Classifying local Artinian Gorenstein algebras

Definition 2.1

Definition 2.2

We define multiplication on \(P\) by

Open image in new window

(3)

In this way \(P\) is a divided power ring.

Example 2.3

Suppose that k is of characteristic 3. Then \(P\) is not isomorphic to a polynomial ring. Indeed, Open image in new window . Moreover Open image in new window is not in the subring generated by \(x_1, \ldots ,x_n\).

Lemma 2.4

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

Open image in new window

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.

Proposition 2.6

Let \(P_{\ge d}\) be the linear span of Open image in new window . Then Open image in new window 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

Open image in new window

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 \(\left( {\begin{array}{c}\mathbf {a}+ \mathbf {b}\\ \mathbf {a}\end{array}}\right) = \prod \frac{(a_i + b_i)!}{a_i!b_i!}\). \(\square \)

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 \(\sigma \circ f\).

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 Open image in new window be defined via

Open image in new window

Then Open image in new window is an isomorphism of rings and an isomorphism of \(S\)-modules.

Proof

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

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

Open image in new window

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 \(g\in P\)

Open image in new window

(6)

Indeed it is enough to check the above equality for Open image in new window and both sides are zero unless \(\mathbf {c}= \mathbf {a}- \mathbf {b}\), thus it is enough to check the case Open image in new window , 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 Open image in new window with non-zero \(\mathbf {c}\). We can use (6) and remove the restriction \(\mathbf {b}\le \mathbf {a}\), obtaining

Open image in new window

\(\square \)

Proposition 2.10

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.

Corollary 2.12

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 \(\mathbf {D}^{(0, b)} = \alpha _1^b\).

Proposition 2.14

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. \(\square \)

Example 2.15

(quadrics) Let \(f\in P_2\) be a quadric of maximal rank. Then Open image in new window 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 Open image in new window 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.

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].

Proposition 2.18

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

Open image in new window

Moreover

Open image in new window

Suppose further that \(f\in P\) is homogeneous of degree d. Then Open image in new window is spanned by homogeneous operators and

Open image in new window

Proof

Let Open image in new window 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 Open image in new window . Now Open image in new window . By Lemma 2.4 we have \(x_i(\sigma \,\lrcorner \,f) \equiv \sigma \,\lrcorner \,(x_i f) \mod Sf\). Thus

Open image in new window

Now let \(\sigma \in S\) be an operator such that Open image in new window . This is equivalent to Open image in new window , which simplifies to \(\sigma \,\lrcorner \,f = 0\) and \((\sigma \mathfrak {m})\,\lrcorner \,(x_i f) = 0\) for all i. We have Open image in new window , 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 Open image in new window . 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.

Proposition 2.20

Let \(f\in P\). Then Open image in new window so that

Open image in new window

If f is homogeneous of degree d then Open image in new window is spanned by homogeneous polynomials and

Open image in new window

Theorem 2.21

For every \(f\in P\) the orbit Open image in new window is closed in \(P\), in both Euclidean and Zariski topologies.

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 Open image in new window 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 Open image in new window is the whole space:

Open image in new window

From the explicit formula in Proposition 2.9 we see that Open image in new window . 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 Open image in new window .

Let Open image in new window 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 Open image in new window and the claim follows.

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

Open image in new window

or, in an invariant form, Open image in new window . Using Proposition 2.9 one can even compute the orbit itself. For example, when \(d = 4\), the orbit is equal to

Open image in new window

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

Open image in new window

so Open image in new window 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 Open image in new window does not hold. This manifests the fact that \({\text {Apolar}}(f)\) is a deformation of \({\text {Apolar}}(x^d + y^d)\).

Proposition 3.3

Let \(f\in P\). Then the top degree form of every element of Open image in new window is equal to the top degree form of f. Moreover,

Open image in new window

(11)

If f is homogeneous, then both sides of (11) are equal to the set of homogeneous elements of Open image in new window .

Proof

Consider the \(S\)-action on \(P_{\le d}\). This action descents to an Open image in new window action. Further in the proof we implicitly replace \(S\) by Open image in new window , thus also replacing Open image in new window and Open image in new window by appropriate truncations. Let Open image in new window . 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 Open image in new window with left inverse \(\log \). Since \(\exp \) is algebraic we see by dimension count that its image is open in Open image in new window . Since \(\log \) is Zariski-continuous, we get that Open image in new window , then Open image in new window 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 Open image in new window 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 Open image in new window is equal to the Open image in new window by Proposition 2.20, and the last claim follows.

For an elementary proof, at least for the subgroup Open image in new window , see [35, Proposition 1.2]. \(\square \)

Corollary 3.4

Let F and f be polynomials. Suppose that the leading form of \(F - f\) lies in Open image in new window . Then there is an element Open image in new window 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 Open image in new window 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 Open image in new window consists precisely of polynomials f having leading form F and such that \(H_{{\text {Apolar}}(f)} = H_{{\text {Apolar}}(F)}\).

Let us first compute Open image in new window . Since F is homogeneous, Proposition 2.20 shows that

Open image in new window

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

Open image in new window

Now we proceed to the description of Open image in new window . It is clear that every Open image in new window 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 Open image in new window , so it lies in Open image in new window . By Corollary 3.4 we may find Open image in new window 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 \(f = F\).

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. 1.

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

    
2. 2.

   we have Open image in new window for all i satisfying \(t \le i \le d-1\).

    

Then there is an element Open image in new window 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 Open image in new window such that \(\deg (uf - F) < \deg (f - F)\).

Let G be the leading form of \(f - F\) and \(t \le r \le d-1\) be its degree. We will now prove that Open image in new window . 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 Open image in new window 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 Open image in new window . Therefore Open image in new window 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 Open image in new window , 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.

Definition 3.8

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.

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.

Remark 3.11

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 Open image in new window -orbit of f contains \(f + P_{\le t+1}\). In particular Open image in new window , so that \({\text {Apolar}}(f) \simeq {\text {Apolar}}(f_{\ge t+2})\).

Proof

First we show that Open image in new window , i.e. that no non-zero operator of order at most \(t+1\) lies in Open image in new window . 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 Open image in new window .

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

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. \(\square \)

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 Open image in new window . 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. \(\square \)

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 Open image in new window . Thus Open image in new window does not contain \(P_{\le 2}\) and so Open image in new window 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 Open image in new window is an example.

A similar example shows that over a field of characteristic three there are compressed quartics which are 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

Open image in new window

(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 Open image in new window is strictly contained in \(F + P_{\le 4}\). Pick any element \(f\in F + P_{\le 4}\) not lying in Open image in new window . 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 \({\text {Apolar}}(f) = S/(\alpha _1^4, \alpha _2^3 - \alpha _1^3\alpha _2)\).

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 Open image in new window . It is enough to show that

Open image in new window

(13)

We will show that Open image in new window . 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 Open image in new window 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 Open image in new window is strictly contained in \(F + P_{<d}\) which is the same as to show that Open image in new window is non-zero.

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 Open image in new window is the image of Open image in new window , 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. \(\square \)

Proposition 3.20

Proof

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

Open image in new window

(14)

Let us now analyse Open image in new window . By Proposition 2.18 the equations of Open image in new window 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 Open image in new window is a non-zero element. Therefore Open image in new window does not contain the whole \(P_{\le d-1}\), which contradicts (14). \(\square \)

Conjecture 3.21

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. 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. 2.

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

    

Then the Open image in new window -orbit of f contains \(f + P_{\le t+1}\). In particular Open image in new window .

Proof

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 \(\delta \,\lrcorner \,f = \ell \). \(\square \)

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 Open image in new window such that \({\varphi }^{\vee }(x^3y) = x^3y + y^3\). Write \(\varphi = gu\), for g linear and Open image in new window . 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 Open image in new window , 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

Open image in new window

Recall that Open image in new window is the product of the group of linear transformations and \(k^*\) acting by multiplication.

The case \(F_1\). Example 3.7 shows that Open image in new window 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 Open image in new window . 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

Open image in new window

We may choose \({\text {span}}\left\langle y^3, y^2z, yz^2\right\rangle \) as the complement of Open image in new window 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 Open image in new window -orbits. We identify the set of Open image in new window -orbits with Open image in new window . We wish to determine isomorphism classes, that is, check which such \(f_{3, *}\) lie in the same Open image in new window -orbit. A little care should be taken here, since Open image in new window -orbits will be bigger than expected.

Recall that Open image in new window 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 Open image in new window 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. 1.

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

    
2. 2.
   for every \(a, b\in k\) with \(b\ne 0\), the map Open image in new window given by

   Open image in new window

   which maps \(F_3\) to \(b^2 F_3\).
    

The action of Open image in new window on Open image in new window 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

Open image in new window

Suppose \(c_1 = 0\). If \(c_2 \ne 0\) then we may choose \(a\), \(b\) and \(\lambda \) so that Open image in new window . 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 Open image in new window action on Open image in new window one checks that the apolar algebras of the above polynomials are pairwise non-isomorphic.

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 Open image in new window -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 Open image in new window -orbit. By Remark 2.17 there is an element of Open image in new window 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 Open image in new window 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 Open image in new window . As in Example 3.27 we see that

Open image in new window

Thus we investigate Open image in new window using Proposition 2.20. Let us suppose it is non-zero and pick a non-zero element Open image in new window . 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 Open image in new window .

Summarizing, Open image in new window 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.

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 Open image in new window 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 Open image in new window 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 Open image in new window 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.