Free extensions and Jordan type

Free extensions and Jordan type

Journal Pre-proof Free extensions and Jordan type Anthony Iarrobino, Pedro Macias Marques, Chris McDaniel PII: S0021-8693(20)30005-3 DOI: https:/...

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Journal Pre-proof Free extensions and Jordan type

Anthony Iarrobino, Pedro Macias Marques, Chris McDaniel

PII:

S0021-8693(20)30005-3

DOI:

https://doi.org/10.1016/j.jalgebra.2020.01.003

Reference:

YJABR 17501

To appear in:

Journal of Algebra

Received date:

19 May 2019

Please cite this article as: A. Iarrobino et al., Free extensions and Jordan type, J. Algebra (2020), doi: https://doi.org/10.1016/j.jalgebra.2020.01.003.

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Free extensions and Jordan type∗ Anthony Iarrobino Department of Mathematics, Northeastern University, Boston, MA 02115, USA.

Pedro Macias Marques Departamento de Matem´ atica, Escola de Ciˆencias e Tecnologia, Centro de Investiga¸c˜ ao em Matem´ atica e Aplica¸c˜ oes, Instituto de Investiga¸c˜ ao e Forma¸c˜ ao Avan¸cada, ´ ´ Universidade de Evora, Rua Rom˜ ao Ramalho, 59, P–7000–671 Evora, Portugal.

Chris McDaniel Department of Mathematics, Endicott College, Beverly, MA 01915, USA.

May 21, 2019, revised October 23, 2019 Abstract Free extensions of graded Artinian algebras were introduced by T. Harima and J. Watanabe, and were shown to preserve the strong Lefschetz property. The Jordan type of a multiplication map m by a nilpotent element of an Artinian algebra is the partition determining the sizes of the blocks in a Jordan matrix for m. We show that a free extension C of the Artinian algebra A with fiber B is a deformation of the usual tensor product. This has consequences for the generic Jordan types of A, B and C: we show that the Jordan type of C is at least that of the usual tensor product in the dominance order (Theorem 2.5). In particular this gives a different proof of the T. Harima and J. Watanabe result concerning the strong Lefschetz property of a free extension. Examples illustrate that a non-strong-Lefschetz graded Gorenstein algebra A with non-unimodal Hilbert function may nevertheless have a non-homogeneous element with strong Lefschetz Jordan type, and may have an A-free extension that is strong Lefschetz. We apply these results to algebras of relative coinvariants of linear group actions on a polynomial ring.

1

Introduction

Let A = (A, m, k) be a local Artinian algebra over a field k = A/m where m is the unique maximum ideal. The Jordan type of an element  ∈ m is the partition P with parts equal to ∗

Keywords: Artinian algebra, coinvariant, deformation, free extension, Hilbert function, invariant, Jordan type, Lefschetz property, tensor product. 2010 Mathematics Subject Classification: Primary: 13E10; Secondary: 13A50, 13D40, 13H10, 14B07, 14C05

1

the sizes of the Jordan blocks of its associated (nilpotent) multiplication map × : A → A. The generic Jordan type of A is defined to be the largest partition PA = P , with respect to the dominance order on partitions, which occurs among the elements  ∈ m. It is known that the Jordan types P are bounded above by the conjugate partitions H(A)∨ of their associated Hilbert functions H(A). We say that an element  ∈ m has strong Lefschetz Jordan type (SLJT) if its Jordan type achieves this bound, i.e. if P = H(A)∨ . We say that the algebra A has strong Lefschetz Jordan type if it admits a SLJT element. Our graded algebras A (we use roman letters to distinguish this case) may or may not be standard graded : that is, generated by A1 over A0 = k. When A is standard graded one typically restricts to linear forms  ∈ A1 , and computes the generic linear Jordan type of A, PA , as the largest Jordan type which occurs among the linear forms  ∈ A1 . We say that a linear form  ∈ A1 is strong Lefschetz (SL) if P = H(A)∨ , the conjugate partition of the Hilbert function of A corresponding to the grading on A. This Hilbert function H(A) may in general be different from the Hilbert function H(A), with respect to the filtration by powers of the maximal ideal mA = ⊕∞ k=1 Ak , where A denotes the related local algebra. The strong Lefschetz property of a linear form  ∈ A1 is equivalent to the familiar condition that the multiplication maps ×k : Ai → Ai+k have maximal rank for all integers i, k. We say that the graded algebra A is strong Lefschetz (SL) if it admits a strong Lefschetz linear element. Let A, B, and C be graded Artinian k-algebras (not necessarily standard-graded). We say that C is a free extension of A with fiber B if there is a map of graded algebras ι : A → C making C into a free A-module and there is an isomorphism of graded algebras B ∼ = C/(ι(mA )). The tensor product algebra A ⊗k B is a free extension of A with fiber B, and one of our principal results is that a free extension may be regarded as a deformation of the tensor product (Theorem 2.1). We use this to show a main goal of this note, that the generic linear Jordan type of a free extension is bounded below by that of its associated tensor product algebra: Theorem. (Theorem 2.5 below) Let A, B, and C be graded Artinian algebras over an infinite field k, and suppose that C is a free extension of A with fiber B. Let PA⊗k B be the generic linear Jordan type of the tensor product algebra, and let PC be the generic linear Jordan type of the A-free extension C. Then in the dominance partial order we have PC ≥ PA⊗k B . This theorem can be viewed as a generalization of a well known result of T. Harima and J. Watanabe, which we derive as a corollary. For a graded algebra jA is the largest degree i for which Ai = 0. Corollary. (Theorem 2.6 below)[HW1, Theorem 6.1] Let the Artinian algebra C be a free extension of A with fiber B. Assume that char k = 0 or char k > jA + jB , that the Hilbert functions of both A and B are symmetric, and that both A and B are strong Lefschetz. Then C is also strong Lefschetz. The invariant theory of finite groups abounds with free extensions, and the above corollary can be used to show that coinvariant algebras associated to certain finite reflection groups are SL. On the other hand, we also find examples of free extensions C over A with fiber B where the inequality in the theorem is strict, i.e. PC > PA⊗k B . Example 3.9 shows that this 2

strict inequality can even occur if the gradings on A, B, and C are all standard, answering a question of J. Watanabe (private communication). For further study of relative coinvariant rings from this viewpoint see [MCIM].

1.1

Jordan Type and Strong Lefschetz

Let A = (A, m, k) be a local Artinian algebra over a field k. Then for any element  ∈ m the multiplication map × : A → A (1.1) is a nilpotent linear transformation of a finite dimensional vector space A. Define the Jordan type of  to be the partition P with parts equal to the block sizes of the Jordan canonical form of the multiplication map (1.1). We recall the dominance order on partitions. Let P = (p1 , . . . , ps ) and P  = (p1 , . . . , pt ) with p1 ≥ · · · ≥ ps and p1 ≥ · · · ≥ pt . Then 

P ≤ P if for all i we have

i  k=1

pk ≤

i 

pk .

(1.2)

k=1

Thus, (2, 2, 1, 1) < (3, 2, 1) but (3, 3, 3) and (4, 2, 2, 1) are incomparable. Given a partition P = (p1 , . . . , pr ), define its conjugate partition by P ∨ = (p∨1 , . . . , p∨s ), p∨i = # {k | pk ≥ i} . A partition can be visualized by a Ferrers diagram, and its conjugate partition is then the Ferrers diagram with the rows and columns interchanged. The following Proposition leads to the definition of generic Jordan type of A: see [IMM1, §2.2, Lemma 2.8] concerning the semicontinuity of Jordan type. Recall that in a local Artinian algebra A, the maximal ideal m is an affine space. We say that the graded algebra A has maximal socle degree j = jA or formal dimension j (as in [MS]) when Aj = 0 but Ai = 0 for i > j.1 Proposition 1.1. Assume that k is an infinite field. In the local (resp. graded) case, there exists a non-empty Zariski open dense set U ⊂ m (respectively U ⊂ A1 ) such that for each pair of elements  ∈ U ,  ∈ m, (of  ∈ U,  ∈ A1 ) we have P ≤ P . Definition 1.2. The generic Jordan type of A, PA (resp. the generic linear Jordan type of A, PA ) is the Jordan type P for any element  ∈ U , the open dense Zariski set of Proposition 1.1 Recall that the Hilbert function for a local Artinian algebra is the tuple of integers   (1.3) H(A) = (h0 , h1 , h2 , . . . , hjA ), hi = dimk mi /mi+1 . If A = A is a graded algebra, it comes with its own Hilbert function H(A) = (h0 , h1 , h2 , . . . , hj ), hi = dimk (Ai ). 1

(1.4)

We follow [H-W, Remark 2.11] in this definition. When A is not standard graded, we take mA = ⊕i≥1 Ai ; 0. then jA might not agree with the largest j such that (mA )j =

3

Note that the Hilbert function defines a partition (after possible rearrangement). We write H(A)∨ or H(A)∨ to mean the conjugate partition of the Hilbert function. For a proof of the following we refer the reader to [IMM1, Theorem 2.23]. Proposition 1.3. The Jordan type of any element  ∈ m is bounded above by the conjugate partition of the Hilbert function of A, i.e. P ≤ H(A)∨ . If A = A is graded, and  ∈ A1 , then the Jordan type is bounded by the conjugate of the graded Hilbert function, i.e. P ≤ H(A)∨ . We say that A has SLJT (respectively, A is SL) if its generic Jordan type (respectively, generic linear Jordan type) achieves this bound, i.e. PA = H(A)∨ , resp. PA = H(A)∨ . One can show, e.g. [IMM1, Proposition 2.37], that in the graded case, A is strong Lefschetz if and only if there exists a linear form  ∈ A1 for which the multiplication maps ×k : Ai → Ai+k have full rank for all integers i, k. Moreover, one can further show that if the Hilbert function H(A) is symmetric, i.e. hi = hjA −i for each i, then SL is in turn equivalent to the condition that the multiplication maps ×jA −2i : Ai → AjA −i are isomorphisms for each i. The following result is well known. For a graded algebra with A0 ∼ = k we set mA = ⊕i≥1 Ai . We denote by k{x, y} the regular local ring in variables x, y over the field k. Lemma 1.4 (Height two Artinian algebras are strong Lefschetz). Let A = k[x, y]/I be standard Artinian graded of socle degree j, or A = k{x, y}/I be local Artinian of socle degree j = jA , and suppose char k = 0 or char k > j. Let  be a general element of mA in the first case, or of mA in the second. Then  has SLJT and A is SL (or A has SLJT). Proof. These statements follow readily from the openness of the set of directions in which there is a standard basis for ideals in the local ring C{x, y} [Bri, Theorem I.2.1], that extends to the case char k = p > j (see [BaI, Theorem 2.16]). 

1.2

Free extensions

Definition 1.5. A. Given graded Artinian algebras A, B, and C, we say that C is a free extension of A with fiber B if there exist graded algebra homomorphisms ι : A → C and π : C → B for which 1. ι is injective and makes C into a free A-module, 2. π is surjective and ker(π) = ι(mA ) · C. Equivalently, C is a free extension of A with fiber B if π is surjective and for some (equivalently, every) k-linear section of π, say s : B → C, the map Φs : A ⊗k B → C is an isomorphism of A-modules. 4

B. Geometrically, as we shall show (Theorem 2.1), a free extension C is a flat deformation of a finite (commutative associative) algebra B over a local algebra (A, m, k): so C is an A-algebra such that the associated map γ : Spec(C) → Spec(A) is finite and flat, and with a closed embedding π : Spec(B) → Spec(C) inducing a cartesian diagram: Spec(B) 



π

/

Spec(C) γ



Spec(k)





/ Spec(A).

Remarks 1.6. 1. Thus, a free extension C can be regarded as a deformed tensor product of A and B, with the deformation occurring in the B-factor. Note that C is isomorphic to the tensor product A ⊗k B as algebras if and only if the k-linear section s : B → C can be chosen as a map of (graded) k-algebras. 2. A related notion is that of coexact sequences which show up in the J. C. Moore related topology literature [Bau, MoSm, Sm1]. Given a sequence of graded Artinian algebras A(n) n ∈ Z, we say that a sequence of maps · · · A(n − 1)

fn−1

/

A(n)

fn

/

A(n + 1) · · ·

is coexact at A(n) if we have ker(fn ) = fn−1 (mA(n−1) ) · A(n). The sequence is coexact if it is coexact at A(n) for every n. Then C is a free extension of A with fiber B if and only if the sequence /A ι /C π /B /k k (1.5) is coexact and ι : A → C makes C into a free A-module. 3. Here is yet another criterion for free extension: C is a free extension of A with fiber B if and only if π is surjective, ker(π) = ι(mA ) · C and dimk (C) = dimk (A) · dimk (B).

2

Free extension as a deformation, and Jordan type

We first will show that an A-free extension C with fiber B is a deformation of the algebra A ⊗ B (Theorem 2.1). We will conclude that the generic Jordan type of C is always greater than or equal to that of the actual tensor product A ⊗ B (Theorem 2.6). Let A, B, and C be graded Artinia k algebras with homogeneous maximal ideals mA , mB and mC , respectively. Suppose that C is a free extension over A with fiber B, meaning that there exist algebra homomorphisms ι : A → C making C into a free A module and π : C → B surjective with ker(π) = mA C (Definition 1.5). We thank the referee for suggesting we show the following stronger result than we had originally shown and outlining the proof given in Remark 2.2 using Rees algebras. We give two different proofs, with the thought that some readers may prefer one. Our first proof was inspired by the discussion in [Ei, §15.8 and Theorem 15.17], which was pointed out by the referee. Theorem 2.1. Assume that k is an infinite field and that C is a free extension over A with fiber B. Then there exists a flat k[t]-algebra C with fibers Cc := C/(t − c)C over c ∈ k satisfying 5

1. C0 := C/(t · C) ∼ = A ⊗k B as A-algebras, and 2. C1 := C/((t − 1) · C) ∼ = C as A-algebras, 3. for every non-zero c ∈ k, we have Cc ∼ = C1 as A-algebras. In other words, C is a flat deformation of A ⊗k B as an A-algebra. Proof. Let b1 , . . . , br ∈ B be a set of homogeneous elements that generate B as a k-algebra: then B has a homogeneous k-linear basis consisting of some monomials in b1 , . . . , br . Let c1 , . . . , cr ∈ C be any homogeneous π-lifts of b1 , . . . , br , respectively. Then since C is a free Amodule with fiber B, C must have an A-linear basis consisting of some monomials in c1 , . . . , cr ; in particular, the elements c1 , . . . , cr must generate C as an A-algebra. Define the surjective A-algebra homomorphism φ : A[y1 , . . . , yr ] → C, φ(yi ) = ci , and let I = ker(φ) ⊂ A[y1 , . . . , yr ]. Since A is Artinian, it is also Noetherian, hence so is A[y1 , . . . , yr ]. Thus I is finitely generated, say I = (g1 , . . . , gr ). Since φ preserves the grading (if we set deg(yi ) = deg(bi )), then we can take the generators g1 , . . . , gr to be homogeneous in A[y1 , . . . , yr ]. Then in A[y1 , . . . , yr , t] define the polynomials   (t) gi := tdeg(gi ) · gi t− deg(y1 ) y1 , . . . , t− deg(yr ) yr and set

A[y1 , . . . , yr , t] C =  (t) . (t)  g1 , . . . , gs

Note that for each c ∈ F we have A[y1 , . . . , yr ] ∼ . =   (t) (t) (c) (c)  g1 , . . . , gs , t − c g1 , . . . , gs A[y1 , . . . , yr , t]

Cc = C/(t − c)C =  (1)

In particular, for c = 1 we have gi

= gi and hence

A[y1 , . . . , yr ] ∼ C1 ∼ = C, = (g1 , . . . , gs ) which shows that (2) holds. Note also that for each non-zero b ∈ k we have a k-algebra isomorphism /

θb : A[y1 , . . . , yr , t]

(t)

A[y1 , . . . , yr , t]

a

/a

yi 

/ bdeg(yi ) yi

t

/ bt

(t)

which satisfies θb (gi ) = bdeg(gi ) · gi , and θb (t − 1) = b(t − 1b ). Hence for a non-zero c ∈ k, the map θc−1 passes to an A-algebra isomorphism on the quotients A[y1 , . . . , yr , t] (t) (t) g1 , . . . , gs , t −

θc−1 : C1 = 

1

A[y1 , . . . , yr , t] (t) (t) (c−1 )deg(g1 ) g1 , . . . , (c−1 )deg(gs ) gs , c−1 (t

→

6

− c)

 = Cc ,

hence (3) is satisfied. To see that (1) holds, note that for each 1 ≤ i ≤ r we have (t)

(0)

gi = gi + t · hi (0)

where gi ∈ k[y1 , . . . , yr ] and hi ∈ A[y1 , . . . , yr , t] is some polynomial with coefficients in mA . Then we have the following string of A-module isomorphisms: A[y1 , . . . , yr , t] (t) (t)  g1 , . . . , gr + (t)

C/(t · C) = 

A[y1 , . . . , yr ] =  (0) (0)  g1 , . . . , gr k[y1 , . . . , yr ] =A ⊗k  (0) (0)  g1 , . . . , gr A[y1 , . . . , yr ] ∼ = A ⊗k (g1 , . . . , gr ) + mA =A ⊗k C/(mA · C) ∼ =A ⊗k B, which implies (1). We will now show that C is free over A[t], hence flat over k[t]. Since  (0) (0)  B∼ = k[y1 , . . . , yr ]/ g1 , . . . , gs there exist monomials m1 , . . . , mb ∈ k[y1 , . . . , yr ] which form a k-linear basis for B. Consequently, the m1 , . . . , mb also form an A basis for C ∼ = C1 , and hence also for Cb for each non-zero b ∈ F (use the isomorphism θb above, and note that θb (mi ) = mi for each i). We claim that m1 , . . . , mb form an A[t]-module basis for C. First, they generate C/mA C ∼ = B ⊗k k[t] as a k[t]-module, hence by Nakayama’s lemma they must also generate C as an A[t]-module. Next, suppose by way of contradiction that there is a dependence relation over A[t], N  b 

aij ti mj = 0, for some aij ∈ A.

i=1 j=1

Then for each c ∈ k, we must have an A-dependence relation in Cc :  N  b   aij ci mj = 0. j=1

i=1

This implies that for each j = 1, . . . , b we must have N 

aij ci = 0

i=0

for every c ∈ k, and some fixed aij ∈ A. But choosing N -different points c1 , . . . , cN ∈ k, which is possible since k is infinite, we can solve the (N + 1) × (N + 1) nonsingular Vandermonde system to show that aij = 0 for all i, j. Therefore the m1 , . . . , mb are A[t]-linearly independent, and hence C is A[t]-free, and in particular k[t]-flat.  7

Remark 2.2 (Proof of Theorem 2.1 using Rees algebra). We now give a second proof, suggested by the referee, that works over any field k; we use also [Ei, §6.5]. Let C be a free extension of A with fibre B as in Theorem 2.1. For each i define the ideal Ii := A≥i · C; these ideals define a decreasing filtration F(C) : I0 = C ⊇ I1 ⊇ I2 ⊇ · · · for which the associated graded algebra is grF (C) (C) = C/I1 ⊕ I1 /I2 ⊕ · · · = ⊕i≥0 Ii /Ii+1 . Note that grF (C) (C) is naturally an algebra over the associated graded algebra grF (A) (A) = A/A≥1 ⊕ A≥1 /A≥2 ⊕ · · · ∼ = A. = A0 ⊕ A1 ⊕ A2 ⊕ · · · ∼ Lemma 2.3. The algebra grF (C) (C) is isomorphic to A ⊗k B as A algebra. Proof. Note that besides the well defined k-algebra map ι : A → grF (C) (C) (which gives grF (C) (C) its A-algebra structure), we also have a well defined k-algebra map s : B → C (since C/I1 = C/mA C ∼ = B). Then by the universal property for tensor products, there is also a well defined A-algebra homomorphism Φ = ι ⊗ s : A ⊗k B → grF (C) (C). If b1 , . . . , br is a homogeneous k-basis for B, and a1 , . . . , as is a homogeneous basis for A, then clearly for each k, {Φ(ai ⊗ bj ) = ai bj + Ik+1 | deg(ai ) = k} is a basis for Ik /Ik+1 , hence Φ is surjective. Finally since dimk (C) = dimk (A) · dimk (B) = dimk (A ⊗k B), the homomorphism Φ must be an A-algebra isomorphism.  Recall that the Rees algebra R(C, F) for the ring C with filtration F = F(C) : I0 ⊇ I1 ⊇ I2 ⊇ · · · , is the following k[t]-subalgebra of C[t, t−1 ] (here we take Ii = R for i < 0) R(C, F) =

∞ 

Ii · t−i ⊂ C[t, t−1 ].

i=−∞

Lemma 2.4.

1. R(C, F) is a flat k[t]-algebra.

2. R(C, F)/t · R(C, F) ∼ = grF (C) as A-algebras. 3. R(C, F)/(t − c) · R(C, F) ∼ = C as A-algebras, for every 0 = c ∈ k. Proof. Note that R(C, F) ⊂ C[t, t−1 ] is k[t]-torsion free, hence it is k[t]-flat, which is (1). To see (2) note that for each integer i there is a natural projection map πi : Ii · t−i → Ii /Ii+1 whose kernel is Ii+1 · t−i ⊂ Ii · t−i ; taking the direct sum of these maps gives the A-algebra map ⊕i∈Z πi : R(C, F) → grF (C) with kernel t · R(C, F) =

∞  i=−∞

8

Ii t−i+1

which gives (2). Finally note that for every non-zero c ∈ k, there is also a natural surjective homomorphism /C ρ : R(C, F)   −i / (xi · t−i )i∈Z  i∈Z xi · c .  Finally note that if i∈Z xi c−i = 0 for some finite sum where each xi ∈ Ii ⊂ Ii−1 , then the difference of Z-tuples   −i   

 xi t i∈Z − xi c−1 t−i+1 i∈Z = (t − c) xi t−i+1 i∈Z is in the kernel and conversely. Therefore ker(ρ) = (t − c)R(C, F), which implies (3).



The above argument shows that C is a flat deformation of A ⊗k B regardless of the field; it also works alike for standard or non-standard graded algebras. We thank the referee for pointing to the stronger statement, and this proof. A key observation is Lemma 2.3. Theorem 2.5. Let A be an Artinian graded algebra, let C be a free extension of A with fiber B, over an infinite field k. Let PA⊗k B be the generic linear Jordan type of the actual tensor product algebra, and let PC be the generic linear Jordan type of the A-free extension C. Then in the dominance partial order we have PC ≥ PA⊗k B . Proof. Let  = A ⊗ 1 + 1 ⊗ B ∈ A ⊗k B be a linear form whose multiplication map has Jordan type partition PA⊗k B . By Theorem 2.1 there is a 1-parameter family Lt : C → C (t ∈ k ∼ = A1 ) of degree one endomorphisms of C such L0 has the same Jordan type as the multiplication map × : A ⊗k B → A ⊗k B, and such that Lt , t = 0 has the same Jordan type as the multiplication map × (ι(A ) + tΛ) : C → C. By the semicontinuity of Jordan type, e.g. Proposition 1.1, there is an open set U ∈ A1 containing t = 0 such that the Jordan type PLt ≥ PL0 for all t ∈ U . Since the generic (linear) Jordan type PC is the maximal Jordan type occurring for  ∈ C1 we have, as desired, PC ≥ PLt ≥ PL0 = PA⊗k B .  We can use Theorem 2.5 to give a new proof of the following result of T. Harima and J. Watanabe, which they show using central simple modules.2 Theorem 2.6. [HW1, Theorem 6.1]. Suppose that C is a free extension of A with fiber B. Assume that char k = 0 or char k > jA + jB , that the Hilbert functions of both A and B are symmetric, and that both A and B are strong Lefschetz. Then C is also strong Lefschetz. 2

We have slightly different notation, our C is the A in [HW1, Theorem 6.1]. The definition in [HW1] of SL is the “narrow strong Lefschetz” of [H-W, Definition 3.18], implying that the Hilbert function is symmetric and unimodal. Their statement of Theorem 6.1 in [HW1] is for char k = 0, but the proof in high enough characteristic p is the same.

9

Proof. Under the assumptions on char k, the Clebsch-Gordan formula applies, and shows that if A, B are strong Lefschetz then the tensor product A ⊗k B is strong Lefschetz.3 Hence the generic linear Jordan type is maximal, i.e. PA⊗B = H(A ⊗k B)∨ . But since C is a free extension of A with fiber B, it must have the same Hilbert function H(C) = H(A ⊗k B), and hence H(C)∨ = H(A ⊗k B)∨ . By Proposition 1.3, we must have PC ≤ H(C)∨ . On the other hand Theorem 2.5 implies that PC ≥ PA⊗k B = H(C)∨ , and hence we must have equality P (C) = H(C)∨ , and hence C must be SL. 

3

Examples

3.1

Free Extensions in Invariant Theory

Let V = kn be a finite dimensional vector space over k, let W ⊂ Gl(V ) be any finite group acting linearly on V , and let R = Sym(V ∗ ) ∼ = k[x1 , . . . , xn ] be the algebra of polynomial ∗ functions on V (here V is the dual). Then W acts on R in the usual way, i.e. (w · f )(v) = f (w−1 (v)), and the set of W -invariant polynomials forms a subalgebra RW ⊂ R called the W -invariant subalgebra. Let h(W ) ⊂ R denote the ideal generated by the non-constant R polynomials in RW . The quotient algebra RW = h(W is called the coinvariant algebra of W ; ) it is always a graded Artinian algebra. K K For any subgroup K ⊂ W , the quotient algebra RW = h(WR)∩RK is called the relative coinvariant algebra for the pair K ⊂ W . The inclusion ˆι : RK → R passes to a well defined K map of quotient algebras ι : RW → RW . We also have an inclusion of ideals h(W ) ⊂ h(K), and hence a natural surjection of quotient algebras π : RW → RK . We say that the invariant subalgebra RW or RK is polynomial if it can be generated as an algebra by algebraically independent polynomials. K Lemma 3.1. If RK is polynomial, then C = RW is a free extension of A = RW with fiber W B = RK via the maps ι and π. Moreover, if R is polynomial and RW is a free extension of K RW with fiber RK via ι and π, then RK must be polynomial. K → RW makes RW Proof. Assume first that RK is polynomial. We must show that ι : RW K K + into a free A = RW module, and that ker(π) = ι(mA ) · RW = (RW ) · RW . Note that an equivalence class f +h(W ) is in ker(π) if and only if f ∈ h(K), which immediately implies that K + ker(π) = (RW ) · RW . To see that RW is a free module, recall, c.f. [Sm2, Corollary 6.7.13], the fact that RK ⊂ R is polynomial if and only if R is a free module over RK . Therefore K ∼ RW ∼ = R ⊗RW k is a free module over RW = RK ⊗RW k. W K Assume R is polynomial but that R is not polynomial. Consider the natural projection π ˆ : R → RK . By Nakayama’s Lemma if ˆs : RK → R is any graded k-linear section for π, ˆ s : RK ⊗k RK → R is surjective; let K be its kernel so that we have a short exact the map Φ sequence of RK modules

0 3

/

K

/

R K ⊗ k RK

Φs

/

R

/0

T. Harima and J. Watanabe’s [HW1, Theorem 3.10], also see [IMM1, Corollary 3.6].

10

(3.1)

Note that since RK is not polynomial, the fact cited above implies that K is non-zero. Applying K k⊗RW − to the exact sequence (3.1), we get another exact sequence of k⊗RW RK ∼ -modules = RW W

TorR 1 (k, R)

/

/

k ⊗R W K

K RW ⊗ k RK

Φs

/

RW

/

0

(3.2)

W

Since RW is polynomial, R is a free RW -module hence TorR 1 (k, R) = 0. Moreover since ∼ K = 0, Nakayama’s Lemma implies that k ⊗RW K = K/hW ∩ K = 0 as well. This implies that the second map Φs = ι ⊗ s in Sequence (3.2) is not an isomorphism. On the other hand, K K + K + we have seen already that ker(π : RW → RW ) = (RW ) · RW and RW /(RW ) · RW ∼ = RK , K K and hence by Nakayama’s Lemma Φs : RW ⊗k RK → RW is a free RW -module cover of RW and so is an isomorphism if and only if RW is free. We conclude that RW cannot be a free K RW -module in this case.  Remark 3.2. In the non-modular case, meaning that |W | ∈ k∗ , we have RW is polynomial if and only if W is generated by pseudo-reflections. A pseudo-reflection is an invertible linear transformation s ∈ Gl(V ) with finite order whose fixed point set is a hyperplane in V . Therefore in the non-modular case, Lemma 3.1 says that for any fixed group W , the coinvariant algebra RW has a free extension structure for each reflection subgroup K ⊂ W . A similar result to Lemma 3.1 was proved by L. Smith in the non-modular case [Sm3, Theorem 1]. Example 3.3. Let k = C, and fix integers r, n ≥ 1. Let W be the pseudo-reflection group G(r, 1, n), i.e. semi-direct product ⎧⎛ ⎪ ⎨ ⎜ W = ⎝ ⎪ ⎩

⎫ ⎞  λ1 · · · 0 ⎪ ⎬  .. . . .. ⎟ λr = 1  S . n . . ⎠ i . ⎪ ⎭  0 · · · λn

In other words W is the group of n × n “r-colored permutation matrices” whose elements are permutation matrices with non-zero entries arbitrary rth roots of unity. W acts on the polynomial ring R = k[x1 , . . . , xn ] and its invariants are generated by the “elementary rsymmetric polynomials” in n-variables, i.e. ei (r, n) = ei (xr1 , . . . , xrn ) where ei (x1 , . . . , xn ) is the ith elementary symmetric polynomial in the variables x1 , . . . , xn . Therefore its coinvariant algebra is the graded Artinian complete intersection RW =

k[x1 , . . . , xn ] . (e1 (r, n), . . . , en (r, n))

As a reflection subgroup take K = G(r, 1, n − 1), i.e. K is the pointwise stabilizer subgroup of the last coordinate function xn . Then the K-coinvariants are RK =

k[x1 , . . . , xn−1 ] k[x1 , . . . , xn ] ∼ . = (e1 (r, n − 1), . . . , en−1 (r, n − 1), xn ) (e1 (r, n − 1), . . . , en−1 (r, n − 1))

Note that for each 1 ≤ i ≤ n − 1 we have the relations ei (r, n) = ei (r, n − 1) + xrn · ei−1 (r, n − 1). 11

Thus, the K ⊂ W relative coinvariants are k[y1 , . . . , yn−1 , xn ] k[e1 (r, n − 1), . . . , en (r, n − 1), xn ] ∼ = r (e1 (r, n), . . . , en (r, n)) (y1 + xn , y2 + xrn y1 , . . . , yn−2 + xrn yn−1 , xrn yn−1 ) K ∼ or RW = k[xn ]/(xnr n ). We can use Theorem 2.5 to see that RW is SL, by induction on n. The base case is n = 2 and we have k[x1 , x2 ] RW = r (x1 + xr2 , xr1 xr2 ) K which is SL by Lemma 1.4. By induction, RK is SL, and RW also obviously is SL. It follows from Corollary 2.6 that RW is SL. K RW =

J. Watanabe et. al. [H-W] used Corollary 2.6 in a different way to show that RW is SL for W = G(r, 1, n). In [McD] Corollary 2.6 was used in conjunction with Schubert calculus to show that RW is SL for every real reflection group. The next example shows that there are complex reflection groups to which the above argument does not apply. Example 3.4. Let k = C and take W ⎧⎛ ⎨ λ1 0 W = ⎝ 0 λ2 ⎩ 0 0

be the complex reflection group G(3, 3, 3), i.e. ⎫ ⎞  0 ⎬  0 ⎠ λ3i = 1, λ1 λ2 λ3 = 1  S3 . ⎭ λ3 

In other words W is the group of 3 × 3 permutation matrices whose non-zero entries are 3rd roots of unity with the additional proviso that their product is equal to one. The coinvariant algebra for W is k[x, y, z] RW = 3 . (x + y 3 + z 3 , x3 y 3 + x3 z 3 + y 3 z 3 , xyz) Take K = G(3, 3, 2), i.e. K is the pointwise stabilizer subgroup of the z-coordinate. The coinvariant algebra for K is RK =

(x3

k[x, y] k[x, y, z] ∼ . = 3 3 + y , xy, z) (x + y 3 , xy)

The relative coinvariant algebra for K ⊂ W is K = RW

k[a, b, c] k[b, c] k[x3 + y 3 , xy, z] ∼ ∼ = = 3 3 3 3 3 3 3 3 3 3 3 3 3 (x + y + z , x y + x z + y z , xyz) (a + c , b + ac , bc) (b − c6 , bc)

K is where deg(a) = 3, deg(b) = 2, and deg(c) = 1. Note that the Hilbert function for RW K K H(RW ) = (1, 1, 2, 1, 2, 1, 1). The non-unimodality of H(RW ) precludes the existence of a SL K element. On the other hand, as a local ring A = RW , and  = b + c ∈ m is a SLJT element, the existence of which is guaranteed by Lemma 1.4.

In case the Hilbert functions of A and B are symmetric, one can show that both A and B are strong Lefschetz if and only if their tensor product A ⊗k B is strong Lefschetz. From this, K one can deduce that in Example 3.4, the tensor product A ⊗k B, for A = RW and B = RK , is not SL. Hence Example 3.4 shows that the inequality in Theorem 2.5 can be strict, i.e. for C = RW , we have PC > PA⊗k B . It is tempting to think that this strictness may be an artifact of the non-standard grading on A. The next Example 3.9 shows that a strict inequality PC > PA⊗k B can also occur where each of A, B, and C has the standard grading, answering a question of J. Watanabe. 12

3.2

Free Extensions via Macaulay Duality

Dual generator of an Artinian Gorenstein algebra. A local Artinian algebra A is called Gorenstein if its socle (0 : m) is a one-dimensional vector space over k. Let R = k{x1 , . . . , xn } be a regular local ring of Krull dimension n. Its divided   [s+t] [s] [t] power algebra is an algebra QR = kDP [X1 , . . . , Xn ] where Xi · Xi = s+t Xi ; R acts s [k ] [k −k] k  for k ≥ k, extended multilinearly. Given on QR by contraction, i.e. xi ◦ Xj = δi,j Xj an element F ∈ QR , define its annihilator ideal Ann(F ) ⊂ R as the ideal consisting of all elements of R which annihilate F . Then the quotient A = R/ Ann(F ) is a local Artinian Gorenstein algebra. The module Aˆ = R ◦ F is the Macaulay dual of A, equivalent to the Macaulay inverse system of A.4 The socle of A is Soc(A) = (0 : mA ) ⊂ A, the unique minimal non-zero ideal of A, and dimk Soc(A) = 1. Letting mA be the maximal ideal of A, we have Soc(A) = mA jA A = 0 and mA jA +1 A = 0. Then we have ([Mac, §60-63], [I, Lemma 1.1], or, in the graded case, [MS, Lemma 1.1.1]) Lemma 3.5. i. Assume that A = R/I is Artinian Gorenstein of socle degree j. Then there is a degree-j element F ∈ QR such that I = Ann F . Furthermore F is uniquely determined up to action of a differential unit u ∈ R: that is Ann F = Ann(u ◦ F ); and Ann F = Ann G ⇔ G = u ◦ F for some unit u ∈ R.

(3.3)

The R-module (Ann F )⊥ = {h ∈ QR |(Ann F ) ◦ h = 0 } satisfies (Ann F )⊥ = R ◦ F . ii. Denote by φ : Soc(A) → k a fixed non-trivial isomorphism, and define the pairing ·, ·φ on A × A by (a, b)φ = φ(ab). Then the pairing (·, ·)φ is an exact pairing on A, for which (mi )⊥ = (0 : mA i ). We have 0 : mA i = Ann(mA i ◦ F ). Also Ann(i ◦ F ) = I : i . When A = R/I is a local ring then in general F is not homogeneous. When A is (possibly non-standard) graded we will write A for A: then the dual generator F ∈ QR may be taken homogeneous, and it is unique up to non-zero scalar multiple. J. Watanabe asked whether there is a converse to Theorem 2.6 if we assume that each of A, B, C are standard graded.5 Question 3.6. Assume that the A-free extension C with fiber B is strong Lefschetz, and that A, B, C are standard graded. Can we conclude that A, B are SL? We will show that the answer to the Question 3.6 is “No” in Example 3.9. In order to show this example we prove a result about freeness of extensions C over the ring A = k[t]/(tm+1 ). Let R = k[x1 , . . . , xn ] be a graded polynomial ring and QR its divided power algebra as above. Let IB ⊂ R be a homogeneous ideal of finite colength such that the quotient B = R/IB is a graded Artinian Gorenstein algebra with socle degree jB . Let FB ∈ (QR )jB be a (homogeneous) Macaulay dual generator for B as in Lemma 3.5. Set A = k[t]/(tm+1 ), standard graded so that deg(t) = 1, and consider FA = T m , a Macaulay dual generator for A in kDP [T ]. Let S = k[x1 , . . . , xn , t] = R[t] be the (graded) polynomial ring with corresponding [s]

F.H.S. Macaulay used the notation x−s for the element Xi in QR . i Recall that T. Harima and J. Watanabe gave a counterexample to the converse of Theorem 2.6 when A is allowed to have non-standard grading [HW2, Example 6.3]. 4 5

13

divided power algebra QS = kDP [X1 , . . . , Xn , T ]. Note that we do not require S to have the standard grading, but we do require deg(t) = 1. Lemma 3.7. Let G ∈ QR be a homogeneous element of degree deg G = jB + m, and under the above assumptions for B, FB , and S consider the element F ∈ QS defined by F = T [m] · FB + G. Then F is the Macaulay dual generator of a free extension C with base A and fiber B if and only if (IB )2 ◦ G = 0. Proof. We first show ⇐, that (IB )2 ◦ G = 0 implies C is an A-free extension. Let C = S/IC , where IC = AnnS F and assume IB2 ◦ G = 0. Consider the projection map π ˆ : S → R defined by xi → xi and t → 0. Let f ∈ IC , where f=

k 

ti fi = f0 + tf1 + · · · + tk fk , with fi ∈ R.

(3.4)

0

We assume k ≥ m, as we do not require fk = 0. Since 0 = f ◦ F = T [m] f0 ◦ FB + T [m−1] f1 ◦ FB + · · · + T fm−1 ◦ FB + fm ◦ FB + f0 ◦ G, we must have f0 ◦ FB = 0, so π ˆ (f ) = f0 ∈ IB . This shows that π ˆ yields a morphism π : C → B, ˆ passes to a map which is surjective, by construction. Also, the natural inclusion ˆι : k[t] → R m+1 ι : A → C, since t ◦ F = 0. We now wish to show that ker π = (ι(A)+ )C. The inclusion (ι(A)+ )C ⊆ ker π is immediate. For the other inclusion, consider f ∈ ker π, with f = f0 + tf1 + · · · + tm fm , fi ∈ R (we do not need higher degree in t, since tm+1 ∈ IC ). We have 0 = π(f) = f 0 ∈ B, which implies f0 ∈ IB . Since (IB )2 ◦ G = 0, we have IB ◦ G ⊆ R ◦ FB , so there exists gm ∈ R such that f0 ◦ G = gm ◦ FB . Let g = tf1 + · · · + tm (fm + gm ) ∈ (t) ⊂ S. Then f − g = f0 − tm gm ∈ IC as f0 ∈ IB , and (f − g) ◦ F = (f0 − tm gm ) ◦ (T [m] FB + G) = T [m] (f0 ◦ FB ) + f0 ◦ G − gm ◦ FB = 0. Hence, as an element of C, we have f ∈ ker π implies f ∈ (ι(A)+ )C, hence the sequence /A ι /C π /B / k is coexact at C; the coexactness at A and B are obvious. k To complete the proof of ⇐ we need to show Claim. We have equality dimk C = dimk (A) · dimk (B) = (m + 1) · dimk (B). This implies that C is a free A module via the inclusion ι : A → C. Proof of Claim. Since C ⊃ tC ⊃ t2 C · · · ⊃ tm C ⊂ tm+1 C = 0 is a filtration of C, we have i i+1 that as vector space C ∼ C). We have shown above that π is surjective with = k ⊕m i=0 t C/(t ker π = tC, and that f = f0 + tf1 + · · · + tm fm satisfies f ∈ ker π implies f0 ∈ IB . Thus we have C/ ker π ∼ = B. Assume by way of induction that for an integer i ∈ [0, m − 1] we have = C/tC ∼ i i+1 ∼ t C/t C = B, and consider the homomorphism mt induced by the multiplication c → t · c mt : ti C/ti+1 C → ti+1 C/ti+2 C. 14

(3.5)

Evidently mt is surjective. We now show mt is injective. Suppose that for an α ∈ ti C/(ti+1 C) we have mt (α) = 0; let α = ti c0 ∈ ti C be a representative of α. Then there exists c1 ∈ C for which ti+1 c = ti+2 c1 ; in particular then we must have tm c0 = 0, which implies that 0 = (tm c0 ) ◦ F = c0 ◦ FB , hence c0 ∈ IB , implying ti c0 ∈ ti+1 C implying α = 0. Hence mt is an isomorphism. This completes the induction step. We have shown equation (3.5) for i = 0, thus, by induction we have dimk ti C/ti+1 C = dimk B for 0 ≤ i ≤ m, implying that dimk C = (m + 1) dimk (B). This completes the proof of the claim, and the proof that (IB )2 ◦ G = 0 implies that C is a free extension of A with fiber B. We prove ⇒. Assume that C is an A-free extension. Then ker π = (ι(A)+ )C. Here know i f = 0 t fi ∈ ker π ⇔ f0 ∈ IB so IB ⊂ tC. So there is h ∈ C so (f0 − th) ◦ F = 0, but (f0 − th) ◦ F = −th ◦ (T [m] FB ) + f0 ◦ G, so we have the highest degree term tm−1 h0 of h satisfies h0 ◦ FB = f0 ◦ G, so IB ◦ (f0 ◦ G) = IB ◦ (h0 ◦ FB ) = 0. Since this occurs for all f0 ∈ IB we have (IB )2 ◦ G = 0. This completes the proof of the Theorem.6  Remark 3.8. Lemma 3.7 might also be derived from results appearing in the papers of J. Elias and M. E. Rossi [ElRo] or J. Jelisiejew [Je]. For example in terms of the paper [ElRo], the sequence Hi = T [i] FB for 0 ≤ i ≤ m − 1 and Hm = T [m] FB + G forms a “truncated” G1 admissible system generating an S-submodule of QS which one could then show corresponds to a flat extension over A = k[t]/(tm+1 ) using an argument similar to the proof of their [ElRo, Theorem 3.8]. We thank the referee for pointing out this connection. We below will use the notation m ⊗ n to denote the generic Jordan type (m + n − 1, m + n − 3, . . .) of k[x, y]/(xm , y n ). Example 3.9 (C is an A-free extension, C is SL, but B is not SL). Let k be an infinite field of characteristic zero or characteristic p ≥ 5. Take R = k[x, y, z, u, v], S = R[t], let A = k[t]/(t2 ) of Hilbert function H(A) = (1, 1) and let B = R/IB , IB = Ann FB , FB = [2] (XU + Y U V + ZV [2] ), an idealization of k[u, v]/(u, v)2 . Then H(B) = (1, 5, 5, 1), IB =   2 (x, y, z) , uy −vz, ux−vy, uz, vx , and it is straightforward to see that the Jordan type of B is JB = (4, 2, 2, 2, 1, 1), so B is not strong Lefschetz, and, since it has symmetric Hilbert function, also cannot have an element of strong Lefschetz Jordan type.7 Let C = S/I, I = Ann F where F = T FB + G, G = X [2] U V + XY V [2] . It is straightforward to verify that (IB )2 ◦ G = 0, so C is an A-free extension with fiber B by Lemma 3.7, and we have H(C) = (1, 6, 10, 6, 1). A calculation in Macaulay2 shows that the (usual) Hessian of F is non-zero, hence for a generic linear form  ∈ S we have the multiplication 2 : C1 → C3 is an isomorphism. Evidently, for a generic , we have 4 = 0.8 Therefore, PC = (5, 35 , 14 ), so C is strong Lefschetz. As an example, we can check that for  = x + u + v, we have 2 x ◦ F = 2X + 2U + 2V 2 z ◦ F = T 2 v ◦ F = 2X + 2Y + U

2 y ◦ F = 2V + 2T 2 u ◦ F = 2X + V + 2T 2 t ◦ F = X + 2Y + Z + 2U

6

This result is generalized in [IMM2, Theorem 2.1]. The cubic defining B was studied by U. Perazzo in 1900, see [Ru, Example 7.5.1, Theorem 7.6.8]. 8 in general, for any standard graded Artinian algebra S of socle degree j and any sufficiently general linear form  we always have j = 0 in S, as the powers of linear forms span Sj when char k = 0 or is greater than j. 7

15

and since these are linearly independent in the dual space, so are 2 x, 2 y, 2 z, 2 u, 2 v, 2 t linearly independent in C3 . Therefore ×2 : C1 → C3 is an isomorphism. Furthermore, 4 = 12x2 uv = 0: these verify that  is a SL element for C. Note that here the Jordan type of A ⊗ B is 2 ⊗ (4 + 23 + 12 ) = 2 ⊗ 4 + 3(2 ⊗ 2) + 2(2 ⊗ 1) = (5, 3) ∪ 3(3, 1) ∪ (2, 2) = (5, 34 , 22 , 13 ), thus PC > PA⊗B and PC covers PA⊗B in the dominance partial order. Acknowledgment. The first and third author appreciate conversations with participants of the informal work group on Jordan type, and others at the workshop “Lefschetz Properties in Geometry, Algebra, and Combinatorics” at Institute Mittag Leffler in July, 2017. We appreciate comments and answers to our questions by Larry Smith, Junzo Watanabe and Alexandra Seceleanu. We thank Oana Veliche, Ivan Martino, Jerzy Weyman, Emre Sen, and Shujian Chen for their comments. We thank the referee for cogent and helpful comments including an improved statement and proof for Theorem 2.1. The second author was partially supported by CIMA – Centro de Investiga¸ca˜o em Mate´ m´atica e Aplica¸co˜es, Universidade de Evora, project UID/MAT/04674/2019 (Funda¸ca˜o para a Ciˆencia e Tecnologia).

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