On the Depth of Cohomology Modules^{∗}

Peter Fleischmann, Gregor Kemper, and R. James Shank

April 14, 2003

Abstract

We study the cohomology modules H^{i}(G, R) of a p-group G acting on a ring R of characteristic p, for i > 0. In particular, we are interested in the Cohen-Macaulay property and the depth of H^{i}(G, R) regarded as an R^{G}-module. We first determine the support of H^{i}(G, R), which turns out to be independent of i. Then we study the Cohen-Macaulay property for H^{1}(G, R). Further results are restricted to the special case that G is cyclic and R is the symmetric algebra of a vector space on which G acts. We determine the depth of H^{i}(G, R) for i odd and obtain results in certain cases for i even. Along the way, we determine the degrees in which the transfer map Tr^{G}: R → R^{G} has non-zero image.

Introduction

Modular invariant theory is the study of invariant rings R^{G}, where G is a finite group acting on a ring R such that the order of G is not invertible in R. The standard situation is the special case where R is the ring of polynomials on a vector space V over a field K with a linear G-action. More precisely, R = S(V^{∗}) is the symmetric algebra of the dual of V. The situation becomes modular if the characteristic of K divides |G|. Since R^{G} coincides with the zeroth cohomology H^{0}(G, R), it is only natural to consider higher cohomology modules, H^{i}(G, R), as well. Each cohomology module is an R^{G}-module, since multiplication with an invariant gives a G-equivariant mapping R → R which induces an endomorphism of H^{i}(G, R). Considering H^{i}(G, R) as an R^{G}-module raises natural questions concerning the Cohen-Macaulay property, the support, and the depth (in the case where R = S(V^{∗}), so we have a grading). Apart from being interesting in themselves, the cohomology modules H^{i}(G, R) have proven to be a sharp tool in the study of the Cohen-Macaulay property and depth of the invariant ring R^{G}. Various authors [6, 9, 12, 13, 14, 15] have used this approach.

Not much is known about the structure of H^{i}(G, R) as an R^{G}-module. Ellingsrud and Skjelbred [6] showed that H^{i} (G, S(V^{∗})) is Cohen-Macaulay if G is cyclic of order p. This result was generalized by Kemper [14, Example 2.14] to all groups G such that p^{2} ∤ |G|. So the first question to ask is: What happens if we consider the group G = Z/p^{2} or, more generally, any cyclic p-group of order greater than p? The answer is that the Cohen-Macaulay property very often fails in these cases (see Theorem 3.1). This observation was the starting point of our investigations, which subsequently included larger classes of groups and actions on non-polynomial rings.

∗This research was supported by EPSRC grant GR/R32055/01.

2000 Mathematics Subject Classification: 20J06, 13A50, 13C14, 13C15.

In the first section of this paper we determine the support of H^{+}(G, R), the positive part of the cohomology, where R is a ring of positive characteristic p. It turns out that this support coincides with the wild branch locus of Spec(R) → Spec(R^{G}) (see Theorem 1.4). This result was obtained by Kemper [14, Theorem 2.10] for a special case. In the second section we restrict our attention to the case of a p-group acting on a commutative ring R of characteristic p. We use localization techniques to show that for i > 0 the support of H^{i}(G, R) does not depend on i. Then we study the first cohomology H^{1}(G, R) and obtain a rather restrictive necessary condition for this module to be Cohen-Macaulay. In the case where R = S(V^{∗}) with V a faithful KG-module, this condition implies that G is generated by elements of order p. In Section 3 we consider the case where G = Z/p^{k+1} is cyclic and R = S(V^{∗}) with V a KG-module. This assumption renders the cohomology modules much more tractable. As a first result we establish that for odd i the depth of H^{i}(G, R) is equal to the dimension of the fixed space V^{G}. The situation turns out to be much more difficult for i even. We first consider the case where V is a free KG-module. In that context we prove a general result, Theorem 3.3, about the image of the transfer for a p-group acting by permutations, and as a corollary we determine the depth of H^{i} (G, S(V^{∗})) for i even and V free. For the further study of H^{i}(G, R) for i even we have to consider the transfer in greater detail. In Section 3.3 we determine the degrees where the transfer Tr^{G}: R → R^{G} has non-zero image (all in the situation G = Z/p^{k+1} and R = S(V^{∗})). Since the transfer map is of utmost importance in modular invariant theory, this should be of some interest in itself. Using the results from Section 3.3, we obtain information about the depth of H^{i}(G, R) in some special cases.

It is clear that our results are far from complete. In fact, we consider this paper to be the beginning of an investigation, which we hope will lead to further developments.

Acknowledgement. Most of this research was done during a stay of the second author at the University of Kent at Canterbury as an EPSRC Visiting Fellow. We thank the EPSRC for funding this visit.

1 The support of cohomology

In this section R is a commutative ring (with unity) of characteristic a prime p, and G is an arbitrary finite group acting on R by ring automorphisms. We first determine the support of the positive cohomology H^{+}(G, R) := ⊕_{i≥1} H^{i}(G, R).

We need two lemmas which are well-known, but we include proofs for completeness.

Lemma 1.1. Let A and B be rings (with unity but not necessarily commutative), let U be a left A-module, V a left A, right B, bimodule and W a left B-module. If U is free of finite rank, then there is a natural isomorphism

Hom_{A}(U, V ⊗_{B} W) ≅ Hom_{A}(U, V) ⊗_{B} W.

Proof. If U is isomorphic to the left regular module _{A}A, then Hom_{A}(U, X) ≅ X for every left A-module X, verifying the lemma immediately. The general case now follows from the additivity of the functors Hom_{A}(·, V ⊗_{B} W) and Hom_{A}(·, V) ⊗_{B} W.

Lemma 1.2. Let A be a commutative ring, G a finite group and M a module over the group ring AG. Furthermore, let S ⊂ A \ {0} be a multiplicative set. Then for every i ≥ 0

H^{i}(G, S^{−1}M) ≅ S^{−1}H^{i}(G, M).

Proof. Choose a resolution F^{∗} → Z of Z as a ZG-module by free ZG-modules F^{i} of finite rank, e.g. the bar resolution. Let Hom_{ZG}(F^{∗}, M) denote the complex with Hom_{ZG}(F^{i}, M) as i-th part. Hence H^{∗}(G, M) = H (Hom_{ZG}(F^{∗}, M)), where the right hand side is the homology of the complex. By Lemma 1.1 we have

Hom_{ZG}(F^{∗}, S^{−1}M) = Hom_{ZG}(F^{∗}, M ⊗_{A} S^{−1}A) ≅ Hom_{ZG}(F^{∗}, M) ⊗_{A} S^{−1}A

and therefore

H^{∗}(G, S^{−1}M) ≅ H (Hom_{ZG}(F^{∗}, M) ⊗_{A} S^{−1}A) ≅ H (Hom_{ZG}(F^{∗}, M)) ⊗_{A} S^{−1}A. (1)

The second isomorphism results from the facts that S^{−1}A is a flat A-module (see Eisenbud [5, Proposition 2.5]) and that application of an exact functor and forming homology commute. The claim follows from (1).

For each subgroup H ≤ G, the abelian group H^{i}(H, R) is an R^{H}-module and hence an R^{G}-module. We define H^{+ev}(G, R) = ⊕_{i≥1} H^{2i}(G, R). For a prime ideal 𝔓 ∈ Spec(R) define the inertia group G_{𝔓} as

G_{𝔓} := {g ∈ G | g(f) − f ∈ 𝔓 for all f ∈ R}.

Lemma 1.3. (a) Supp_{R^{G}}(Im res_{G→H} (H^{+ev}(G, R))) ⊆ Supp_{R^{G}} (H^{+ev}(G, R)).

(b) For 𝔭 ∈ Spec(R^{G}), suppose 𝔓 ∈ Spec(R) with 𝔓 ∩ R^{G} = 𝔭. Then

𝔭 ∈ Supp_{R^{G}} (Im res_{G→⟨g⟩} (H^{+ev}(G, R)))

for each g ∈ G_{𝔓} with |g| = p.

Proof. (a) Suppose 𝔭 ∉ RHS, then by exactness of localization:

Im res_{G→H} (H^{+ev}(G, R))_{𝔭} = Im res_{G→H} (H^{+ev}(G, R)_{𝔭}) = 0,

so 𝔭 ∉ LHS.

(b) Suppose g ∈ G_{𝔓} with |g| = p, H := ⟨g⟩, α ∈ H^{2i}(G, 𝔽_{p}) with i > 0, such that 0 ≠ β := res_{G→H}(α) (which exists by Benson [2, Theorem 4.1.3 and its proof]) and let β̃ denote the image of β under the map H^{∗}(H, 𝔽_{p}) → H^{∗}(H, R), induced by the inclusion 𝔽_{p} ↪ R. Then β̃ ∈ Im res_{G→H} (H^{+ev}(G, R)). Let f ∈ R^{G} with fβ̃ = 0. We have H^{2i}(H, R) ≅ R^{H}/(g − 1)^{p−1}R. The induced map

H^{2}(H, 𝔽_{p}) ≅ 𝔽_{p} = 𝔽_{p}^{H}/(g − 1)^{p−1}𝔽_{p} → R^{H}/(g − 1)^{p−1}R ≅ H^{2i}(H, R)

takes β = λ ∈ 𝔽_{p} to β̃ = λ+(g−1)^{p−1}R. Since λ ∈ 𝔽_{p}\{0} we have λ ∉ 𝔓 (as 𝔓 is a proper ideal). Since g ∈ G_{𝔓}, (g − 1)^{p−1}R ⊆ 𝔓, hence fλ ∈ 𝔓. It follows that f ∈ 𝔓 ∩ R^{G} = 𝔭. We have shown that every element from R^{G} which sends β̃ to zero lies in 𝔭. This means that β̃ remains non-zero in the localization H^{+}(G, R)_{𝔭}, so

𝔭 ∈ Supp_{R^{G}} (Im res_{G→H} (H^{+ev}(G, R))).

Theorem 1.4.

Supp_{R^{G}}(H^{+ev}(G, R)) = Supp_{R^{G}}(H^{+}(G, R)) = V(Ann_{R^{G}}(H^{+}(G, R)))
= V(Tr^{G}(R)) = {𝔓 ∩ R^{G} | 𝔓 ∈ Spec(R), |G_{𝔓}| ≡ 0 mod p}.

Proof. The last equality is Lorenz and Pathak [15, Lemma 1.1]. To prove the remaining equalities, we first show V(Ann_{R^{G}}(H^{+}(G, R))) ⊆ V(Tr^{G}(R)): Let 𝔭 ∉ V(Tr^{G}(R)), then for some r ∈ R, f := Tr^{G}(r) ∉ 𝔭. By Lorenz and Pathak [15, Lemma 1.3], f annihilates ker res_{G→1}(H^{∗}(G, R)) = H^{+}(G, R). Hence 𝔭 ∉ V(Ann_{R^{G}}(H^{+}(G, R))).

Now assume 𝔭 ∉ Supp_{R^{G}}(H^{+ev}(G, R)); then 𝔭 ∉ Supp_{R^{G}} (Im res_{G→H} (H^{+ev}(G, R))) for each H ≤ G. From Lemma 1.3 (b), we see that G_{𝔓} has no elements of order p. Thus

V(Ann_{R^{G}}(H^{+}(G, R))) ⊆ V(Tr^{G}(R)) ⊆ Supp_{R^{G}}(H^{+ev}(G, R)) ⊆ Supp_{R^{G}}(H^{+}(G, R))

and the theorem follows from the standard inclusion

Supp_{R^{G}}(H^{+}(G, R)) ⊆ V(Ann_{R^{G}}(H^{+}(G, R))).

Corollary 1.5.

\sqrt{Ann_{R^{G}}(H^{+}(G, R))} = \sqrt{Tr^{G}(R)}.

2 Cohomology modules for p-groups

In this section R is again a commutative ring of characteristic p, but now G is a finite pgroup acting on R by ring automorphisms. Using Corollary 1.5, we will first determine the support of the cohomology modules H^{i}(G, R). This implies knowledge about the dimension of H^{i}(G, R). Then we derive rather restrictive necessary conditions for H^{1}(G, R) to be Cohen-Macaulay.

Lemma 2.1. With the notation introduced at the beginning of Section 2 we have

Supp_{R^{G}} (H^{i}(G, R)) = Supp_{R^{G}} (H^{j}(G, R))

for all positive integers i and j.

Proof. Take 𝔭 ∈ Spec(R^{G}). Using Lemma 1.2 we obtain

𝔭 ∈ Supp_{R^{G}} (H^{i}(G, R)) ⇔ H^{i}(G, R_{𝔭}) ≠ 0.

Since G is a p-group, H^{i}(G, R_{𝔭}) ≠ 0 is equivalent to H^{j}(G, R_{𝔭}) ≠ 0 (see Brown [3, Theorem VI.8.5] or Benson [1, Proposition 3.14.4]). The result follows.

Now Corollary 1.5 and Lemma 2.1 determine the support of the cohomology for a pgroup, in fact the following theorem tells us that the support of H^{i}(G, R) is equal to the image of the branch locus under the categorical quotient, independent of i.

Theorem 2.2. For every positive integer i we have

Supp_{R^{G}} (H^{i}(G, R)) = Supp_{R^{G}} (H^{+}(G, R)) = {R^{G} ∩ 𝔓 | 𝔓 ∈ Spec(R), G_{𝔓} ≠ {1}}.

As a special case we obtain:

Corollary 2.3. Let G be a finite p-group acting linearly on a finite-dimensional vector space V over a field K of characteristic p. Then

dim (H^{i} (G, S(V^{∗}))) = max {dim_{K}(V^{σ}) | σ ∈ G \ {1}} .

We obtain the following, somewhat technical, necessary condition for H^{1}(G, R) to be Cohen-Macaulay.

Theorem 2.4. Assume that R^{G} is Noetherian and R is finitely generated as a module over R^{G}. Define H := ⟨G_{𝔓} | 𝔓 ∈ Spec(R)⟩ and

H_{min} := ⟨G_{𝔓} | 𝔓 ∈ Spec(R) is minimal with G_{𝔓} ≠ {1}⟩ .

If H^{1}(G, R) is Cohen-Macaulay as an R^{G}-module, then H ⊆ H_{min} · Φ(G), where Φ(G) is the Frattini subgroup.

Proof. First observe that H^{1}(G, R) is the first homology of a complex of finitely generated R^{G}-modules, and is therefore itself finitely generated over R^{G}.

Assume that H ⊈ H_{min} · Φ(G). This implies H · Φ(G)/H_{min} · Φ(G) ≠ {1}. But this factor group is a subgroup of G/H_{min} · Φ(G), which is elementary abelian since G is a p-group (see Huppert [11, Chapt. III, Satz 3.14(a)]). It follows that there exists a maximal subgroup of G/H_{min} · Φ(G) which does not contain H · Φ(G)/H_{min} · Φ(G). This means that we have a normal subgroup N ⊴ G of index p such that H_{min} ⊆ N but H ⊈ N. Thus there exists a 𝔔 ∈ Spec(R) such that G_{𝔔} ⊈ N. Choose σ ∈ G_{𝔔} \ N and write σ̄ := σN ∈ G/N. Then G/N = ⟨σ̄⟩. Consider the class α ∈ H^{1}(G/N, R^{N}) = ker(Tr_{N}^{G})/(σ̄ − 1)R^{N} given by α = 1 + (σ̄ − 1)R^{N}. The fact that σ lies in G_{𝔔} implies (σ̄ − 1)R^{N} ⊆ 𝔔 and therefore α ≠ 0.

Let β := inf(α) be the image of α under the inflation map inf : H^{1}(G/N, R^{N}) → H^{1}(G, R). Then β ≠ 0 (see Evens [7, Corollary 7.2.3]). Consider the ideal I := Tr_{N}^{G}(R^{N}) ⊆ R^{G}. We have I = (σ̄−1)^{p−1}R^{N} and therefore I ⊆ Ann_{R^{G}}(α). It follows that I ⊆ Ann_{R^{G}}(β), which implies

grade (I, H^{1}(G, R)) ∈ {0, ∞}  (2)

(using the convention of Bruns and Herzog [4, Definition 1.2.6] that grade(I, M) = ∞ if IM = M). On the other hand, Bruns and Herzog [4, Proposition 1.2.10(a)] (which is applicable since R^{G} is Noetherian and H^{1}(G, R) is finitely generated over R^{G}) yields

grade (I, H^{1}(G, R)) = min {depth (H^{1}(G, R)_{𝔭}) | 𝔭 ∈ Spec(R^{G}), I ⊆ 𝔭} .  (3)

The hypothesis that H^{1}(G, R) is Cohen-Macaulay implies that

depth (H^{1}(G, R)_{𝔭}) = dim (H^{1}(G, R)_{𝔭})

for every 𝔭 ∈ Spec(R^{G}) (see Bruns and Herzog [4, Theorem 2.1.3(b)]), where we use the convention dim(0) = ∞. Thus in the right hand side of (3) we can substitute the depth by the dimension. Now it follows from (2) that there exists 𝔭 ∈ Spec(R^{G}) with I ⊆ 𝔭 and H^{1}(G, R)_{𝔭} = 0 or dim (H^{1}(G, R)_{𝔭}) = 0. Choose a 𝔓 ∈ Spec(R) with R^{G} ∩ 𝔓 = 𝔭. Then I ⊆ 𝔓, which by Lorenz and Pathak [15, Lemma 1.1] implies

N_{𝔓} ⫋ G_{𝔓},  (4)

hence G_{𝔓} ≠ {1}. By Theorem 2.2 this implies H^{1}(G, R)_{𝔭} ≠ 0. Hence dim (H^{1}(G, R)_{𝔭}) = 0, which means that 𝔭 is a minimal element of Supp_{R^{G}} (H^{1}(G, R)). But then 𝔓 must also be minimal with G_{𝔓} ≠ {1}, since by Theorem 2.2, an ideal 𝔓′ ⫋ 𝔓 with G_{𝔓′} ≠ {1} would yield 𝔭′ ⫋ 𝔭 with 𝔭′ ∈ Supp_{R^{G}} (H^{1}(G, R)) (see Eisenbud [5, Corollary 4.18]). By the definition of H_{min}, this means that G_{𝔓} ⊆ H_{min}, implying G_{𝔓} ⊆ N and contradicting (4).

Remark. The hypotheses in Theorem 2.4 requiring R^{G} to be Noetherian and R to be finitely generated over R^{G}, may seem a bit awkward. In particular, the hypotheses imply that R is Noetherian, so one might wonder if it suffices to assume that R is Noetherian. However, there exist examples of a Noetherian ring R and a finite group G ≤ Aut(R) such that R^{G} is not Noetherian and R is not finitely generated over R^{G} (see Montgomery [16, Example 5.5]). ◁

The following corollaries are less technical. We consider the special case where G (still a p-group) acts linearly on a finite-dimensional vector space V over a field K of characteristic p. Then G also acts on R := S(V^{∗}), the symmetric algebra of the dual of V.

Corollary 2.5. In the above situation define F := {V^{σ} | σ ∈ G \ {1}}. If H^{1}(G, R) is Cohen-Macaulay (as a module over R^{G}), then

G = ⟨σ ∈ G \ {1} | V^{σ} is maximal in F⟩ .

The following elementary lemma is needed for the proof.

Lemma 2.6. Let 𝔓 = ((V^{σ})^{⊥}) be the ideal in R generated by all linear forms in V^{∗} vanishing on V^{σ}. Then

(a) σ ∈ G_{𝔓};

(b) if σ ∈ G_{𝔔} for 𝔔 ∈ Spec(R), then 𝔓 ⊆ 𝔔.

Proof. (a) Let l ∈ V^{∗} be a linear form. Then for v ∈ V^{σ} we have

((σ − 1)l) (v) = l (σ^{−1}(v) − v) = l(0) = 0,  (5)

so (σ − 1)l ∈ (V^{σ})^{⊥} ⊂ 𝔓. Assume that we have shown (σ − 1)f ∈ 𝔓 for an f ∈ R. Then

(σ − 1)(lf) = σ(l) · (σ − 1)f + f · (σ − 1)l ∈ 𝔓.

By induction, (σ − 1)R ⊆ 𝔓 now follows.

(b) Equation (5) shows that (σ − 1)V^{∗} ⊆ (V^{σ})^{⊥}. The rank of σ − 1 on V^{∗} equals the rank on V, which in turn is equal to dim_{K}(V) − dim_{K}(V^{σ}) = dim_{K} ((V^{σ})^{⊥}). Thus (σ − 1)V^{∗} = (V^{σ})^{⊥}, so for l ∈ (V^{σ})^{⊥} there exists h ∈ V^{∗} with (σ − 1)h = l. Hence l ∈ (σ − 1)R ⊆ 𝔔. Since 𝔓 is generated by (V^{σ})^{⊥}, the result follows.

Proof of Corollary 2.5. The hypotheses of Theorem 2.4 are satisfied, so we need to analyse the conclusion in the context of the corollary. For 𝔔 := R^{+}, the direct sum of all S^{i}(V), i > 0, we have 𝔔 ∈ Spec(R) and G_{𝔔} = G. Therefore the subgroup H from Theorem 2.4 coincides with G. Hence Theorem 2.4 says that H_{min} = G. In other words, G is generated by σ ∈ G_{𝔓} \ {1} with 𝔓 ∈ Spec(R) minimal with the property that G_{𝔓} ≠ {1}.

Take such a σ and assume, by way of contradiction, that V^{σ} is not maximal in F. Hence there exists a τ ∈ G \ {1} with V^{σ} ⫋ V^{τ}. Let 𝔓′ := ((V^{τ})^{⊥}) be the ideal in R generated by the linear forms vanishing on V^{τ}. Then τ ∈ G_{𝔓′} by Lemma 2.6(a), hence G_{𝔓′} ≠ {1}. We claim that 𝔓′ ⫋ 𝔓, which will contradict the minimality of 𝔓. Indeed, take l ∈ (V^{τ})^{⊥}. Then V^{σ} ⊆ V^{τ} implies l ∈ (V^{σ})^{⊥} and so, by Lemma 2.6(b), l ∈ 𝔓. This yields 𝔓′ ⊆ 𝔓. Since V^{σ} is proper in V^{τ} there exists an l ∈ (V^{σ})^{⊥} \(V^{τ})^{⊥}. As above, it follows that l ∈ 𝔓, but l ∉ 𝔓′ by the construction of 𝔓′. Therefore 𝔓′ ⫋ 𝔓, which completes the proof.

Corollary 2.7. In the situation of Corollary 2.5, assume that the action of G on V is faithful and H^{1}(G, R) is Cohen-Macaulay as an R^{G}-module. Then G is generated by elements of order p.

Proof. By Corollary 2.5, G is generated by σ ∈ G\{1} such that V^{σ} is maximal among the V^{τ}, τ ∈ G \ {1}. But it can be seen from the Jordan canonical form of σ that V^{σ} ⫋ V^{σ^{p}}. Hence σ^{p} = 1.

Remark. Corollary 2.7 is in sharp contrast with Example 2.14 from Kemper [14], where it is shown that H^{i}(G, R) is Cohen-Macaulay for all i > 0 if the order of G is not divisible by p^{2}. ◁

3 Cohomology modules for cyclic p-groups

In this section G = ⟨σ⟩ is a cyclic group of order p^{k+1} acting on a finite-dimensional vector space V over a field K of characteristic p. We write R = S(V^{∗}) for the symmetric algebra of the dual of V. We will analyse the R^{G}-modules H^{j}(G, R). The depth of H^{0}(G, R) is known by Ellingsrud and Skjelbred [6]. We are interested in the depth of H^{j}(G, R) for j > 0 as R^{G}-modules. By Kemper [14, Theorem 2.13(a)] we know that

depth (H^{j}(G, R)) ≥ dim_{K}(V^{G}).  (6)

To obtain more information we use the special form of the group. In fact, we have

H^{i}(G, V) = \begin{cases}V^{G}/ Tr^{G}(V) & if i > 0 is even, \\ ker(Tr^{G})/(σ − 1)V & if i is odd \end{cases}  (7)

for any KG-module V (see, for example, Evens [7, p. 6]). Hence it suffices to consider j = 1 and j = 2, where the case j = 1 is fairly straightforward and will be discussed in the next subsection. After that we discuss the case j = 2 with V being a free KG-module. The case j = 2 and V not free is more subtle and requires detailed information on the transfer ideal Tr^{G}(R). These results, some of which are of independent interest, will be collected in the third subsection. Throughout this section we use V_{m}, for m ≤ |G|, to denote the indecomposable KG-module of dimension m.

3.1 The depth of H^{1}(G, R)

In this section we determine the depth of H^{1}(G, R) when G a cyclic p-group.

Theorem 3.1. With the notation introduced at the beginning of Section 3 we have

depth (H^{1}(G, R)) = dim_{K}(V^{G}).  (8)

In particular, H^{1}(G, R) is Cohen-Macaulay if and only if |G| = p.

Proof. V^{G} is a submodule of V on which G acts trivially. Moreover, H^{1}(G, R) is a module over End_{KG}(R). Thus Theorem 1.5 of Kemper [14] yields

depth (H^{1}(G, R)) = dim_{K}(V^{G}) + grade (𝔦, H^{1}(G, R)) ,

where 𝔦 is the intersection with R^{G} of the ideal in R generated by all linear forms vanishing on V^{G}. But 𝔦 coincides with the radical ideal of the image of the relative transfer Tr_{N}^{G} : R^{N} → R^{G} for N := ⟨σ^{p}⟩ (see Fleischmann [8]). Thus

depth (H^{1}(G, R)) = dim_{K}(V^{G}) + grade (Tr_{N}^{G}(R^{N}), H^{1}(G, R)) .

Take any f ∈ Tr_{N}^{G}(R^{N}). Then f = Tr_{N}^{G}(g) = (σ − 1)^{p−1}(g) with g ∈ R^{N}. Let α := 1 + (σ − 1)R be the class of 1 in H^{1}(G, R). Then

fα = (σ − 1)^{p−1}(g) + (σ − 1)R = 0.

Thus every element in Tr_{N}^{G}(R^{N}) is a zero-divisor on H^{1}(G, R), and, therefore, we obtain grade (Tr_{N}^{G}(R^{N}), H^{1}(G, R)) = 0. This yields (8).

The additional statement follows from comparing (8) with Corollary 2.3.

3.2 The depth of H^{2}(G, S(V^{∗})) for V free

We now consider H^{2}(G, R) when V is a free G-module, i.e., V = mV_{p^{k+1}}. In this case V ≅ V^{∗} is a permutation module and, choosing a permutation basis for V^{∗}, the group G permutes monomials in R = S(V^{∗}), i.e. R^{G} is what is called a ring of permutation invariants. We will now prove a general result on permutation invariants of p-groups that contains the required information on H^{2}(G, R) in the special case where G is cyclic.

The following lemma, which will also be used later in the paper, tells how to recognize free summands using the transfer. Although it is well known in representation theory, we will provide a short proof:

Lemma 3.2. Let P be a p-group and V a finitely generated KP -module. Then V contains a free direct summand if and only if Tr^{P} (V) ≠ 0 and V is free if and only if Tr^{P} (V) = V^{P}.

Proof. First assume that w := Tr^{P} (v) ≠ 0 for v ∈ V. The homomorphism

ϑ : KP → V, f ↦ fv

maps the one-dimensional socle soc(KP) = K · Tr^{P} (1_{P}) onto the line K · w ⊆ V^{P}. Hence ϑ is a monomorphism. Since KP is a finite-dimensional Frobenius-algebra, the notions of finitely generated projective, injective and free modules coincide. Hence ϑ splits, i.e. V ≅ KP ⊕ W. For the second claim we can assume that V ≠ 0 is indecomposable. If V is not free, then by the previous argument Tr^{P} (V) must be zero, whereas V^{P} is nonzero. If V is free, then V ≅ KP and clearly V^{P} = Tr^{P} (V).

Theorem 3.3. Let P be a p-group acting on the finite set Ω and let K^{Ω} denote the corresponding KP -permutation module over the field K of characteristic p. Let Z ◁ P be a normal subgroup of order p with quotient \overline{P} := P/Z; furthermore let Ω/Z denote the set of Z-orbits on Ω. Then one has for R := S(K^{Ω}):

R^{P} / Tr^{P} (R) ≅ S(K^{Ω/Z})\overline{P} ⊕ (Tr^{Z}(R))\overline{P} / Tr^{P} (R).

Moreover (Tr^{Z}(R))\overline{P} = Tr^{P} (R) if and only if Tr^{Z}(R) is free as K\overline{P} -module. In particular, this is true if P is cyclic, i.e., in this case R^{P} / Tr^{P} (R) ≅ S(K^{Ω/Z})\overline{P} .

Proof. Note that the action of P on Ω induces an action of \overline{P} on Ω/Z in a natural way. Let V := K^{Ω} with basis \{X_{ω} | ω ∈ Ω\}. Then R = ⊕_{α: Ω→ℕ_{0}} K\underline{X}^{α} with \underline{X}^{α} := ∏_{ω∈Ω} X_{ω}^{α_{ω}} and P acts on R by g(\underline{X}^{α}) = \underline{X}^{α∘g^{−1}}. For every subgroup H ≤ P the invariant ring R^{H} has a K-basis consisting of H-orbit sums, i.e. relative transfers Tr_{H_{α}}^{H}(\underline{X}^{α}), where H_{α} := \{h ∈ H | α ∘ h = α\} is the stabiliser of \underline{X}^{α}. For each Z-orbit O ∈ Ω/Z let N_{O} = ∏_{ω∈O} X_{ω} be the orbit product in R^{Z}. A monomial \underline{X}^{α} is stabilised by Z if and only if α is constant on the orbits in Ω/Z, and hence if and only if it is a monomial in the N_{O}’s. These orbit products generate a polynomial subalgebra B of R^{Z}, which is isomorphic to S(K^{Ω/Z}) as a K\overline{P} -module. All the other Z-orbit sums in R^{Z} are absolute transfers, because Z has no nontrivial proper subgroups. It is now easy to see that R^{Z} = Tr^{Z}(R)⊕B as K\overline{P} -modules. Moreover

R^{P} = (R^{Z})\overline{P} = Tr^{Z}(R)\overline{P} ⊕ B\overline{P}

and Tr^{P} (R) = Tr_{Z}^{P}(Tr^{Z}(R)) = Tr\overline{P} (Tr^{Z}(R)) ⊆ Tr^{Z}(R)\overline{P} . This, together with Lemma 3.2, proves the first two statements.

Now assume that P is cyclic. Then R = X ⊕ Y, where X is spanned by the \underline{X}^{α} with P_{α} = 1 and Y is spanned by the \underline{X}^{β} with Z ≤ P_{β}. It follows that Tr^{Z}(R) = Tr^{Z}(X). But X is a direct sum of regular KP -modules, hence it is free in each fixed degree. Let Z^{+} := Tr^{Z}(1) ∈ KP; then

Tr^{Z}(KP) = Z^{+}KP = Z^{+}KZ ⊗_{KZ} K\overline{P} ≅ KZ^{+} ⊗_{KZ} K\overline{P} ≅ K\overline{P} .

Hence in each degree Tr^{Z}(R) is free as K\overline{P} -module.

Returning to the standard situation of this section, let G be cyclic of order p^{k+1} and R = S(V^{∗}). Assume that V := mV_{p^{k+1}} is the direct sum of m copies of the regular module KG and let Z ≤ G be the unique minimal subgroup. The corresponding set Ω is the union of m copies of G and Ω/Z is the union of m copies of \overline{G} = G/Z. Hence K^{Ω/Z} is the sum of m copies of the regular module \overline{G} and we get

H^{2}(G, R) ≅ R^{G}/ Tr^{G}(R) ≅ S(mK\overline{G})\overline{G}.

Thus applying Ellingsrud and Skjelbred [6] gives:

Corollary 3.4. In the above situation we have

depth (H^{2}(G, R)) = min{2 + m, mp^{k}}.

In particular, H^{2}(G, R) is Cohen-Macaulay if and only if

m(p^{k} − 1) ≤ 2.

Proof. We only need to prove the statement on the Cohen-Macaulay property. From Corollary 2.3 we see that dim (H^{2}(G, R)) = dim_{K}(V^{Z}) = mp^{k}. Thus H^{2}(G, R) is Cohen-Macaulay if and only if 2 + m ≥ mp^{k}.

Remark 3.5. Note that the isomorphism B ≅ S(K^{Ω/Z}) does not preserve degrees. In fact, for O ∈ Ω/Z the ‘variable’ N_{O} ∈ B has degree |O| rather than degree one. Let P be cyclic of order p^{2} and Z the subgroup of order p. Then Theorem 3.3 gives

S(V_{p^{2}}^{∗})^{Z/p^{2}}/ Tr^{Z/p^{2}}(S(V_{p^{2}}^{∗})) ≅ S (V_{p}^{∗})^{Z/p} .

From Shank and Wehlau [17, Theorem 6.2], S (V_{p}^{∗})^{Z/p} contains an indecomposable invariant of degree 2p − 3. Taking into account the ‘blow-up’ of degrees in the isomorphism B ≅ S(K^{Ω/Z}), we see that the quotient ring S(V_{p^{2}}^{∗})^{Z/p^{2}}/ Tr^{Z/p^{2}}(S(V_{p^{2}}^{∗})) contains an indecomposable element of degree p(2p−3) = 2p^{2}−3p > p^{2} (for p > 3). It had been conjectured that all indecomposable homogeneous modular invariants of degree larger than |P | lie in the transfer ideal. The above arguments show that this is not the case, if P is cyclic of order p^{2}.

3.3 The lowest degree in the image of the transfer for a cyclic p-group

Next we consider H^{2}(G, R) in greater generality. For that purpose we study the image of the transfer in R^{G}. We are particularly interested in the smallest degree for which the transfer is non-zero. From Lemma 3.2, we know that Tr^{G}(R)_{d} ≠ 0 is equivalent to the condition that the KG-module R_{d} contains a free direct summand. Recall that R = S(V^{∗}) for some KG-module V. Hence R_{d} ≅ S^{d}(V^{∗}) is a quotient of the d-fold tensor space (V^{∗})^{⊗d}. Hence Tr^{G}(R)_{d} ≠ 0 also implies that (V^{∗})^{⊗d} contains a free summand. It is somewhat surprising that the reverse turns out to be true for a cyclic group G.

If V is not faithful, then the minimal subgroup Z ≤ G acts trivially on R and Tr^{G}(R) = Tr_{Z}^{G}(Tr^{Z}(R)) = 0. Thus we can restrict attention to faithful modules. For any KG-module V we define mt(V) := min{d ∈ ℕ | Tr^{G}(S^{d}(V^{∗})) ≠ 0} with the convention min ∅ = ∞. Recall that the indecomposable modules for G (the cyclic group of order p^{k+1}) are given by V_{n} (1 ≤ n ≤ p^{k+1}), where n = dim(V_{n}), and a generator of G acts as a full Jordan block. The following is the main result of this subsection:

Theorem 3.6. Let G ≅ Z/p^{k+1} and V a KG-module.

i) Let 0 ≤ n = ∑_{i=0}^{k} n_{i}p^{i} < p^{k+1} with 0 ≤ n_{i} < p; if n_{k} > 0, let p − 1 = dn_{k} + r with 0 ≤ r < n_{k}. Then

mt(V_{n+1}) = \begin{cases}∞ & if n_{k} = 0 \\ d & if r = 0 and n ≥ n_{k}(1 + p + ⋯ + p^{k}) \\ d + 1 & otherwise. \end{cases}

ii) Suppose that V ≅ V_{m_{1}} ⊕ V_{m_{2}} ⊕ ⋯ ⊕ V_{m_{l}} with 0 ≤ m_{1} ≤ m_{2} ≤ ⋯ ≤ m_{l} ≤ p^{k+1}. Then mt(V) = mt(V_{m_{l}}).

iii) For any non-negative integer d the following are equivalent:

(a) the free module V_{p^{k+1}} appears as a direct summand of V^{⊗d};

(b) the free module V_{p^{k+1}} appears as a direct summand of S^{d}(V);

(c) d ≥ mt(V).

The proof will be given below, after a series of technical lemmas. We will make use of the representation ring (also called the Green ring) R_{KG}. As a Z-module R_{KG} is a free module with the isomorphism classes of indecomposable KG-modules as basis elements, and the multiplication is given by the tensor product. We will consider the elements χ_{i} := V_{p^{i}+1} − V_{p^{i}−1} ∈ R_{KG} (0 ≤ i ≤ k), where formally we set V_{0} := 0. We will use the following formulas of Green [10, Theorem 3], which hold for 0 ≤ i ≤ k.

χ_{i}V_{r} = \begin{cases}V_{r+p^{i}} − V_{p^{i}−r} & if 1 ≤ r ≤ p^{i}, \\ V_{r+p^{i}} + V_{r−p^{i}} & if p^{i} < r ≤ (p − 1)p^{i}, \\ V_{r−p^{i}} + 2V_{p^{i+1}} − V_{2p^{i+1}−p^{i}−r} & if (p − 1)p^{i} < r ≤ p^{i+1}. \end{cases}  (9)

Lemma 3.7. Let n be an integer with 0 ≤ n < p^{k+1} and write n = ∑_{i=0}^{k} n_{i}p^{i} with 0 ≤ n_{i} < p. There exists a unique polynomial f_{n} ∈ Z[X_{0}, …, X_{k}] with deg_{X_{i}}(f_{n}) < p for all i, such that

V_{n+1} = f_{n}(χ_{0}, …, χ_{k}).

Moreover, using the lexicographical monomial ordering with X_{k} > X_{k−1} > ⋯ > X_{0}, f_{n} has the leading monomial X_{k}^{n_{k}}X_{k−1}^{n_{k−1}} ⋯ X_{0}^{n_{0}} with leading coefficient 1.

Proof. We first remark that the existence of the f_{n} will imply uniqueness. In fact, if B is the set of all polynomials f ∈ Z[X_{0}, …, X_{k}] with deg_{X_{i}}(f) < p for all i, then the existence will provide an epimorphism B → R_{KG} of Z-modules. However B and R_{KG} are both free of rank p^{k+1}, therefore the epimorphism must in fact be an isomorphism.

We prove the existence of f_{n} and the statement on the leading monomial and coefficient by induction on n. For n = 0 we have f_{0} = 1. So assume n > 0 and let j be maximal with n_{j} ≠ 0. We have 1 ≤ n + 1 − p^{j} ≤ (p − 1)p^{j}. Hence Green’s formulas (9) yield

χ_{j}V_{n+1−p^{j}} = \begin{cases}V_{n+1} − V_{2p^{j}−n−1} & if n + 1 ≤ 2p^{j}, \\ V_{n+1} + V_{n+1−2p^{j}} & if n + 1 > 2p^{j}. \end{cases}

We have 2p^{j} − n − 1 < n + 1, hence in each case we obtain V_{n+1} = χ_{j}V_{n+1−p^{j}} ± V_{m+1} for some m < n. By induction, this yields

V_{n+1} = χ_{j}f_{n−p^{j}}(χ_{0}, …, χ_{k}) ± f_{m}(χ_{0}, …, χ_{k}),

so with f_{n} := X_{j}f_{n−p^{j}} ± f_{m} we have V_{n+1} = f_{n}(χ_{0}, …, χ_{k}). Also by induction, the leading monomial of f_{m} is less than X_{j}^{n_{j}}X_{j−1}^{n_{j−1}} ⋯ X_{0}^{n_{0}}, and the leading term of f_{n−p^{j}} is X_{j}^{n_{j}−1}X_{j−1}^{n_{j−1}} ⋯ X_{0}^{n_{0}}. This yields the statement on the leading monomial and coefficient of f_{n}. We also get from the induction hypothesis that deg_{X_{i}}(f_{n}) < p for i ≠ j, and it follows from the description of the leading monomial of f_{n} that deg_{X_{j}}(f_{n}) = n_{j} < p.

By Lemma 3.7, we have an epimorphism Z[X_{0}, …, X_{k}] → R_{KG} by sending each X_{i} to χ_{i}. We denote the kernel of this epimorphism by I.

Lemma 3.8. There exist polynomials r_{0}, …, r_{k} ∈ I with the leading monomial LM(r_{i}) = X_{i}^{p}. Moreover, the leading coefficient of r_{i} is 1.

Proof. Take an integer i with 0 ≤ i ≤ k. By (9) we have

χ_{i}V_{(p−1)p^{i}+1} = V_{(p−2)p^{i}+1} + 2V_{p^{i+1}} − V_{p^{i+1}−1},

so r_{i} := X_{i}f_{(p−1)p^{i}} − f_{(p−2)p^{i}} − 2f_{p^{i+1}−1} + f_{p^{i+1}−2} lies in I. By Lemma 3.7 this has leading monomial X_{i}^{p} with coefficient 1.

Proposition 3.9. Let V = V_{m_{1}+1}⊕⋯⊕V_{m_{l}+1} be a finitely generated KG-module and let d be a non-negative integer. Write n := max{m_{1}, …, m_{l}} as n = ∑_{i=0}^{k} n_{i}p^{i} with 0 ≤ n_{i} < p. If there exists a j ∈ {0, …, k} such that n_{i} = (p − 1)/d for j < i ≤ k but n_{j} < (p − 1)/d, then Tr^{G} (V^{⊗d}) = 0.

Proof. By Lemma 3.2 we have to show that under the assumption of the proposition, no V_{p^{k+1}} occurs as an indecomposable summand in V^{⊗d}. With f := f_{m_{1}} + ⋯ + f_{m_{l}}, Lemma 3.7 tells us that V = f(χ_{0}, …, χ_{k}), and LM(f) = X_{k}^{n_{k}}X_{k−1}^{n_{k−1}} ⋯ X_{0}^{n_{0}}. Hence V^{⊗d} = f^{d}(χ_{0}, …, χ_{k}), and

LM(f^{d}) = X_{k}^{dn_{k}}X_{k−1}^{dn_{k−1}} ⋯ X_{0}^{dn_{0}} = X_{k}^{p−1} ⋯ X_{j+1}^{p−1}X_{j}^{dn_{j}}X_{j−1}^{dn_{j−1}} ⋯ X_{0}^{dn_{0}} < (X_{k} ⋯ X_{0})^{p−1} .

Let g ∈ Z[X_{0}, …, X_{k}] be a normal form of f^{d} with respect to the set {r_{0}, …, r_{k}} from Lemma 3.8. Then LM(g) ≤ LM(f^{d}) and g − f^{d} ∈ I, so V^{⊗d} = g(χ_{0}, …, χ_{k}). But we have deg_{X_{i}}(g) < p for every i, since LM(r_{i}) = X_{i}^{p}. Therefore g is the unique polynomial with deg_{X_{i}}(g) < p and V^{⊗d} = g(χ_{0}, …, χ_{k}). But if V^{⊗d} had V_{p^{k+1}} as an indecomposable summand, this unique polynomial would have the leading monomial (X_{k} ⋯ X_{0})^{p−1} by Lemma 3.7. However, the leading monomial of g is smaller than this.

Corollary 3.10. With the setting and assumptions of Proposition 3.9, there exists no nonzero element of degree d in Tr^{G} (S(V)). In particular, the numbers given in Theorem 3.6(i) and (ii) are lower bounds for mt(V).

Proof. We have to show that Tr^{G} (S^{d}(V)) = 0, where S^{d}(V) is the d-th symmetric power. But there exists natural epimorphism φ: V^{⊗d} → S^{d}(V) of KG-modules. Therefore

Tr^{G} (S^{d}(V)) = Tr^{G} (φ (V^{⊗d})) = φ (Tr^{G} (V^{⊗d})) = 0

by Proposition 3.9.

So far we have looked at conditions ensuring that V^{⊗d} and therefore S^{d}(V) do not contain a free summand. We now look at the opposite, namely conditions such that S^{d}(V) and therefore V^{⊗d} do contain such a summand. We will use Lemma 3.2 as our main tool and therefore we need some results on transfers.

3.3.1 Computing transfers

Lemma 3.11. Let X be an arbitrary finite group, acting on a ring A. Suppose that H is a subgroup of X, g ∈ A and h ∈ A^{H}. Further suppose f := Tr^{H}(g) ∈ A^{X}. Then Tr^{X}(gh) = f Tr_{H}^{X}(h).

Proof. Using the factorisation Tr^{X} = Tr_{H}^{X} ∘ Tr^{H} gives Tr^{X}(gh) = Tr_{H}^{X}(Tr^{H}(gh)). Since h ∈ A^{H}, we have Tr^{H}(gh) = h Tr^{H}(g). By hypothesis Tr^{H}(g) ∈ A^{X}. Thus

Tr_{H}^{X}(h Tr^{H}(g)) = Tr_{H}^{X}(h) Tr^{H}(g) = f Tr_{H}^{X}(h)

as required.

Lemma 3.12. For an integer 0 ≤ ℓ < p we have

∑_{c∈F_{p}} c^{ℓ} = \{ \begin{matrix}0 & if 0 ≤ ℓ < p − 1; \\ −1 & if ℓ = p − 1. \end{matrix}

Defining Δ := σ − 1 and applying the binomial theorem gives

σ^{i} = ∑_{j=0}^{i} \binom{i}{j}Δ^{j}.

Over a field of characteristic p this gives σ^{p^{i}} = 1 + Δ^{p^{i}}. Suppose G = ⟨σ⟩ ≅ Z/p^{k+1} and α is an element in a KG-module with KGα ≅ V_{n+1}. Then comparing binomial coefficients gives

σ^{i}α = ∑_{j=0}^{n} \binom{i}{j}Δ^{j}α.

If n = p^{i}m + r with 0 ≤ r < p^{i} and H := ⟨σ^{p^{i}}⟩ then KHα ≅ V_{m+1} and the corresponding fixed point is Δ^{mp^{i}}α.

Theorem 3.13. Let H = Z/p and suppose that V is a KH-module and α_{1}, α_{2}, …, α_{ℓ} is a sequence of not necessarily distinct elements of S(V^{∗}). Define m_{i} by KH · α_{i} ≅ V_{m_{i}+1} and define β_{i} by Δ^{m_{i}}α_{i} = β_{i}. If ∑_{i=1}^{ℓ} m_{i} = p − 1 then

Tr^{H} (∏_{i=1}^{ℓ} α_{i}) = − ∏_{i=1}^{ℓ} \frac{β_{i}}{m_{i}!}.

Proof. Observe that σ^{c}(α_{i}) = ∑_{j=0}^{m_{i}} \binom{c}{j}Δ^{j}α_{i}. Thus

Tr^{Z/p} (∏_{i=1}^{ℓ} α_{i}) = ∑_{c∈F_{p}} ∏_{i=1}^{ℓ} ∑_{j=0}^{m_{i}} \binom{c}{j}Δ^{j}α_{i}

= ∑_{c∈F_{p}} ∏_{i=1}^{ℓ} (α_{i} + ⋯ + \binom{c}{m_{i}}β_{i})

= ∑_{c∈F_{p}} (∏_{i=1}^{ℓ} α_{i} + ⋯ + ∏_{i=1}^{ℓ} \binom{c}{m_{i}}β_{i}) .

Note that the degree of \binom{c}{j} as a polynomial in c is j. Thus ∏_{i=1}^{ℓ} \binom{c}{m_{i}} has degree ∑_{i=1}^{ℓ} m_{i} = p − 1. All other terms in the sum have coefficients which, as polynomials in c, have degree less than p − 1. Therefore, using Lemma 3.12, these terms don’t contribute. Thus

Tr^{Z/p} (∏_{i=1}^{ℓ} α_{i}) = ∑_{c∈F_{p}} ∏_{i=1}^{ℓ} \binom{c}{m_{i}}β_{i}

= (∏_{i=1}^{ℓ} \frac{β_{i}}{m_{i}!}) ∑_{c∈F_{p}} ∏_{i=1}^{ℓ} c(c − 1) ⋯ (c − m_{i} + 1).

A further application of Lemma 3.12 gives

∑_{c∈F_{p}} ∏_{i=1}^{ℓ} c(c − 1) ⋯ (c − m_{i} + 1) = ∑_{c∈F_{p}}(c^{p−1} + ⋯) = −1

as required.

In the following we write x_{n+1} for a linear form in V^{∗} which generates a KG-module of type V_{n+1}. Furthermore, for r ≤ n, we write x_{r} := Δ^{n+1−r}(x_{n+1}).

Corollary 3.14. Suppose n + 1 ≤ p and nd = p − 1. Then

Tr^{Z/p}(x_{n+1}^{d}) = − (\frac{x_{1}}{n!} )^{d} .

Proof. Apply Theorem 3.13 with α_{i} = x_{n+1}, β_{i} = x_{1}, m_{i} = n and ∑ m_{i} = dn = p − 1.

Lemma 3.15. Suppose t and d are positive integers with td = p − 1 and let n := t(1 + p + p^{2} + ⋯ + p^{k}) = (p^{k+1} − 1)/d. Then

Tr^{G}(x_{n+1}^{d}) = \frac{x_{1}^{d}}{(−(t!)^{d})^{k+1}} .

Proof. The proof is by induction on k. For k = 0, the result is given by Corollary 3.14. For k > 0, let N := ⟨σ^{p}⟩ ≅ Z/p^{k}. The N-module generated by x_{n+1} is isomorphic to V_{n′+1} with n′ = (p^{k} − 1)/d and the corresponding fixed point is x_{t+1}. Therefore, using the induction hypothesis, Tr^{N}(x_{n+1}^{d}) = (−(t!)^{−d})^{k}x_{t+1}^{d}. However Tr^{G} = Tr_{N}^{G} ∘ Tr^{N} = Tr^{G/N} ∘ Tr^{N}. Thus Tr^{G}(x_{n+1}^{d}) = (−(t!)^{−d})^{k} Tr^{G/N}(x_{t+1}^{d}). The G/N-module generated by x_{t+1} is isomorphic to V_{t+1} and the corresponding fixed point is x_{1}. Therefore, applying Corollary 3.14 gives Tr^{G/N}(x_{t+1}^{d}) = −(t!)^{−d}x_{1}^{d} and the result follows.

Lemma 3.16. Suppose t ≤ p−1 is a positive integer. Divide t into p−1 to get p−1 = td+r with 0 ≤ r < t. Then

Tr^{G} (x_{tp^{k}+1}^{d}x_{(r+1)p^{k}}) = \frac{−x_{1}^{d+1}}{(t!)^{d}r!} .

Proof. Let H := ⟨σ^{p^{k}}⟩ ≅ Z/p. The H-module generated by x_{tp^{k}+1} is isomorphic to V_{t+1} with corresponding fixed point x_{1}. Since (r + 1)p^{k} = rp^{k} + (p^{k} − 1) + 1, the H-module generated by x_{(r+1)p^{k}} is isomorphic to V_{r+1} with corresponding fixed point x_{p^{k}}. Thus applying Theorem 3.13 gives

Tr^{H} (x_{tp^{k}+1}^{d}x_{(r+1)p^{k}}) = \frac{−x_{1}^{d}x_{p^{k}}}{(t!)^{d}r!} .

The result then follows from the factorisation Tr^{G} = Tr_{H}^{G} ∘ Tr^{H} and the fact that Tr_{H}^{G}(x_{p^{k}}) = Tr^{Z/p^{k}}(x_{p^{k}}) = x_{1}.

3.3.2 Proof of Theorem 3.6

For brevity we write I^{G} := Tr^{G}(R). Assume p^{k} ≤ n < p^{k+1} and V = V_{n+1}. Write n = ∑_{i=0}^{k} n_{i}p^{i} with 0 ≤ n_{i} ≤ p − 1. Divide n_{k} into p − 1 to get p − 1 = dn_{k} + r with 0 ≤ r < n_{k}. Using Lemma 3.16, we see that x_{1}^{d+1} ∈ I^{G}. If r > 0, applying Proposition 3.9 with j = k shows that I^{G} is zero in degrees less than or equal to d. If r = 0 and n_{k}(p^{j} + ⋯ + p^{k}) ≤ n < n_{k}(p^{j−1} + ⋯ + p^{k}) then again Proposition 3.9 tells us that I^{G} is zero in degrees less than or equal to d. However if n ≥ n_{k}(1 + p + ⋯ + p^{k}), then from Lemma 3.15, x_{1}^{d} ∈ I^{G}. A further application of Proposition 3.9 shows that, in this case, I^{G} is zero in degrees less than or equal to d − 1. Note that n necessarily falls into one of three cases:

(i) r > 0, (ii) r = 0 and n_{k}(p^{j} + ⋯ + p^{k}) ≤ n < n_{k}(p^{j−1} + ⋯ + p^{k}), (iii) r = 0 and n ≥ n_{k}(1 + p + ⋯ + p^{k}). In the first two cases the minimal non-zero degree for I^{G} is d + 1 while in the third case the minimal non-zero degree for I^{G} is d. In all three cases there is a power of x_{1} in the first non-zero degree.

This proves part i) of Theorem 3.6 for indecomposable KG-modules. Suppose V is arbitrary KG-module. Since G is a p-group, there is an invariant 0 ≠ v ∈ S(V^{∗})^{G} of degree one. Suppose d > mt(V) and X ≅ V_{p^{k+1}} is a submodule of S^{mt(V)}(V^{∗}). Then V_{p^{k+1}} ≅ v^{d−mt(V)} ·X ≤ S^{d}(V^{∗}), which splits off, because X is projective and hence injective in the category of KG-modules. This shows that the implications (c) ⇔ (b) ⇒ (a) hold for arbitrary V. Together with the proof so far we see that (a), (b) and (c) are equivalent if V is indecomposable.

Suppose V is decomposed as in part ii) of the theorem. Then S^{mt(V_{m_{l}})}(V_{m_{l}}) appears as a submodule of (S^{mt(V_{m_{l}}} (V)), hence mt(V) ≤ mt(V_{m_{l}}). Now assume that some summand

F ≅ V_{p^{k+1}} appears in a submodule S of S^{d}(V) ≅ (⊗_{i=1}^{l} S(V_{m_{i}}))_{d}. We can choose S of the form S^{s_{1}}(V_{m_{1}})⊗⋯⊗S^{s_{l}}(V_{m_{l}}) with s_{1} +s_{2} +⋯+s_{k} = d = mt(V). Hence F also appears in V_{m_{1}}^{⊗s_{1}} ⊗⋯⊗V_{m_{l}}^{⊗s_{l}}. Since m_{1} ≤ m_{2} ≤ ⋯ ≤ m_{l}, we know that V_{m_{1}}^{⊗s_{1}} ⊗⋯⊗V_{m_{l}}^{⊗s_{l}} is isomorphic to a submodule of V_{m_{l}}^{⊗s_{1}} ⊗ ⋯ ⊗ V_{m_{l}}^{⊗s_{l}} ≅ V_{m_{l}}^{⊗mt(V)}. Hence F splits off V_{m_{l}}^{⊗mt(V)}. Since V_{m_{l}} is indecomposable, we can use iii) (c) to conclude mt(V) ≥ mt(V_{m_{l}}). This finishes the proof of Theorem 3.6.

3.4 On the depth of even cohomology

We keep the notation introduced at the beginning of Section 3, and we continue to write I^{G} = Tr^{G}(R). For a subgroup H ≤ G we write I_{H}^{G} = Tr_{H}^{G}(R^{H}) for the image of the relative transfer.

Theorem 3.17. Suppose that H is a subgroup of G and (I^{H} \ I^{G}) ∩ R^{G} is non-empty. Then depth(R^{G}/I^{G}) ≤ dim(R^{G}/I_{H}^{G}).

Proof. Choose f ∈ (I^{H} \ I^{G}) ∩ R^{G}. By Lemma 3.11, f + I^{G} is a nonzero element of M := R^{G}/I^{G} which annihilates I_{H}^{G} + I^{G} ⊂ R^{G}/I^{G}. Thus every element of I_{H}^{G} acts as a zero divisor on M and therefore I_{H}^{G} is contained in some associated prime 𝔭 of M. Now the ‘graded version’ of Bruns and Herzog [4, Proposition 1.2.13] shows that the depth of M is less than or equal to dim(R^{G}/𝔭) which in turn is less than or equal to dim(R^{G}/I_{H}^{G}).

Theorem 3.18. Suppose td = p − 1 and t(p^{i} + p^{i+1} + ⋯ + p^{k}) ≤ n < t(p^{i−1} + p^{i} + ⋯ p^{k}). Then depth(H^{2}(G, S(V_{n+1}^{∗}))) ≤ p^{i−1} and, if i ≤ k, H^{2}(G, S(V_{n+1}^{∗})) is not Cohen-Macaulay.

Proof. Let H := ⟨σ^{p^{i}}⟩ ≅ Z/p^{k+1−i}, m := t(p^{i} +p^{i+1} +⋯+p^{k}) and let m′ := m/p^{i}. The Hmodule generated by x_{m+1} is isomorphic to V_{m′+1} with corresponding fixed point x_{1}. Thus applying Lemma 3.15 gives Tr^{H}(x_{m+1}^{d}) = (−1(t!)^{−d})^{k+1−i}x_{1}^{d} ∈ R^{G}. By Corollary 3.10 we know that I^{G} is zero in degree d thus x_{1}^{d} ∉ I^{G}. Therefore x_{1}^{d} ∈ (I^{H} \ I^{G}) ∩ R^{G}. Applying Theorem 3.17 gives depth(H^{2}(G, R)) ≤ dim(R^{G}/I_{H}^{G}). However dim(R^{G}/I_{H}^{G}) is the dimension of the subspace of V_{n+1} fixed by σ^{p^{i−1}}. Thus depth(H^{2}(G, R)) ≤ p^{i−1}. Furthermore dim(H^{2}(G, R)) = p^{k}. Therefore, as long as k ≥ i, H^{2}(G, R) is not Cohen-Macaulay.

Theorem 3.19. Suppose td = p − 1 and t(p + p^{2} + ⋯ + p^{k}) ≤ n < t(1 + p + p^{2} + ⋯ p^{k}). Then depth(H^{2}(G, S(V_{n+1}^{∗})) = 1.

Proof. Applying Theorem 3.18 with i = 1 gives depth(H^{2}(G, R)) ≤ 1 and from 6, we have depth(H^{2}(G, S(V_{n+1}^{∗})) ≥ 1.

Theorem 3.20. Let V = V_{m_{1}+1} ⊕ ⋯ ⊕ V_{m_{ℓ}+1} be a finitely generated KG-module and define n := max{m_{1}, …, m_{ℓ}}. Suppose that td = p − 1 and t(p + p^{2} + ⋯ + p^{k}) ≤ n < t(1 + p + p^{2} + ⋯ p^{k}). Then depth(H^{2}(G, S(V^{∗})) = ℓ.

Proof. It follows from Corollary 3.10 that x_{1}^{d} + I^{G} is a non-zero element of H^{2}(G, S(V^{∗})). We choose a homogeneous system of parameters for R^{G} consisting of the norms of the terminal variables and elements from the image of the relative transfer and then proceed as in the proof of Theorem 3.19.

The following gives an ‘if and only if’ - criterion for the hypothesis in Theorem 3.17, in the special case where H is the maximal subgroup of G.

Lemma 3.21. If N ◁ G, the equation

I^{G} = I^{N} ∩ R^{G}

holds if and only if I^{N} is free as a K(G/N)-module.

Proof. Let Ḡ := G/N and note that I^{G} = TrḠ(I^{N}), whereas I^{N} ∩ R^{G} = (I^{N})Ḡ. Now the statement follows immediately from Lemma 3.2 with P = Ḡ.

Lemma 3.22. Let N < G be the maximal subgroup and S := ⊕_{m=1}^{p^{k+1}} n_{m}V_{m}. Then Tr^{N}(S) is free as KḠ-module if and only if n_{m} = 0 for all m satisfying p^{k+1} − p + 1 ≤ m < p^{k+1}.

Proof. Note that V_{m} viewed as K[Δ]-module satisfies: V_{m} ≅ K[Δ]/(Δ^{m}) and hence

Δ^{i}(V_{m}) ≅ Δ^{i} · K[Δ]/(Δ^{m}) ≅ K[Δ]/(Δ^{m−i}) ≅ V_{m−i},

with V_{j} := 0 for j ≤ 0. Therefore one has

Tr^{N}(S) = (σ^{p} − 1)^{p^{k}−1}(S) = Δ^{p^{k+1}−p}(S) = ⊕_{ℓ=0}^{p−1} n_{p^{k+1}−ℓ}(Δ^{p^{k+1}−p}(V_{p^{k+1}−ℓ}))
≅ ⊕_{ℓ=0}^{p−1} n_{p^{k+1}−ℓ}(V_{p−ℓ}) ∈ KḠ − mod.

Now the statement in the lemma follows from the fact that V_{p−ℓ} is a free KḠ - module if and only if ℓ = 0.

Using (6), Kemper [14, Theorem 1.5] and the preceding lemmas, we obtain:

Proposition 3.23. Let N < G be the maximal subgroup. Then we have I^{G} ⫋ I^{N} ∩ R^{G} if and only if the KG-module R has a direct summand V_{m} with p(p^{k} − 1) < m < p^{k+1}. In this case

grade (\sqrt{Tr_{N}^{G}(R^{N})} , H^{2}(G, R)) = 0

and

depth(H^{2}(G, R)) = dim(V^{G}).

We think that it is quite common to find a V_{m} with p(p^{k} − 1) < m < p^{k+1} in the symmetric powers of V^{∗}. Unfortunately we were not able to prove any concrete instances of this, apart from the ones found in Theorems 3.19 and 3.20.

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P. Fleischmann and R.J. Shank  G. Kemper
Institute of Mathematics and Statistics  Technische Universität München
University of Kent  Zentrum Mathematik - M11
Canterbury, CT2 7NF  Boltzmannstr. 3
UK  85 748 Garching

  P.Fleischmann@ukc.ac.uk  Germany
  R.J.Shank@ukc.ac.uk                 kemper@ma.tum.de