# The Cohen-Macaulay property of

separating invariants of finite groups

###### Abstract.

In the case of finite groups, a separating algebra is a subalgebra of the ring of invariants which separates the orbits. Although separating algebras are often better behaved than the ring of invariants, we show that many of the criteria which imply the ring of invariants is non Cohen-Macaulay actually imply that no graded separating algebra is Cohen-Macaulay. For example, we show that, over a field of positive characteristic , given sufficiently many copies of a faithful modular representation, no graded separating algebra is Cohen-Macaulay. Furthermore, we show that, for a -group, the existence of a Cohen-Macaulay graded separating algebra implies the group is generated by bireflections. Additionally, we give an example which shows that Cohen-Macaulay separating algebras can occur when the ring of invariants is not Cohen-Macaulay.

## 1. Introduction

Let be a finite group, and let be a finite dimensional representation of over a field of characteristic . We say is a modular representation if divides . We write for the symmetric algebra on the vector space dual of . It is a polynomial ring with the standard grading. The action of on induces an action on : for and , the action of is given by . The ring of invariants, denoted , is the ring formed by the elements of fixed by . Since the group action preserves degree, is a graded subalgebra of .

Let be an algebraic closure of the field , and let . As , any in can be considered as a function . The action of on extends to an action of on , and so .

By definition, elements of are constant on -orbits. Accordingly, if an invariant takes distinct values on elements , then these elements belong to distinct orbits, and we say
*separates* and . A *geometric separating set* is then a set of invariants which separates exactly the same points of as the whole ring of invariants. As is finite, the ring of invariants separates orbits in [11, Lemma 2.1]. Hence, a geometric separating set is a set of invariants which separates the orbits of the -action on .

The study of separating invariants, which has become quite popular in recent
years, was initiated by Derksen and Kemper ([8, Section 2.3]
and [24]). They defined a *separating set* as a set separating the same orbits in as the whole ring of invariants [8, Definition 2.3.8].
For algebraically closed, geometric separating sets coincide with separating sets, but in general, a separating set is not always a geometric separating set (see Example 2.5). On the other hand, is a geometric separating algebra if and only if is a separating algebra. Moreover, if is graded, so is , and the two rings have the same dimension and depth. Furthermore, the extension is integral if and only if is integral. Thus, it often suffices to write proofs for separating sets over algebraically closed fields. Note that this works only because we are interested in geometric separating algebras.

Many defects of invariant rings disappear when one considers separating invariants. The ring of invariants is not always finitely generated (for non-reductive groups) [8, Example 2.1.4, due to Nagata], but there always exist finite geometric separating sets [8, Theorem 2.3.15]. Over algebraically closed fields, there is an upper bound on the size of minimal separating sets, depending only on the dimension of the representation [12, Proposition 5.1.1]. Polarization, a classical method for obtaining vector invariants in characteristic zero, extends, for separating invariants, to all characteristics [11, 10]. For finite, the Noether bound holds for separating invariants in all characteristics: although they may not generate the ring of invariants, the invariants of degree at most always form a geometric separating set [8, Section 3.9].

Graded separating algebras are very closely related to the ring of invariants:

###### Proposition 1.1.

If is a graded geometric separating algebra, then is an integral extension, and is a finitely generated -algebra.

###### Proposition 1.2.

Suppose . If is a graded subalgebra, then is a geometric separating algebra if and only if is the purely inseparable closure of in , that is,

###### Proof.

For algebraically closed, see [9, Remark 1.3]. The proof is an application of a result of van der Kallen [28, Sublemma A.5.1] (see the extended proof in [27]). A variation of the same argument first appeared in [19, Theorem 6]. For and arbitrary, we have that for some . The stability of the rank of matrices under field extensions implies .

Kemper [24] exploited this close relationship to compute the invariants of reductive groups in positive characteristic. On the other hand, Dufresne [13] showed that the existence of polynomial or complete intersection separating algebras imposes strong conditions on the representation. The present paper is in the latter vein. We show that, in many instances, conditions which ensure that the ring of invariants is non Cohen-Macaulay, in fact imply that no graded geometric separating algebra is Cohen-Macaulay. We thus provide a (partial) negative answer to Kemper who asked if Cohen-Macaulay separating algebras should always exist [25]. Notably, we show:

###### Theorem 1.4.

If is faithful and modular, then there exists such that, for all , every graded geometric separating algebra in has Cohen-Macaulay defect at least . In particular, for , no graded geometric separating algebra in is Cohen-Macaulay.

An element of acts as a
*bireflection* on if its fixed space is of codimension at most in .

###### Theorem 1.5.

Let be a -group. If there exists a graded geometric separating algebra in which is Cohen-Macaulay, then is generated by elements acting as bireflections.

Theorem 1.5 fits well with [13, Theorem 1.3]. In the important special case of -groups, we obtain that is generated by bireflections from a much weaker hypothesis: the existence of a Cohen-Macaulay rather than a complete intersection graded geometric separating algebra. This mirrors the situation for invariant rings ([22, Corollary 3.7] and [21, Theorem A]). Example 2.10 shows that not only the converse of Theorem 1.5, but also the converses of [13, Theorem 1.1 and Theorem 1.3], are not true.

In Section 2, we extend the methods introduced in [22] to prove our main results. Section 3 concentrates on the alternating group . We conclude in Section 4 with a discussion of the general situation and examples which show that the depth of graded geometric separating algebras can be both larger and smaller than that of the corresponding invariant ring.

###### Acknowledgements. ()

This paper was prepared during visits of the first and second authors to TU München, and of the second and third authors to Universität Heidelberg. We thank Gregor Kemper, B. Heinrich Matzat, and MATCH for making these visits possible. Finally, we thank the anonymous referees for helpful suggestions, in particular for pointing out an error in our original proof of Lemma 2.2.

## 2. The Cohen-Macaulay Defect of Separating Algebras

Let be a finitely generated graded subalgebra, and let
denote its maximal homogeneous ideal. Homogeneous elements
in form a *partial homogeneous system of
parameters (phsop)* if they generate an ideal of height in . If
additionally , then they form a *homogeneous system of
parameters (hsop)*. Noether’s normalization theorem guarantees that a hsop
always exists. If, for , the element is not a zero
divisor in , then the elements
form a *regular sequence*. Every regular sequence is a phsop. We say
is *Cohen-Macaulay* when every phsop is a regular sequence. The depth of a
homogeneous ideal , written , is the maximal length of a
regular sequence in . Note that the height of , , is equal to the maximal length of a phsop in . We write , and define the *Cohen-Macaulay defect* of to be . Thus, is Cohen-Macaulay precisely when .

In Theorem 2.1, we relate the Cohen-Macaulay defect of graded geometric separating algebras to the -th cohomology group . For the theory of these groups for arbitrary , we refer to [2, 29]. Since we have in most applications of this theorem, we construct this group now.

Let be a representation of the group over the field . A
*1-cocycle* is a map
such that , for all
. We write for the additive group of all
1-cocycles. For each , the map given by is a 1-cocycle, which is called a *1-coboundary*. The coboundaries form the subgroup
of . The *first cohomology group of with
coefficients in * is the quotient group
. A cocycle is *nontrivial* if
and only if its *cohomology class* is nonzero in
. Similar definitions hold for . We will sometimes abuse
notation by using the same symbols for cocycles and cohomology classes.

The group has a natural graded -module structure. For a homogeneous , its annihilator in ,

is a homogeneous ideal. If , the -fold Frobenius homomorphism , induces a map . This map is a homomorphism of abelian groups, but not of -vector spaces. We write for the image of an element under this map. In particular, for , the cohomology class is given by the cocycle .

Over fields of characteristic zero, and in the non-modular case in general, the ring of invariants is always Cohen-Macaulay [20]. In particular, there will always be a Cohen-Macaulay geometric separating algebra. Accordingly, from now on we assume is modular.

Our most general statement generalizes [22, Corollary 1.6] (see also [23, Proposition 6]) to the case of separating algebras:

###### Theorem 2.1.

Let be the smallest integer such that there exists a homogeneous such that is nonzero for every . If is a graded geometric separating algebra in , then , the annihilation ideal of in , has depth at most . Furthermore, if has height , then has Cohen-Macaulay defect at least .

Since the case suffices for most of our applications, we give an additional more elementary proof of the first part. Without loss of generality, we assume in both arguments.

###### Proof of the case ..

By Proposition 1.2, there exists a -power such that . Suppose, for a contradiction, that is at least . Hence, there exists an -regular sequence in . Since , there are such that

Set , for . For all , is invariant, and so belongs to . Since forms an -regular sequence [14, Corollary 17.8 (a)], and since

it follows that . Thus, there exist such that

As is a phsop in , it is also a phsop in its integral extension , and thus are coprime in . From , it follows that divides in . Therefore, there exists such that . Hence, for every , we have

that is, , a contradiction since is nonzero. Thus, .

###### Proof of the general case..

For some , we have . For each , is finitely generated as a -module. Therefore, for each , there exists some such that for all . Set , where is the maximum of . Assume, for a contradiction, that , and let be an -regular sequence in . Consider the bar resolution of as a -module:

where , and is the augmentation map. The cohomology class is represented by a such that , that is, (the notation will become clear later). For each , , so there is such that . We next define, for each , and for each ordered -subset of , a homomorphism such that , that is, . For , the definition of implies , so there is a map satisfying . We now define , and thus , by induction on :

(1) |

Next, we show that for . For , we get

For , we obtain similarly:

(2) |

since the middle term is

which equals zero. When , . It follows that , which implies ( is the neutral element). Therefore, for each , we have . The second equality in (2) is also valid for (), that is,

As is -regular, . Thus there exist (in particular, ) such that . Substituting (1) for yields

Since is a phsop in , it is also a phsop in , and so . It follows that

for some . We next apply to this expression. For , we have , or alternatively , a contradiction to . Now assume that . Applying leads to

(3) |

For , we prove by reverse induction that there exist elements such that

(4) |

The case is covered by (3). Suppose for some . Using (1), we obtain

and rearranging yields

Since is a phsop for , we have

for some . Here we have used that is a free -module. Applying to this expression gives us

which implies (4), as required.

When , Equation (4) reads , where . Substituting (1) for , we obtain . For the same reasons as before, we have for some . Applying gives us . Hence, , a contradiction to . Therefore, .

Finally, if forms a phsop in , then forms a phsop in , and so has height at least . The graded analogue of [5, Exercise 1.2.23] implies the Cohen-Macaulay defect of is at least .

###### Lemma 2.2.

If is nonzero, then is nonzero for all .

###### Proof.

For , this is clear since elements of are group homomorphisms . For arbitrary , by the Universal Coefficient Theorem [18, page 30], . We have for some and . Without loss of generality, we can assume that the set is -linearly independent and is nonzero for all . Then , since the -fold Frobenius homomorphism induces the identity map on , and thus also on . Therefore, as is still -linearly independent, is still nonzero.

###### Remark 2.3.

For , Theorem 2.1 is new even in the case .

###### Example 2.4.

Let be a finite nontrivial subgroup. Consider the threefold sum of the -dimensional representation of over given by and for the induced -action. The map yields a nonzero element in . For all , we have , that is, is trivial. Since form a phsop in , Theorem 2.1 implies that no graded geometric separating algebra in is Cohen-Macaulay. . Write

The following example shows that Theorem 2.1 applies only to
graded *geometric* separating algebras:

###### Example 2.5.

Let be the permutation representation of the cyclic group of order over the field . Consider the -invariants , , , and . The action of on partitions its elements into orbits, which one can check are separated by , , , and . As form a hsop in , the subalgebra is a polynomial graded (non geometric) separating algebra. In particular, it is Cohen-Macaulay.

On the other hand, if is the nontrivial cocycle given by , then , , and , that is, . Since form a hsop in , by Theorem 2.1, no graded geometric separating algebra in is Cohen-Macaulay.

Using Theorem 2.1, we can quickly generalize several results which were consequences of its analogue [22, Corollary 1.6]. A first consequence is the following generalization of [6, Corollary 21]:

###### Corollary 2.6.

Assume contains a normal subgroup of index (for example, is a -group). Then for any faithful representation , every graded geometric separating algebra in has Cohen-Macaulay defect at least .

###### Proof.

Since , there is a nonzero element in . As is faithful, the fixed subspaces of nonidentity elements in have codimension at least in . The result follows from Lemma 2.7.

###### Lemma 2.7.

Suppose is faithful. If the fixed subspace of every element of order in has codimension at least in , then for any homogeneous , the ideal has height at least .

Therefore, if there exists a homogeneous satisfying the hypotheses of Theorem 2.1, then every graded geometric separating algebra in has Cohen-Macaulay defect at least .

###### Proof.

We may assume . By Kemper [22, Lemma 2.1] there exist (in general, non-homogeneous) elements such that , for all , and . It follows that has height at least .

###### Proof of Theorem 1.4.

Next, we generalize three results of [22]. Note that since, for us, elements acting trivially are bireflections, we do not need to assume that is faithful.

###### Corollary 2.8.

If has a normal subgroup of index which contains all elements acting as bireflections on , then no graded geometric separating algebra in is Cohen-Macaulay.

###### Proof of Theorem 1.5.

For -groups, if the elements acting as bireflections generate a proper subgroup, then this subgroup lies in a normal subgroup of index .

###### Proposition 2.9.

Suppose has a normal subgroup with factor group an elementary abelian -group. If there exists whose fixed space in is not contained in the fixed space of any bireflection in , then no graded geometric separating algebra in is Cohen-Macaulay.

###### Proof.

Without loss of generality, assume . As is an elementary abelian -group, there is a with kernel . The proof of [22, Theorem 3.9] provides a phsop in , and a homogeneous invariant such that , for . Hence, for some , the invariants annihilate . Thus, the annihilator ideal has height at least . By Theorem 2.1, it now suffices to show that is nonzero for all . By [22, Proposition 3.5], we have . Therefore, if is zero, then annihilates , and so , a contradiction.

###### Example 2.10.

Let be a finite field. For , set , and consider the group formed by the matrices of the form

where , and denotes the identity matrix. The group is a -group, and is generated by reflections, that is, by elements whose fixed space has codimension at most 1 in . Example 3.10 in [22] shows that the hypotheses of Proposition 2.9 are satisfied, with the subgroup formed by the elements such that , and the element such that for all . Hence, no graded geometric separating algebra in is Cohen-Macaulay.

We end this section with a generalization of [22, Theorem 2.7].

###### Theorem 2.11.

Let be the regular representation of over . If divides , then every graded geometric separating algebra in has Cohen-Macaulay defect at least . For , this number is at least one.

The regular representation of was studied in Example 2.5. For , and , the invariant ring is Cohen-Macaulay [22, Theorem 2.7]. Thus, these are the only groups such that there exists a Cohen-Macaulay graded geometric separating algebra in .

###### Lemma 2.12.

Let be a permutation representation of . If in is nonzero, then is nonzero for all . If, in addition, is faithful, then every graded geometric separating algebra in has Cohen-Macaulay defect at least .

###### Proof.

As is a permutation representation, there is a set of monomials such that . Thus, if , there is a (finite) decomposition , where each is in . For , we have the decomposition , where is in , and is the unique element of . If is nonzero, then is nonzero for some . As and are isomorphic permutation representations of , the element is also nonzero. The additional statement follows by Lemma 2.7.