (1, 3, 3, 3, 1), obtaining eleven isomorphism types, see Example 3.27. We also discuss the case of (1, 3, 4, 3, 1), show that there are infinitely many isomorphism types and classify algebras which are not canonically graded, see Example 3.28. Non-canonically graded algebras with Hilbert function (1, 3, 4, 3, 1) are investigated independently in [34].

• 5. We classify algebras with Hilbert function (1, 2, 2, 2, 1, 1, 1) obtaining $|k| + 1$ isomorphism types, see Example 3.29. This example was worked out by Elias and Valla [18].

We now explain our approach. In short, the classification problem may be reduced to representation theory, see [19]. 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 $S = k[[\alpha _1, \ldots ,\alpha _n]]$. The ring S acts on a polynomial ring $P= k[x_1, \ldots ,x_n]$ by letting $\alpha _i$ act as the partial derivative with respect to $x_i$. It turns out that $A = S/{\text {Ann}}\left( f \right)$ for an element $f\in P$, 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 $f\in P$. There is a certain group acting on $P$ such that $S/{\text {Ann}}\left( f \right) \simeq S/{\text {Ann}}\left( g \right)$ if and only if f and g lie in the same -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 -orbit and its tangent space, thus we are able to compute dimensions of orbits. We also find the unipotent radical of 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 Sect. 2.4. Finally, in Sect. 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 [39]. The -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 -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 $\mathbb {N}_{0}$ 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 $\mathfrak {m}$ be its maximal ideal. By ${\text {ord}}\left( \sigma \right)$ we denote the order of a non-zero $\sigma \in S$ i.e. the largest i such that $\sigma \in \mathfrak {m}^i$. Then ${\text {ord}}\left( \sigma \right) = 0$ if and only if $\sigma$ is invertible. Let ${S}^{\vee } = {\text {Hom}}_{k}(S, k)$ be the space of functionals on $S$. We denote the pairing between $S$ and ${S}^{\vee }$ by

$$\begin{aligned} \left\langle -, -\right\rangle : S\times {S}^{\vee } \rightarrow k. \end{aligned}$$

Definition 2.1

The dual space $P\subset {S}^{\vee }$ is the linear subspace of functionals eventually equal to zero:

$$\begin{aligned} P= \left\{ f\in {S}^{\vee }\ |\ \forall _{D\gg 0}\ \left\langle \mathfrak {m}^D, f\right\rangle = 0\right\} . \end{aligned}$$

On $P$ we have a structure of $S$-module via precomposition: for every $\sigma \in S$ and $f\in P$ the element $\sigma \,\lrcorner \,f\in P$ is defined via the equation

$$\begin{aligned} \left\langle \tau , \sigma \,\lrcorner \,f\right\rangle = \left\langle \tau \sigma , f\right\rangle \quad \hbox {for every}\ \tau \in S. \end{aligned}$$

(1)

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 $L:S\rightarrow S$ be a k-linear map. Assume that there exists an integer s such that $L(\mathfrak {m}^{i}) \subset \mathfrak {m}^{i+s}$ for all i. Then the dual map ${L}^{\vee }:{S}^{\vee }\rightarrow {S}^{\vee }$ restricts to ${L}^{\vee }:P\rightarrow P$. Explicitly, ${L}^{\vee }$ is given by the equation

$$\begin{aligned} \left\langle \tau , {L}^{\vee }(f)\right\rangle = \left\langle L(\tau ), f\right\rangle \quad \hbox {for every}\ \tau \in S,\ f\in P. \end{aligned}$$

(2)

To obtain contraction with respect to $\sigma$ we use $L(\tau ) = \sigma \tau$, the multiplication by $\sigma$. 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 $\alpha _1, \ldots ,\alpha _n\in S$ so that $S= k[[\alpha _1, \ldots ,\alpha _n]]$. Write $\alpha ^{\mathbf {a}}$ to denote $\alpha _1^{a_1} \ldots \alpha _n^{a_n}$. For every $\mathbf {a}\in \mathbb {N}_{0}^{n}$ there is a unique element dual to $\alpha ^{\mathbf {a}}$, given by

Additionally, we define $x_i$ as the functional dual to $\alpha _i$, so that with one on i-th position.

Let us make a few remarks:

• 1. The functionals form a basis of $P$.

• 2. The contraction action is given by the formula

  

  Therefore our definition agrees with the standard, see [29, Definition 1.1, p. 4].

We say that has degree $\sum a_i$. 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-indices $\mathbf {a}, \mathbf {b}\in \mathbb {N}_{0}^{n}$ we define $\mathbf {a}! = \prod (a_i!)$, $\sum \mathbf {a}= \sum a_i$ and $\left( {\begin{array}{c}\mathbf {a}+ \mathbf {b}\\ \mathbf {a}\end{array}}\right) = \prod _i \left( {\begin{array}{c}a_i + b_i\\ a_i\end{array}}\right) = \left( {\begin{array}{c}\mathbf {a}+ \mathbf {b}\\ \mathbf {b}\end{array}}\right)$.

Definition 2.2

We define multiplication on $P$ by

(3)

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 [13, §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, . Moreover is not in the subring generated by $x_1, \ldots ,x_n$.

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 $\sigma \in S$ and denote by $\sigma ^{(i)}$ its i-th partial derivative. For every $f\in P$ we have

$$\begin{aligned} \sigma \,\lrcorner \,(x_i\cdot f) - x_i\cdot (\sigma \,\lrcorner \,f) = \sigma ^{(i)}\,\lrcorner \,f. \end{aligned}$$

(4)

Proof

Since the formula is linear in $\sigma$ and f we may assume these are monomials. Let $\sigma = \alpha _i^{r}\tau$, where $\alpha _i$ does not appear in $\tau$. Then $\sigma ^{(i)} = r\alpha _i^{r-1}\tau$. Moreover $\tau \,\lrcorner \,(x_i \cdot f) = x_i\cdot (\tau \,\lrcorner \,f)$, thus we may replace f by $\tau \,\lrcorner \,f$ and reduce to the case $\tau = 1$, $\sigma = \alpha _i^{r}$.

Write where g is a monomial in variables other than $x_i$. Then according to (3). If $s+1 < r$ then both sides of (4) are zero. Otherwise

so Eq. (4) is valid in this case also. $\square$

Remark 2.5

Lemma 2.4 applied to $\sigma = \alpha _i$ shows that $\alpha _i\,\lrcorner \,(x_i\cdot f) - x_i\cdot (\alpha _i\,\lrcorner \,f) = f$. This can be rephrased more abstractly by saying that $\alpha _i$ 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 Propositions 2.6 and 2.8 below.

Proposition 2.6

Let $P_{\ge d}$ be the linear span of . Then is an ideal of $P$, for all d. Let k be a field of characteristic p. The ring $P/P_{\ge p}$ is isomorphic to the truncated polynomial ring. In fact

$$\begin{aligned} \Omega :P/P_{\ge p}\rightarrow k[x_1, \ldots ,x_n]/(x_1, \ldots ,x_n)^p \end{aligned}$$

defined by

is an isomorphism.

Proof

Since $\Omega$ maps a basis of $P/I_p$ to a basis of $k[x_1, \ldots ,x_n]/(x_1, \ldots ,x_n)^p$, it is clearly well defined and bijective. The fact that $\Omega$ is a k-algebra homomorphism reduces to the equality $\left( {\begin{array}{c}\mathbf {a}+ \mathbf {b}\\ \mathbf {a}\end{array}}\right) = \prod \frac{(a_i + b_i)!}{a_i!b_i!}$. $\square$

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 $k[x_1, \ldots ,x_n]$ be a polynomial ring. There is a (unique) action of $S$ on $k[x_1, \ldots ,x_n]$ such that the element $\alpha _i$ acts a $\frac{\partial }{\partial x_i}$. For $f\in k[x_1, \ldots ,x_n]$ and $\sigma \in S$ we denote this action as $\sigma \circ f$.

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 $k[x_1, \ldots ,x_n]$ be a polynomial ring with $S$-module structure as defined in Definition 2.7. Let be defined via

Then is an isomorphism of rings and an isomorphism of $S$-modules.

Proof

The map is an isomorphism of k-algebras by the same argument as in Proposition 2.6. We leave the check that is a $S$-module homomorphism to the reader. $\square$

Summarising, we get the following corresponding notions.

Arbitrary characteristic	Characteristic zero
Divided power series ring $P$	Polynomial ring $k[x_1, \ldots ,x_n]$
$S$-action by contraction (precomposition) denoted $\sigma \,\lrcorner \,f$	$S$ action by derivations denoted $\sigma \circ f$
	$\mathbf {x}^{\mathbf {a}}/\mathbf {a}!$
	$x_i$

2.2 Automorphisms and derivations of the power series ring

Let as before $S= k[[\alpha _1, \ldots ,\alpha _n]]$ be a power series ring with maximal ideal $\mathfrak {m}$. This ring has a huge automorphism group: for every choice of elements $\sigma _1, \ldots ,\sigma _n\in \mathfrak {m}$ whose images span $\mathfrak {m}/\mathfrak {m}^2$ there is a unique automorphism $\varphi :S\rightarrow S$ such that $\varphi (\alpha _i) = \sigma _i$. Note that $\varphi$ preserves $\mathfrak {m}$ and its powers. Therefore the dual map ${\varphi }^{\vee } : {S}^{\vee }\rightarrow {S}^{\vee }$ restricts to . The map is defined (using the pairing of Definition 2.1) via the condition

(5)

Now we will describe this action explicitly.

Proposition 2.9

Let $\varphi :S\rightarrow S$ be an automorphism. Let $D_i = \varphi (\alpha _i) - \alpha _i$. For $\mathbf {a}\in \mathbb {N}_{0}^{n}$ denote $\mathbf {D}^{\mathbf {a}} = D_1^{a_1} \ldots D_n^{a_n}$. Let $f\in P$. Then

Proof

We need to show that $\left\langle \sigma , {\varphi }^{\vee }(f)\right\rangle = \left\langle \varphi (\sigma ), f\right\rangle$ for all $\sigma \in S$. Since $f\in P$, it is enough to check this for all $\sigma \in k[\alpha _1, \ldots ,\alpha _n]$. By linearity, we may assume that $\sigma = \alpha ^{\mathbf {a}}$.

For every $g\in P$ let $\varepsilon (g) = \left\langle 1, g\right\rangle \in k$. We have

$$\begin{aligned} \left\langle \varphi (\sigma ), f\right\rangle= & {} \left\langle 1, \varphi (\sigma )\,\lrcorner \,f\right\rangle = \varepsilon \left( \varphi (\sigma )\,\lrcorner \,f \right) = \varepsilon \left( \sum _{\mathbf {b}\le \mathbf {a}} \left( {\begin{array}{c}\mathbf {a}\\ \mathbf {b}\end{array}}\right) (\alpha ^{\mathbf {a}- \mathbf {b}}\mathbf {D}^{\mathbf {b}}) \,\lrcorner \,f \right) \\= & {} \sum _{\mathbf {b}\le \mathbf {a}} \varepsilon \left( \left( {\begin{array}{c}\mathbf {a}\\ \mathbf {b}\end{array}}\right) \alpha ^{\mathbf {a}- \mathbf {b}}\,\lrcorner \,(\mathbf {D}^{\mathbf {b}}\,\lrcorner \,f) \right) . \end{aligned}$$

Consider a term of this sum. Observe that for every $g\in P$

(6)

Indeed it is enough to check the above equality for and both sides are zero unless $\mathbf {c}= \mathbf {a}- \mathbf {b}$, thus it is enough to check the case , which is straightforward. Moreover note that if $\mathbf {b}\not \le \mathbf {a}$, then the right hand side is zero for all g, because $\varepsilon$ is zero for all with non-zero $\mathbf {c}$. We can use (6) and remove the restriction $\mathbf {b}\le \mathbf {a}$, obtaining

$\square$

Consider now a derivation $D:S\rightarrow S$. Here and elsewhere a derivation is a k-linear map satisfying the Leibnitz rule. The derivation D gives rise to a dual map ${D}^{\vee }:P\rightarrow P$. We wish to describe it explicitly.

Proposition 2.10

Let $D:S\rightarrow S$ be a derivation and $D_i := D(\alpha _i)$. Let $f\in P$. Then

$$\begin{aligned} {D}^{\vee }(f) = \sum _{i=1}^n x_i\cdot (D_i\,\lrcorner \,f). \end{aligned}$$

Proof

The proof is similar, though easier, to the proof of Proposition 2.9. $\square$

Remark 2.11

Suppose $D:S\rightarrow S$ is a derivation such that $D(\mathfrak {m}) \subseteq \mathfrak {m}^2$. Then $\deg ({D}^{\vee }(f)) < \deg (f)$. We say that D lowers the degree.

Characteristic zero case. Let k be a field of characteristic zero. By $\mathbf {x}^{\mathbf {a}}$ we denote the monomial $x_1^{a_1} \ldots x_n^{a_n}$ in the polynomial ring $k[x_1, \ldots ,x_n]$. Then, in the notation of Proposition 2.8, we have

Clearly, an automorphism of $S$ gives rise to an linear map $k[x_1, \ldots ,x_n]\rightarrow k[x_1, \ldots ,x_n]$. We may restate Propositions 2.9 and 2.10 as

Corollary 2.12

Let $\varphi :S\rightarrow S$ be an automorphism. Let $D_i = \varphi (\alpha _i) - \alpha _i$. For $\mathbf {a}\in \mathbb {N}_{0}^{n}$ denote $\mathbf {D}^{\mathbf {a}} = D_1^{a_1} \ldots D_n^{a_n}$. Let $f\in k[x_1, \ldots ,x_n]$. Then

$$\begin{aligned} {\varphi }^{\vee }(f) = \sum _{\mathbf {a}\in \mathbb {N}_{0}^n} \frac{\mathbf {x}^\mathbf {a}}{\mathbf {a}!} \left( \mathbf {D}^{\mathbf {a}} \circ f \right) = f + \sum _{i=1}^n x_i (D_i \circ f) + \ldots . \end{aligned}$$

Let $D:S\rightarrow S$ be a derivation and $D_i := D(\alpha _i)$. Then

$$\begin{aligned} {D}^{\vee }(f) = \sum _{i=1}^n x_i(D_i\circ f). \end{aligned}$$

Example 2.13

Let $n = 2$, so that $S= k[[\alpha _1, \alpha _2]]$ and consider a linear map $\varphi :S\rightarrow S$ given by $\varphi (\alpha _1) = \alpha _1$ and $\varphi (\alpha _2) = \alpha _1 + \alpha _2$. Dually, ${\varphi }^{\vee }(x_1) = x_1 + x_2$ and ${\varphi }^{\vee }(x_2) = x_2$. Since $\varphi$ is linear, ${\varphi }^{\vee }$ is an automorphism of $k[x_1, x_2]$. Therefore ${\varphi }^{\vee }(x_1^3) = (x_1 + x_2)^3$. Let us check this equality using Proposition 2.12. We have $D_1 = \varphi (\alpha _1) - \alpha _1 = 0$ and $D_2 = \varphi (\alpha _2) - \alpha _2 = \alpha _1$. Therefore $\mathbf {D}^{(a, b)} = 0$ whenever $a > 0$ and $\mathbf {D}^{(0, b)} = \alpha _1^b$.

We have

$$\begin{aligned} {\varphi }^{\vee }\left( x_1^3 \right)= & {} \sum _{(a, b)\in \mathbb {N}^2} \frac{x_1^{a}x_2^{b}}{a!b!}(\mathbf {D}^{(a, b)} \circ x_1^3) = \sum _{b\in \mathbb {N}} \frac{x_2^{b}}{b!}(\alpha _1^{b} \circ x_1^3) = x_1^3 + \frac{x_2}{1} \cdot (3x_1^2) + \frac{x_2^2}{2} \cdot (6x_1) \\&+\, \frac{x_2^3}{6}\cdot (6) = (x_1 + x_2)^3, \end{aligned}$$

which indeed agrees with our previous computation.

When $\varphi$ is not linear, ${\varphi }^{\vee }$ is not an endomorphism of $k[x_1, x_2]$ and computing it directly from definition becomes harder. For example, if $\varphi (\alpha _1) = \alpha _1$ and $\varphi (\alpha _2) = \alpha _2 + \alpha _1^2$, then

$$\begin{aligned} {\varphi }^{\vee }(x_1) = x_1,\quad {\varphi }^{\vee }(x_1^4) = x_1^4 + 12x_1^2x_2 + 12x_2^2. \end{aligned}$$

2.3 Local Artinian Gorenstein algebras

Let $(A, \mathfrak {m}_A, k)$ be a local Artinian Gorenstein algebra. Recall that the socle degree of A is defined as $d =\max \{ i\ |\ \mathfrak {m}_A^i\ne 0\}$. The embedding dimension of A is $H_A(1) = \dim \mathfrak {m}_A/\mathfrak {m}_A^2$. Note that A may be presented as a quotient of a power series ring $S$ over k if and only $\dim S\ge H_A(1)$. 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 $f\in P$ such that $I = {\text {Ann}}\left( f \right)$, see [28, 29], [13, Chapter 21] and [33, 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 $f\in P$ we denote

$$\begin{aligned} {\text {Apolar}}(f) = S/{\text {Ann}}\left( f \right) , \end{aligned}$$

which is called the apolar algebra of f and ${\text {Ann}}\left( f \right)$ 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 ${\text {gr }} A$, see [28, Chapter 1]. We may compare A and f as follows:

Algebra A	Module $Sf$
$S$-module A	$\simeq$ $Sf$
$\dim _k A$	= $\dim _k Sf$
Socle degree of A	= $\deg (f)$
Certain quotient of ${\text {gr }} A$	$\simeq {\text {Apolar}}(f_{\deg (f)})$
If for all i we have $H_A(d-i) = H_A(i)$ then ${\text {gr }} A$	$\simeq {\text {Apolar}}(f_{\deg (f)})$

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. [19]. Let $S^*$ denote the group of invertible elements of $S$ and let

be the group generated by and $S^*$ in the group of linear operators on $S$. As the notation suggests, the group is a semidirect product of those groups: indeed $\varphi \circ \mu _s \circ \varphi ^{-1} = \mu _{\varphi (s)}$, where $\varphi$ is an automorphism, $s\in S$ is invertible and $\mu _{s}$ denotes multiplication by s. By Eq. (2) we have an action of on $P$. Here $S^*$ acts by contraction and 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 $f, g\in P$ so that $I = {\text {Ann}}\left( f \right)$ and $J = {\text {Ann}}\left( g \right)$. The following conditions are equivalent:

• 1. A and B are isomorphic,

• 2. there exists an automorphism $\varphi :S\rightarrow S$ such that $\varphi (I) = J$,

• 3. there exists an automorphism $\varphi :S\rightarrow S$ such that ${\varphi }^{\vee }(f) = \sigma \,\lrcorner \,g$, for an invertible element $\sigma \in S$.

• 4. f and g lie in the same -orbit of $P$.

Proof

Taking an isomorphism $A \simeq B$, one obtains $\varphi ':S\rightarrow B = S/I$, which can be lifted to an automorphism of $S$ by choosing lifts of linear forms. This proves $1.\iff 2.$

$2.\iff 3.$ Let $\varphi$ be as in 2. Then ${\text {Ann}}\left( {\varphi }^{\vee }(f) \right) = \varphi ({\text {Ann}}\left( f \right) ) = \varphi (I) = J$. Therefore the principal $S$-submodules of $P$ generated by ${\varphi }^{\vee }(f)$ and g are equal, so that there is an invertible element $\sigma \in S$ such that ${\varphi }^{\vee }(f) = \sigma \,\lrcorner \,g$. The argument can be reversed.

Finally, 4. is just a reformulation of 3. $\square$

The invertible element in Point 3 cannot be discarded, see Examples 2.16 and 3.27. The outcome of this theorem is that we are interested in the orbits of elements of $P$ under the action.

Example 2.15

(quadrics) Let $f\in P_2$ be a quadric of maximal rank. Then 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.

Let $f = f_3 + f_{\le 2}\in P_{\le 3}$ be an element of degree three such $H_{{\text {Apolar}}(f)} = (1, n, n, 1)$, where $n = H_{S}(1)$. Such a polynomial f is called compressed, see [17, 27] or Sect. 3.2 for a definition.

We claim that there is an element such that ${\varphi }^{\vee }(f_3) = f$. This implies that ${\text {Apolar}}(f) \simeq {\text {Apolar}}(f_{3}) = {\text {gr}}\,{\text {Apolar}}(f)$. We say that the apolar algebra of f is canonically graded.

Let $A = {\text {Apolar}}(f_3)$. Since $H_A(2) = n = H_{S}(1)$, every linear form in $P$ may be obtained as $\delta \,\lrcorner \,f$ for some operator $\delta \in S$ of order two, see Remark 3.11 below. We pick operators $D_1, \ldots ,D_n$ so that $\sum x_i \cdot (D_i \,\lrcorner \,f_3) = f_2 + f_1$. Explicitly, $D_i$ is such that $D_i\,\lrcorner \,f_3 = (\alpha _i \,\lrcorner \,f_2)/2 + \alpha _i\,\lrcorner \,f_1$. Here we use the assumption on the characteristic.

Let $\varphi :S\rightarrow S$ be an automorphism defined via $\varphi (\alpha _i) = \alpha _i + D_i$. Since $(D_i D_j)\,\lrcorner \,f = 0$ by degree reasons, the explicit formula in Proposition 2.9 takes the form

$$\begin{aligned} {\varphi }^{\vee }(f_3) = f_3 + \sum x_i \cdot (D_i \,\lrcorner \,f_3) = f_3 + f_2 + f_1. \end{aligned}$$

The missing term $f_0$ is a constant, so that we may pick an order three operator $\sigma \in S$ with $\sigma \,\lrcorner \,{\varphi }^{\vee }(f_3) = f_0$. Then $(1 + \sigma )\,\lrcorner \,({\varphi }^{\vee }(f_3)) = f_3 + f_2 + f_1 + f_0 = f$, as claimed.

The isomorphism ${\text {Apolar}}(f) \simeq {\text {Apolar}}(f_3)$ was first proved by Elias and Rossi, see [16, Thm 3.3]. See Example 2.22 for a more conceptual proof for $k = \mathbb {C}$ and Sect. 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 $f, g\in P$. Then Condition 1. is equivalent to the fact that f and g lie in the same ${\text {GL}}(\mathfrak {m}/\mathfrak {m}^2)$-orbit. The proof of Proposition 2.14 easily restricts to this case, see [22].

Now we turn our attention to derivations. The notation and motivation come from Lie theory, see section on characteristic zero below. Let denote the space of derivations of $S$ preserving $\mathfrak {m}$, i.e. derivations such that $D(\mathfrak {m}) \subseteq \mathfrak {m}$. Let $S\subset {\text {Hom}}_{k}(S, S)$ be given by sending $\sigma \in S$ to multiplication by $\sigma$. Let

where the sum is taken in the space of linear maps from $S$ to $S$. Then acts on $P$ as defined in Eq. (2). The space is naturally the tangent space to the group . Similarly, is naturally the tangent space of the orbit and is the tangent space to the orbit for every $f\in P$.

Sometimes it is more convenient to work with $S$ than with $P$. For each subspace $W\subseteq P$ we may consider the orthogonal space $W^{\perp } \subseteq S$. Below we describe the linear space . For $\sigma \in S$ by $\sigma ^{(i)}$ we denote the i-th partial derivative of $\sigma$. We use the convention that $\deg (0) < 0$.

Proposition 2.18

(Tangent space description) Let $f\in P$. Then

Moreover

Suppose further that $f\in P$ is homogeneous of degree d. Then is spanned by homogeneous operators and

Proof

Let and $D_i := D(\alpha _i)$. By Proposition 2.10 we have ${D}^{\vee }(f) = \sum _{i=1}^n x_i\cdot (D_i\,\lrcorner \,f)$. For any $\delta \in \mathfrak {m}$ we may choose D so that $D_i = \delta$ and all other $D_j$ are zero. This proves the description of . Now . By Lemma 2.4 we have $x_i(\sigma \,\lrcorner \,f) \equiv \sigma \,\lrcorner \,(x_i f) \mod Sf$. Thus

Now let $\sigma \in S$ be an operator such that . This is equivalent to , which simplifies to $\sigma \,\lrcorner \,f = 0$ and $(\sigma \mathfrak {m})\,\lrcorner \,(x_i f) = 0$ for all i. We have , thus we get equivalent conditions:

$$\begin{aligned} \sigma \,\lrcorner \,f = 0\quad \hbox {and} \quad \mathfrak {m}\,\lrcorner \,({\sigma ^{(i)}} \,\lrcorner \,f) = 0, \end{aligned}$$

and the claim follows. Finally, if f is homogeneous of degree d and $\sigma \in S$ is homogeneous of degree at most d then $\sigma ^{(i)}\,\lrcorner \,f$ has no constant term and so $\deg (\sigma ^{(i)}\,\lrcorner \,f) \le 0$ implies that $\sigma ^{(i)}\,\lrcorner \,f = 0$. $\square$

Remark 2.19

Let $f\in P$ be homogeneous of degree d. Let $i \le d$ and . Proposition 2.18 gives a useful connection of $K_i$ with the conormal sequence. Namely, let $I = {\text {Ann}}\left( f \right)$ and $B = {\text {Apolar}}(f) = S/I$. We have $(I^2)_i \subseteq K_{i}$ and the quotient space fits into the conormal sequence of $S\rightarrow B$, making that sequence exact:

$$\begin{aligned} 0\rightarrow (K/I^2)_i\rightarrow (I/I^2)_i \rightarrow (\Omega _{S/k}\otimes B)_i\rightarrow (\Omega _{B/k})_{i}\rightarrow 0. \end{aligned}$$

(7)

This is expected from the point of view of deformation theory. Recall that by [24, Theorem 5.1] the deformations of B over $k[\varepsilon ]/\varepsilon ^2$ 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 [24, Prop 3.10] into the sequence

$$\begin{aligned} 0\rightarrow {\text {Hom}}_{B}(\Omega _{B/k}, B) \rightarrow {\text {Hom}}_{B}(\Omega _{S/k}\otimes B, B)\rightarrow {\text {Hom}}_{B}(I/I^2, B) \rightarrow T^1(B/k, B)\rightarrow 0. \end{aligned}$$

Since B is Gorenstein, ${\text {Hom}}_{B}(-, B)$ is exact and we have $T^1(B/k, B)_i \simeq {\text {Hom}}(K/I^2, B)_i$ for all $i \ge 0$. The restriction $i\ge 0$ appears because ${\text {Hom}}(K/I^2, B)$ 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 of . It plays an important part in characteristic zero, because -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 . Let us denote by the group of automorphisms which act as identity on the tangent space: . We have the following sequence of groups:

(8)

We define

Note that we have the following exact sequence:

(9)

Correspondingly, let denote the space of derivations preserving $\mathfrak {m}$, i.e. derivations such that $D(\mathfrak {m}) \subseteq \mathfrak {m}$. Let denote the space of derivations such that $D(\mathfrak {m}) \subseteq \mathfrak {m}^2$. Denoting by the space of linear endomorphisms of $\mathfrak {m}/\mathfrak {m}^2$, we have we following sequence of linear spaces:

(10)

We define

Following the proof of Proposition 2.18 we get the following proposition.

Proposition 2.20

Let $f\in P$. Then so that

If f is homogeneous of degree d then is spanned by homogeneous polynomials and

2.4 Characteristic zero

For simplicity we restrict to $k = \mathbb {C}$ in this section, to freely speak about Lie groups and algebras. As a general reference we suggest [21]. One technical problem is that the group is infinite-dimensional. In this section we implicitly replace $S$ with $S/\mathfrak {m}^{D}$ and with the group ${\text {Aut}}(S/\mathfrak {m}^{D})$, where $D\gg 0$ is larger than the degree of any element of $P$ considered. Then and become Lie groups.

Note that the group is unipotent: the map $\varphi - {\text {id}}$ is nilpotent for every . Therefore, we get the following theorem.

Theorem 2.21

For every $f\in P$ the orbit is closed in $P$, in both Euclidean and Zariski topologies.

Classically, the Lie algebra corresponding to is . Moreover by the exact sequences (8) and (10), the Lie algebra of is . Finally, the Lie algebra of is and the Lie algebra of is . In particular

We now give another proof of Elias–Rossi theorem on canonically graded algebras, in the following Example 2.22.

Example 2.22

(compressed cubics, using Lie theoretic ideas) Assume that $k = \mathbb {C}$.

Let $f = f_3 + f_{\le 2}\in P_{\le 3}$ be an element of degree three such $H_{{\text {Apolar}}(f)} = (1, n, n, 1)$, where $n = H_{S}(1)$. Such a polynomial f is called compressed, see [17, 27] and Sect. 3.2 for a definition. Then $H_{{\text {Apolar}}(f)}$ is symmetric, thus $H_{{\text {Apolar}}(f_3)} = (1, n, n, 1)$ as explained in Sect. 2.3.

We claim that there is an element $\varphi$ of such that ${\varphi }^{\vee }(f_3) = f$. This proves that ${\text {Apolar}}(f) \simeq {\text {Apolar}}(f_{3}) = {\text {gr}}\,{\text {Apolar}}(f)$. We say that the apolar algebra of f is canonically graded.

In fact, we claim that already is the whole space:

From the explicit formula in Proposition 2.9 we see that . 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 .

Let be non-zero. By Proposition 2.20 we get that $\sigma \,\lrcorner \,f_3 = 0$ and $\sigma ^{(i)} \,\lrcorner \,f_3 = 0$ for all i. Then there is a degree one operator annihilating $f_3$. But this contradicts the fact that $H_{{\text {Apolar}}(f_3)}(1) = n = H_{S}(1)$. Therefore and the claim follows.

3 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 Sect. 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 [6, 32] 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 $\mathbb {C}$ and $\mathbb {R}$ 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 Eq. (9), the group is build from , and $k^*$. We will be mostly concerned with , 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 is not sufficient and the difference between this group and becomes the substance of the result. This is the case for algebras having Hilbert function (1, 3, 3, 3, 1). The group 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.

3.1 Basic examples and tools

In this section we analyse the -orbits of “easy” homogeneous polynomials: those with small Waring and border Waring rank.

Example 3.1

(Rank one) The simplest example seems to be the case $f = \ell ^d$ for some linear form $\ell$. We may assume $\ell = x_1$ and apply Proposition 2.20 to see that

or, in an invariant form, . Using Proposition 2.9 one can even compute the orbit itself. For example, when $d = 4$, the orbit is equal to

This example together with Corollary 3.4 plays an important role in the paper [6], compare [6, Lemma 4.2, Example 4.4, Proposition 5.13].

Example 3.2

(Border rank two) Consider $f = x^{d-1}y$. Assume $S= k[[\alpha , \beta ]]$ and $P= k_{dp}[x, y]$. As above, the apolar ideal of f is monomial, equal to $(\alpha ^d, \beta ^2)$. Using Proposition 2.20 we easily get that

so is spanned by monomials $x^{a}y^{b}$, where $b\le 2$ and $a + b < d$. Note that in contrast with Example 3.1, the equality does not hold. This manifests the fact that ${\text {Apolar}}(f)$ is a deformation of ${\text {Apolar}}(x^d + y^d)$.

Before we analyse the rank two polynomials i.e. $x^d + y^d$, we need a few more observations.

Let for every $g\in P$ the symbol ${\text {tdf}}(g)$ denote the top degree form of g, so that for example ${\text {tdf}}(x_1^3 + x_2^2x_3 + x_4^2) = x_1^3 + x_2^2x_3$.

Proposition 3.3

Let $f\in P$. Then the top degree form of every element of is equal to the top degree form of f. Moreover,

(11)

If f is homogeneous, then both sides of (11) are equal to the set of homogeneous elements of .

Proof

Consider the $S$-action on $P_{\le d}$. This action descents to an action. Further in the proof we implicitly replace $S$ by , thus also replacing and by appropriate truncations. Let . Since $({\text {id}} - \varphi )\left( \mathfrak {m}^i \right) \subseteq \mathfrak {m}^{i+1}$ for all i, we have $({\text {id}} - \varphi )^{d+1} = 0$. By our global assumption on the characteristic $D := \log (\varphi )$ is well-defined and $\varphi = \exp (D)$. We get an injective map with left inverse $\log$. Since $\exp$ is algebraic we see by dimension count that its image is open in . Since $\log$ is Zariski-continuous, we get that , then is an isomorphism.

Therefore

$$\begin{aligned} {\varphi }^{\vee }(f) = f + \sum _{i=1}^{d} \frac{\left( {D}^{\vee }\right) ^i(f)}{i!} = f + {D}^{\vee }(f) + \left( \sum _{i=1}^{d-1} \frac{\left( {D}^{\vee }\right) ^i}{(i+1)!}\right) {D}^{\vee }(f). \end{aligned}$$

By Remark 2.11 the derivation lowers the degree, we see that ${\text {tdf}}({\varphi }^{\vee }f) = {\text {tdf}}(f)$ and ${\text {tdf}}({\varphi }^{\vee }f - f) = {\text {tdf}}({D}^{\vee }(f))$. This proves (11). Finally, if f is homogeneous then is equal to the by Proposition 2.20, and the last claim follows.

For an elementary proof, at least for the subgroup , see [35, Proposition 1.2]. $\square$

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 . Then there is an element such that $\deg ({\varphi }^{\vee }f - F) < \deg (f - F)$.

Proof

Let G be the leading form of $f - F$ and e be its degree. By Proposition 3.3 we may find such that ${\text {tdf}}({\varphi }^{\vee }(F) - F) = -G$, so that ${\varphi }^{\vee }(F) \equiv F - G \mod P_{\le e-1}$. By the same proposition we have $\deg ({\varphi }^{\vee }(f - F) - (f - F)) < \deg (f - F) = e$, so that ${\varphi }^{\vee }(f - F) \equiv f - F \mod P_{\le e-1}$. Therefore ${\varphi }^{\vee }(f) -F = {\varphi }^{\vee }(F) + {\varphi }^{\vee }(f - F) -F \equiv f - G - F \equiv 0 \mod P_{\le e-1}$, as claimed. $\square$

Example 3.5

(Rank two)

Let $P = k_{dp}[x, y]$, $S = k[[\alpha , \beta ]]$ and $F = x^d + y^d$ for some $d\ge 2$. Then $H_{{\text {Apolar}}(F)} = (1, 2, 2, 2, \ldots , 2, 1)$. We claim that the orbit consists precisely of polynomials f having leading form F and such that $H_{{\text {Apolar}}(f)} = H_{{\text {Apolar}}(F)}$.

Let us first compute . Since F is homogeneous, Proposition 2.20 shows that

Since ${\text {Ann}}\left( F \right) = (\alpha \beta , \alpha ^d - \beta ^d)$, we see that

Now we proceed to the description of . It is clear that every satisfies the conditions given. Conversely, suppose that $f \in P$ satisfies these conditions: the leading form of f is F and $H_{{\text {Apolar}}(f)} = H_{{\text {Apolar}}(F)}$. If $d = 2$, then the claim follows from Example 2.15, so we may assume $d\ge 3$. By applying $\alpha ^{d-2}$ and $\beta ^{d-2}$ to f, we see that $x^2$ and $y^2$ are leading forms of partials of f. Since $H_{{\text {Apolar}}(f)}(2) = 2$, 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 $x^ay^b$, where $a, b \ge 2$. Then $\alpha ^{a-1}\beta ^{b-1}\,\lrcorner \,f = xy + l$, where l is linear, then we get a contradiction with the conclusion of the previous paragraph.

Since G contains no monomials of the form $x^ay^b$ with $a, b \ge 2$, we see that G is annihilated by , so it lies in . By Corollary 3.4 we may find such that $\deg (u f - F) < \deg (f - F)$. 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 $f\in P$ be a polynomial with leading form F. Let $I = {\text {Ann}}\left( F \right)$. Fix an integer $t\ge 0$ and assume that

• 1. $\dim _k\,{\text {Apolar}}(f) = \dim _k\,{\text {Apolar}}(F)$.

• 2. we have for all i satisfying $t \le i \le d-1$.

Then there is an element such that $\deg (g) < t$. Equivalently, ${\text {Apolar}}(f) \simeq {\text {Apolar}}(F + g)$ for some polynomial g of degree less than t.

Proof

We apply induction with respect to $\deg (f - F)$. If $\deg (f - F) < t$ then we are done. Otherwise, it is enough to find such that $\deg (uf - F) < \deg (f - F)$.

Since ${\text {Apolar}}(F)$ is a quotient of ${\text {gr}}\,{\text {Apolar}}(f)$, the first condition implies that

$$\begin{aligned} {\text {Apolar}}(F) \simeq {\text {gr}}\,{\text {Apolar}}(f), \end{aligned}$$

Thus for every element $i\in I = {\text {Ann}}\left( F \right)$ we may find $\sigma \in S$ such that $(i + \sigma )\,\lrcorner \,f = 0$ and ${\text {ord}}\left( \sigma \right) > {\text {ord}}\left( i\right)$.

Let G be the leading form of $f - F$ and $t \le r \le d-1$ be its degree. We will now prove that . 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 $i, j\in I$ such that $\deg (ij) = r$ we have

$$\begin{aligned} (ij)\,\lrcorner \,G = 0. \end{aligned}$$

Take $\sigma \in S$ such that $(j - \sigma )\,\lrcorner \,f = 0$. Then ${\text {ord}}\left( i\sigma \right) > {\text {ord}}\left( ij\right) = r = \deg (f-F)$, thus $(i\sigma )\,\lrcorner \,f = (i\sigma )\,\lrcorner \,F = 0$. Therefore,

$$\begin{aligned} (ij)\,\lrcorner \,G = (ij)\,\lrcorner \,(F + G) = (ij)\,\lrcorner \,f = i(j - \sigma )\,\lrcorner \,f = 0. \end{aligned}$$

and the claims follow. $\square$

Example 3.7

(Rank n) Let $f\in P= k_{dp}[x_1, \ldots ,x_n]$ be a polynomial of degree $d\ge 4$ with leading form $F = x_1^{d} + \cdots + x_{n}^d$ and Hilbert function $H_{{\text {Apolar}}(f)} = (1, n, n, n, \ldots , n, 1)$. 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 $I = {\text {Ann}}\left( F \right)$. Then the generators of $I_{<d}$ are $\alpha _i \alpha _j$ for $i\ne j$, thus the ideals $I_{<d}$ and $(I^2)_{<d}$ are monomial. Also the ideal 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 $\alpha _i^{d-1}\alpha _j$ and these do not lie in . Therefore for all $4\le i\le d-1$. 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 $(1, n, n, \ldots , n, 1)$, so the above considerations give a full classification of polynomials in $P$ with this Hilbert function and leading form F.

Since , we may also assume that $f = F + \sum _{i<j<l} \lambda _{ijl} x_ix_jx_l$ for some $\lambda _{ijl}\in k$, see also Example 3.27.

3.2 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 [17, Remark 3.6].

Example 2.16 is concerned with a degree three polynomial f such that the Hilbert function of ${\text {Apolar}}(f)$ 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 d (see Sect. 2.3) is called compressed if

$$\begin{aligned} H_A(i)= & {} \min \left( H_{S}(i), H_{S}(d - i)\right) = \min \left( \left( {\begin{array}{c}i+n-1\\ i\end{array}}\right) , \left( {\begin{array}{c}d-i+n-1\\ d-i\end{array}}\right) \right) \\&\quad \hbox {for all}\ \ i=0, 1, \ldots , d. \end{aligned}$$

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 $t\ge 1$. Then A is called t-compressed if the following conditions are satisfied:

• 1. $H_A(i) = H_{S}(i) = \left( {\begin{array}{c}i+n-1\\ i\end{array}}\right)$ for all $0\le i \le t$,

• 2. $H_{A}(d-1) = H_{S}(1)$.

Example 3.9

Let $n = 2$. Then $H_A = (1, 2, 2, 1, 1)$ is not t-compressed, for any t. The function $H_A = (1, 2, 3, 2, 2, 2, 1)$ is 2-compressed. For any sequence $*$ the function $(1, 2, *, 2, 1)$ is 1-compressed.

Note that it is always true that $H_{A}(d-1) \le H_{A}(1) \le H_{S}(1)$, thus both conditions above assert that the Hilbert function is maximal possible. Therefore they are open in $P_{\le d}$.

Remark 3.10

The maximal value of t, for which t-compressed algebras exists, is $t = \left\lfloor d/2\right\rfloor$. Every compressed algebra is t-compressed for $t = \left\lfloor d/2 \right\rfloor$ but not vice versa. If A is graded, then $H_A(1) = H_A(d-1)$, so the condition $H_{A}(d-1) = H_{S}(1)$ is satisfied automatically.

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 $A = {\text {Apolar}}(f)$ be a t-compressed algebra with maximal ideal $\mathfrak {m}_A$. We have $\dim P_{\le 1} = H_{A}(d-1) + H_{A}(d) = \dim \mathfrak {m}_A^{d-1}/\mathfrak {m}_A^d + \dim \mathfrak {m}_A^d = \dim \mathfrak {m}_A^{d-1}$. Moreover $\mathfrak {m}_A^{d-1} \simeq \mathfrak {m}^{d-1} f$ as linear spaces and $\mathfrak {m}^{d-1} f \subseteq P_{\le 1}$. Thus

$$\begin{aligned} \mathfrak {m}^{d-1} f = P_{\le 1}. \end{aligned}$$

The definition of t-compressed algebras explains itself in the following Proposition 3.12.

Proposition 3.12

Let $f\in P$ be a polynomial of degree $d\ge 3$ and A be its apolar algebra. Suppose that A is t-compressed. Then the -orbit of f contains $f + P_{\le t+1}$. In particular , so that ${\text {Apolar}}(f) \simeq {\text {Apolar}}(f_{\ge t+2})$.

Proof

First we show that , i.e. that no non-zero operator of order at most $t+1$ lies in . Pick such an operator. By Proposition 2.20 it is not constant. Let $\sigma '$ be any of its non-zero partial derivatives. Proposition 2.20 asserts that $\deg (\sigma '\,\lrcorner \,f)\le 1$. Let $\ell := \sigma '\,\lrcorner \,f$. By Remark 3.11 every linear polynomial is contained in $\mathfrak {m}^{d-1} f$. Thus we may choose a $\delta \in \mathfrak {m}^{d-1}$ such that $\delta \,\lrcorner \,f = \ell$. Then $(\sigma ' - \delta )\,\lrcorner \,f = 0$. Since $d\ge 3$, we have $d - 1 > \left\lfloor d/2 \right\rfloor \ge t$, so that $\sigma - \delta$ is an operator of order at most t annihilating f. This contradicts the fact that $H_{A}(i) = H_{S}(i)$ for all $i\le t$. Therefore .

Second, pick a polynomial $g\in f + P_{\le t+1}$. We prove that by induction on $\deg (g - f)$. The top degree form of $g - f$ lies in . Using Corollary 3.4 we find such that $\deg ({\varphi }^{\vee }(g) - f) < \deg (g - f)$. $\square$

For completeness, we state the following consequence of the previous result.

Corollary 3.13

Let $f\in P$ be a polynomial of degree $d\ge 3$ and A be its apolar algebra. Suppose that A is compressed. Then $A \simeq {\text {Apolar}}(f_{\ge \left\lfloor d/2 \right\rfloor + 2})$.

Proof

The algebra A is $\left\lfloor d/2 \right\rfloor$-compressed and the claim follows from Proposition 3.12. $\square$

As a corollary we reobtain the result of Elias and Rossi, see [17, Thm 3.1].

Corollary 3.14

Suppose that A is a compressed Artinian Gorenstein local k-algebra of socle degree $d\le 4$. Then A is canonically graded i.e. isomorphic to its associated graded algebra ${\text {gr}} A$.

Proof

The case $d\le 2$ is easy and left to the reader. We assume $d\ge 3$, so that $3\le d\le 4$.

Fix $n = H_A(1)$ and choose $f\in P= k_{dp}[x_1, \ldots ,x_n]$ such that $A \simeq {\text {Apolar}}(f)$. This is possible by the existence of standard form, see [28, Thm 5.3AB] or [30]. Let $f_d$ be the top degree part of f. Since $\left\lfloor d/2 \right\rfloor + 2 = d$, Corollary 3.13 implies that . Therefore the apolar algebras of f and $f_d$ are isomorphic. The algebra ${\text {Apolar}}(f_d)$ is a quotient of ${\text {gr}} {\text {Apolar}}(f)$. Since $\dim _k\,{\text {gr}}{\text {Apolar}}(f) = \dim _k\,{\text {Apolar}}(f) = \dim _k\,{\text {Apolar}}(f_d)$ it follows that

$$\begin{aligned} {\text {Apolar}}(f) \simeq {\text {Apolar}}(f_{d}) \simeq {\text {gr}}{\text {Apolar}}(f), \end{aligned}$$

which was to be proved. $\square$

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 $f_3\in P_{3}$ be a cubic form such that $H_{{\text {Apolar}}(f_3)} = (1, n, n, 1)$ and $\alpha _1^2\,\lrcorner \,f_3 = 0$. Then there is a degree three polynomial f with leading form $f_3$, whose apolar algebra is compressed but not canonically graded.

Indeed, take $\sigma = \alpha _1^2$. Then all derivatives of $\sigma$ are zero because the characteristic is two. By Proposition 2.18 the element $\sigma$ lies in . Thus does not contain $P_{\le 2}$ and so does not contain $f_3 + P_{\le 2}$. Taking any $f\in f_3 + P_{\le 2}$ outside the orbit yields the desired polynomial. In fact, one can explicitly check that 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 $d \ge 5$ in two variables. As noted in [17, 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

($H_A = (1, 2, 3, 3, 2, 1)$, special) Let $n = 2$, $P= k_{dp}[x_1, x_2]$ and $S= k[[\alpha _1, \alpha _2]]$. Take $F = x_1^3x_2^2\in P$ and $A = {\text {Apolar}}(F) = S/(\alpha _1^4, \alpha _2^3)$. Since $x_1^2, x_1x_2, x_2^2$ are all partials of F we have $H_{A}(2) = 3$ and the algebra A is compressed. It is crucial to note that

(12)

Indeed, the only nontrivial derivative of this element is $4\alpha _2^3\in {\text {Ann}}\left( F \right)$, so (12) follows from Proposition 2.18. Therefore is strictly contained in $F + P_{\le 4}$. Pick any element $f\in F + P_{\le 4}$ not lying in . The associated graded of ${\text {Apolar}}(f)$ is equal to ${\text {Apolar}}(F)$ but ${\text {Apolar}}(f)$ is not isomorphic to ${\text {Apolar}}(F)$, thus ${\text {Apolar}}(f)$ is not canonically graded. In fact one may pick $f = F + x_2^4$, so that ${\text {Apolar}}(f) = S/(\alpha _1^4, \alpha _2^3 - \alpha _1^3\alpha _2)$.

On the positive side, Corollary 3.13 implies that every compressed algebra of socle degree 5 is isomorphic to ${\text {Apolar}}(f)$ 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 $f\in P= k_{dp}[x_1, x_2]$ be a general polynomial of degree five with respect to the natural affine space structure on $k_{dp}[x_1, x_2]_{\le 5}$. Then ${\text {Apolar}}(f)$ and ${\text {Apolar}}(f_5)$ are compressed by [27]. The ideal ${\text {Ann}}\left( f_5 \right)$ 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 ${\text {Ann}}\left( f_5 \right)$ is not a power of a linear form.

We claim that ${\text {Apolar}}(f)$ is canonically graded. Equivalently, we claim that . It is enough to show that

(13)

We will show that . This space is spanned by homogeneous elements. Suppose that this space contains a non-zero homogeneous $\sigma$. Proposition 2.20 implies that $\sigma$ is non-constant and that all partial derivatives of $\sigma$ annihilate $f_5$. These derivatives have degree at most three, thus they are multiples of the cubic generator c of ${\text {Ann}}\left( f_5 \right)$. This implies that all partial derivatives of $\sigma$ are proportional, so that $\sigma$ 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 exists. Now, the Eq. (13) follows from Corollary 3.4.

Example 3.18

(general polynomials in two variables of large degree) Let $F\in P= k_{dp}[x_1, x_2]$ be a homogeneous form of degree $d\ge 9$ and assume that no linear form annihilates F. The ideal ${\text {Ann}}\left( F \right) = (q_1, q_2)$ is a complete intersection. Let $d_i := \deg q_i$ for $i=1, 2$ then $d_1 + d_2 = d + 2$. Since $d_1, d_2 \ge 2$, we have $d_1, d_2 \le d$.

We claim that the apolar algebra of a general (in the sense explained in Example 3.17) polynomial $f\in P$ with leading form F is not canonically graded. Indeed, we shall prove that is strictly contained in $F + P_{<d}$ which is the same as to show that is non-zero.

The sequence (7) from Remark 2.19 becomes

$$\begin{aligned} 0 \rightarrow (K/I^2)_{d-1} \rightarrow \left( A[-d_1] \oplus A[-d_2]\right) _{d-1}\rightarrow \left( A\oplus A\right) _{d-2}, \end{aligned}$$

where the rightmost map has degree $-1$. Note that

$$\begin{aligned}&\dim \left( A[-d_1] \oplus A[-d_2]\right) _{d-1} = (d-1-d_1 + 1) + (d-1-d_2+1)\\&\quad = 2d - (d_1 + d_2) = d - 2. \end{aligned}$$

By assumption $d - 2 > 6 = \dim \left( A\oplus A\right) _{d-2}$, so that $(K/I^2)_{d-1} \ne 0$ and our claim follows.

The set of canonically graded algebras is constructible but not necessarily open nor closed.

Again consider $P_{\le d}$ 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 [17, 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 $\mathcal {G} \subset P_{\le d}$ 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 is the image of , thus irreducible and constructible. Examples 3.16 and 3.17 together show that for $n=2$, $d = 5$ this set is not closed. Example 3.18 shows that for $n=2$, $d \ge 9$ this set is not open. $\square$

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 $P_{\le d}$ 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.

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 $P_{\le d}$ is canonically graded. Then (n, d) belongs to the following list:

• 1. $d\le 4$ and n arbitrary,

• 2. $d = 5$ and $n\le 6$,

• 3. $d = 6$ and $n = 2$,

• 4. d arbitrary and $n = 1$.

Proof

Fix n, d outside the above list and suppose that is dense in $P_{\le d}$. Since we are over a field of characteristic zero, the tangent map at a general point (g, F) is surjective. Then by -action, this map is surjective also at (1, F). Its image is . Thus

(14)

Let us now analyse . By Proposition 2.18 the equations of are given by $\sigma \in S$ satisfying $\sigma \,\lrcorner \,F = 0$ and

$$\begin{aligned} \sigma ^{(i)}\,\lrcorner \,F = 0\quad \hbox {for all}\quad i=1, \ldots ,n. \end{aligned}$$

(15)

Consider $\sigma \in S$ homogeneous of degree $d-1$. The space ${\text {Ann}}\left( F \right) _{d-2}$ is of codimension $H_{{\text {Apolar}}(F)}(d-2) = H_{{\text {Apolar}}(F)}(2) \le \left( {\begin{array}{c}n+1\\ 2\end{array}}\right)$. Thus for fixed i the condition $\sigma ^{(i)}\,\lrcorner \,F = 0$ amounts to at most $\left( {\begin{array}{c}n+1\\ 2\end{array}}\right)$ linear conditions on the coefficients of $\sigma$. Summing over $i=1, \ldots ,n$ we get $n\cdot \left( {\begin{array}{c}n+1\\ 2\end{array}}\right)$ conditions. Now, if (n, d) is outside the list, then

$$\begin{aligned} n\cdot \left( {\begin{array}{c}n+1\\ 2\end{array}}\right) < \left( {\begin{array}{c}n-1+d-1\\ d-1\end{array}}\right) , \end{aligned}$$

so that there exists a non-zero $\sigma$ satisfying (15). Since $\sigma = (d-1)^{-1}\sum \alpha _i\sigma ^{(i)}$ also $\sigma \,\lrcorner \,F = 0$. Thus is a non-zero element. Therefore does not contain the whole $P_{\le d-1}$, which contradicts (14). $\square$

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 $d\le 4$ it is the result of Elias and Rossi reproved in Corollary 3.14. We put forward the following conjecture.

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 $P_{\le d}$ is canonically graded are precisely the pairs listed below:

• 1. $d\le 4$ and n arbitrary,

• 2. $d = 5$ and $n\le 6$,

• 3. $d = 6$ and $n = 2$.

• 4. d arbitrary and $n = 1$.

Here $n = \dim S$ is the number of variables.

We have some supporting evidence in all cases $d = 5$, $n\le 6$ and $d = 6$, $n=2$. Namely, for a pseudo-randomly chosen form F a machine computation of via Proposition 2.18 reveals that

Thus arguing by semicontinuity we get that the map 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 $\Delta _{r}$ of the Hilbert function, see [6, 28] or [1, Section 2]. In short, the Hilbert function H of an apolar algebra of socle degree d admits a canonical decomposition $H = \sum _{i=0}^{d} \Delta _i$, where $\Delta _i$ is a vector of length $d-i$ which is symmetric: $\Delta _i(j) = \Delta _i(d-i-j)$ for all j.

Proposition 3.22

Let $f\in P$ be a polynomial of degree $d\ge 3$ and A be its apolar algebra. Let $\Delta _{\bullet }$ be the symmetric decomposition of $H_A$. Suppose that

• 1. $H_A(r) = H_{S}(r) = \left( {\begin{array}{c}r + n - 1\\ r\end{array}}\right)$ for all $0\le r \le t$.

• 2. $\Delta _{r}(1) = 0$ for all $d - 1 - t \le r$.

Then the -orbit of f contains $f + P_{\le t+1}$. In particular .

Proof

In the notation of [1, Section 2] or [6] we see that

$$\begin{aligned} \Delta _a(1) = \dim C_{a, 1}(1)/C_{a-1, 1}(1),\quad \hbox {where}\quad C_{a, 1}(1) = \{\mathrm{linear\,polynomials\,in\,}\mathfrak {m}^{d - 1 - a}f\}. \end{aligned}$$

Moreover, the assumption $H_A(1) = n$ guarantees that $C_{d-2, 1}(1) = P_{\le 1}$ is the full space of linear polynomials. The assumption $\Delta _{r}(1) = 0$ for all $d - 1 - t\le r \le d - 2$ shows that

$$\begin{aligned} P_{\le 1} = C_{d-2, 1}(1) = C_{d-3, 1}(1)= \cdots = C_{d-2 - t, 1}(1)= \mathfrak {m}^{t+1} f. \end{aligned}$$

(16)

Thus, for every $\ell \in P_{\le 1}$ we have a $\delta$ of order greater than t such that $\delta \,\lrcorner \,f = \ell$.

Now we repeat the proof of Proposition 3.12 with one difference: instead of referring to Remark 3.11 we use Eq. (16) to obtain, for a given linear form $\ell$, an element $\delta \in S$ of order greater than t and such that $\delta \,\lrcorner \,f = \ell$. $\square$

3.3 Further examples

Below we present some more involved examples, which employ the tools developed in the previous section.

Example 3.23

(Hilbert function (1, 2, 2, 1)) By the result of Elias and Rossi (Proposition 3.14) every apolar algebra with Hilbert function (1, 2, 2, 1) is canonically graded. Then it is isomorphic to ${\text {Apolar}}(x^3 + y^3)$ or ${\text {Apolar}}(x^2y)$ and these algebras are not isomorphic.

The example was treated in [7, p. 12] and [16, Prop 3.6].

Example 3.24

(Hilbert function (1, 2, 2, 2, 1)) Let $f\in P_{\le 4}$ be a polynomial such that $H_{{\text {Apolar}}(f)} = (1, 2, 2, 2, 1)$. If $f_4 = x^4 + y^4$ then if fact ${\text {Apolar}}(f) \simeq {\text {Apolar}}(f_4)$ by Example 3.5, so we may assume $f_4 = x^3y$. Since f is 1-compressed we may also assume $f_{\le 2} = 0$, whereas by Corollary 3.4 together with Example 3.2 we may assume $f_3 = cy^3$ for $c\in k$. Thus

$$\begin{aligned} {\text {Apolar}}(f) \simeq {\text {Apolar}}(x^3y + cy^3). \end{aligned}$$

By multiplying variables by suitable constants, we may assume $c = 0$ or $c =1$ and obtain three possibilities:

$$\begin{aligned} f = x^4 + y^4,\qquad f = x^3y,\qquad f = x^3y + y^3. \end{aligned}$$

Note that the three types appearing above are pairwise non-isomorphic. Indeed, an isomorphism may only occur between ${\text {Apolar}}(x^3y)$ and ${\text {Apolar}}(x^3y + y^3)$. Suppose that such exists, so that there is such that ${\varphi }^{\vee }(x^3y) = x^3y + y^3$. Write $\varphi = gu$, for g linear and . Then g preserves $x^3y$ and a direct check shows that g is diagonal. Then ${u}^{\vee }(x^3y) = x^3y + cy^3$ for some non-zero $c\in k$ and , by Proposition 3.3. This is a contradiction with Example 3.2.

This example was analysed, among others, in [5], where Casnati classifies all Artinian Gorenstein algebras of length at most 9. See especially [5, Theorem 4.4].

Example 3.25

(Hilbert function (1, 2, 2, ..., 2, 1), general) Consider the set of polynomials $f\in P= k_{dp}[x, y]_{\le d}$ such that $H_{{\text {Apolar}}(f)} = (1, 2, \ldots , 2, 1)$ with $d-1$ twos occurring. This set is irreducible in Zariski topology, by [6, Proposition 4.8] which uses [25, Theorem 3.13]. The leading form of a general element of this set has, up to a coordinate change, the form $x^d + y^d$. Example 3.5 shows that the apolar algebra of such an element is canonically graded.

Remark 3.26

In general, there are $d-1$ isomorphism types of almost-stretched algebras of socle degree d and with Hilbert function $(1, 2, 2, \ldots , 2, 1)$ as proved in [5, Theorem 4.4]; see also [18, Remark 5, p. 447]. The claim is recently generalised by Elias and Homs, see [15].

Example 3.27

(Hilbert function (1, 3, 3, 3, 1)) Consider now a polynomial $f\in P= k_{dp}[x, y, z]$ whose Hilbert function is (1, 3, 3, 3, 1). Let F denote the leading form of f. By [32] or [6, Prop 4.9] the form F is linearly equivalent to one of the following:

$$\begin{aligned} F_1 = x^4 + y^4 + z^4,\quad F_2 = x^3y + z^4,\quad F_3 = x^2(xy + z^2). \end{aligned}$$

Since ${\text {Apolar}}(f)$ is 1-compressed, we have ${\text {Apolar}}(f) \simeq {\text {Apolar}}(f_{\ge 3})$; 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 is the product of the group of linear transformations and $k^*$ acting by multiplication.

The case $F_1$. Example 3.7 shows that is spanned by $\alpha \beta \gamma$. Therefore we may assume $f = F_1 + c\cdot xyz$ for some $c\in k$. 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:

$$\begin{aligned} f_{1,0} = x^4 + y^4 + z^4,\quad f_{1,1} = x^4 + y^4 + z^4 + xyz. \end{aligned}$$

Note that $f_{1, 0}$ is canonically graded, whereas $f_{1,1}$ is a complete intersection.

The case $F_2$. We have ${\text {Ann}}\left( F_2 \right) _2 = (\alpha \gamma , \beta ^2, \beta \gamma )$, so that . Thus we may assume $f = F_2 + c_1 y^3 + c_2 y^2z$. As before, multiplying x, y and z by suitable constants we may assume $c_1, c_2\in \{0, 1\}$. We get four isomorphism types:

$$\begin{aligned}&f_{2, 00} = x^3y + z^4,\quad f_{2, 10} = x^3y + z^4 + y^3,\quad f_{2, 01} = x^3y + z^4 + y^2z,\\&\quad f_{2, 11} = x^3y + z^4 + y^3 + y^2z. \end{aligned}$$

To prove that the apolar algebras are pairwise non-isomorphic one shows that the only linear maps preserving $F_2$ are diagonal and argues as an in Example 3.24 or as described in the case of $F_3$ below.

The case $F_3$. We have ${\text {Ann}}\left( F_3 \right) _2 = (\beta ^2,\, \beta \gamma ,\,\alpha \beta - \gamma ^2)$ and

We may choose ${\text {span}}\left\langle y^3, y^2z, yz^2\right\rangle$ as the complement of in $P_3$. Therefore the apolar algebra of each f with top degree form $F_3$ is isomorphic to the apolar algebra of

$$\begin{aligned} f_{3, *} = x^3y + x^2z^2 + c_1y^3 + c_2y^2z + c_3yz^2 \end{aligned}$$

and two distinct such polynomials $f_{3, *1}$ and $f_{3, *2}$ lie in different -orbits. We identify the set of -orbits with . We wish to determine isomorphism classes, that is, check which such $f_{3, *}$ lie in the same -orbit. A little care should be taken here, since -orbits will be bigger than expected.

Recall that preserves the degree. Therefore, it is enough to look at the operators stabilising $F_3$. These are $c\cdot g$, where $c\in k^*$ is a constant and stabilises ${\text {span}}\left\langle F_3\right\rangle$, i.e. ${g}^{\vee }({\text {span}}\left\langle F_3\right\rangle ) = {\text {span}}\left\langle F_3\right\rangle$. Consider such a g. It is a linear automorphism of $P$ and maps ${\text {Ann}}\left( F \right)$ into itself. Since $\beta (\lambda _1\beta + \lambda _2\gamma )$ for $\lambda _i\in k$ are the only reducible quadrics in ${\text {Ann}}\left( F_3 \right)$ we see that g stabilises ${\text {span}}\left\langle \beta , \gamma \right\rangle$, so that ${g}^{\vee }(x) = \lambda x$ for a non-zero $\lambda$. Now it is straightforward to check directly that the group of linear maps stabilising ${\text {span}}\left\langle F_3\right\rangle$ is generated by the following elements

• 1. homotheties: for a fixed $\lambda \in k$ and for all linear $\ell \in P$ we have ${g}^{\vee }(\ell ) = \lambda \ell$.

• 2. for every $a, b\in k$ with $b\ne 0$, the map given by

  

  which maps $F_3$ to $b^2 F_3$.

The action of on in the basis $(y^3, y^2z, yz^2)$ is given by the matrix

$$\begin{aligned} \begin{pmatrix} b^6 &{}\quad 0 &{}\quad 0\\ -6ab^5 &{}\quad b^5 &{}\quad 0\\ \frac{27}{2}a^2b^4 &{}\quad -\frac{9}{2}ab^4 &{}\quad b^4\\ \end{pmatrix} \end{aligned}$$

Suppose that $f_{3, *} = x^3y + x^2z^2 + c_1y^3 + c_2y^2z + c_3yz^2$ has $c_1 \ne 0$. The above matrix shows that we may choose $a$ and $b$ and a homothety h so that

Suppose $c_1 = 0$. If $c_2 \ne 0$ then we may choose $a$, $b$ and $\lambda$ so that . Finally, if $c_1 = c_2 = 0$, then we may choose $a= 0$ and $b$, $\lambda$ so that $c_3 = 0$ or $c_3 = 1$. We get at most five isomorphism types:

$$\begin{aligned}&f_{3, 100} = x^3y + x^2z^2 + y^3,\quad f_{3, 101} = x^3y + x^2z^2 + y^3 + yz^2,\\&\quad f_{3, 010} = x^3y + x^2z^2 + y^2z,\quad f_{3, 001} = x^3y + x^2z^2 + yz^2,\\&\quad f_{3, 000} = x^3y + x^2z^2. \end{aligned}$$

By using the explicit description of the action on one checks that the apolar algebras of the above polynomials are pairwise non-isomorphic.

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:

Orbit	Dimension	Orbit	Dimension
	29		27
	28		26
	28		26
	27		25
	27		24
	26

The closure of the orbit of $f_{1, 1} = x^4 + y^4 + z^4 + xyz$ is contained in ${\text {GL}}_3(x^4 + y^4 + z^4) + P_{\le 3}$, 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 ${\text {GL}}_3(x^4 + y^4 + z^4) + P_{\le 3}$. Moreover, the set ${\text {GL}}_{3}(x^4 + y^4 + z^4)$ is dense inside the set $\sigma _3$ 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 ${\text {GL}}_3$-orbit of $x^3y + x^2z^2$ has smaller dimension than the ${\text {GL}}_3$-orbit of $x^3y + z^4$. Thus $x^3y + x^2z^2 + y^3 + yz^2$ does not specialize to $x^3y+z^4$ even though its -orbit has higher dimension.

Example 3.28

(Hilbert function (1, 3, 4, 3, 1))

Consider now a polynomial $f\in k_{dp}[x, y, z]$ 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 ${\text {Ann}}\left( F \right)$ 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 $V = V(Q) \subset \mathbb {P}^2$ be the zero set of Q inside the projective space with coordinates $\alpha , \beta , \gamma$.

Infinite family of isomorphism types. Consider

$$\begin{aligned} F = F_{\lambda } = \lambda _1 x^4 + \lambda _2 y^4 + \lambda _3 z^4 + \lambda _4(x+y+z)^4. \end{aligned}$$

(17)

for some non-zero numbers $\lambda _{i}$. Then the Hilbert function of ${\text {Apolar}}(f)$ is (1, 3, 4, 3, 1) and V(Q) is a set of four points, no three of them collinear. Suppose that $F_{\lambda }$ and $F_{\lambda '}$ are in the same -orbit. By Remark 2.17 there is an element of mapping $F_{\lambda }$ to $F_{\lambda '}$. Such element stabilizes V(Q), which is a set of four points, no three of them collinear. But the only elements of stabilizing such set of four points are the scalar matrices. Therefore we conclude that the set of isomorphism classes of $F_{\lambda }$ is the set of quadruples $\lambda _{\bullet }$ up to homothety. This set is in bijection with $(k^*)^4/k^* \simeq (k^*)^3$, thus infinite. It fact the set of $F_{\lambda }$ with $\lambda _4 = 1$ is a threefold in the moduli space of finite algebras with a fixed basis, see [38] 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 . As in Example 3.27 we see that

Thus we investigate using Proposition 2.20. Let us suppose it is non-zero and pick a non-zero element . Then $\sigma ^{(i)}\in {\text {Ann}}\left( F \right) _2 = Q$ for all i. Since Q is two-dimensional we see that the derivatives of $\sigma$ are linearly dependent. Thus up to coordinate change we may assume that $\sigma \in k[\alpha , \beta ]$. If $\sigma$ has one-dimensional space of derivatives, then $\sigma = \alpha ^3$ up to coordinate change and $\frac{1}{3}\sigma ^{(1)} = \alpha ^2$ annihilates Q. If $\sigma$ has two-dimensional space of derivatives Q, then Q intersects the space of pure squares in an non-zero element $\alpha ^2$, thus $\alpha ^2$ annihilates F in this case also. Conversely, if $\alpha ^2$ annihilates F, then $\alpha ^3$ lies in .

Summarizing, if and only if no square of a linear form annihilates F. For example, apolar algebras of all polynomials with leading form $F_{\lambda }$ from (17) are canonically graded. On the other hand, for $F = xz^3 + z^2y^2 + y^4$ and $f_{\lambda } = F + ax^3$ the apolar algebra of $f_{\lambda }$ is canonically graded if and only if $a = 0$.

Non-canonically graded algebras with Hilbert function (1, 3, 4, 3, 1) are investigated independently in [34], where more generally Hilbert functions (1, n, m, n, 1) are considered.

Any discussion of isomorphism types would be incomplete without tackling the example of Hilbert function (1, 2, 2, 2, 1, 1, 1), which is the smallest example where infinitely many isomorphism types appear. It is also instructive as a non-homogeneous example. Strangely enough, the argument is similar to the previous examples. We will use some standard tools to deal with the dual socle generator, see [6, Chap 3].

Example 3.29

(Hilbert function (1, 2, 2, 2, 1, 1, 1)) Consider any $f\in P$ such that

$$\begin{aligned} H_{{\text {Apolar}}(f)} = (1, 2, 2, 2, 1, 1, 1). \end{aligned}$$

Then f is of degree 6. Using the standard form, see [28, Theorem 5.3] or [6, Section 3], we may assume, after a suitable change of f, that

$$\begin{aligned} f = x^6 + f_{\le 4}. \end{aligned}$$

If $y^4$ appears with non-zero coefficient in $f_{\le 4}$, we conclude that arguing similarly as in Example 3.5. Otherwise, we may assume that $\beta ^2$ is a leading form of an element of an annihilator of f, so that the only monomials in $f_4$ are $x^4$, $x^3y$ and $x^2y^2$. By subtracting a suitable element of and rescaling by homotheties, we may assume

$$\begin{aligned} f = x^6 + c x^2y^2 + f_{\le 3}, \end{aligned}$$

for $c = 0$ or $c = 1$. If $c = 0$, then the Hilbert function of f is $(1, *, *, 1, 1, 1, 1)$, which is a contradiction; see [28, Lem 1.10] or [31, Lem 4.34] for details. Thus $c = 1$.

Consider elements of with order at most three. Every non-zero partial derivative $\tau$ of such an element $\sigma$ has order at most two and satisfies $\deg (\tau \,\lrcorner \,f) \le 1$. This is only possible if $\tau _2 = \beta ^2$, so that $\sigma _3 = \beta ^3$ up to a non-zero constant. Therefore we may assume

$$\begin{aligned} f = x^6 + x^2y^2 + \lambda y^3 + f_{\le 2}\quad \hbox {for some}\quad \lambda \in k. \end{aligned}$$

The symmetric decomposition of ${\text {Apolar}}(f)$ is $\Delta _0 = (1, 1, 1, 1, 1, 1, 1)$, $\Delta _1 = \mathbf {0}$, $\Delta _2 = (0, 1, 1, 1, 0)$, $\Delta _3 = \mathbf {0}$, $\Delta _4 = \mathbf {0}$. In particular $\Delta _4(1) = 0$, so that by Proposition 3.22 we may assume $f_{\le 2} = 0$ and

$$\begin{aligned} f = f_{\lambda } = x^6 + x^2y^2 + \lambda y^3. \end{aligned}$$

Similarly as in the final part of Example 3.24 we see that for every $\lambda \in k$ we get a distinct algebra $A_{\lambda } = {\text {Apolar}}(f_{\lambda })$ where $A_{\lambda } \simeq A_{\lambda '}$ if and only if $\lambda = \lambda '$. Thus in total we get $|k| + 1$ isomorphism types. Note that the |k| types $f_{\lambda }$ form a curve in the moduli space of finite algebras with a fixed basis, see [38] for information on this space. The point corresponding to $y^6 + x^4$ does not seem to be related to this curve.