# Inverse Systems of Local Rings

Chapter
Part of the Lecture Notes in Mathematics book series (LNM, volume 2210)

## Abstract

Matlis duality and the particular case of Macaulay correspondence provide a dictionary between the Artin algebras and their inverse systems. Inspired in a result of Emsalem we translate the problem of classification of Artin algebras to a problem of linear system of equations on the inverse systems.

The main purpose of these notes is to use this result to classify Artin Gorenstein algebras with Hilbert function {1, 3, 3, 1}, level algebras and compressed algebras. The main results presented in these notes were obtained in collaboration with M.E. Rossi.

## 2.1 Introduction

These notes are based on a series of lectures given by the author at the Vietnam Institute for Advanced Study in Mathematics, Hanoi, during the period February 8–March 7, 2014. The aim of these three lectures was to present some recent results on the classification of Artin Gorenstein and level algebras by using the inverse system of Macaulay. These notes are not a review on the known results of Macaulay’s inverse systems. See [12, 20, 21, 22, 23] and  for further details on inverse systems.

Let be the ring of the formal series and let S = k[y1, …, yn] be a polynomial ring. Macaulay established a one-to-one correspondence between the Gorenstein Artin algebras A = RI and cyclic submodules 〈F〉 of the polynomial ring S. This correspondence is a particular case of Matlis duality because the injective hull of k as R-module is isomorphic to S. The structure of S as R-module is defined, depending on the characteristic of the residue field k, by derivation or by contraction. Macaulay’s correspondence establish a dictionary between the algebraic-geometric properties of Artin Gorenstein algebras A and the algebraic properties of its inverse system F or the geometric properties of the variety defined by F ≡ 0. See  for the extension to higher dimensions of Macaulay’s correspondence.

In the second chapter we review the main results on injective modules. We prove the existence on the injective hull of a ring and we prove Matlis’ duality for a complete ring. The main references used in this chapter are:  and .

In the third chapter we study Macaulay’s correspondence that is a particular case of Matlis’ duality. In the main result of this we prove that S is the injective hull of the residue field of the R-module k. From this result and Matlis’ duality we deduce Macaulay’s correspondence. We end the chapter computing the Hilbert function of a quotient A = RI in terms of its inverse system. The main references used in this chapter are: [18, 20, 21, 22, 23] and .

The fourth chapter is devoted to give a quick introduction to Artin Gorenstein, level and compressed algebras. We only quote the results needed to achieve the main goal of these notes. The main references used in this chapter are:  and .

The fifth chapter is the core of these notes. We present the main results obtained in collaboration with M.E. Rossi on the classification of Artinian Gorenstein algebras, level algebras and compressed algebras,  and . After a short review of the classification of Artin algebra we show the difficulty of the problem of the classification of Artin algebras recalling some results obtained in collaboration with Valla,  and .

Inspired in a result of Emsalem, , we translate the problem of classification of Artin algebras to a problem of linear systems of equations. The study of these systems of equations permits to establish the main result of this paper, Theorem 2.5.10. We end the chapter by giving a complete analytic classification of Artin Gorenstein algebras with Hilbert function {1, 3, 3, 1} by using the Weierstrass form of an elliptic plane curve. The main references used in this chapter are: [12, 20] and .

In Sect. 2.6 we consider the problem of computing the Betti numbers of an ideal I by considering only its inverse system without computing the ideal I. The main open problem is to characterize the complete intersection ideals in terms of their inverse systems. In this chapter we focus the study on the computation of the last Betti number (i.e. the Cohen-Macaulay type) and the first Betti number (i.e. the minimal number of generators)

In the last chapter we show that some results of the chapter four cannot be generalized and we present several explicit computations of the minimal number of generators of some families of Artin Gorenstein and level algebras.

In these notes we omit reviewing some recent interesting results on the rationality of the Poincaré series of an Artin Gorenstein algebra, on the smoothability of the Artinian algebras, and the applications of these results to the study of the geometric properties of Hilbert schemes, see for instance [4, 5] and their reference’s list.

The examples of this paper are done by using the Singular library [7, 8], and Mathematica.

## 2.2 Injective Modules: Matlis’ Duality

Given a commutative ring R we denote by R_mod, resp. R_mod.Noeth, R_mod.Artin, the category of R-modules, resp. category of Noetherian R-modules, Artinian R-modules.

### Definition 2.2.1 (Injective Module)

Let R be a commutative ring and let E be an R-module. E is injective if and only if HomR(⋅, E) is an exact functor.

Since for all R-module E the contravariant functor HomR(⋅, E) is right exact, we have that E is injective if and only for all injective morphism h : MN and for all morphism f : ME, where M and N are R-modules, there exists a morphism g : NE making the following diagram commutative:

In the following result we collect some basic properties of injective modules.

### Proposition 2.2.2

1. (i)
If a R-module E is injective, then every short exact sequence splits:
\displaystyle \begin{aligned}0\longrightarrow E \longrightarrow M \longrightarrow N \longrightarrow 0 \end{aligned}

2. (ii)

If an injective module E is a submodule of a module M, then E is a direct summand of M, in other words, there is a complement S with M = S  E.

3. (iii)

If (Ej)J is a family of injective R-modules, thenjJEj is also an injective module.

4. (iv)

Every direct summand of an injective R-module is injective.

5. (v)

A finite direct sum of injective R-modules is injective.

Now that we have showed some of the properties of the injective modules, we need to find an easier way to check the injectivity of a module. This criterion is the following:

### Proposition 2.2.3 (Baer’s Criterion)

A R-module E is injective if and only if every homomorphism f : I  E, where I is an ideal of R, can be extended to R.

### Proof

First, if E is injective, then, as I is a submodule of R, the existence of an extension g of f is just a a straight consequence of the injectivity of E.

Consider that we have the following diagram, where M is a submodule of a R-module N:
We may assume that M is a submodule of N. Let us consider the set X = {(M′, g′)|M ⊂ M′⊂ N, g′|M = f}. Note that X ≠ ∅ because (M, f) ∈ X. Now we put a partial order in X, (M′, g′) ≼ (M″, g″), which means that M′⊂ M″ and g″ extends g′. It is easy to see that any chain in X has an upper bound in X (just take the union). By Zorn’s Lemma we have that there is a maximal element (M0, g0) of X. If M0 = N we are done, so we can assume that there is some b ∈ N that is not in M0. Define I = {r ∈ R : r.b ∈ M0}, which is clearly an ideal of R. Now define h : I → E by h(r) = g0(r.b). By hypothesis, there is a map h extending h. Finally define M1 = M0 + 〈b〉 and g1 : M1 → E by
\displaystyle \begin{aligned} g_1(a_0+br)=g_0(a_0)+r\cdot h^*(1), \end{aligned}
where a0 ∈ M0 and r ∈ R. Notice that if $$a_0+r.b=a_0^{\prime }+r'.b$$ then $$(r-r')b=a_0^{\prime }-a_0\in M_0$$ and (r − r′) ∈ I. Therefore, g0((r − r′)b) and h(r − r′) are defined and we have:
\displaystyle \begin{aligned}g_0(a_0^{\prime}-a_0)=g_0((r-r')b)=h(r-r')=h^*(r-r')=(r-r')\cdot h^*(1). \end{aligned}
Thus, $$g_0(a_0^{\prime })-g_0(a_0)=r\cdot h^*(1)-r'\cdot h^*(1)$$ and this shows that $$g_0(a_0^{\prime })+r'\cdot h^*(1)=g_0(a_0)+r\cdot h^*(1)$$.

Clearly, g1(a0) = g0(a0) for all a0 ∈ M0, so that the map g1 extends g0. We conclude that (M0, g0) ≼ (M1, g1) and M0 ≠ M1, contradicting the maximality of (M0, g0). Therefore, M0 = N, the map g0 is a lifting of f and then E is injective.

### Proposition 2.2.4

If R is a Noetherian ring and (Ej)jJ is a family of injective R-modules, thenjJEj is an injective R-module.

### Proof

By the Baer criterion, it suffices to complete the diagram
where I is an ideal of R. If x ∈⊕jEj, then x = (ej), where ej ∈ Ej. Since R is noetherian, I is finitely generated. There exists a finite set S such that Im(f) ⊂⊕sSEs. But we already know that the finite direct sums are injective. Hence, there is a homomorphism g′ : R →⊕sSEs. Finally, composing g′ with the inclusion of ⊕sSEs into ⊕jJEj completes the given diagram.

Next step is to show that any R-module is a sub-module of an injective module, for this end we have to recall the basics of divisible modules.

### Definition 2.2.5 (Divisible Modules)

Let M be an R-module over a ring R and let r ∈ RZ(R) and m ∈ M. We say that m is divisible by r if there is some m′∈ M with m = rm′. In general, we say that M is a divisible module if for all r ∈ RZ(R) and for all m ∈ M we have that m is divisible by r.

### Proposition 2.2.6

Every injective module E is divisible.

### Proof

Assume that E is injective. Let e ∈ E and a ∈ RZ(R), we must find x ∈ E with e = ax. Define f : (a) → E by f(ra) = rm. Observe that this map is well defined because a is not a zero divisor. Since E is injective we have the following diagram:
where $$\overline {f}$$ extends f. In particular, $$m=f(a)=\overline {f}(a)=a\overline {f}(1)$$. So, the x that we need is $$x=\overline {f}(1)$$.

### Proposition 2.2.7

Let R be a principal ideal domain and M an R-module. Then we have that M is divisible if and only if M is injective.

### Proof

We are going to use Baer’s criterion. Assume that f : I → E is a homomorphism where I is a non zero ideal. By hypothesis, I = (a) for some non zero a ∈ I. Since E is divisible, there is some e ∈ E with f(a) = ae. Define h : R → E by h(s) = se. It is easy to check that h is a homomorphism, moreover, it extends f. That is, if s = ra ∈ I, we have that h(s) = h(ra) = rae = rf(a) = f(ra). Therefore, by Baer’s criterion, E is injective.

### Lemma 2.2.8

Let R be a ring. Then:
1. (i)

For all G abelian groups, $$\mathrm {Hom}_{\mathbb {Z}}(R,G)$$ is an R-module.

2. (ii)

If G is injective as a $$\mathbb {Z}$$ -module, then $$\mathrm {Hom}_{\mathbb {Z}}(R,G)$$ is R-injective.

### Proof

1. (i)

This statement is clear, because the addition is as usual, and with the multiplication by elements of R, we define (rf)(x) = f(rx) if r ∈ R and $$f\in \mathrm {Hom}_{\mathbb {Z}}(R,G)$$.

2. (ii)

If we have a monomorphism g : M1 → M2 and a homomorphism $$f:M_1\rightarrow \mathrm {Hom}_{\mathbb {Z}}(R,G)$$, we have to find an extension from M2 to $$\mathrm {Hom}_{\mathbb {Z}}(R,G)$$. But if we have that f, we can also define a homomorphism f′ between M1 and G in the following way, f′(m1) = (f(m1))(1). Is an homomorphism because f is also an homomorphism. So, as G is injective, we can find an extension of f′, namely $$\overline {f'}$$. With this map, we can define the extension we wanted $$\overline {f}(m_2):R\rightarrow G$$ where $$\overline {f}(m_2)(r)=\overline {f'}(rm_2)$$. The way that we constructed the map assure us that is an homomorphism and that extends f.

### Theorem 2.2.9

Let R be a ring and M an R-module. Then there exists an R-injective module E and a monomorphism f : M  E. In other words, any module M can be embedded as a submodule of an injective module.

### Proof

Since M is a $$\mathbb {Z}$$-module we have that $$M\cong \mathbb {Z}^{(I)}/H$$ for a suitable subgroup H of $$\mathbb {Z}^{(I)}$$. Notice that $$\mathbb {Z}^{(I)}\subset \mathbb {Q}^{(I)}$$ as abelian groups, so $$M\subset G=\mathbb {Q}^{(I)}/H$$. But as $$\mathbb {Q}$$ is divisible, we have that also G is divisible. Hence MG, where G is an injective abelian group. So from the last Lemma we deduce that $$\mathrm {Hom}_{\mathbb {Z}}(R,G)$$ is an R-injective module. Then we have the exact sequence of R-modules
\displaystyle \begin{aligned}0\longrightarrow \mathrm{Hom}_{\mathbb{Z}}(R,M) \longrightarrow E=\mathrm{Hom}_{\mathbb{Z}}(R,G). \end{aligned}
Next step is to embed M in E; it is enough to show that the linear map $$f:M\rightarrow \mathrm {Hom}_{\mathbb {Z}}(R,M)$$, defined by f(m)(r) = rm if r ∈ R, is injective. If f(m)(r) = 0 for all r ∈ R, we have that f(m)(1) = m = 0.

### Definition 2.2.10 (Proper Essential Extensions)

Let R be a ring and let N ⊂ M be R-modules. We say that M is an essential extension of N if for any non-zero submodule U of M one has U ∩ N ≠ 0. An essential extension M of N is called proper if N ≠ M.

### Proposition 2.2.11

Let R be a ring.
1. (i)

An R-module N is injective if and only if it has no proper essential extensions.

2. (ii)

Let N  M be an essential extension. Let E be an injective module containing N. Then there exists a monomorphism ϕ : ME extending the inclusion N  M.

### Proof

1. (i)

Let’s assume that N is injective and let N ⊂ M be an essential extension. Since N is injective, N is a direct summand of M, Proposition 2.2.2. Let S be the complement of N in M, Proposition 2.2.2. Then N ∩ S = 0 and so, the extension N ⊂ M is essential, so S = 0 and N = M. Conversely, suppose that N has no proper essential extensions. Let E be an injective module containing N, Theorem 2.2.9. Let us consider the set of submodules M ⊂ E such that M ∩ N = 0. This set is not empty 0 ∈ X and it is inductively ordered. By Zorn’s Lemma there is a maximal element L ∈ X, so NN + LL ⊂ EL. This extension is essential. Let K be an R-module L ⊂ K ⊂ E such that KL ∩ (N + L)∕L = 0. Hence K ∩ (N + L) = 0, so K ∩ N = 0. From the maximality of L we deduce K = L. Since N has no proper essential extensions we obtain E = N + L. On the other hand we have L ∩ N = 0, so E = N ⊕ L. From Proposition 2.2.2 we get that N is injective.

2. (ii)

Since E is injective there exists a homomorphism ϕ : ME extending the inclusion N ⊂ M. If $$\operatorname {ker}(\phi )\neq 0$$ then $$\operatorname {ker}(\phi )\cap M\neq 0$$ because the extension N ⊂ M is essential. Let $$0\neq x\in \operatorname {ker}(\phi )\cap M$$ then we get a contradiction: x = ϕ(x) = 0.

### Definition 2.2.12

Let be R a ring and M an R-module. An injective hull of M is an injective module ER(M) such that M ⊂ ER(M) is an essential extension.

### Proposition 2.2.13

Let R be a ring and let M be an R-module.
1. (i)

M admits an injective hull. Moreover, if M  I and I is injective, then a maximal essential extension of M in I is an injective hull of M.

2. (ii)
Let E be an injective hull of M, let I be an injective R-module, and α : M  I a monomorphism. Then there exists a monomorphism φ : E  I such that the following diagram is commutative, where i is the inclusion:
In other words, the injective hulls of M are the “minimal” injective modules in which M can be embedded.

3. (iii)
If E and E′ are injective hulls of M, then there exists an isomorphism φ : E  E′ such that the following diagram commutes:

### Proof

1. (i)

We know by Theorem 2.2.9 that we can embed M into an injective module I. Now consider $$\mathscr {S}$$ to be the set of all essential extensions N with M ⊂ N ⊂ I. Applying Zorn’s Lemma to this set yields to a maximal essential extension M ⊂ E such that E ⊂ I. We claim that E has no proper essential extensions and because of Proposition 2.2.11 we can say that E will be injective and therefore it will be the injective hull we are looking for. Assume that E has a proper essential extension E′. Since I is injective, there exists ψ : E′→ I extending the inclusion E ⊂ I. Suppose $$\operatorname {ker} \psi =0$$; then Imψ ⊂ I is an essential extension of M (in I) properly containing E, which contradicts the fact that E is maximal. On the other hand, since ψ extends the inclusion E ⊂ I we have $$E\cap \operatorname {ker}\psi =0$$. But this contradicts with the essentiality of the extension E ⊂ E′. And then we have the result we were looking for.

2. (ii)

Since I is injective, α can be extended to an homomorphism φ : E → I. We have that φ|M = α, and so $$M\cap \operatorname {ker} \varphi =\operatorname {ker} \alpha =0$$. Thus, since the extension M ⊂ E is essential, we even have $$\operatorname {ker} \phi =0$$ and therefore φ is a monomorphism.

3. (iii)

By (ii) there is a monomorphism ϕ : E → E′ such that ϕ|M equals the inclusion M ⊂ E′. Then, as ImϕE because of the injectivity, Imϕ is also injective and hence a direct summand of E′. However, since the extension M ⊂ E′ is essential, ϕ is exhaustive because there can’t be direct summands different than the total. Therefore, ϕ is an isomorphism.

### Remark

We can use this proposition to build an injective resolution, $$E_R^*(M)$$ of a module M. We let E0(M) = ER(M) and denote the embedding by −1. Now suppose that the injective resolution has been constructed till the i-th step:
We define then Ei+1 = ER(Coker i−1), and i is defined as the inclusion.

### Definition 2.2.14

Let $$(R,\mathfrak {m},{\mathbf {k}})$$ be a local ring. Given an R-module M the Matlis dual of M is M = HomR(M, ER(k)). We write (−) = HomR(−, ER(k)), which is a contravariant exact functor from the category R_mod into itself.

### Proposition 2.2.15

Let $$(R,\mathfrak {m},{\mathbf {k}})$$ be a local ring. Then (−) is a faithful functor. Furthermore, if M is a R-module of finite length, then ℓR(M) = ℓR(M). If R is in addition an Artin ring then ℓR(ER(k)) = ℓR(R) < ∞.

### Proof

We have to show that if M is a nonzero R-module then M is nonzero. Let’s take a non-zero cyclic submodule $$R/\mathfrak {a}$$ of M. Since $$\mathfrak {a}\subset \mathfrak {m}$$ we have the maps
\displaystyle \begin{aligned} M\hookleftarrow R/\mathfrak{a} \twoheadrightarrow R/\mathfrak{m}\cong {\mathbf{k}}. \end{aligned}
Notice that k = HomR(k, ER(k))≅k. Applying the functor (−) to this diagram we get
\displaystyle \begin{aligned}M^\vee \twoheadrightarrow (R/\mathfrak{a})^\vee\hookleftarrow {\mathbf{k}}^\vee\cong{\mathbf{k}}, \end{aligned}
implying that M is nonzero.
Let M be a finite length R-module, we use induction on (M) to prove R(M) = R(M). If R(M) = 1, then M is a simple R-module and thus $$M\cong R/\mathfrak {m}={\mathbf {k}}$$. Thus R(M)≅R(k) = 1. For the general case, pick a simple submodule S ⊂ M. We apply (−) to the short exact sequence:
\displaystyle \begin{aligned}0\longrightarrow S \longrightarrow M \longrightarrow M/S \longrightarrow 0 \end{aligned}
Since Sk, we have (S) = 1. Now, by induction, R((MS)) = R(MS) = R(M) − 1. We conclude then R(M) = R(M).

Let us assume that R is Artin, so R(R) < . From the first part we get R(ER(k)) = R(R) < .

### Proposition 2.2.16

Let R be a ring, $$\mathfrak {a}$$ an ideal of R and M a R-module annihilated by $$\mathfrak {a}$$. Then, if E = ER(M):
\displaystyle \begin{aligned}E_{R/\mathfrak{a}}(M)=\{e\in E\ :\ \mathfrak{a}e=0\}=(0:_E\mathfrak{a})\end{aligned}

### Proof

Both M and $$(0:_E\mathfrak {a})$$ are annihilated by $$\mathfrak {a}$$ and thus can be thought as $$R/\mathfrak {a}$$-modules. Clearly $$M\subset (0:_E\mathfrak {a})\subset E$$. Since all $$R/\mathfrak {a}$$-submodule of $$(0:_E\mathfrak {a})$$ is also a R-submodule of E, necessarily $$(0:_E\mathfrak {a})$$ is an essential extension on M. So now we need to check that $$(0:_E\mathfrak {a})$$ is injective. So let us consider a diagram of $$R/\mathfrak {a}$$-modules:
We have to prove that there is $$g:B\longrightarrow 0:_E\mathfrak {a})$$ such that f = g ∘ i. But as we can think these modules as R-modules, we can replace $$(0:_E\mathfrak {a})$$ by E and, since E is injective, we can extend the diagram and make the diagram commutative. But this commutativity implies that $$\text{Im}(g)\subset (0:_E\mathfrak {a})$$ and therefore the original diagram also commutes.

### Corollary 2.2.17

Let $$(R,\mathfrak {m},{\mathbf {k}})$$ be a local ring and E = ER(k). Let $$\mathfrak {a}$$ be an ideal of R. Then:
1. (i)

$$E_{R/\mathfrak {a}}({\mathbf {k}})=(0:_E\mathfrak {a})$$

2. (ii)

$$E=\bigcup _{t\ge 1}E_{R/\mathfrak {m}^t}({\mathbf {k}})$$

Now it’s time to prove some technical results with the assumption that we need, the completeness of the Noetherian local ring.

### Lemma 2.2.18

Let $$(R,\mathfrak {m},{\mathbf {k}})$$ be a complete Noetherian local ring and E = ER(k). Then:
1. (i)

RE and ER.

2. (ii)

For every R-module M the natural map M  M∨∨ induce isomorphisms R  R∨∨ and E  E∨∨.

### Proof

1. (i)

It is well known that R = HomR(R, E)≅E. Now let’s prove ER. Assume first that R is Artinian. Consider the map θ : R → E = HomR(E, E) which sends an element r ∈ R to the homothety defined by r. Since (R) = (E), Proposition 2.2.15, we only need to prove that θ is injective. Suppose that rE = 0. Then, by the last Corollary, ER∕(r)(k) = (0 :E(r)) = E, and, by the same argument, (E) = (R∕(r)). This implies that (R) = (R∕(r)), then r = 0.

Assume now that R is Noetherian and complete. We consider the map θ : R → E = HomR(E, E) as above, we will prove that θ is an isomorphism. Let’s write $$R_t=R/\mathfrak {m}^t$$ for each t. By the last corollary $$E_t:=E_{R_t}({\mathbf {k}})=(0:_E\mathfrak {m}^t)$$. Let φ ∈HomR(E, E) = E. It is clear that φ(Et) ⊂ Et and thus $$\varphi \in \mathrm {Hom}_{R_t}(E_t,E_t)$$. Since Rt is Artinian we have φ is a homothety defined by an element rt ∈ Rt. The fact Et ⊂ Et+1 implies that $$r_t-r_{t+1}\in \mathfrak {m}^t$$ for all t ≥ 1. In consequence, $$r=(r_t)_t\in \hat {R}=R$$ and $$r_t=r+\mathfrak {m}^t$$ for all t ≥ 1. We claim that φ is given by multiplication by r. This follows from the fact that E = ∪tEt and that φ(e) = rte for all e ∈ Et. Moreover, r is uniquely determined by φ, and we conclude that θ is bijective.

2. (ii)

We consider the natural homomorphism γ : M → M∨∨ = HomR(HomR(M, E), E) given by γ(m)(φ) = φ(m). Fisrt we prove that γ : R → R∨∨ is an isomorphism. This map is the composition of the two isomorphisms given in part (i) RE≅(R). In fact, if r ∈ R, the map RE sends r to multiplication by r, hr : E → E. Now the map E≅(R) sends hr to αr defined by αr(φ) = hr(φ(1)) = φ(r), so αr = γ(r). The case of E is analogous to this one.

### Proposition 2.2.19

Let $$(R,\mathfrak {m},{\mathbf {k}})$$ be a complete Noetherian local ring and E = ER(k).
1. (i)

There is an order-reversing bijectionbetween the set of R-submodules of E and the set of ideals of R given by: if M is a submodule of E then (EM)M = (0 :RM), and (RI)I = (0 :EI) for an ideal I  R,

2. (ii)

E is an Artinian R-module,

3. (iii)

an R-module is Artinian if and only if it can be embedded in En for some $$n\in \mathbb {N}$$.

### Proof

(i) Since M ⊂ M⊥⊥ we have to prove that M⊥⊥⊂ M. Consider the exact sequence
\displaystyle \begin{aligned}0\longrightarrow M \longrightarrow E \stackrel{\pi}{\longrightarrow} E/M \longrightarrow 0, \end{aligned}
dualizing with respect E, we get an injective homomorphism, Lemma 2.2.18,
\displaystyle \begin{aligned}0 \longrightarrow (E/M)^\vee \stackrel{\pi^\vee}{\longrightarrow} E^\vee\stackrel{\theta^{-1}}{\cong} R. \end{aligned}
Hence every g ∈ (EM) is mapped to an r ∈ R such that (θ−1 ∘ π)(g) = r, or equivalently g ∘ π = π(g) = hr = θ(r) where hr : EE is the homothety defined by r. Since g ∘ π(M) = g(0) = 0 we get rM = 0, so (EM)⊂ M. On the other hand if r ∈ M then we can consider the map g : EME such that $$g(\overline {x})= r x$$ for all x ∈ E. It is easy to see that (θ−1 ∘ π)(g) = r, so $$(E/M)^\vee \stackrel {\theta ^{-1} \pi ^\vee }{\cong }M^\perp$$. Let x ∈ E ∖ M then there is g ∈ (EM) such that $$g(\overline {x})\neq 0$$, Lemma 2.2.18. From the above isomorphism we deduce that there is r ∈ M such that rx ≠ 0. This shows that M∨∨⊂ M and then M = M∨∨.
Let I be an ideal of R. As in the previous case we have I ⊂ I⊥⊥. From the natural exact sequence
\displaystyle \begin{aligned}0\longrightarrow I \longrightarrow R \stackrel{\pi}{\longrightarrow} R/I \longrightarrow 0, \end{aligned}
we get an injective homomorphism, Lemma 2.2.18,
\displaystyle \begin{aligned}0 \longrightarrow (R/I)^\vee \stackrel{\pi^\vee}{\longrightarrow} R^\vee\stackrel{\theta^{-1}}{\cong} E. \end{aligned}
As in the previous case θ−1 ∘ π maps (RI) to I. Let r ∈ R ∖ I then there is g ∈ (RI) such that $$g(\overline {r})\neq 0$$, Lemma 2.2.18. Hence $$x=g(\overline {1})\in I^\perp$$ and rx ≠ 0, i.e. r∉(0 :Rx). Since $$I^{\perp \perp }=\bigcap _{x\in I^\perp } (0:_R x)$$ we get I⊥⊥⊂ I and then I = I⊥⊥.

(ii) Since R is Noetherian, by (i) we get that E is Artinian.

(iii) We consider the set X of kernels of all homomorphisms F : MEn, for all $$n\in \mathbb N$$. This is a set of submodules of M. Since M is Artininan there is a minimal element $$\operatorname {ker}(F)$$ of X, where F : MEn for some $$n\in \mathbb N$$. Assume that $$\operatorname {ker}(F)\neq 0$$ and pick $$0\neq x \in \operatorname {ker}(F)$$. From Proposition 2.2.15 there is σ : ME such that σ(x) ≠ 0. Let us consider F : MEn+1 defined by F(y) = (F(y), σ(y)). Since $$\operatorname {ker}(F^*)\varsubsetneq \operatorname {ker}(F)$$ we get a contradiction with the minimality of $$\operatorname {ker}(F)$$.

Assume that M is a submodule of En for some integer n. From (ii) we get that M is an Artin module.

In the next result we will prove Matlis’ duality, see  Theorem 5.20.

### Theorem 2.2.20 (Matlis Duality)

Let $$(R,\mathfrak {m},{\mathbf {k}})$$ be a complete Noetherian local ring, E = ER(k) and let M be a R-module. Then:
1. (i)

If M is Noetherian then M is Artinian.

2. (ii)

If M is Artinian then M is Noetherian.

3. (iii)

If M is either Noetherian or Artinian then M∨∨M.

4. (iv)

The functor (−) is a contravariant, additive and exact functor.

5. (v)

The functor (−) is an anti-equivalence between R_mod.Noeth and R_mod.Artin (resp. between R_mod.Artin and R_mod.Noeth). It holds (−)∘ (−) is the identity functor of R_mod.Noeth (resp. R_mod.Artin).

### Proof

1. (i)
Let’s consider a presentation of M
Since (−) is exact, it induces an exact sequence:
Thus M can be seen as a submodule of (Rn)≅(R)nEn, Lemma 2.2.18. Since E is Artinian as we saw in the previous corollary, so is En and hence also M. Applying the functor (−) again we get a commutative diagram:
whose rows are exact. Since we proved that in this context R → R∨∨ is an isomorphism, MM∨∨

2. (ii)
We proved that MEn for some $$n\in \mathbb {N}$$. Since E is Artinian, so is EnM and thus EnMEm for some $$m\in \mathbb {N}$$. In consequence, we have an exact sequence:
As before, if we apply (−) we have an exact sequence:
and M can be seen as a quotient of (En)≅(E)nRn, where the isomorphism is the one we proved in Lemma 2.2.18. This implies that M is Noetherian.

3. (iii)
Finally, we apply the functor (−) to the last exact sequence we obtain the commutative diagram
And again, since E → E∨∨ is an isomorphism, MM∨∨

4. (iv)

This is a consequence of the previous statements.

## 2.3 Macaulay’s Correspondence

Let k be an arbitrary field. Let be the ring of the formal series with maximal ideal $${\mathfrak m} =(x_1,\cdots ,x_n)$$ and let S = k[y1, …, yn] be a polynomial ring, we denote by μ = (x1, …, xn) the homogeneous maximal ideal of S.

It is well known that R is an S-module with the standard product. On the other hand, S can be considered as R-module with two linear structures: by derivation and by contraction.

If char(k) = 0, the R-module structure of S by derivation is defined by
\displaystyle \begin{aligned}\begin{array}{cccc} R\times S & \longrightarrow & S & \\ (x^\alpha, y^\beta) & \mapsto & x^\alpha\circ y^\beta = & \left\{ \begin{array}{ll} \frac{\beta!}{(\beta-\alpha)!}y^{\beta-\alpha} & \beta \ge \alpha \\ \\ 0, & {\text{otherwise}} \end{array} \right. \end{array} \end{aligned}
where for all $$\alpha , \beta \in \mathbb N^n$$, $$\alpha !=\prod _{i=1}^n \alpha _i!$$
If char(k) ≥ 0, the R-module structure of S by contraction is defined by:
\displaystyle \begin{aligned}\begin{array}{cccc} R\times S & \longrightarrow & S & \\ (x^\alpha, y^\beta) & \mapsto & x^\alpha \circ y^\beta = & \left\{ \begin{array}{ll} y^{\beta-\alpha} & \beta \ge \alpha \\ \\ 0, & otherwise \end{array} \right. \end{array} \end{aligned}
$$\alpha , \beta \in \mathbb N^n$$

### Proposition 2.3.1

For any field k there is a R-module homomorphism
\displaystyle \begin{aligned}\begin{array}{cccc} \sigma : &(S,der) & \longrightarrow & (S,cont) \\ &y^\alpha& \mapsto& \alpha! \; y^\alpha \end{array} \end{aligned}

If char(k) = 0 then σ is an isomorphism of R-modules.

### Proof

For proving the first statement it is enough to show that
\displaystyle \begin{aligned}\sigma(x^\alpha \circ y^\beta)=x^\alpha \sigma( y^\beta). \end{aligned}
This is easy:
\displaystyle \begin{aligned}\sigma(x^\alpha \circ y^\beta)=\sigma\left(\frac{\beta!}{(\beta-\alpha)!} y^{\beta-\alpha}\right)=\frac{\beta!}{(\beta-\alpha)!}( (\beta-\alpha)! y^{\beta-\alpha}) \end{aligned}
\displaystyle \begin{aligned}= \beta! \; y^{\beta-\alpha}= x^{\alpha} \circ \sigma(y^\beta) \end{aligned}
If char(k) = 0 then the inverse of σ is yα→(1∕α!)yα

Given a family of polynomials Fj, j ∈ J, we denote by 〈Fj, j ∈ J〉 the submodule of S generated by Fj, j ∈ J, i.e. the k-vector subspace of S generated by xα ∘ Fj, j ∈ J, and $$\alpha \in \mathbb N^n$$. We denote by 〈Fj, jJk the k-vector space generated by Fj, j ∈ J.

In the next result we compute the injective hull of the residue field of a power series ring, [18, 25].

### Theorem 2.3.2

Let be the n-dimensional power series ring over a field k . If k is of characteristic zero then
\displaystyle \begin{aligned}E_R({\mathbf{k}})\cong (S, der) \cong (S,cont). \end{aligned}
If k is of positive characteristic then
\displaystyle \begin{aligned}E_R({\mathbf{k}})\cong (S,cont). \end{aligned}

### Proof

We write E = ER(k). From Corollary 2.2.17 we get
\displaystyle \begin{aligned}E=\bigcup_{i\ge 0} (0:_E {\mathfrak m}_R^i)=\bigcup_{i\ge 0} E_{R/{\mathfrak m}_R^i}({\mathbf{k}}) \end{aligned}
Hence the problem is reduced to the computation of $$E_{R/{\mathfrak m}_R^i}({\mathbf {k}})\subset E$$.

Notice that Si−1 := {f ∈ S∣deg(f) ≤ i − 1}⊂ S is an sub-R-module of S, with respect to the derivation or contraction structure of S, and that Si−1 is annihilated by $${\mathfrak m}_R^i$$. Hence Si−1 is an $$R/ {\mathfrak m}_R^i$$-module. For any characteristic of the ground field k the extension k ⊂ Si−1 := {f ∈ S∣deg(f) ≤ i − 1} is essential. In fact, let 0 ≠ M ⊂ Si−1 be a sub-$$R/{\mathfrak m}_R^i$$-module then it holds 1 ∈ M.

From Theorem 2.2.13 there exists $$L\cong E_{R/{\mathfrak m}_R^i}({\mathbf {k}})$$ such that
\displaystyle \begin{aligned}{\mathbf{k}} \subset S_{\le i-1} \subset L\cong E_{R/{\mathfrak m}_R^i}({\mathbf{k}}). \end{aligned}
Since, Proposition 2.2.15,
\displaystyle \begin{aligned} \begin{array}{rcl} {\mathrm{Length}}_{R/{\mathfrak m}_R^i}(E_{R/{\mathfrak m}_R^i}({\mathbf{k}})) &\displaystyle =&\displaystyle {\mathrm{Length}}_{R/{\mathfrak m}_R^i}(R/{\mathfrak m}_R^i)\\ &\displaystyle =&\displaystyle {\mathrm{Length}}_{R/{\mathfrak m}_R^i}(S_{\le i-1}) \end{array} \end{aligned}
from the last inclusions we get $$S_{\le i-1} \cong E_{R/{\mathfrak m}_R^i}({\mathbf {k}})$$. Hence
\displaystyle \begin{aligned}E_R({\mathbf{k}})\cong\bigcup_{i\ge 0}S_{\le i-1}=S. \end{aligned}

From the previous results we can recover the classical result of Macaulay, , for the power series ring, see [16, 21].

If I ⊂ R is an ideal, then (RI) is the sub-R-module of S that we already denote by I, see Proposition 2.2.19,
\displaystyle \begin{aligned} {I^{\perp}}=\{ g \in S\ |\ I \circ g = 0 \}, \end{aligned}
this is the Macaulay’s inverse system of I. Given a sub-R-module M of S then dual M is an ideal of R that we already denote by (SM), see Proposition 2.2.19,
\displaystyle \begin{aligned} M^\perp= \{ f \in R \ \mid \ f \circ g= 0 \ \text{ for all }\ g \in M\}. \end{aligned}
We will write sometimes this module as $$M^\perp =\operatorname {Ann}_R(M)$$.

### Proposition 2.3.3 (Macaulay’s Duality)

Let be the n-dimensional power series ring over a field k. There is a order-reversing bijectionbetween the set of finitely generated sub-R-submodules of and the set of $${\mathfrak m}$$-primary ideals of R given by: if M is a submodule of S then M = (0 :RM), and I = (0 :SI) for an ideal I  R.

### Proof

The one-to-one correspondence is a particular case of Proposition 2.2.19. Theorem 2.2.20 gives the one-to-one correspondence between finitely generated sub-R-submodules of S and $${\mathfrak m}$$-primary ideals of R.

### Remark

Macaulay proved more as we will see later on. Trough this correspondence Macaulay proved that Artin Gorenstein k-algebras A = RI of socle degree s correspond to R-submodules of S generated by a polynomial F of degree s, see Proposition 2.4.4.

Let A = RI be an Artin quotient of R, we denote by $${\mathfrak n}={\mathfrak m}/I$$ the maximal ideal of A. The socle of A is the colon ideal $$\operatorname {Soc}(A) = 0 :_A {\mathfrak n}$$, notice that $$\operatorname {Soc}(A)$$ is a k-vector space subspace of A. We denote by s(A) the socle degree of A, that is the maximum integer j such that $${\mathfrak n}^j \neq 0.$$ The (Cohen-Macaulay) type of A is $$t(A) := \operatorname {dim}_{{\mathbf {k}}} \operatorname {Soc}(A)$$.

The Hilbert function of A = RI is by definition
\displaystyle \begin{aligned} \operatorname{HF}_A(i) = \operatorname{dim}_{{\mathbf{k}}} \left(\frac{{\mathfrak n}^i}{{\mathfrak n}^{i+1}}\right), \end{aligned}
the multiplicity of A is the integer $$e(A):=\operatorname {dim}_{{\mathbf {k}}} (A) = \operatorname {dim}_{{\mathbf {k}}} I^{\perp }$$, Propositions 2.3.3 and 2.2.19. Notice that s(A) is the last integer such that $$\operatorname {HF}_A(i)\neq 0$$ and that $$e(A)=\sum _{i=0}^s \operatorname {HF}_A(i)$$. The embedding dimension of A is $$\operatorname {HF}_A(1)$$.

### Example 2.3.4

Let be a polynomial. We consider the R-module structure of S = k[x, y] defined by the contraction ∘. Then 〈F〉 = 〈F, y2 + x, y + x, x, 1〉k and $$\operatorname {dim}_{{\mathbf {k}}}(\langle F\rangle )=5$$. We have that $$I=\operatorname {Ann}_R(\langle F\rangle )=(xy-y^3, x^2-xy)$$, i.e. I is a complete intersection ideal of R. The Hilbert function of A is $$\operatorname {HF}_A=\{1,2,1,1\}$$, so e(A) = 5 and s(A) = 3

The associated graded ring to A is the graded k-algebra ring $$gr_{{\mathfrak n}}(A)= \oplus _{i\ge 0} {\mathfrak n}^i/{\mathfrak n}^{i+1}$$. Notice that the Hilbert function of A and its associated graded ring $$gr_{{\mathfrak n}}(A)$$ agrees. We denote by I the homogeneous ideal of S generated by the initial forms of the elements I. It is well known that $$gr_{{\mathfrak n}}(A)\cong S/I^*$$ as graded k-algebras, in particular $$gr_{{\mathfrak n}}(A)_i\cong (S/I^*)_i$$ for all i ≥ 0.

We denote by Si (resp. S<i, resp. Si), $$i\in \mathbb N$$, the k-vector space of polynomials of S of degree less or equal (resp. less, resp. equal to) to i, and we consider the following k-vector space
\displaystyle \begin{aligned} (I^{\perp})_i := {\frac{ I^{\perp} \cap S_{\le i} + S_{< i}}{ S_{< i}}}. \end{aligned}

### Proposition 2.3.5

For all i ≥ 0 it holds
\displaystyle \begin{aligned}\operatorname{HF}_{A}(i) = \operatorname{dim}_{{\mathbf{k}}} (I^{\perp})_i. \end{aligned}

### Proof

Let’s consider the following natural exact sequence of R-modules
\displaystyle \begin{aligned}0 \longrightarrow \frac{{\mathfrak n}^i}{{\mathfrak n}^{i+1}} \longrightarrow \frac{A}{{\mathfrak n}^{i+1}} \longrightarrow \frac{A}{{\mathfrak n}^{i}} \longrightarrow 0. \end{aligned}
Dualizing this sequence we get
\displaystyle \begin{aligned}0 \longrightarrow (I+{\mathfrak m}^{i})^\perp \longrightarrow (I+{\mathfrak m}^{i+1})^\perp \longrightarrow \left(\frac{{\mathfrak n}^i}{{\mathfrak n}^{i+1}}\right)^\vee \longrightarrow 0 \end{aligned}
so we get the following sequence of k-vector spaces:
\displaystyle \begin{aligned}\left(\frac{{\mathfrak n}^i}{{\mathfrak n}^{i+1}}\right)^\vee \cong \frac{(I+{\mathfrak m}^{i+1})^\perp}{(I+{\mathfrak m}^{i})^\perp} = \frac{I^\perp \cap S_{\le i}}{I^\perp \cap S_{\le i-1}}\cong {\frac{ I^{\perp} \cap S_{\le i} + S_{< i}}{ S_{< i}}}. \end{aligned}
From Proposition 2.2.15 we get the claim.
Consider the map
\displaystyle \begin{aligned}\begin{array}{cccc} \langle \mid \rangle :& R\times S & \longrightarrow & {\mathbf{k}} \\ & (F,G) & \mapsto & (F\circ G)(0) \end{array} \end{aligned}
In the next result we collect some results on 〈∣〉 that we will use later on.

### Proposition 2.3.6

1. 1.

〈∣〉 is a bilinear non-degenerate map of k-vector spaces.

2. 2.
If I is an ideal of R then
\displaystyle \begin{aligned}I^\perp =\{G\in S\mid \langle I \mid G\rangle =0\} \end{aligned}

3. 3.
〈∣〉 induces a bilinear non-degenerate map of k-vector spaces
\displaystyle \begin{aligned}\overline{\langle \mid \rangle}:\frac{R}{I}\times I^\perp \longrightarrow {\mathbf{k}} \end{aligned}

4. 4.
We have an isomorphism of k-vector spaces:
\displaystyle \begin{aligned}\left(\frac{S}{I^*}\right)_i\cong (I^\perp)_i \end{aligned}
for all i ≥ 0.

We will denote by ∗ the duality defined by exact pairing $$\overline {\langle \mid \rangle }$$, notice that (RI)I.

If $${ \underline i}=(i_1,\cdots , i_n)\in \mathbb N^n$$ is a integer n-pla we denote by $$\partial _{{ \underline i}}(G)$$, G ∈ S, the derivative of G with respect to $$y_1^{i_1}\cdots y_n^{i_n}$$, i.e. $$\partial _{{ \underline i}}(G)=(x_1^{i_1}\cdots x_n^{i_n})\circ G$$.

Let Ω = {ωi} be the canonical basis of $$R/{\mathfrak m}^{s+1}$$ as a k-vector space consisting of the standard monomials xα ordered by the deg-lex order with x1 > ⋯ > xn and, then the dual basis with respect to ∗ is the basis $$\varOmega ^*=\{ \omega _i^* \}$$ of Sj where
\displaystyle \begin{aligned} (x^{\alpha})^* = \frac 1 {\alpha !} y^{\alpha}, \end{aligned}
in fact $$\omega _j \circ \omega _i^* = \overline {\langle \omega _j \mid \omega _i^*\rangle }=\delta _{ij}$$, where δij = 0 if i ≠ j and δii = 1.

## 2.4 Gorenstein, Level and Compressed Algebras

### Definition 2.4.1

An Artin ring A is Gorenstein if t(A) = 1; A is an Artin level algebra if $$\operatorname {Soc}(A)={\mathfrak m}^s$$, where s is the socle degree of A.

### Proposition 2.4.2

Let A = RI be an Artin ring, the following conditions are equivalent:
1. (i)

A is Gorenstein,

2. (ii)

AEA(k) as R-modules,

3. (iii)

A is injective as A-module.

### Proof

Assume (i). Since the extension $${\mathbf {k}}=\operatorname {Soc}(A) \subset A$$ is essential we have the A-module extensions, Proposition 2.2.11 (ii),
\displaystyle \begin{aligned} {\mathbf{k}}=\operatorname{Soc}(A) \subset A \subset E_A({\mathbf{k}}). \end{aligned}
so A = EA(k), Proposition 2.2.15. Since SEk(k) is an injective R-module, (ii) implies (iii).
Assume that A is injective as A-module. From Proposition 2.2.13 (ii) we get the A-module extensions
\displaystyle \begin{aligned} {\mathbf{k}} \subset E_A({\mathbf{k}}) \subset A, \end{aligned}
from Proposition 2.2.15 we get (i).

Given an R-module M we denote by μ(M) the minimal number of generators of M.

### Proposition 2.4.3

Let A = RI be an Artinian local ring. Then
\displaystyle \begin{aligned}\operatorname{Soc}(A)^\vee=\frac{I^\perp}{ {\mathfrak m} \circ I^\perp}. \end{aligned}
In particular the Cohen-Macaulay type of A is
\displaystyle \begin{aligned}t(A)=\operatorname{dim}_{{\mathbf{k}}}(I^\perp/{\mathfrak m}\circ I^\perp)=\mu_R(I^\perp). \end{aligned}

### Proof

Let’s consider exact sequence of R-modules
\displaystyle \begin{aligned}0\longrightarrow \operatorname{Soc}(A)=(0:_A {\mathfrak n} )\longrightarrow A \stackrel{(x_1,\cdots,x_n)}{\longrightarrow} A^n, \end{aligned}
dualizing this sequence we get
\displaystyle \begin{aligned}(I^\perp)^n\stackrel{\sigma}{\longrightarrow} I^\perp \longrightarrow \operatorname{Soc}(A)^\vee \longrightarrow 0 \end{aligned}
where $$\sigma (f_1,\cdots ,f_n)=\sum _{i=1}^n x_i\circ f_i$$. Hence
\displaystyle \begin{aligned}\operatorname{Soc}(A)^\vee=\frac{I^\perp}{ (x_1,\dots,x_n)\circ I^\perp}=\frac{I^\perp}{ {\mathfrak m}\circ I^\perp} \end{aligned}
Since $$t(A)=\operatorname {dim}_{{\mathbf {k}}}(\operatorname {Soc}(A))=\operatorname {dim}_{{\mathbf {k}}}(\operatorname {Soc}(A)^\vee )=\mu (I^\perp )$$, Proposition 2.2.15.

Given a polynomial F ∈ S of degree r we denote by top(F) the degree r form of F where r = deg(F).

### Proposition 2.4.4

Let I be an $${\mathfrak m}$$-primary ideal of R. The quotient A = RI is an Artin level algebra of socle degree s and Cohen-Macaulay type t if and only if I is generated by t polynomials F1, ⋯ , Ft ∈ S such that deg(Fi) = s, i = 1, ⋯ , t, and top(F1), ⋯ , top(Ft) are k-linear independent forms of degree s. In particular, A = RI is Gorenstein of socle degree s if and only if I is a cyclic R-module generated by a polynomial of degree s.

### Proof

Assume that A is an Artin level algebra of socle degree s and Cohen-Macaulay type t. In particular $$\operatorname {Soc}(A)={\mathfrak n}^s={\mathfrak m}^s+I/I$$ so
\displaystyle \begin{aligned}\operatorname{Soc}(A)^\vee= \frac{I^\perp}{I^\perp\cap S_{\le s-1}}. \end{aligned}
From the last result we get
\displaystyle \begin{aligned}{\mathfrak m}\circ I^\perp= I^\perp\cap S_{\le s-1}. \end{aligned}
From this identity we deduce that I is generated by t polynomials F1, ⋯ , Ft of degree s and top(F1), ⋯ , top(Ft) are k-linear independent.
Assume that I = 〈F1, ⋯ , Ft〉 such that deg(Fi) = s, i = 1, ⋯ , t, and that top(F1), ⋯ , top(Ft) are k-linear independent forms of degree s. Hence F1, ⋯ , Ft is a minimal system of generators of I, in particular μR(I) = t and from the last result we have that t is the Cohen-Macaulay type of A. Furthermore, since deg(Fi) = s, i = 1, ⋯ , t, we have
\displaystyle \begin{aligned}{\mathfrak m}\circ I^\perp= I^\perp\cap S_{\le s-1}. \end{aligned}
From the last result we deduce $$\operatorname {Soc}(A)={\mathfrak n}^s$$, i.e. A is Artin level of socle degree s.

In the last section we will prove the following result, see Proposition 2.6.3,

### Corollary 2.4.5

Let A = RI be an Artin algebra of embedding dimension two. Then
\displaystyle \begin{aligned}\mu(I)=t(R/I)+1. \end{aligned}

A is Gorenstein if and only if I is a complete intersection.

The initial degree of A = RI is the integer r such that $$I \subseteq {\mathfrak m}^r$$ and $$I\nsubseteq {\mathfrak m}^{r+1}$$. The socle type of A is the sequence σ(A) = (0, …, σr−1, σr, …, σs, 0, 0, … ), s is the socle degree of A, with
\displaystyle \begin{aligned}\sigma_i := \operatorname{dim}_{{\mathbf{k}}} \left(\frac{(0: {\mathfrak n}) \cap {\mathfrak n}^i}{ (0:{\mathfrak n}) \cap {\mathfrak n}^{i+1}}\right). \end{aligned}
Notice that σs > 0 and σj = 0 for j > s, . See  for some conditions on a sequence of integers to be the socle type of an Artin algebra

### Remark

An Artin algebra of socle degree s and Cohen-Macaulay type t is level if and only if σj = 0 for j ≠ s and σs = t. The Artin algebra is Gorenstein if and only if σj = 0 for j ≠ s and σs = 1 .

We say that the Hilbert function $$\operatorname {HF}_A$$ is maximal in the class of Artin level algebras of given embedding dimension and socle type, if for each integer i, $$\operatorname {HF}_A(i) \ge \operatorname {HF}_{B}(i)$$ for any other Artin algebra B in the same class. The existence of a maximal $$\operatorname {HF}_A$$ was shown for graded algebras by Iarrobino . In the general case by Fröberg and Laksov , by Emsalem , by Iarrobino and the author of this notes in  in the local case.

### Definition 2.4.6

An Artin algebra A = RI of socle type σ is compressed if and only if it has maximal length $$e(A)= \operatorname {dim}_{{\mathbf {k}}} A$$ among Artin quotients of R having socle type σ and embedding dimension n.

The maximality of the Hilbert function characterizes compressed algebras as follows. If A is an Artin algebra of socle type σ, it is known that for i ≥ 0,
\displaystyle \begin{aligned}\operatorname{HF}_A(i) \le \min \{\operatorname{dim}_{{\mathbf{k}} } S_i, \sigma_i \operatorname{dim}_{{\mathbf{k}} } S_0 + \sigma_{i+1} \operatorname{dim}_{{\mathbf{k}} } S_1 + \dots + \sigma_s \operatorname{dim}_{{\mathbf{k}} } S_{s-i} \}.\end{aligned}
Accordingly with , Definition 2.4. B, we can rephrase the previous definition in terms of the Hilbert function.

### Definition 2.4.7

A local k-algebra A of socle degree s, socle type σ and initial degree r is compressed if
\displaystyle \begin{aligned}\operatorname{HF}_A(i) = \left\{ \begin{array}{ll} \sum_{u = i}^s \sigma_u ( \operatorname{dim}_{{\mathbf{k}}} S_{u-i}) & \text{ if } i \ge r \\ \\ \operatorname{dim}_{{\mathbf{k}}} S_i & \text{ otherwise.} \end{array} \right. \end{aligned}
In particular a level algebra A of socle degree s, type t and embedding dimension n is compressed if
\displaystyle \begin{aligned}\operatorname{HF}_A(i) = \min \left\{ \binom{n+i-1}{i}, \ t \binom{n+s-i-1}{s-i}\right\}.\end{aligned}
If t = 1 and the above equality holds then A is called compressed Gorenstein algebra or also extremal Gorenstein algebra.
It is clear that compressed algebras impose several restrictive numerical conditions on the socle sequence σ (see [20, Definition 2.2]). For instance if r is the initial degree of A, then
\displaystyle \begin{aligned} \sigma_{r-1} = {\mathrm{max}} \{0, \operatorname{dim}_{{\mathbf{k}}} S_{r-1} - \sum_{u \ge r}(\sigma_u \operatorname{dim}_{{\mathbf{k}}} S_{u-(r-1)})\}. \end{aligned}
(2.1)
If s ≥ 2(r − 1), then it is easy to see that σr−1 = 0 because $$\operatorname {dim}_{{\mathbf {k}}} S_{s-(r-1)} \ge \operatorname {dim}_{{\mathbf {k}}} S_{r-1}$$. This is the case if A is Gorenstein.

The following result was proved in [20, Proposition 3.7 and Corollary 3.8].

### Proposition 2.4.8

A compressed local algebra A = RI whose dual module I is generated by F1, …, Ft of degrees d1, …, dt has a compressed associated graded ring $$gr_{{\mathfrak n}}(A)$$ whose dual module is generated by the leading forms of F1, …, Ft. Conversely if $$gr_{{\mathfrak n}}(A)$$ is compressed, then A is compressed and $$\sigma (A)=\sigma (gr_{{\mathfrak n}}(A)).$$

It is well known that if $$gr_{{\mathfrak n}}(A)$$ is Gorenstein then A is Gorenstein. On the other hand, if A is Gorenstein then $$gr_{{\mathfrak n}}(A)$$ is no longer Gorenstein. In order to study the associated graded ring to A Iarrobino considered the following construction. For a = 0, ⋯ , s − 1, s = s(A), consider the homogeneous ideals of $$gr_{{\mathfrak n}}(A)$$
\displaystyle \begin{aligned}C(a)=\bigoplus_{i\ge 0} C(a)_i \end{aligned}
\displaystyle \begin{aligned}C(a)_i=\frac{(0:_A {\mathfrak n}^{s+1-a-i})\cap {\mathfrak n}^i}{(0:_A {\mathfrak n}^{s+1-a-i})\cap {\mathfrak n}^{i+1}} \subset gr_{{\mathfrak n}}(A)_i \end{aligned}
This defining a decreasing filtration of ideals of $$gr_{{\mathfrak n}}(A)$$
\displaystyle \begin{aligned}gr_{{\mathfrak m}_A}(A)=C(0)\supseteq C(1)\supseteq \cdots \supseteq C(s)=0 \end{aligned}
Notice that if a ≥ 1 then C(a)i = 0 for all i ≥ s − a and C(0)i = 0 for all i ≥ s + 1

### Definition 2.4.9 (Iarrobino’s Q-Decomposition of $$gr_{{\mathfrak n}}(A)$$)

For all a = 0, ⋯ , s − 1 we consider the $$gr_{{\mathfrak m}_A}(A)$$-module
\displaystyle \begin{aligned}Q(a)=C(a)/C(a+1). \end{aligned}
Since the Hilbert function of A and $$gr_{{\mathfrak n}}(A)$$ agree we have the Iarrobino’s Shell decomposition of $$\operatorname {HF}_A$$:
\displaystyle \begin{aligned}\operatorname{HF}_{A}=\sum_{a=0}^{s-1} \operatorname{HF}_{Q(a)} \end{aligned}

### Proposition 2.4.10

If A is Artin Gorenstein then Q(a) is a reflexive $$gr_{{\mathfrak n}}(A)$$-module:
\displaystyle \begin{aligned}\mathrm{Hom}_{{\mathbf{k}}}(Q(a)_i,{\mathbf{k}})\cong Q(a)_{s-a-i} \end{aligned}

i = 0, ⋯ , s  a. In particular, $$\operatorname {HF}_{Q(a)}$$ is a symmetric function w.r.t $$\frac {s-a}{2}$$.

### Example 2.4.11 (Shell Decomposition)

Assume that $$\operatorname {HF}=\{1,m,n,1\}$$ is the Hilbert function of an Artin Gorenstein algebra A = RI The Shell decomposition of $$\operatorname {HF}$$ is, s = 3,
so m ≥ n. In fact, all function {1, m, n, 1}, m ≥ n, is the Hilbert function of an Artin Gorenstein algebra Theorem 2.5.11. Notice that from Macaulay’s characterization of Hilbert functions we get that {1, m, n, 1} is the Hilbert function of an Artin algebra iff $$1\le n\le \binom {m+1}{2}$$, [3, 30].

The following result is due to De Stefani, , it is a generalization of some results of Iarrobino.

### Proposition 2.4.12

Let A = RI be an Artin level algebra of socle degree s and Cohen-Macaulay type t. Then
1. (i)

$$Q(0)= gr_{{\mathfrak n}}(A)/C(1)$$ is the unique (up to iso) graded level quotient of $$gr_{{\mathfrak n}}(A)$$ with socle degree s and Cohen-Macaulay type t.

2. (ii)

Let F1, ⋯ , Ft ∈ S be generators of I such that such that deg(Fi) = s, i = 1, ⋯ , t, and top(F1), ⋯ , top(Ft) are k-linear independent forms of degree s, Proposition 2.4.4. Then Q(0)≅R∕〈top(F1), ⋯ , top(Ft)〉.

3. (iii)

The associated graded ring $$gr_{{\mathfrak n}}(A)$$ is an Artin level algebra of socle degree s and Cohen-Macaulay type t iff $$gr_{{\mathfrak n}}(A)\cong Q(0)$$.

As corollary we get:

### Proposition 2.4.13

Let A = RI be an Artin Gorenstein algebra of socle degree s. Then the following conditions are equivalent:
1. (i)

$$gr_{{\mathfrak n}}(A)$$ is Gorenstein,

2. (ii)

$$gr_{{\mathfrak n}}(A)=Q(0)$$ ,

3. (iii)

$$\operatorname {HF}_A$$ is symmetric.

## 2.5 Classification of Artin Rings

It is known that there are a finite number of isomorphism classes for e ≤ 6. J. Briançon  proved this result for n = 2, $${\mathbf {k}}=\mathbb C$$; G. Mazzola  for $${\mathbf {k}}=\bar {\mathbf {k}}$$ and char(k) ≠ 2, 3; finally B. Poonen  proved the finiteness for any $${\mathbf {k}}=\bar {\mathbf {k}}$$. On the other hand D.A. Suprunenko  proved that if k infinite, there are infinite number of isomorphism classes for e ≥ 7.

The problem of classification is in general very hard. For instance, before the paper , an open problem was the classification of Artin algebras with Hilbert function {1, m, n, 1}, even if A is Gorenstein.

Other families that has been classified are the almost stretched algebras, [14, 15]. We say that a Artin Gorenstein algebra A = RI is Almost Stretched if $${\mathfrak m}^2$$ is minimally generated by two elements or equivalently, the Hilbert function of A is
\displaystyle \begin{aligned}\operatorname{HF}_A=\{1,n, \stackrel{t-1}{\overbrace{2,\cdots,2}},\stackrel{s-t}{\overbrace{1,\cdots,1}}\} \end{aligned}
We assume that 3 ≤ t + 1 ≤ s. We say that A is of type (s, t).

In the following result we present the possible analytic types of almost stretched algebras, [14, 15] and . In fact, we proved more: we determined the pairwise analytic types of almost stretched algebras. We omit describing it here.

### Theorem 2.5.1

Let A = RI be an Almost Stretched algebra of type (s, t) with 3 ≤ t + 1 ≤ s.

If there is not r such that 2(r + 1) = s  t + 1 or s ≥ 3t − 1 then I is isomorphic to one of the following ideals:
\displaystyle \begin{aligned}I_{0,1}, I_{1,1},\dots , I_{\min\{t-1, s-t\},1}. \end{aligned}
Assume that s ≤ 3t − 2 and let r be the integer such that 2(r + 1) = s  t + 1, then I is isomorphic to one of the following ideals:
\displaystyle \begin{aligned}I_{0,1},\dots, I_{r-1,1},I_{r+1,1},\dots , I_{\min\{t-1, s-t\},1}\end{aligned}
\displaystyle \begin{aligned}\{I_{r,a}\}_{a\in {\mathbf{k}}^*},\{I_{r,a+x_1}\}_{a\in {\mathbf{k}}^*},\dots,\{I_{r,a+x_1^{t-r-2}}\}_{a\in {\mathbf{k}}^*}\end{aligned}

Where I p, z is the ideal generated by

$$\{x_ix_j\}_{1\le i< j\le n, (i,j)\neq (1,2)}, \{x_j-x_1^s\}_{3\le j\le n}, x_2^2-x_1^{p+1}x_2- zx_1^{s-t+1}, x_1^tx_2$$

### Example 2.5.2 ()

Let A be an Artin Gorenstein algebra with Hilbert function {1, 2, 2, 2, 1, 1, 1}. Then the analytic types are represented by
1. 1.

I1 = (y2 − xy − x4, x3y)

2. 2.

I2 = (y2 − x3y − x4, x3y)

3. 3.

Ic = (y2 − x2y − cx4, x3y), c ∈k

The moduli space, see  has two isolated points and a punctured affine line.

The main result of this section shows that some Artin algebras are isomorphic to their associated graded ring. J. Emsalem called these algebras “canonically graded”.

### Definition 2.5.3 (Emsalem)

An Artin local algebra A = RI is canonically graded if A is analytically isomorphic to $$gr_{{\mathfrak n}}(A)\cong R/I^*R$$.

Notice that there are non-canonically graded algebras, for instance:

### Example 2.5.4 ()

Let A be an Artin Gorenstein algebras with $$\operatorname {HF}_A=\{1,2,3,2,1\}$$ then A is is isomorphic to one and only one of the following quotients of R = k[[x1, x2]]:
1. 1.

$$I_1=(x_1^4,x_2^2)$$,

2. 2.

$$I_2=(x_1^4, x_1^2+x_2^2)$$, and

3. 3.

$$I_3=(x_1^4, x_2^2-x_1^3)$$.

Notice: $$I_3^*=(x_1^4,x_2^2)=I_1$$ and $$I_1\ncong I_3$$, i.e. RI3 is not canonically graded.

From now on we assume that the ground field k is of characteristic zero.

If L is a submodule of S generated by a sequence $$\underline G := G_1, \dots , G_t$$ of polynomials of S, then we will write
\displaystyle \begin{aligned} A_{\underline G}= R/ \operatorname{Ann}(L) . \end{aligned}

Given a form G of degree s and an integer q ≤ s, we denote by Δq(G) the $$\binom {n-1+s-q}{n-1}\times \binom {n-1+q}{n-1}$$ matrix whose columns are the coordinates of $$\partial _{{ \underline i}}(G)$$, $$|{ \underline i}|=q$$, with respect to $$(x^{L})^* = \frac 1 {L !} y^{L}$$, |L| = s − q. We will denote by $$(L, { \underline i})$$ the corresponding position in the matrix Δq(G). In the following $$L + { \underline i}$$ denotes the sum in $$\mathbb N^n.$$

### Proposition 2.5.5 ()

Let G  S = k[y1, ⋯ , yn] be a form of degree s. Then
\displaystyle \begin{aligned} \operatorname{HF}_{A_G}(s-i)={\mathrm{rank}}\,(\varDelta^{i}(G)) \le \min \left\{ \binom{n-1+s-i}{n-1}, \binom{n-1+i}{n-1}\right\} \end{aligned}

for i = 0, ⋯ , s. The equality holds if and only if AG is compressed.

Given an integer i  s, then
\displaystyle \begin{aligned} \varDelta^i(G)= ^\tau\varDelta^{s-i}(G) \end{aligned}

where τ denotes the transpose matrix.

Notice that from the last result and Proposition 2.3.5 it is easy to deduce an alternative proof of the fact that a graded Gorenstein algebra AG has symmetric Hilbert function.

Let $$A_{ \underline {G}}$$ be a graded level algebra. We can define for all integers i ≤ s the block matrix which is a $$t \binom {n-1+s-i}{n-1} \times \binom {n-1+i}{n-1}$$ matrix. We get the following result.

### Proposition 2.5.6

Let $$A=A_{ \underline G}$$ be a compressed algebra of socle degree s and Cohen-Macaulay type type t. Then for every i = 1, …, s
\displaystyle \begin{aligned} \operatorname{HF}_A(i) = rank ( \varDelta^i (\underline{G}[s] ))= \min \left\{\binom{n-1+i}{n-1}, t \binom{n-1+s-i}{n-1}\right\}. \end{aligned}

### Proof

By Proposition 2.4.8 we know that grμ(A) is level compressed of socle degree s and type t. Since grμ(A) is level if and only if $$gr_{\mu }(A) \simeq Q(0)= S/\langle {\mathrm {top}}( \underline {G})\rangle ^\perp ,$$ the result follows by Proposition 2.5.5.

Given a k-algebra C, quotient of R, we will denote by Aut(C) the group of the automorphisms of C as a k-algebra and by Autk(C) as a k-vector space. Since R is complete φ ∈ Aut(R) is determined by
\displaystyle \begin{aligned} \varphi(x_i)\in {\mathfrak m} \end{aligned}
i = 1, ⋯ , n, i.e. φ acts by substitution of xi by φ(xi).
For any $$\varphi \in Aut_{{\mathbf {k}}}(R/{{\mathfrak m} }^{s+1})$$ we may associate a matrix M(φ) with respect to the basis Ω of size $$r= dim_K (R/{{\mathfrak m}}^{s+1}) = {{n+s} \choose s}$$ already defined at the end of Sect. 2.5. Given I and J ideals of R such that $${\mathfrak m}^{s+1}\subset I, J,$$ there exists an isomorphism of k-algebras
\displaystyle \begin{aligned} \varphi : R/I \to R/J \end{aligned}
if and only if φ is canonically induced by a k-algebra automorphism of $$R/{{\mathfrak m}}^{s+1}$$ sending $$I/{{\mathfrak m}}^{s+1}$$ to $$J/{{\mathfrak m}}^{s+1}.$$ In particular φ is an isomorphism of k-vector spaces. Dualizing
\displaystyle \begin{aligned} \varphi^* : (R/J)^* \to (R/I)^* \end{aligned}
is an isomorphism of the k-vector subspaces where (RI)≃ I and (RJ)≃ J of Ss according to the exact paring (2.3.6). Hence τM(φ) is the matrix associated to φ with respect to the basis Ω of Ss.

We denote by $${\mathscr R}$$ the subgroup of Autk(Ss) (automorphisms of Ss as a k-vector space) represented by the matrices τM(φ) of Glr(k) with $$\varphi \in Aut (R/{{\mathfrak m}}^{s+1})$$. For all p ≥ 1, Ip denotes the identity matrix of order $$\binom {n+p-1}{p}.$$ By Emsalem, [16, Proposition 15], the classification, up to analytic isomorphism, of the Artin local k-algebras of multiplicity e, socle degree s and embedding dimension n is equivalent to the classification, up to the action of $${\mathscr R},$$ of the k-vector subspaces of Ss of dimension e, stable by derivations and containing the k-vector space S≤1.

Let $${ \underline F} = F_1, \dots , F_t,$$ respectively $${ \underline G} = G_1, \dots , G_t,$$ be polynomials of degree s. Let $$\varphi \in Aut (R/ {\mathfrak m}^{s+1}),$$ from the previous facts we have
\displaystyle \begin{aligned} \varphi(A_{\underline F})=A_{\underline G} \ \ \text{ if}\\text{and }\\text{only}\\text{if } \ (\varphi^*)^{-1}(\langle {\underline F} \rangle_R)=\langle {\underline G} \rangle_R. \end{aligned}
(2.3)
If $$F_i = b_{i1} \omega _1^*+ \dots b_{ir} \omega _r^* \in S_{\le s},$$ then we will denote the row vector of the coefficients of the polynomial with respect to the basis Ω by
\displaystyle \begin{aligned}{}[F_i]_{\varOmega^*} = (b_{i1}, \dots, b_{ir}). \end{aligned}
If there exists $$\varphi \in Aut (R/ {\mathfrak m}^{s+1})$$ such that
\displaystyle \begin{aligned}{}[ G_i]_{\varOmega^*} M(\varphi ) = [ F_i]_{\varOmega^*}, {\text{for }\text{every }} i=1, \dots, t, {\text{ then }} \varphi(A_{\underline F})=A_{\underline G} \end{aligned}
(2.4)
Let φsp be an automorphism of $$R/{\mathfrak m}^{s+1}$$ such that φsp = Id modulo $${\mathfrak m}^{p+1}$$, with 1 ≤ p ≤ s, that is
\displaystyle \begin{aligned} \varphi_{s-p}(x_j)=x_j+ \sum_{|{\underline i}|=p+1} a_{{\underline i}}^j x^{{\underline i}}+ \text{higher terms} \end{aligned}
(2.5)
for j = 1, …, n and $$a_{{ \underline i}}^j \in {\mathbf {k}}$$ for each n-uple $${ \underline i}$$ such that $$|{ \underline i}|=p+1.$$ In the following we will denote $${ \underline a}: = (a_{{ \underline i}}^1, |{ \underline i}|=p+1 ; \cdots ; a_{{ \underline i}}^n, |{ \underline i}|=p+1 )\in {\mathbf {k}}^{n \binom {n+p}{n-1}}.$$
The matrix associated to φsp, say M(φsp), is an element of Glr(k), $$r=\binom {n+s}{s+1}$$, with respect to the basis Ω of $$R/{\mathfrak m}^{s+1}$$. We write M(φsp) = (Bi,j)0≤i,js where Bi,j is a $$\binom {n+i-1}{i}\times \binom {n+j-1}{j}$$ matrix of the coefficients of monomials of degree i appearing in $$\varphi (x^{{ \underline j}})$$ where $${ \underline j}=(j_1, \dots , j_n)$$ such that $$|{ \underline j}|=j.$$ It is easy to verify that:
\displaystyle \begin{aligned}B_{i,j}= \left\{ \begin{array}{ll} 0, & 0\le i < j \le s, \text{ or } j=1, i=1,\cdots, s,\\ \\ I_i,& i=j=0,\cdots, s,\\ \\ 0, & j=s-p,\cdots, s-1, i=j+1,\cdots, s, \text{ and } (i,j)\neq (s,s-p). \end{array} \right. \end{aligned}
The matrix M(φsp) has the following structure
The entries of Bp+1,1, Bp+2,2, …, Bs,sp are linear forms in the variables $$a_{{ \underline i}}^j$$, with $$|{ \underline i}|=p+1$$, j = 1, ⋯ , n. We are mainly interested in Bs,sp which is a $$\binom {n+s-1}{s}\times \binom {n+s-p-1}{s-p}$$ matrix whose columns correspond to xW with |W| = s − p and the rows correspond to the coefficients of xL with |L| = s in φ(xW).

Let F, G be polynomials of degree s of P and let φsp be a k-algebra isomorphism of type (2.5) sending AF to AG. We denote by F[j] (respectively G[j]) the homogeneous component of degree j of F (respectively of G), that is F = F[s] + F[s − 1] + … (G = G[s] + G[s − 1] + …).

By (2.4) we have
\displaystyle \begin{aligned}{}[G]_{\varOmega^*} M(\varphi_{s-p}) = [F]_{\varOmega^*}, \end{aligned}
(2.6)
in particular we deduce
\displaystyle \begin{aligned}{}[F[j]]_{\varOmega^*}= \left\{ \begin{array}{ll} [G[s-p]]_{\varOmega^*}+ [G[s]]_{\varOmega^*}B_{s,s-p},& j=s-p, \\ \\ {[G[j]]}_{\varOmega^*}, & j=s-p+1,\cdots, s. \end{array} \right. \end{aligned}
(2.7)
We are going to study $$[G[s]]_{\varOmega ^*}B_{s,s-p}.$$ Let $$[\alpha _{ \underline i} ]$$ be the vector of the coordinates of G[s] w.r.t. Ω, i.e.
\displaystyle \begin{aligned}G[s]= \sum_{|{\underline i}|=s} \alpha_{\underline i} \; \frac{1}{{\underline i} !} y^{\underline i}; \end{aligned}
the entries of $$[G[s]]_{\varOmega ^*}B_{s,s-p}$$ are bi-homogeneous forms in the components of $$[\alpha _{ \underline i} ]$$ and $${ \underline a}=(a_{{ \underline i}}^1, \dots , a_{{ \underline i}}^n)$$ such that $$|{ \underline i}|=p+1$$ of bi-degree (1, 1). Hence there exists a matrix M[sp](G[s]) of size $$\binom {n-1+s-p}{n-1} \times n \binom {n+p}{n-1}$$ and entries in the $${\mathbf {k}}[\alpha _{ \underline i} ]$$ such that where denotes the transpose of the row-vector $${ \underline a}.$$ We are going to describe the entries of M[sp](G[s]). We label the columns of M[sp](G[s]) with the set of indexes $$(j,{ \underline i})$$, j = 1, ⋯ , n, $$|{ \underline i}|=p+1$$, corresponding to the entries of $${ \underline a} = (a_{{ \underline i}}^1, |{ \underline i}|=p+1 ; \cdots ; a_{{ \underline i}}^n, |{ \underline i}|=p+1 )\in {\mathbf {k}}^{n \binom {n+p}{n-1}}$$.

For every i = 1, ⋯ , n, we denote $$S^i_p$$ the set of monomials xα of degree p such that xα ∈ xi(xi, ⋯ , xn)p−1, hence $$\#(S^i_p)={p-1+n-i \choose p-1}.$$

### Lemma 2.5.7

The matrix M[sp](G[s]) has the following upper-diagonal structure
where Mj is a matrix of size $$\binom {s-p-1+n-j}{s-p-1}\times \binom {n+p}{n-1},$$ j = 1, ⋯ , n, defined as follows: the entries of Mj are the entries of M[sp](G[s]) corresponding to the rows $$W\in \log (S^j_{s-p})$$ and columns $$(j,{ \underline i})$$, $$|{ \underline i}|=p+1$$. We label the entries of Mj with respect to these multi-indices. Then it holds:
1. (i)
for all $$W=(w_1,\cdots ,w_n)\in \log (S^1_{s-p})$$ and $${ \underline i}$$ , $$|{ \underline i}|=p+1$$ ,
\displaystyle \begin{aligned}w_1 \varDelta^{p+1}(G[s])_{(W-\delta_1,{\underline i})} ={M_1}_{(W,(1,{\underline i}))}, \end{aligned}

2. (ii)
for all j = 1, ⋯ , n − 1, $$W\in \log (S_{s-p}^{j+1})$$,
\displaystyle \begin{aligned}M_{j+1, (W,(j+1,*))}= w_{j+1} M_{j,(L,(j,*))} \end{aligned}
with L = δj + W  δj+1,

From the last result we get the key result of this chapter.

### Corollary 2.5.8

If s ≤ 4 then rank (M[sp](G[s])) is maximal if and only if rank (Δp+1(G[s])) is maximal.

### Proof

Notice that M[sp](G[s]) has an upper-diagonal structure where the rows of the diagonal blocks Mj are a subset of the rows of the first block matrix M1. Let us assume that the number of rows of M1 is not bigger than the number of columns of M1, as a consequence the same holds for Mj with j > 1. Then we can compute the rank of M[sp](G[s]) by rows, so rank (M[sp](G[s])) is maximal if and only if rank (Δp+1(G[s])) is maximal. Since M1 is a $$\binom {s-p-2+n}{s-p-1}\times \binom {n+p}{n-1}$$ matrix, if $$\binom {n+s-p-2}{s-p-1}=\binom {n+s-p-2}{n-1}\le \binom {n+p}{n-1}$$ we get the result. This inequality is equivalent to n + s − p − 2 ≤ n + p, i.e. s ≤ 2p + 2, since p ≥ 1 we get that s ≤ 4.

We may generalize the previous facts to a sequence $$\underline G=G_1, \dots , G_t$$ of polynomials of degree s of S. Let φsp be a k-algebra isomorphism of type (2.5) sending $$A_{ \underline F}$$ to $$A_{ \underline G}$$ where $${ \underline F}=F_1, \dots ,F_t .$$ In particular we assume that, as in (2.6),
\displaystyle \begin{aligned}{}[G_r]_{\varOmega^*} M(\varphi_{s-p}) = [F_r]_{\varOmega^*},\end{aligned}
for every r = 1, …, t. We deduce the analogues of (2.7) and we restrict our interest to
\displaystyle \begin{aligned}{}[\underline G[s]]_{\varOmega^*}B^{\oplus t}_{s,s-p}\end{aligned}
where
obtained by gluing t times the matrix Bs,sp and where $$[ \underline G[s]]_{\varOmega ^*}$$ is the row $$([G_r[s]]_{\varOmega ^*} : r=1, \dots , t).$$ In accordance with (2.8), it is defined the matrix M[sp](Gr[s]) of size $$\binom {n-1+s-p}{n-1} \times n \binom {n+p}{n-1}$$ and entries depending on $$[ \underline G[s]]_{\varOmega ^*}$$ such that
\displaystyle \begin{aligned} ^\tau([G_r[s]]_{\varOmega^*} B_{s,s-p}) = M^{[s-p]}(G_r[s]) \ ^\tau {\underline a} \end{aligned}
If we define which is a $$t \binom {n-1+s-p}{n-1} \times n \binom {n+p}{n-1}$$ matrix, we get The matrix $$M ^{[s-p]}({ \underline G}[s])$$ has the same shape of M[sp](G[s]), already described in Lemma 2.5.7 and its blocks correspond to suitable submatrices of $$(\varDelta ^{p+1}( \underline G[s]))$$ (see (2.2)). Hence we have an analogue to (2.7) for the level case
\displaystyle \begin{aligned}{}[F_r[j]]_{\varOmega^*}= \left\{ \begin{array}{ll} [G_r[s-p]]_{\varOmega^*}+ {\underline a} \; \; ^\tau(M^{[s-p]}({\underline G}[s])),& j=s-p, \\ \\ {[G_r[j]]}_{\varOmega^*}, & j=s-p+1,\cdots, s. \end{array} \right. \end{aligned}
(2.11)
for all r = 1, …, t.

In the next result we generalize the main result of .

### Theorem 2.5.9

Let A be an Artin compressed Gorenstein local k-algebra. If s ≤ 4 then A is canonically graded.

### Proof

Let A be an Artin compressed Gorenstein local k-algebra of socle degree s ≥ 2 and embedding dimension n. Then A = AG with G ∈ S a polynomial of degree s and $$gr_{\mu }(A)= S/\operatorname {Ann}(G[s])$$ is a compressed Gorenstein graded algebra of socle degree s ≥ 2 and embedding dimension n (see Proposition 2.4.8).

The main result of  shows that if s ≤ 3 then A is canonically graded. Let us assume s = 4, then the Hilbert function is $$\{ 1, n, {n+1 \choose 2}, n, 1\}.$$ Because AG is a compressed Gorenstein algebra with the same Hilbert function of A, we may assume G = G + G. In fact S1, S2 ⊆〈G〉R because of (2.5.6) and, as a consequence, it is easy to see that 〈G + G〉R = 〈G + G + G + …〉R. We have to prove that there exists an automorphism $$\varphi \in Aut(R/{\mathfrak m}^{5})$$ such that
\displaystyle \begin{aligned}A_{G} \simeq A_{G}.\end{aligned}
We consider for every j = 1, …, n
\displaystyle \begin{aligned}\varphi_3(x_j)=x_j+ \sum_{|{\underline i}|=2} a_{{\underline i}}^j x^{{\underline i}}+ \text{higher terms} \end{aligned}
If $$A_F=\varphi ^{-1}_{3}(A_G),$$ then from (2.7) and (2.8) we get where $${ \underline a}=(a_{{ \underline i}}^1, \dots , a_{{ \underline i}}^n).$$ By Proposition 2.5.6 and Corollary 2.5.8, we know that the matrix M(G) has maximal rank and it coincides with the number of rows, so there exists a solution $${ \underline a} \in {\mathbf {k}}^n$$ of (2.12) such that F = 0 and F = G.

The aim is now to list classes of local compressed algebras of embedding dimension n, socle degree s and socle type σ = (0, …, σr−1, σr, …, σs, 0, 0, … ) which are canonically graded. Examples will prove that the following result cannot be extended to higher socle degrees. This result extends the main result of  and .

### Theorem 2.5.10

Let A = RI be an Artin compressed k-algebra of embedding dimension n, socle degree s and socle type σ. Then A is canonically graded in the following cases:
1. (1)

s ≤ 3,

2. (2)

s = 4 and e4 = 1,

3. (3)

s = 4 and n = 2.

### Proof

Since a local ring with Hilbert function {1, n, t} is always graded, we may assume s ≥ 3.

If s = 3 and A is a compressed level algebra, then A is canonically graded by De Stefani . If A is not necessarily level, but compressed, then by (2.1) the socle type is {0, 0, σ2, σ3} and the Hilbert function is {1, n, h2, σ3} where $$h_2 = \min \{ \operatorname {dim}_{{\mathbf {k}}} R_2, \sigma _2+ \sigma _3 n\}.$$ Then we may assume that in any system of coordinates I is generated by e2 quadratic forms and e3 polynomials $$G_1, \dots , G_{\sigma _3}$$ of degree 3. Then the result follows because $$R/\operatorname {Ann}_R(G_1, \dots , G_{\sigma _3})$$ is a 3-level compressed algebra of type σ3 and hence canonically graded.

Let us assume s = 4 and σ4 = 1. We recall that if A is Gorenstein, then the result follows by Theorem 2.5.9. Since A is compressed, then by (2.1) the socle type is (0, 0, 0, σ3, 1). This means that I is generated by e3 polynomial of degree 3 and one polynomial of degree 4. Similarly to the above part, because S≤2 ⊆ (I), I can be generated by σ3 forms of degree 3 and one polynomial of degree 4. As before the problem is reduced to the Gorenstein case with s = 4 and the result follows.

Assume s = 4 and n = 2. If σ4 = 1, then we are in case (2). If σ4 > 1, because A is compressed, the possible socle types are: σi = (0, 0, 0, 0, i) with i = 2, ⋯ , 5 and since A is compressed, the corresponding Hilbert function is {1, 2, 3, 4, i}. In each case A is graded because the Hilbert function forces the dual module to be generated by forms of degree four.

As a corollary of the last result we get .

### Theorem 2.5.11

Let A be an Artinian Gorenstein k-algebra with Hilbert function {1, n, m, 1}. Then the following conditions are equivalent:
1. (i)

2. (ii)

m = n,

3. (iii)

A is compressed.

From this result we can deduce

### Corollary 2.5.12 ()

The classification of Artinian Gorenstein local k -algebras with Hilbert function $$\operatorname {HF}_A=\{1,n,n,1\}$$ is equivalent to the projective classification of the hypersurfaces $$V(F)\subset \mathbb {P}^{n-1}_{{\mathbf {k}}}$$ where F is a degree three non degenerate form in n variables.

Next we will recall the classification of the Artin Gorenstein algebras for n = 1, 2, 3, .

If n = 1, then it is clear that Ak[[x]]∕(x4), so there is only one analytic model. If n = 2 we have the following result:

### Proposition 2.5.13 ()

Let A be an Artinian Gorenstein local K-algebra with Hilbert function $$\operatorname {HF}_A=\{1,2,2,1\}$$. Then A is isomorphic to one and only one of the following quotients of R = K[[x1, x2]]:
Finally, for n = 3 first we have to study with detail the classification of plane curves, in particular, the elliptic curves, see for instance . Any plane elliptic cubic curve $$C\subset \mathbb P^{2}_{{\mathbf {k}}}$$ is defined, in a suitable system of coordinates, by a Weierstrass’ equation, ,
\displaystyle \begin{aligned} W_{a,b}:y_2^2 y_3=y_1^3+a y_1 y_3^2+ b y_3^3 \end{aligned}
with a, b ∈k such that 4a3 + 27b2 ≠ 0. The j invariant of C is
\displaystyle \begin{aligned} j(a,b)=1728\frac{4a^3}{4a^3+27b^2} \end{aligned}
It is well known that two plane elliptic cubic curves $$C_i=V(W_{a_i,b_i})\subset \mathbb P^{2}_{{\mathbf {k}}}$$, i = 1, 2, are projectively isomorphic if and only if j(a1, b1) = j(a2, b2).
For elliptic curves the inverse moduli problem can be done as follows. We denote by W(j) the following elliptic curves with j as moduli : $$W(0)=y_2^2y_1+y_2y_3^2-y_1^3$$, $$W(1728)= y_2^2y_3-y_1y_3^2-y_1^3$$, and for j ≠ 0, 1728
\displaystyle \begin{aligned} W(j)=(j-1728) (y_2^2 y_3+y_1 y_2 y_3 -y_1^3) + 36 y_1 y_3^2 +y_3^3. \end{aligned}

We will show by using the library INVERSE-SYST.LIB that:

### Proposition 2.5.14

Let A be an Artin Gorenstein local k-algebra with Hilbert function $$\operatorname {HF}_A=\{1,3,3,1\}$$. Then A is isomorphic to one and only one of the following quotients of R = k[[x1, x2, x3]] :

with:

$$H_j=6jx_1x_2- 144(j-1728)x_1x_3+72(j-1728)x_2x_3- (j-1728)^2 x_3^2$$ , and

$$G_j=jx_1^2-12(j-1728)x_1x_3+ 6(j-1728)x_2x_3+ 144(j-1728)x_3^2$$ ;

I(j1)≅I(j2) if and only if j1 = j2.

### Proof

Let us assume that F is the product of the linear forms l1, l2, l3. If l1, l2, l3 are k-linear independent we get the first case. On the contrary, if these linear forms are k-linear dependent, we deduce that F is degenerate. Let us assume that F is the product of a linear form l and an irreducible quadric Q. According to the relative position of V (l) and V (Q) we get the second and the third case.

Let F be a degree three irreducible form. The first seven models can be obtained from the corresponding inverse system F by using the command idealAnn of . For the last case see .

## 2.6 Computation of Betti Numbers

In this chapter we address the following problem: How can we compute the Betti numbers of I in terms of its Macaulay’s inverse system L = I without computing the ideal I? This is a longstanding problem in commutative algebra that has been considered by many authors, see for instance , Chap. 9, Problem L. For instance, if A = RI is an Artin Gorenstein local ring then its inverse system is a polynomial F on the variables x1, …, xn of degree the socle degree of A. In this chapter we compute the Betti numbers of A in terms of the polynomial F instead of computing I and then to compute the Betti numbers of A = RI.

Let I be an $${\mathfrak m}$$-primary ideal of R. Let $$\mathbb F_{\bullet }$$ be a minimal free resolution of the R-module RI
\displaystyle \begin{aligned} \begin{array}{rcl} \mathbb F_{\bullet}\qquad0\longrightarrow \mathbb F_n=R^{\beta_n}\longrightarrow \dots \longrightarrow \mathbb F_1=R^{\beta_1} \longrightarrow \mathbb F_0=R \longrightarrow R/I \longrightarrow 0, \end{array} \end{aligned}
the p-th Betti number of RI is $$\beta _p(R/I)={\mathrm {rank}}\,_R(\mathbb F_p)$$, 1 ≤ p ≤ n. Tensoring $$\mathbb F_{\bullet }$$ by the R-module k we get the complex
\displaystyle \begin{aligned} \begin{array}{rcl} \mathbb F_{\bullet}\otimes_R {\mathbf{k}}\qquad \qquad \qquad0\longrightarrow {\mathbf{k}}^{\beta_n}\longrightarrow \dots \longrightarrow {\mathbf{k}}^{\beta_1} \longrightarrow \mathbb {\mathbf{k}} ={\mathbf{k}} \longrightarrow 0 .\vspace{-3pt} \end{array} \end{aligned}
Since $$\mathbb F_{\bullet }$$ is a minimal resolution we get that the morphisms of $$\mathbb F_{\bullet }\otimes _R {\mathbf {k}}$$ are zero, so
\displaystyle \begin{aligned}\beta_p(R/I)=\operatorname{dim}_{{\mathbf{k}}}(\operatorname{Tor}_p^R(R/I, {\mathbf{k}})), \end{aligned}
p = 1, …, n. Let us now consider Koszul’s resolution of R defined by the regular sequence x1, …, xn
\displaystyle \begin{aligned} \begin{array}{rcl} \mathbb K_{\bullet}\qquad0\stackrel{d_{n+1}}{\longrightarrow} \mathbb K_n=\bigwedge^n R^n \stackrel{d_{n}}{\longrightarrow} \dots \longrightarrow \mathbb K_1=\bigwedge^1 R^n \stackrel{d_{1}}{\longrightarrow} \mathbb K_0=R \longrightarrow {\mathbf{k}} \longrightarrow0. \end{array} \end{aligned}
We consider the R-basis of Rn: ei = (0, …, 1(i, …, 0) ∈ Rn, i = 1, …, n; for all 1 ≤ i1 < ⋯ < ip ≤ n we set $$e_{i_1,\dots , i_p}=e_{i_1}\wedge \dots e_{i_p}\in \bigwedge ^p R^n$$. Since the set $$e_{i_1,\dots , i_p}$$, 1 ≤ i1 < ⋯ < ip ≤ n, form a R-basis of $$\mathbb K_p$$ we define the morphism
\displaystyle \begin{aligned}d_p: \mathbb K_p\longrightarrow \mathbb K_{p-1} \end{aligned}
by $$d_p(e_{i_1,\dots , i_p})=\sum _{j=1}^p (-1)^{j-1} x_{i_j} e_{i_1,\dots ,i_{j-1},i_{j+1},\dots , i_p}\in \mathbb K_{p-1}$$. Notice that the basis $$e_{i_1,\dots , i_p}$$, 1 ≤ i1 < ⋯ < ip ≤ n, this defines an isomorphism of R-modules $$\bigwedge ^p R^n\stackrel {\phi _p}{\cong }R^{{n \choose p}}$$, such that $$\phi _p(e_{i_1,\dots , i_p})=v_{i_1,\dots , i_p}$$, 1 ≤ i1 < ⋯ < ip ≤ n, is the element of $$R^{{n \choose p}}$$ with all entries zero but the (i1, …, ip)-th that it is equal to 1. We denote by Δp the associated matrix to dp with respect the above bases of $$R^{{n \choose p}}$$ and $$R^{{n \choose p-1}}$$, notice that the entries of Δp are zero or ± xi, i = 1, …, n.
We can compute $$\operatorname {Tor}_p^R(R/I, {\mathbf {k}})$$ by considering the complex $$R/I \otimes _R \mathbb K_{\bullet }$$
\displaystyle \begin{aligned} \begin{array}{rcl} R/I \otimes_R \mathbb K_{\bullet}\qquad \quad0\longrightarrow (R/I)^{{n \choose n}}\longrightarrow \dots \longrightarrow (R/I)^{{n \choose 1}} \longrightarrow R/I \longrightarrow {\mathbf{k}} \longrightarrow0. \end{array} \end{aligned}
We denote again by dp the morphism $$\operatorname {Id}_{R/I}\otimes _R d_p$$ then
\displaystyle \begin{aligned}\operatorname{Tor}_p^R(R/I, {\mathbf{k}})=H_p(R/I \otimes_R \mathbb K_{\bullet})= \frac{\operatorname{ker} (d_p)}{\mathrm{im} (d_{p+1})} \end{aligned}
for p = 1, …, n. If we consider the dual of $$R/I \otimes _R \mathbb K_{\bullet }$$ with respect to E we get, L = I,
\displaystyle \begin{aligned} \begin{array}{rcl} (R/I \otimes_R \mathbb K_{\bullet})^* \qquad \qquad \qquad \quad0\longrightarrow {\mathbf{k}} \stackrel{d_0^*}{\longrightarrow} L \stackrel{d_1^*}{\longrightarrow} L^{{n \choose 1}} \longrightarrow \dots L^{{n \choose n}}\stackrel{d_{n+1}^*}{\longrightarrow} 0. \end{array} \end{aligned}
Notice that, if $$\underbar {h}=(h_{i_1,\dots , i_p}, 1\le i_1<\dots <i_{p-1}\le n)\in L^{{n \choose p-1}}$$ then

### Proposition 2.6.1

Let L  S be a finitely generated sub-R-module of S of dimension $$e=\operatorname {dim}_{{\mathbf {k}}}(L)$$. If I = AnnR(L) ⊂ R then
\displaystyle \begin{aligned}\beta_p(R/I)=e {n \choose p} - \operatorname{dim}_{{\mathbf{k}}}(\mathrm{im}( d_p^*))- \operatorname{dim}_{{\mathbf{k}}}(\mathrm{im} ( d_{p+1}^*)). \end{aligned}

for p = 1, …, n.

### Proof

Since L is a finitely dimensional k-vector space and the duallzing functor is exact and additive we get
\displaystyle \begin{aligned}\beta_p(R/I)=\operatorname{dim}_{{\mathbf{k}}}\left(\frac{\operatorname{ker} (d_{p+1}^*)}{\mathrm{im} (d_p^*)}\right)= \operatorname{dim}_{{\mathbf{k}}}\operatorname{ker} (d_{p+1}^*)-\operatorname{dim}_{{\mathbf{k}}}\mathrm{im} (d_p^*). \end{aligned}
On the other hand $$\operatorname {dim}_{{\mathbf {k}}}(\operatorname {ker} (d_{p+1}^*))= e {n \choose p} - \operatorname {dim}_{{\mathbf {k}}}(\mathrm {im} (d_{p+1}^*))$$, so from these identities we get the claim.
Next step is to compute the Betti numbers effectively. For all t ≥ 0 let Wt be the set of standard monomials xα, $$\alpha \in \mathbb N^n$$, of degree at most t ordered by the local deg-rev-lex ordering with xn < ⋯ < x1. For instance, for n = 3 and t = 2, $$W_2=\{ x_3^2, x_3 x_2, x_3 x_1, x_2^2, x_2x_1,x_1^2, x_3, x_2, x_1, 1\}$$. For all p = 1, …, n we consider the following set $$\mathscr M_{s,p}$$ of linearly independent elements of $$R^{n \choose p}$$
\displaystyle \begin{aligned}m_{\alpha; i_1,\dots,i_p}:= \left\{ \begin{array}{ll} x^{\alpha} & \text{in the }(i_1,\dots ,i_p)\text{-th component,}\\ \\ 0 & \text{otherwise,} \end{array} \right. \end{aligned}
deg(α) ≤ s, 1 ≤ i1 < ⋯ < ip ≤ n. Notice that $$\# (\mathscr M_{s,p})= {n+s\choose n}{n \choose p}$$.
Assume that s = deg(L). Given a k-basis w1, …, we of L we consider the following k-basis, say Bp, of $$L^{{n \choose p}}$$:
\displaystyle \begin{aligned}w_{i_1,\dots,i_p}^i:= \left\{ \begin{array}{ll} w_i & \text{in the }(i_1,\dots ,i_p)\text{-th component,}\\ \\ 0 & \text{otherwise,} \end{array} \right. \end{aligned}
i = 1, …, e, 1 ≤ i1 < ⋯ < ip ≤ n. We denote by $$\varDelta ^+_p(L)$$ the matrix such that the columns are the coordinates of $$d_p^*(w_{i_1,\dots ,i_p}^i)$$ with respect the basis $$\mathscr M_{s,p-1}$$. This is a matrix of $${n+s\choose n}{n \choose p}$$ rows and $$e{n\choose p-1}$$ columns and the entries are zero or ± xi, i = 1, …, n. Then we have:

### Proposition 2.6.2

For any finitely generated R-module L and 1 ≤ p  n we have
\displaystyle \begin{aligned}\operatorname{dim}_{{\mathbf{k}}}(\mathrm{im}(d_p^*))={\mathrm{rank}}\,(\Delta^+_p(L)). \end{aligned}
If $$e=\operatorname {dim}_{{\mathbf {k}}}(L)$$ and I = AnnR(L) ⊂ R then
\displaystyle \begin{aligned}\beta_p(R/I)=e {n \choose p} - {\mathrm{rank}}\,(\varDelta^+_p(L))- {\mathrm{rank}}\,(\varDelta^+_{p+1}(L)). \end{aligned}

for p = 1, …, n.

From this result we get that the determination of the Betti number βp(RI) involves the computation of the rank of $$\varDelta ^+_p(L)$$, p = 1, …, n. Recall that these matrices are huge, see the comments before last result, so they are difficult to manage. Moreover, this method of computation of Betti numbers implies the computation or election of a k-basis of L. This is not possible if we want to consider a general L or the deformations of L, see Example 2.6.7.

In the next result we compute the Cohen-Macaulay type of RI and we partially recover the classical result of Macaulay. In the second part, case n = 2, we prove a well known result of Serre that can be deduced from Hilbert-Burch structure theorem, i.e. the class of codimension two complete intersection ideals coincides with the class of codimension two Gorenstein ideals.

### Proposition 2.6.3

Let L be a finitely generated R-module of S of dimension e. Then the Cohen-Macaulay type of RI, I = AnnR(L), is
\displaystyle \begin{aligned}t(R/I)=\operatorname{dim}_{{\mathbf{k}}}(L/{\mathfrak m}\circ L)=\mu_R(L). \end{aligned}
In particular, for n = 2 then
\displaystyle \begin{aligned}\mu(I)=t(R/I)+1. \end{aligned}

In particular, I is a complete intersection if and only if RI is Gorenstein.

### Proof

The first result is Proposition 2.4.3.

Assume that n = 2. Then the complex $$(R/I \otimes _R \mathbb K_{\bullet })^*$$ is
\displaystyle \begin{aligned}0\longrightarrow {\mathbf{k}} \stackrel{d_0^*}{\longrightarrow} L \stackrel{d_1^*}{\longrightarrow} L^{2} \stackrel{d_2^*}{\longrightarrow} L\stackrel{d_{3}^*}{\longrightarrow} 0, \end{aligned}
and $$\mathrm {im} ( d_{2}^*)={\mathfrak m} \circ L$$. From Corollary 2.6.1 with p = 1 we get
\displaystyle \begin{aligned}\mu(I)=2e-(e-1)-\operatorname{dim}_{{\mathbf{k}}}({\mathfrak m}\circ L)=\operatorname{dim}_{{\mathbf{k}}}(L/{\mathfrak m}\circ L)+1=t(R/I)+1, \end{aligned}
so I is a complete intersection ideal, i.e. μ(I) = 2, if and only if t(RI) = 1, i.e. RI is Gorenstein.
Given a finitely generated sub-R-module L of S we denote by $$L :_S {\mathfrak m}$$ the sub-R-module of S formed by the polynomials h ∈ S such that $${\mathfrak m} \circ h\subset L$$. Notice that if L ⊂ Ss then $$L:_S {\mathfrak m}\subset S_{\le s+1}$$ and, in particular, $$\operatorname {dim}_{{\mathbf {k}}}(L:_S {\mathfrak m})<\infty$$. We consider the k-vector space morphism induced by $$d_{1}^*$$
\displaystyle \begin{aligned}d_{1,s}^*:S_{\le s+1} \longrightarrow S_{\le s}^n \end{aligned}
with
\displaystyle \begin{aligned}d_{1,s}^*(h)=(x_1\circ h,\dots, x_n \circ h) \end{aligned}
for all h ∈ Ss+1. It is easy to prove that $$L:_S {\mathfrak m}=(d_{1,s}^{*})^{-1}(L^n ).$$

### Proposition 2.6.4

Let L  S be a finitely generated sub-R-module of S of dimension $$e=\operatorname {dim}_{{\mathbf {k}}}(L)$$ and degree s = deg(L). If I = AnnR(L) ⊂ R then
\displaystyle \begin{aligned}\mu(I)= e (n-1) + {n+s+1 \choose n} - \operatorname{dim}_{{\mathbf{k}}} (\mathrm{im}(d_{1,s}^{*}) +L^n). \end{aligned}

### Proof

If we write V = Ln then we have
\displaystyle \begin{aligned}\begin{array}{ccl} \operatorname{dim}_{{\mathbf{k}}}(L:_S {\mathfrak m})&=& \operatorname{dim}_{{\mathbf{k}}}((d_{1,s}^{*})^{-1}(V))\\ \\ &=& \operatorname{dim}_{{\mathbf{k}}}((d_{1,s}^{*})^{-1}(V\cap \mathrm{im} ( d_{1,s}^{*})) =\operatorname{dim}_{{\mathbf{k}}}(V\cap \mathrm{im} (d_{1,s}^{*}))+1 \end{array} \end{aligned}
because $$\operatorname {dim}_{{\mathbf {k}}}(\operatorname {ker} (\phi _s))=1$$, so
\displaystyle \begin{aligned}\begin{array}{ccl} \operatorname{dim}_{{\mathbf{k}}}(L:_S {\mathfrak m})& = & \operatorname{dim}_{{\mathbf{k}}}(V) + \operatorname{dim}_{{\mathbf{k}}}(\mathrm{im} (d_{1,s}^{*})) - \operatorname{dim}_{{\mathbf{k}}}(\mathrm{im} (d_{1,s}^{*})+V)+1\\ \\ & = & n.e + dim_{{\mathbf{k}}}(S_{\le s+1})-1 - \operatorname{dim}_{{\mathbf{k}}}(\mathrm{im} (d_{1,s}^{*})+V)+1 \\ \\ & = & n.e + {n+s+1 \choose n} - \operatorname{dim}_{{\mathbf{k}}}(\mathrm{im} (d_{1,s}^{*})+V). \end{array} \end{aligned}
Claim $$\mu (I)=\operatorname {dim}_{{\mathbf {k}}}\left (L :_S {\mathfrak m} /L\right )$$.
Proof of the Claim Let us consider the exact sequence of R-modules
\displaystyle \begin{aligned}0\longrightarrow \frac{I}{{\mathfrak m} I} \longrightarrow \frac{R}{{\mathfrak m} I} \longrightarrow \frac{R}{I} \longrightarrow 0 \end{aligned}
dualizing this sequence we get the exact sequence on S-modules
\displaystyle \begin{aligned}0\longrightarrow L \longrightarrow ({\mathfrak m} I)^{\perp} \longrightarrow \left( \frac{I}{{\mathfrak m} I}\right)^* \longrightarrow 0.\end{aligned}
Hence
\displaystyle \begin{aligned}\begin{array}{ccl} \operatorname{dim}_{{\mathbf{k}}}\left( \frac{I}{{\mathfrak m} I}\right)^* & = & \operatorname{dim}_{{\mathbf{k}}}(({\mathfrak m} I)^{\perp})-\operatorname{dim}_{{\mathbf{k}}}(L) \\ \\ & = & dim_{{\mathbf{k}}}(R/{\mathfrak m} I)-dim_{{\mathbf{k}}}(R/I) \\ \\ & = & dim_{{\mathbf{k}}}(I/{\mathfrak m} I)=\mu(I) \end{array}\end{aligned}
by Nakayama’s lemma. In particular, we get $$\mu (I)=\operatorname {dim}_{{\mathbf {k}}}\left (({\mathfrak m} I)^{\perp }/L\right ).$$ Last step is to prove that $$({\mathfrak m} I)^{\perp }= L:_S {\mathfrak m}$$. Given a polynomial h ∈ S then $$h\in ({\mathfrak m} I)^{\perp }$$ if and only if $$0=({\mathfrak m} I)\circ h= I\circ ({\mathfrak m} \circ h)$$, so $$({\mathfrak m} I)^{\perp }$$ is the set of polynomial h such that $${\mathfrak m} \circ h \subset L$$, i.e. $$({\mathfrak m} I)^{\perp }= L:_S {\mathfrak m}$$.
From the Claim we get
\displaystyle \begin{aligned}\mu(I) = \operatorname{dim}_{{\mathbf{k}}}(L:_S {\mathfrak m})-\operatorname{dim}_{{\mathbf{k}}}(L) = n.e + {n+s+1 \choose n} - \operatorname{dim}_{{\mathbf{k}}} (\mathrm{im}(d_{1,s}^{*}) +V)- e,\end{aligned}
so
\displaystyle \begin{aligned}\mu(I) = (n-1)e + {n+s+1 \choose n} - \operatorname{dim}_{{\mathbf{k}}} (\mathrm{im}(d_{1,s}^{*}) + V).\end{aligned}

Next we will compute $$\operatorname {dim}_{{\mathbf {k}}}(\mathrm {im}(d_{1,s}^{*}) +L^n)$$ by considering a matrix that we are going to define. We denote by $$\mathbb M_s$$ the $$n {n+s \choose n} \times \left ( {n+s+1 \choose n}-1 \right )$$-matrix such that the i-th column, $$i\in [1,{n+s+1 \choose n}-1]$$, consists in the coordinates of xn ∘ xα,…, x1 ∘ xα with respect the base Ws, where xα is the i-th monomial of Ws+1.

Let L ⊂ S be a finitely generated R-module of dimension e and degree s. We pick a basis w1, …, we of L and we consider the following basis, say B, of Ln: $$(0,\dots , \stackrel {j}{w_i},\dots , 0)\in L^n$$ for j = 1, …, n, i = 1, …, e. We denote by $$\mathbb B(L)$$ the $$n {n+s \choose n} \times ( n.e )$$-matrix, such that the columns consists of the coordinates of the elements of B with respect $$\mathscr M_{s,1}$$. Finally, $$\mathbb M(L)$$ is the $$n {n+s \choose n} \times \left ( {n+s+1 \choose n}-1 + n.e \right )$$ block matrix
\displaystyle \begin{aligned}\mathbb M(L)=(\mathbb M_s\mid \mathbb B(L)); \end{aligned}
notice that
\displaystyle \begin{aligned}\operatorname{dim}_{{\mathbf{k}}}(\mathrm{im}(d_{1,s}^{*}) +L^n)={\mathrm{rank}}\,(\mathbb M(L)). \end{aligned}
If we want to consider a general L ⊂ S, see for instance Example 2.6.7, we have to avoid considering a basis of L. Let F1, …, Fr be a system of generators of L as R-module. Then consider the following system of generators of L as k-vector space xα ∘ Fi for all $$\alpha \in \mathbb N^n$$ of degree less or equal to s = deg(L) and for all i = 1, …, r. We consider now the following system of generators, say B+, of Ln:
\displaystyle \begin{aligned}(0,\dots, \stackrel{j}{x^{\alpha} \circ F_i},\dots , 0)\in L^n \end{aligned}
for j = 1, …, n, $$\alpha \in \mathbb N^n$$ with deg(α) ≤ s. We denote by $$\mathbb B^{+}(L)$$ the $$n {n+s \choose n} \times r {n+s\choose n }$$-matrix, such that the columns are the coordinates of the system of generators B+ with respect $$\mathscr M_{s,1}$$. This (lazy) method generates a matrix
\displaystyle \begin{aligned}\mathbb L(L)=(\mathbb M_s\mid \mathbb B^{+}(L)) \end{aligned}
with $${n+s+1\choose n}-1+ n r {n+s\choose n}$$ columns and $$n {n+s\choose n}$$ rows. Notice that the rank of $$\mathbb M(L)$$ and $$\mathbb L(L)$$ agree. Since the rank of $$\mathbb M_s$$ is $${n+s+1\choose n}-1$$ there is a dimension $${n+s+1\choose n}-1$$ square invertible matrix Gs such that
where Id is the identity matrix of dimension $${n+s+1\choose n}-1$$ and Z is the $$\left (n {n+s \choose n}- {n+s+1\choose n}+1\right )\times ({n+s+1\choose n}-1)$$ zero matrix. We denote by $$\mathbb L^*(L)$$, resp. $$\mathbb M^*(L)$$, the sub-matrix of $$G_s\mathbb L(L)$$, resp. $$G_s\mathbb M(L)$$, consisting of the last $$n {n+s\choose n}- {n+s+1\choose n}+1$$ rows and the last $$n r {n+s\choose n}$$, resp. n.e, columns. Hence we have

### Proposition 2.6.5

Let L be a degree s finitely generated sub-R-module of S. Then
1. (i)

$${\mathrm {rank}}\, (\mathbb M(L))= {\mathrm {rank}}\, (\mathbb L(L))$$ and $${\mathrm {rank}}\, (\mathbb M^*(L))={\mathrm {rank}}\, (\mathbb L^*(L))$$ ,

2. (ii)

$${\mathrm {rank}}\, (\mathbb M(L))={\mathrm {rank}}\, (\mathbb M^*(L))+{n+s+1\choose n}-1$$.

### Remark

Recall that $$\varDelta ^+_1(L)$$ is a matrix of $$n {n+s\choose n}$$ rows and e columns. If we mimic the construction of the matrix $$\mathbb M^*(L)$$ in the definition of $$\varDelta ^+_1(L)$$, i.e. considering a system of generators of L instead a k-basis of L, we get a matrix with $$n{n+s\choose n}$$ rows and $$n r {n+s \choose n}$$ columns. Notice that $$\mathbb M^*(L)$$ is a smaller matrix: has $$n{n+s\choose n}- {n+s+1\choose n}+1$$ rows and $$n r {n+s \choose n}$$ columns.

In the next result we compute more efficiently the minimal number of generators of an ideal by considering the matrix $$\mathbb M^*(L)$$.

### Theorem 2.6.6

Let L  S be a finitely generated sub-R-module of S of dimension $$e=\operatorname {dim}_{{\mathbf {k}}}(L)$$ and degree s = deg(L). If I = AnnR(L) ⊂ R then
\displaystyle \begin{aligned}\mu(I)= e (n-1) +1 - {\mathrm{rank}}\, (\mathbb M^*(L)). \end{aligned}

In particular, I is a complete intersection if and only if $${\mathrm {rank}}\, (\mathbb M^*(L))=(e-1) (n-1)$$.

### Proof

The statement follows from Propositions 2.6.4 and 2.6.5.

### Example 2.6.7

Let n = 2 and consider a general polynomial of degree two
\displaystyle \begin{aligned}F=c_{6}+c_{5} x_{1}+c_{4} x_{2}+c_{3} x_{1}^{2}+c_{2} x_{1}x_{2}+c_{1} x_{2}^{2}. \end{aligned}
We assume that A = RI, I = Ann(〈F〉), is an Artinian Gorenstein ring of embedding dimension two, in particular μ(I) = 2. Hence the Hilbert function of A is {1, 2, 1}. $$\mathbb L(F)$$ is the 12 × (9 + 2.6)-matrix:
The matrix $$\mathbb L^*(F)$$ is
\displaystyle \begin{aligned}\mathbb L^*(F)= \left( \begin{array}{*{12}{c}} -c_{5} & c_{3} & -c_{2} & 0 & 0 & 0 & c_{4} & c_{2} & c_{1} & 0 & 0 & 0 \\ -c_{2} & 0 & 0 & 0 & 0 & 0 & c_{1} & 0 & 0 & 0 & 0 & 0 \\ -c_{3} & 0 & 0 & 0 & 0 & 0 & c_{2} & 0 & 0 & 0 & 0 & 0 \end{array} \right) \end{aligned}
after elementary transformations, the rank of $$\mathbb L^*(F)$$ agrees with the rank of
\displaystyle \begin{aligned}\left( \begin{array}{*{5}{c}} c_{5} & c_{4}& c_{3} & c_{2} & c_{1} \\ c_{2} & c_{1}& 0 & 0 & 0 \\ c_{3} & c_{2}& 0 & 0 & 0 \end{array} \right) \end{aligned}
Since the embedding dimension of A is two the rank of this matrix is 3. Hence $${\mathrm {rank}}\,(\mathbb M^*(F))=3$$ and by Proposition 2.6.6 μ(I) = 4 − 3 + 1 = 2, as expected, Proposition 2.4.3.

## 2.7 Examples

In this chapter we present several explicit examples proving that some results cannot be improved. We also give some explicit computations of the matrices introduced in chapter 5 and some explicit commutations of the minimal number of generators following the results of Chapter 6.

The following example shows that Theorem 2.5.9 fails if A is Gorenstein of socle degree s = 4, but not compressed, i.e. the Hilbert function is not maximal.

### Example 2.7.1 ()

Let A be an Artin Gorenstein local k-algebra with Hilbert function $$\operatorname {HF}_A=\{1,2,2,2,1\}$$. The local ring is called almost stretched and a classification can be found in . In this case A is isomorphic to one and only one of the following rings :
1. (a)

A = RI with $$I=(x_1^4, x_2^2) \subseteq R={\mathbf {k}}[[x_1, x_2]],$$ and $$I^{\perp }=\langle y_1^3 y_2\rangle .$$ In this case A is canonically graded,

2. (b)

A = RI with $$I=(x_1^4, -x_1^3 + x_2^2)\subseteq R={\mathbf {k}}[[x_1, x_2]],$$ and $$I^{\perp }=\langle y_1^3 y_2+ y_2^3\rangle$$. The associated graded ring is of type (a) and it is not isomorphic to RI. Hence A is not canonically graded.

3. (c)

A = RI with $$I=(x_1^2+x_2^2, x_2^4) \subseteq R={\mathbf {k}}[[x_1, x_2]],$$ and $$I^{\perp }=\langle y_1 y_2(y_1^2-y_2^2) \rangle .$$ In this case A is graded.

The following example shows that Theorem 2.5.9 cannot be extended to compressed Gorenstein algebras of socle degree s = 5.

### Example 2.7.2 ()

Let us consider the ideal
\displaystyle \begin{aligned}I=(x_1^4, x_2^3 - 2 x_1^3 x_2)\subset R={\mathbf{k}}[[x_1,x_2]]. \end{aligned}
The quotient A = RI is a compressed Gorenstein algebra with $$\operatorname {HF}_{A}=\{1,2,3,3,2,1\}$$, $$I^*=(x_1^4,x_2^3)$$ and $$I^{\perp }=\langle y_1^3 y_2^2 + y_2^4\rangle$$. Assume that there exists an analytic isomorphism φ of R mapping I into I. It is easy to see that the Jacobian matrix of φ is diagonal because $$(I^*)^{\perp } = \langle y_1^3 y_2^2\rangle .$$ We perform the computations modulo (x1, x2)5, so we only have to consider the following coefficients of φ
\displaystyle \begin{aligned}\left\{ \begin{array}{ll} \varphi(x_1)= a x_1 + \dots \\ \\ \varphi(x_2)= b x_2 + i x_1^2 + j x_1 x_2 + k x_2^2+ \dots \end{array} \right. \end{aligned}
where a, b are units, i, j, k ∈k. After the isomorphism x1 → 1∕ax1, x2 → 1∕bx2, we may assume a = b = 1. Then we have
\displaystyle \begin{aligned}I^*= \varphi(I )= (x_1^4, x_2^3 -2 x_1^3 x_2 + 3 i x_1^2 x_2^2 + 3 j x_1 x_2^3 + 3 k x_2^4) \text{\quad modulo } (x_1,x_2)^5. \end{aligned}
Hence there exist α ∈ K, β ∈ R such that
\displaystyle \begin{aligned}x_2^3 -2 x_1^3 x_2 + 3 i x_1^2 x_2^2 + 3 j x_1 x_2^3 + 3 k x_2^4 = \alpha x_1^4 + \beta x_2^3 \text{\quad modulo } (x_1,x_2)^5. \end{aligned}
From this equality we deduce α = 0 and
\displaystyle \begin{aligned}2 x_1^3 x_2 = x_2^2( x_2 + 3 i x_1^2 + 3 j x_1 x_2 + 3 k x_2^2 - \beta x_2 ) \text{ \quad modulo } (x_1,x_2)^5, \end{aligned}
a contradiction, so I is not isomorphic to I.
Let φ as above sending I into I. If we denote by (zi)i=1,…,6 the coordinates of a homogeneous form G of degree 5 in y1, y2 with respect to Ω, then the matrix M(G) (s = 5, p = 1) has the following shape
\displaystyle \begin{aligned}\left( \begin{array}{cccccc} 4 z_1 & 4 z_2 & 4 z_3 & 0 & 0 & 0 \\ 3 z_2 & 3 z_3 & 3 z_4 & z_1 & z_2 & z_3 \\ 2 z_3 & 2 z_4 & 2 z_5 & 2 z_2 & 2 z_3 & 2 z_4 \\ z_4 & z_5 & z_6 & 3 z_3 & 3 z_4 & 3 z_5 \\ 0 & 0 & 0 & 4 z_4 & 4 z_5 & 4 z_6 \end{array} \right) \end{aligned}
In our case $$G=y_1^3 y_2^2$$, so all zi are zero but z3 = 12, hence the above matrix has rank 4 and it has not maximum rank given by Corollary 2.5.8. Since all the rows are not zero except the last one, it is easy to see that $$F = y_2^4$$ is not in the image of M(G), as (2.7) requires.

The following example shows that Theorem 2.5.10 cannot be extended to compressed type 2 level algebras of socle degree s = 4.

### Example 2.7.3 ()

Let us consider the forms $$G_1=y_1^2 y_2 y_3$$, $$G_2=y_1 y_2^2 y_3+ y_2y_3^3$$ in S = k[y1, y2, y3] of degree 4 and define in R = k[[x1, x2, x3]] the ideal
\displaystyle \begin{aligned}I= Ann (G_1+y_3^3, G_2).\end{aligned}
Then A = RI is a compressed level algebra with socle degree 4, type 2 and Hilbert function $$\operatorname {HF}_A=\{1,3,6,6,2\}$$. We prove that A is not canonically graded.

We know that I = Ann(G1, G2) and we prove that A and $$gr_{{\mathfrak n}}(A)$$ are not isomorphic as k-algebras. Let φ an analytic isomorphism sending I to I, then it is easy to see that φ = I3 modulo (x1, x2, x3)2. The matrix M(G1, G2) is of size 20 × 18 and, accordingly with (2.7), we will show that $$y_3^3$$ is not in the image of M(G1, G2).

Let F1, F2 be two homogeneous forms of degree 4 of R = k[y1, y2, y3]. We denote by $$(z_i^j)_{i=1,\dots , 15}$$ the coordinates of Fj with respect the basis Ω, j = 1, 2. Then the 20 × 18 matrix M(F1, F2) has the following shape, see (2.9),
It is enough to specialize the matrix to our case for proving that $$y_3^3$$ is not in the image of M(G1, G2).

Next we will show how to apply the main result of the chapter six, Theorem 2.6.6. We assume that the ground field k is infinite.

### Example 2.7.4

Artin Graded Level algebras of type 2.

Let F, G be two forms of degree three of S = k[x1, x2, x3]. We write $$I=\operatorname {Ann}_R(\langle F, G \rangle )$$. Then $$\mathbb L^*(\langle F, G\rangle )$$ is a 26 × 120 matrix in the coefficients of F, c1, …, c10, and the coefficients of G, c11, …, c20. This matrix has rank 17 considered as matrix with entries in the field K of fractions of c1, …, c20. This means that for generic c1, …, c20 the matrix $$\mathbb L^*(\langle F, G\rangle )$$ has rank 17. Moreover, there is a 17 × 17 submatrix of $$\mathbb L^*(\langle F , G\rangle )$$ whose determinant is non-zero in K
\displaystyle \begin{aligned}D_1=c_1 (c_5 c_7-c_3 c_8) (c_1 c_{12}-c_2 c_{11}) G_4 G_8\end{aligned}
where G4 is a form of degree 4 on c1, …, c10 and G8 is a form of degree 8 on c11, …, c20. The condition c1c12 − c2c11 ≠ 0 implies that F, G are linearly independent over k, so A = RI is an Artin level algebra of socle degree three and type 2. If the determinant, say D2, of the matrix
\displaystyle \begin{aligned}\left( \begin{array}{ccc} c_9 & c_8 & c_6\\ c_7 & c_6 & c_3 \\ c_6 & c_5 & c_2 \\ \end{array} \right) \end{aligned}
is non-zero, the embedding dimension of A is three.
Let now consider the k-vector space V  generated by xi ∘ F, i = 1, 2, 3; xi ∘ G, i = 1, 2, 3. The dimension of V  equals $$\operatorname {HF}_A(2)$$ and agrees with the rank of the following matrix
\displaystyle \begin{aligned}\left( \begin{array}{cccccc} c_{10} & c_9 & c_8 & c_7 & c_6 & c_5 \\ c_{20} & c_{19} & c_{18} & c_{17} & c_{16} & c_{15} \\ c_{9} & c_7 & c_6 & c_4 & c_3 & c_2 \\ c_{19} & c_{17} & c_{16} & c_{14} & c_{13} & c_{12} \\ c_{8} & c_6 & c_5 & c_3 & c_2 & c_1 \\ c_{18} & c_{16} & c_{15} & c_{13} & c_{12} & c_{11} \\ \end{array} \right) \end{aligned}
If the determinant D3 of this matrix is non-zero then the Hilbert function of A is {1, 3, 6, 2}. Hence, if D1D2D3 ≠ 0 then A is a compressed Artin level algebra of type 2, socle degree 3, embedding dimension 3 and Hilbert function {1, 3, 6, 2}. From Theorem 2.6.6 we get
\displaystyle \begin{aligned} \mu(I)= 2. 12 - {\mathrm{rank}}\,(\mathbb L^*(\langle F, G\rangle))+1=8 \end{aligned}
as Böij conjecture predicts, [1, Section 3.2].

Let $$\mathbb P_{{\mathbf {k}}}^9 \times \mathbb P_{{\mathbf {k}}}^9$$ be the space parameterizing the pairs (F, G) up to scalars in each component. Since D1D2D3 is bi-homogeneous form of degree 26 on (c1, …, c10) and (c11, …, c20), in this example we have shown a principal non-empty subset $$U=\mathbb P_{{\mathbf {k}}}^9 \times \mathbb P_{{\mathbf {k}}}^9\setminus V(D_1 D_2 D_3)$$ parameterizing a family of compressed Artin level algebra of type 2, socle degree 3, embedding dimension 3 and Hilbert function {1, 3, 6, 2}.

### Example 2.7.5

Artin Gorenstein algebras with Hilbert function {1, 4, 4, 1}.

Let us consider a general polynomial F of degree 3 of . We write I = 〈F. Then $$\mathbb L^*(F)$$ is a 71 × 140 matrix in the coefficients of F, say c1, …, c35. This matrix has rank 25 considered as matrix with coefficients in the field K of fractions of c1, …, c35. Hence for generic values of c1, …, c35 the ring A = RI is a compressed Gorenstein algebra with Hilbert function {1, 4, 4, 1}, and the matrix $$\mathbb L^*(F)$$ has rank 25 so
\displaystyle \begin{aligned}\mu(I)= 3. 10- 25 +1= 6 \end{aligned}
as it was expected, .

### Example 2.7.6

Artin Gorenstein algebras with Hilbert function {1, 3, 3, 1}.

In this example we assume that the ground field k is algebraically closed. In  we prove that all Artin Gorenstein algebra A = RI with Hilbert function {1, n, n, 1} is isomorphic to its associated graded ring. Hence in the case n = 3 we may assume that I is generated by a form F in x1, x2, x3 of degree three. In [12, Proposition 3.7 ] we classify such algebras in terms of the geometry of the projective plane cubic C defined by F. Next we will compute the minimal number of generators of I in the case that C is non-singular by using the main theorem of this paper.

Let is consider the Legendre form attached to C, λ ≠ 0, 1,
\displaystyle \begin{aligned}U_{\lambda}=x_2^2x_3- x_1(x_1-x_2)(x_1- \lambda x_3)\end{aligned}
the j-invariant of C is
\displaystyle \begin{aligned}j(\lambda)=2^8\frac{(\lambda^2-\lambda+1)^3}{\lambda^2(\lambda-1)^3}. \end{aligned}
$$\mathbb L^*(U_{\lambda })$$ is a 20 × 24-matrix, after elementary transformations we get that the rank of $$\mathbb L^*(U_{\lambda })$$ is 10 plus the rank of the 4 × 4 square matrix
The determinant of W is
\displaystyle \begin{aligned}\det(K)=(\lambda^2-\lambda+1)^2. \end{aligned}
Hence if j(λ) ≠ 0 then μ(I) = 2.8 − (10 + 4) + 1 = 3, i.e. I is a complete intersection, as we get in [12, Proposition 3.7]. If j(λ) = 0, i.e. C is the elliptic Fermat curve, then one gets rank (W) = 2 for all roots of $$\det (W)=0$$. Hence μ(I) = 2.8 − (10 + 2) + 1 = 5 as we get in [12, Proposition 3.7].

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