Article Submitted to Journal of Symbolic Computation

Computing Modular Invariants of p-Groups

R. James Shank^{1} and David L. Wehlau^{∗}^{2}

1Institute of Mathematics & Statistics, University of Kent at Canterbury, CT2 7NF, United Kingdom, R.J.Shank@ukc.ac.uk 2Department of Mathematics & Computer Science, Royal Military College, Kingston, Ontario, Canada K7K 7B4, wehlau@rmc.ca

Abstract

Let V be a finite dimensional representation of a p-group, G, over a field, k, of characteristic p. We show that there exists a choice of basis and monomial order for which the ring of invariants, k[V]^{G}, has a finite SAGBI basis. We describe two algorithms for constructing a generating set for k[V]^{G}. We use these methods to analyse k[2V_{3}]^{U_{3}} where U_{3} is the p-Sylow subgroup of GL_{3}(F_{p}) and 2V_{3} is the sum of two copies of the canonical representation. We give a generating set for k[2V_{3}]^{U_{3}} for p = 3 and prove that the invariants fail to be Cohen-Macaulay for p > 2. We also give a minimal generating set for k[mV_{2}]^{Z/p} were V_{2} is the two dimensional indecomposable representation of the cyclic group Z/p.

1. Introduction

Let V be a finite dimensional vector space over a field k. We choose a basis, {x_{1}, …, x_{n}}, for the dual, V^{∗}, of V. Consider a subgroup G of GL(V). The action of G on V induces an action on V^{∗} which extends to an action by algebra automorphisms on the symmetric algebra of V^{∗}, S = k[x_{1}, …, x_{n}]. Specifically, for g ∈ G, f ∈ S and v ∈ V, (g · f)(v) = f(g^{−1} · v). The ring of invariants of G is the subring of S given by

S^{G} := {f ∈ S | g · f = f for all g ∈ G}.

For an introduction to the invariant theory of finite groups see [3] or [29].

If G is a finite group and |G| is not invertible in k then we say the representation of G on V is modular. If |G| is invertible in k then V is called a non-modular representation. E. Noether proved in [21; 22] that S^{G} is always a finitely generated algebra. In [21] she showed that if the characteristic of k is zero then S^{G} is

∗Research partially supported by grants from ARP and NSERC

R.J. Shank & D.L. Wehlau: Invariants of p-Groups 2

always generated by the invariant polynomials in S of degree less than or equal to |G|. Recently this result has been extended independently by Fleischmann [11] and Fogarty [12] to the general case of a non-modular representation. The result does not hold for modular representations. In fact, as illustrated by the vector invariants of the regular representation of Z/2 over a field of characteristic 2 (see [23] or [5]), no function depending solely on the order of the group can serve as an upper bound on the degrees of the generators.

The central problem of invariant theory is to find generators for the algebra S^{G}. In practice, this problem is much harder in the modular setting. In this paper we describe various methods for computing generators of S^{G} for modular representations. We especially consider the case where G is a p-group and k is a field of characteristic p.

A SAGBI basis for a subalgebra of S is the analog of a Gröbner basis and as such is a particularly nice generating set. SAGBI bases were introduced independently by Robbiano & Sweedler [24] and Kapur & Madlener [16]. Unfortunately, even a finitely generated subalgebra does not necessarily have a finite SAGBI basis. In fact even the ring of invariants of a finite group may fail to have a SAG-BI basis (see [13, Lemma 2.1], [14] or [30, Example 11.2]). The characterization of subalgebras which admit a finite SAGBI basis is an important open problem. In Section 3 we show that for any representation of p-group over a field of characteristics p, there is a choice of basis and monomial order for which the ring of invariants has a finite SAGBI basis. In fact our result applies to any triangular representation.

In Section 5 we give a number of criteria for determining whether an algebra consisting of invariants is in fact the entire ring of invariants S^{G}. We also give an algorithm for constructing a generating set for the ring of invariants of a p-group or, more generally, a triangular representation. The algorithm makes use of the theory of SAGBI bases, in particular the computation of syzygy modules for subalgebras, and exploits the fact that S^{G} is integrally closed.

Suppose that N ◁ G is a normal subgroup of G. Then G acts on k[V]^{N} and k[V]^{G} = (k[V]^{N})^{G} = (k[V]^{N})^{G/N}. Thus we may reduce the problem of computing S^{G} to two smaller problems: computing invariants first under the subgroup N and then under the quotient group G/N. However computing the G/N-invariants is considerably complicated by the fact that the algebra, k[V]^{N}, on which G/N is acting is not, in general, a polynomial ring. One solution to this difficulty is to construct a G/N-module W together with a G/N-equivariant surjection ρ : k[W] → S^{N}. In the non-modular case the restriction of this homomorphism is a surjection ρ^{G} : k[W]^{G/N} → S^{G}. For non-modular representations this technique, called a ladder, is one of the most effective for computing rings of invariants (see for example [31]). However, in the modular setting the induced map ρ^{G} is not, in general, surjective. In Section 7 we describe how group cohomology may be used to overcome this difficulty. If G is a p-group this provides a method to compute S^{G} by computing the Z/p-invariants of a number of Z/prepresentations together with a number of group cohomology computations. In particular, one must be able to compute rings of invariants for modular Z/prepresentations.

Attempts to apply the ladder technique to modular representations of p-groups emphasize the importance of being able to construct manageable generating sets for rings of invariants for representations of Z/p. However, for most such representations, this is quite difficult. Hughes & Kemper [15] have given an upper bound on the degrees of the generators for any representation of Z/p. Therefore by taking all homogeneous invariants with degree less than or equal to the upper bound we do get a finite generating set. However such generating sets are far from manageable. Throughout the paper we use V_{n}, for n ≤ p, to denote the unique indecomposable modular representation of Z/p with dimension n. Minimal generating sets for k[V_{2}]^{Z/p} and k[V_{3}]^{Z/p} can be found in Dickson’s Madison Colloquium [7]. Finite SAGBI bases for k[V_{4}]^{Z/p} and k[V_{5}]^{Z/p} can be found in [26]. The problem of finding a nice generating set for k[V_{n}]^{Z/p} for n > 5 remains open. Even when the invariants of the indecomposable summands are understood, it can be difficult to construct generating sets for decomposable representations. Campbell & Hughes, in [5], manage to describe a generating set for k[mV_{2}]^{Z/p}. In Section 4 we refine their solution giving a minimal generating set for this ring. This result is used in Section 8. For the special case of p = 2, every representation is of the form mV_{2} ⊕ ℓV_{1}. Since k[mV_{2} ⊕ ℓV_{1}]^{Z/p} = k[mV_{2}]^{Z/p} ⊗ k[ℓV_{1}], we therefore obtain a minimal generating set for the ring of invariants of every finite dimensional modular representation of Z/2.

Suppose that R is a graded subalgebra of S and M is an R-module. Let R_{+} denote the augmentation ideal of R, i.e., the ideal generated by the homogeneous elements of positive degree. A sequence of homogeneous elements h_{1}, …, h_{k} in R_{+} is regular on M if, for each i ≤ k, h_{i} is not a zero-divisor on M/(h_{1}, …, h_{i−1})M. The depth of M is the length of the longest regular sequence on M. The depth of a ring is bounded above by its Krull dimension. A ring is Cohen-Macaulay if the depth equals the dimension. For a detailed discussion of depth and dimension see [8]. For a non-modular representation, the ring of invariants is always Cohen–Macaulay. However, when the characteristic of k divides the order of the group, the invariants often fail to be Cohen–Macaulay. Characterizing the modular representations which have a Cohen-Macaulay ring of invariants is an interesting and important problem. In [17], Kemper proved that if G is a p-group and S^{G} is Cohen-Macaulay then G is generated by a set of bi-reflections, i.e., by elements which fix pointwise a subspace of codimension 1 or 2. In particular, this means that if k[V ⊕ V]^{G} is Cohen-Macaulay then the action of G on V must be generated by reflections, i.e., by elements which fix pointwise a subspace of codimension 1. In Section 8 we analyze the invariants of U_{3}, the p-Sylow subgroup of GL_{3}(F_{p}), acting on V_{3} ⊕ V_{3}. The action of U_{3} on V_{3} is generated by reflections and the ring of invariants is a polynomial algebra, i.e., there are no relations among the generators. Therefore k[2V_{3}]^{U_{3}} passes Kemper’s criteria and could be Cohen-Macaulay. In fact, for p = 2, the invariant ring is Cohen-Macaulay. However, using the ladder technique, we are able to show that for p > 2 the invariants fail to be Cohen-Macaulay.

2. Preliminaries

The transfer is defined by:

\begin{aligned}T r^{G} : & k[V] & ⟶ & k[V]^{G} \\ & f & ⟼ & ∑_{g∈G} g · f \end{aligned}

and is a homomorphism of k[V]^{G}-modules. For non-modular representations, Tr^{G} is surjective. For modular representations, the image of the transfer, Im Tr^{G}, is a proper non-zero ideal of k[V]^{G}. For proofs of this fact and other general properties of the modular transfer see [27].

If a is an element of a set on which the finite group G acts, we write G · a = {g · a | g ∈ G} for the G orbit of a. For f ∈ k[V], we define the norm of f by N_{G}(f) = N(f) = ∏_{h∈G·f} h.

We will consider representations of Z/p, the cyclic group of order p, in some detail. Let σ denote a fixed generator of Z/p. Define Δ := σ−1 and Tr := ∑_{i=1}^{p} σ^{i} in the group ring of Z/p. There are exactly p distinct inequivalent indecomposable representations of Z/p, one of each dimension 1, 2, …, p. We will denote the indecomposable representation of Z/p of dimension n by V_{n}. There exists a basis, {e_{1}, …, e_{p}}, of V_{p}, with Δe_{1} = 0 and, for i > 1, Δe_{i} = e_{i−1}. The vector space spanned by {e_{1}, …, e_{n}} is a Z/p-submodule isomorphic to V_{n}. There are Z/p-equivariant inclusions: V_{1} ⊂ V_{2} ⊂ ⋯ ⊂ V_{p}, and V_{n}^{Z/p} is isomorphic to V_{1}.

Consider the vector space of linear functionals V_{n}^{∗}. Since V_{n}^{∗} is an indecomposable Z/p-module, V_{n}^{∗} and V_{n} are isomorphic. We will call an element, z, of V_{n}^{∗} a distinguished variable for V_{n} if z is a generator of the cyclic Z/p-module V_{n}^{∗}. Equivalently z is a distinguished variable if z restricted to V_{n}^{Z/p} is not identically zero. For any distinguished variable z there is a triangular basis, {z, Δz, Δ^{2}z, …, Δ^{n−1}z}, of V_{n}^{∗}. For any f ∈ k[V_{n}], let deg_{z}(f) denote the degree of f as a polynomial in z with coefficients in k[Δz, Δ^{2}z, …, Δ^{n−1}z]. The special property of the distinguished variable z, and the corresponding triangular basis, which we will exploit, is the fact that deg_{z}(σ(f)) = deg_{z}(f).

Consider a Z/p-module W. Decompose W into a direct sum of indecomposable Z/p-summands:

W = ⨁_{i=1}^{t} W_{i}

where W_{i} ≅ V_{dim(W_{i})} for all i. For each i choose a distinguished variable z_{i} ∈ W_{i}^{∗} and use the corresponding triangular basis for W_{i}^{∗}. Let N_{i} denote the norm of z_{i}. Thus N_{i} = z_{i} if W_{i} ≅ V_{1} and N_{i} := ∏_{j=1}^{p} σ^{j}(z_{i}), otherwise.

Let f ∈ k[W]^{Z/p}. Since N_{1}, considered as a polynomial in z_{1}, is monic we may divide N_{1} into f to obtain the unique decomposition f = f_{1}N_{1} + r_{1} where the remainder r_{1} has degree at most p−1 in the variable z_{1}. Next we divide r_{1} by N_{2} to obtain a decomposition: f = f_{1}N_{1}+f_{2}N_{2}+r_{2} where deg_{z_{1}}(f_{2}) < p, deg_{z_{1}}(r_{2}) < p and deg_{z_{2}}(r_{2}) < p. Continuing in this manner we obtain a decomposition

f = f_{1}N_{1} + f_{2}N_{2} + … + f_{t}N_{t} + r

where deg_{z_{i}}(f_{j}) < p for all i < j and deg_{z_{i}}(r) < p for all i. Note that r is the normal form of f with respect to the Gröbner basis {N_{1}, N_{2}, …, N_{t}} of the ideal (N_{1}, N_{2}, …, N_{t})k[W]. Furthermore the decomposition f = f_{1}N_{1} + f_{2}N_{2} + … + f_{t}N_{t} + r is a normal decomposition of f with respect to this Gröbner basis. We will call this the norm decomposition of f. Note that the norm decomposition depends upon the choice of the z_{i} but is otherwise unique.

Let k[W]^{♭} := {r ∈ k[W] | deg_{z_{i}}(r) < p for all i = 1, 2, …, t}. Thus k[W]^{♭} is the set of functions f having all coefficients f_{i} = 0 in its norm decomposition. Note that k[W]^{♭} is Z/p-stable.

The ring k[W] has a multi-grading given by the degrees in each W_{i}, that is, induced by k[W] ≅ k[W_{1}] ⊗ k[W_{2}] ⊗ ⋯ ⊗ k[W_{t}]. The action of Z/p preserves this grading and thus k[W]^{Z/p} and k[W]^{♭} inherit this grading.

3. SAGBI bases

We use the convention that a monomial is a product of variables and that a term is a monomial with a non-zero coefficient. We direct the reader to [6, Chapter 2] for a detailed discussion of monomial orders. For f ∈ S we use LT(f) to denote the lead term of f and LM(f) to denote the lead monomial of f.

Suppose that R is a subalgebra of S. Let LT(R) denote the vector space spanned by the lead terms of elements of R. Then LT(R) is a subalgebra of S. If C is a subset of R then let LM(C) denote the set of lead monomials of elements of C. If C is a subset of R such that LM(C) generates the algebra LT(R) then C generates R and C is called a SAGBI basis for R. For a detailed discussion of SAGBI bases see [24], [16] or [30, Chapter 11].

Taking C = R gives a SAGBI basis for R. Thus every subalgebra has a SAGBI basis. However, subalgebras of S are not necessarily finitely generated. If LT(R) is not finitely generated then R does not have a finite SAGBI basis (at least using the given monomial order). Even if R is finitely generated, LT(R) may fail to be finitely generated. In fact, as shown by Göbel [13, Lemma 2.1], the ring of invariants of the permutation representation of the alternating group on three letters does not have a finite SAGBI basis using the lexicographic order. Although the characterization of subalgebras which admit a finite SAGBI basis remains an important open problem, there are some circumstances which guarantee the existence of a finite SAGBI basis.

Lemma 3.1: Suppose {h_{1}, …, h_{n}} is a homogeneous system of parameters for k[V] with LM(h_{i}) = x_{i}^{d_{i}}. If A ⊆ k[V] is a subalgebra with {h_{1}, …, h_{n}} ⊆ A, then A has a finite SAGBI basis.

Proof: Since {h_{1}, …, h_{n}} ⊆ A and LM(h_{i}) = x_{i}^{d_{i}}, {x_{1}^{d_{1}}, …, x_{n}^{d_{n}}} ⊆ LT(A). Therefore k[x_{1}^{d_{1}}, …, x_{n}^{d_{n}}] is contained in LT(A). Furthermore the set {x_{1}^{d_{1}}, …, x_{n}^{d_{n}}} is a homogeneous system of parameters for k[V]. Thus LT(A) is a submodule of the finitely generated module k[V] over the Noetherian algebra k[x_{1}^{d_{1}}, …, x_{n}^{d_{n}}]. Hence LT(A) is a finite module over k[x_{1}^{d_{1}}, …, x_{n}^{d_{n}}]. Since LT(A) is generated by monomials, we may choose the module generators to be monomials. For each module generator α ∈ LT(A) choose an element f ∈ A with LT(f) = α. These elements along with h_{1}, …, h_{n} form a SAGBI basis for A. □

Choose an order with x_{1} < x_{2} < ⋯ < x_{n}. We call the representation of G triangular if LM(g(x_{i})) = x_{i} for every g ∈ G. If we view the variables as column vectors then the elements of G are upper-triangular matrices.

Theorem 3.2: If the representation of G is triangular then k[V]^{G} has a finite SAGBI basis.

Proof: Since LM(g(x_{i})) = x_{i}, we see that LM(N(x_{i})) = x_{i}^{d_{i}} where d_{i} is the index of the isotropy subgroup G_{x_{i}}. Thus {N(x_{1}), …, N(x_{n})} ⊂ k[V]^{G} is a homogeneous system of parameters for k[V] satisfying the hypotheses of Lemma 3.1. Therefore k[V]^{G} has a finite SAGBI basis. □

Corollary 3.3: Suppose G is a p-group and k has characteristic p. Then there is a choice of basis and monomial order with respect to which k[V]^{G} has a finite SAGBI basis.

Proof: Under these conditions G is conjugate to a subgroup of the upper-triangular matrices. If k is finite then the set of upper-triangular matrices with 1’s along the diagonal form a p-Sylow subgroup of GL(n, k). In this case G is clearly conjugate to a subgroup of this p-Sylow group. For more general fields, first observe that G has a composition series, {e} = G_{0} ≤ G_{1} ≤ ⋯ ≤ G_{m} = G, whose factors are all isomorphic to Z/p. As a simple consequence of the Jordan canonical form of a generator, every representation of Z/p over a field of characteristic p has a fixed line. Thus, for any representation of G over a field of characteristic p, say W, using the fact that (W^{G_{i}})^{G_{i+1}/G_{i}} = W^{G_{i+1}}, we conclude that dim(W^{G}) ≥ 1. The proof now proceeds by induction on dim(W) − dim(W^{G}). Clearly if dim(W) − dim(W^{G}) = 0, any basis is triangular. Suppose dim(W) − dim(W^{G}) > 0 and consider the G-module W/W^{G}. Since dim(W/W^{G}) − dim ((W/W^{G})^{G}) ≤ dim(W) − dim(W^{G}) − 1, the induction hypothesis gives a triangular basis for W/W^{G}. Lifting the elements of this basis to W and adjoining elements which form a basis for W^{G} gives a triangular basis for W. □

4. The vector invariants of V_{2}

Let k be field of characteristic p and let V_{n} denote the n dimensional indecomposable representation of Z/p. The ladder technique for computing rings of invariants of p-groups described in Section 7 relies heavily upon computing Z/p-invariants. One step in this method requires the construction of a surjection from a polynomial ring, A := k[a_{1}, …, a_{t}], onto a ring of invariants, k[V]^{Z/p}. In order to minimize the complexity of the ladder computation it is desirable to minimize the Krull dimension of A and this usually means having a minimal set of generators for k[V]^{Z/p}.

As discussed in Section 1, the problem of constructing a manageable generating set for k[V]^{Z/p} is, in general, quite difficult. If V = mV_{2}⊕ℓV_{1} then, since k[mV_{2}⊕ ℓV_{1}]^{Z/p} ≅ k[mV_{2}]^{Z/p} ⊗ k[ℓV_{1}], the problem reduces to constructing a generating set for k[mV_{2}]^{Z/p}. This was done by Campbell & Hughes in [5]. However the generating set given in [5] is usually not a minimal set. The current section is devoted to identifying a minimal generating set for k[mV_{2}]^{Z/p}. The results for m = 3 will play a role in Section 8.

Choose a basis {x_{i}, y_{i} | i = 1, …, m} for mV_{2}^{∗} with Δ(y_{i}) = x_{i}. Define N_{i} := N(y_{i}) and, for i < j, define u_{ij} := x_{j}y_{i} − x_{i}y_{j}. For m = 2, k[mV_{2}]^{Z/p} is generated by x_{1}, x_{2} N_{1}, N_{2} and u_{12}. This is clearly a minimal generating set. For m > 2 the generating set must include some elements from the image of the transfer, Im Tr^{Z/p}. In particular Tr^{Z/p}(y_{1}^{p−1} ⋯ y_{m}^{p−1}) is not contained in the subalgebra generated by invariants of lower degree (see [23] or [5]) and m(p − 1) is the least upper bound on the degrees of a generating set. Using the homogeneous system of parameters consisting of x_{i}, N_{i}, we see that the factors of (y_{1}y_{2} ⋯ y_{m})^{p−1} generate k[mV_{2}] as a k[mV_{2}]^{Z/p}-module and, therefore, the ideal Im Tr^{Z/p} is generated by

{Tr^{Z/p} (y_{1}^{e_{1}} ⋯ y_{m}^{e_{m}}) | e_{1} ≤ p − 1, e_{2} ≤ p − 1, …, e_{m} ≤ p − 1}.

Campbell & Hughes showed that this set of transfers together with the x_{i}, N_{i} and u_{ij} generate the ring of invariants, k[mV_{2}]^{Z/p}.

Suppose E = (e_{1}, …, e_{m}) is a sequence of non-negative integers. Let x^{E} := x_{1}^{e_{1}} ⋯ x_{m}^{e_{m}}, y^{E} := y_{1}^{e_{1}} ⋯ y_{m}^{e_{m}}, and |E| := e_{1} + ⋯ + e_{m}. If J = (j_{1}, …, j_{m}) is a second sequence of non-negative integers then we say that J ≤ E if j_{i} ≤ e_{i} for i = 1, …, m and we denote \binom{E}{J} := ∏_{i=1}^{m} \binom{e_{i}}{j_{i}}. Thus

Tr^{Z/p}(y^{E}) = ∑_{c∈F_{p}} ∏_{i=1}^{m} (y_{i} + cx_{i})^{e_{i}}

= ∑_{c∈F_{p}} ∏_{i=1}^{m} ∑_{j_{i}=0}^{e_{i}} c^{j_{i}} \binom{e_{i}}{j_{i}}x_{i}^{j_{i}}y_{i}^{e_{i}−j_{i}}

= ∑_{J≤E} (∑_{c∈F_{p}} c^{|J|}) \binom{E}{J}x^{J}y^{E−J}.  (1)

Recall that

∑_{c∈F_{p}} c^{|J|} = \{ \begin{matrix}−1 & if |J| = k(p − 1) for some k > 0; \\ 0 & otherwise. \end{matrix}

Thus Tr^{Z/p}(y^{E}) = 0 if |E| < p − 1. Introduce a bidegree on k[mV_{2}] by taking x_{i} to have bidegree (1, 0) and y_{i} to have bidegree (0, 1). If |E| < 2(p − 1), then Tr^{Z/p}(y^{E}) is homogeneous with bidegree (p − 1, |E| − p + 1). If |E| = 2(p − 1) then Tr^{Z/p}(y^{E}) + x^{E} is homogeneous with bidegree (p − 1, p − 1).

Let R denote the subalgebra of k[mV_{2}] generated by x_{i}, N_{i} and u_{ij}. If we use a graded reverse lexicographic order with x_{i} < y_{i} and x_{i} < x_{i+1} then the only non-trivial tête-a-têtes^{†} are of the form u_{ij}^{p} − x_{j}^{p}N_{i}. These tête-a-têtes subduct to zero using the relation u_{ij}^{p} − x_{j}^{p}N_{i} + x_{i}^{p}N_{j} − (x_{i}x_{j})^{p−1}u_{ij}. Thus {x_{i}, N_{i}, u_{ij} | i = 1, …, m and i < j ≤ m} is a SAGBI basis for R.

Lemma 4.1: If f ∈ R is homogeneous of bidegree (i, j) with j < p, then f is in the subalgebra generated by {x_{i}, u_{ij} | i = 1, …, m and i < j ≤ m}.

Proof: Suppose f is a minimal counter-example where minimal is defined using the partial order induced on R by the monomial order. Using the SAGBI basis for R, the lead monomial of f is of the form LM(x^{I}u^{J}N^{K}). However LM(f) has bidegree (i, j) with j < p. Thus K = 0. Furthermore, x_{i} and u_{ij} are homogeneous with respect to the bidegree. Thus f − x^{I}u^{J} is still a homogeneous element of R with bidegree (i, j). Clearly f > f − x^{I}u^{J}, contradicting the minimality hypothesis. □

The following lemma shows that the transfers of the form Tr^{Z/p}(y^{E}) with |E| ≤ 2(p − 1) are not required as generators of k[mV_{2}]^{Z/p}.

Lemma 4.2: The algebras k[mV_{2}]^{Z/p} and R agree in degrees less than or equal to 2(p − 1), i.e., (k[mV_{2}]^{Z/p})_{i} = R_{i} for i = 0, 1, …, 2(p − 1).

Proof: The proof is by induction on degree. Clearly the two algebras agree in degree zero. If |E| < p − 1, then Tr^{Z/p}(y^{E}) = 0. If |E| = p − 1, then Tr^{Z/p}(y^{E}) = −x^{E} ∈ R. Therefore the algebras agree in degrees less than or equal to p − 1. Consider Tr^{Z/p}(y^{E}) = Tr^{Z/p}(y_{1}^{e_{1}} ⋯ y_{ℓ}^{e_{ℓ}}) with e_{ℓ} ≠ 0 and p − 1 < |E| ≤ 2(p − 1). Work modulo the ideal (x_{ℓ})k[mV_{2}]. Using the definition of Tr^{Z/p},

Tr^{Z/p}(y_{1}^{e_{1}} ⋯ y_{ℓ}^{e_{ℓ}}) ≡ y_{ℓ}^{e_{ℓ}} Tr^{Z/p}(y_{1}^{e_{1}} ⋯ y_{ℓ−1}^{e_{ℓ−1}}) (mod (x_{ℓ})k[mV_{2}]).

By the induction hypothesis, Tr^{Z/p}(y_{1}^{e_{1}} ⋯ y_{ℓ−1}^{e_{ℓ−1}}) lies in R. Furthermore, since

†Given a set of algebra generators, say C, and a pair of polynomials, f and h, each given by a product of an element of k with elements of C, then if LT(f) = LT(h) the polynomial f − h is called a tête-a-tête. If no element of C divides both f and h, then the tête-a-tête is said to be non-trivial. See [24, p. 71]

e_{1} + ⋯ + e_{ℓ−1} < 2(p − 1), Tr^{Z/p}(y_{1}^{e_{1}} ⋯ y_{ℓ−1}^{e_{ℓ−1}}) is homogeneous of bidegree (p − 1, |E|−(p−1)−e_{ℓ}). Thus by Lemma 4.1 Tr^{Z/p}(y_{1}^{e_{1}} ⋯ y_{ℓ−1}^{e_{ℓ−1}}) lies in the subalgebra generated by x_{i} and u_{ij}. Since 2(p−1) ≥ |E|, we have p−1−e_{ℓ} ≥ |E|−(p−1)−e_{ℓ} and each monomial in Tr^{Z/p}(y_{1}^{e_{1}} ⋯ y_{ℓ−1}^{e_{ℓ−1}}) has at least e_{ℓ} more x’s than y’s. Thus

Tr^{Z/p}(y_{1}^{e_{1}} ⋯ y_{ℓ−1}^{e_{ℓ−1}}) = ∑_{|I|=e_{ℓ}} x^{I}f_{I}

with I = (i_{1}, …, i_{ℓ−l}) and f_{I} in the subalgebra generated by x_{i} and u_{ij}. Let u_{ℓ}^{I} := ∏_{j=1}^{ℓ−1} u_{jℓ}^{i_{j}}. Then u_{ℓ}^{I} ≡ y_{ℓ}^{e_{ℓ}}x^{I} (mod (x_{ℓ})k[mV_{2}]) and

Tr^{Z/p}(y^{E}) ≡ ∑_{I} u_{ℓ}^{I}f_{I} (mod (x_{ℓ})k[mV_{2}]).

Thus

Tr^{Z/p}(y^{E}) = ∑_{I} u_{ℓ}^{I}f_{I} + x_{ℓ}h

for some h ∈ k[mV_{2}]. However x_{ℓ}h = Tr^{Z/p}(y^{E}) − ∑_{I} u_{ℓ}^{I}f_{I} ∈ k[mV_{2}]^{Z/p}. Hence h ∈ k[mV_{2}]^{Z/p}. Furthermore the degree of h is |E| − 1. Thus by the induction hypothesis, h ∈ R. Therefore Tr^{Z/p}(y^{E}) ∈ R. □

The next lemma shows that each of the transfers Tr^{Z/p}(y^{E}) with e_{i} ≤ p − 1 and |E| > 2(p − 1) where E = (e_{1}, …, e_{m}) is required in our minimal generating set.

Lemma 4.3: If E = (e_{1}, …, e_{m}), e_{i} ≤ p − 1 and |E| > 2(p − 1) then Tr^{Z/p}(y^{E}) is indecomposable.

Proof: We use the graded reverse lexicographic order with x_{1} < x_{2} < ⋯ < x_{m} < y_{1} < ⋯ < y_{m}. Suppose, by way of contradiction, that Tr^{Z/p}(y^{E}) = c_{1}m_{1} + c_{2}m_{2} +⋯+c_{r}m_{r} were each m_{i} is a non-trivial product of generators, each c_{i} ∈ k and LM(m_{i}) ≥ LM(m_{i+1}). Either LM(m_{1}) = LM(Tr^{Z/p}(y^{E})) or LM(m_{1}) = LM(m_{2}) > LM(Tr^{Z/p}(y^{E})). From Equation 1 we see that LM(Tr^{Z/p}(y^{E})) = x^{I}y^{E−I} where x^{I} = max{x^{J} | |J| = p − 1 and J ≤ E}.

We first show that LM(m_{1}) ≠ LM(Tr^{Z/p}(y^{E})). Since e_{i} ≤ p − 1, LM(N_{i}) = y_{i}^{p} does not divide x^{I}y^{E−I}. Note that x^{I}y^{E−I} has bidegree (p − 1, |E| − (p + 1)) and |E| − (p − 1) > p − 1. Therefore x^{I}y^{E−I} does not factor using only elements from {x_{i}, LM(u_{ij}) | i = 1, …, m and i < j ≤ m}. If we try to factor x^{I}y^{E−I} using LM(Tr^{Z/p}(y^{F})) with F = (f_{1}, …, f_{m}), f_{i} ≤ p − 1 and |F | ≥ p − 1, then the complement is y^{F −E}. However, since f_{i} − e_{i} ≤ p − 1, this is not the lead term of a product of generators. Thus LM(m_{1}) ≠ LM(Tr^{Z/p}(y^{E})).

Now suppose LM(m_{1}) = LM(m_{2}) > LM(Tr^{Z/p}(y^{E})). This means that m_{1} and m_{2} form a tête-a-tête. However, for every non-trivial tête-a-tête formed from the generators, the leading monomial has bidegree (d_{1}, d_{2}) with d_{1} > p − 1. Thus LM(m_{1}) < x^{I}y^{E−I} giving the required contradiction. □

Putting together Lemma 4.2 and Lemma 4.3 we obtain the following corollary.

Corollary 4.4: The set

{x_{i}, N_{i}, u_{ij} | i = 1, … m and i < j ≤ m}

⋃ {Tr^{Z/p} (y_{1}^{e_{1}} ⋯ y_{m}^{e_{m}}) | e_{i} ≤ p − 1 and e_{1} + ⋯ + e_{m} > 2(p − 1)}

is a minimal generating set for k[mV_{2}]^{Z/p}.

5. Localization and normalization

Let R be a finitely generated algebra. Throughout this section we will further suppose that R contains no zero-divisors. We denote by R_{f} the localization of R with respect to the multiplicative set generated by f.

The following theorem is essentially [25, 15.11]. See also [31, Lemma 4.6.10].

Theorem 5.1: Suppose that A is a subalgebra of R and that f_{1}, f_{2} is a regular sequence in A such that A_{f_{1}} = R_{f_{1}} and A_{f_{2}} = R_{f_{2}}. Then A = R.

Proof: Take h ∈ R. Since R ⊆ R_{f_{i}} = A_{f_{i}} we may write h = a_{1}/f_{1}^{n} and h = a_{2}/f_{2}^{m} for some a_{1}, a_{2} ∈ A and n, m ∈ Z. Thus a_{1}f_{2}^{m} = a_{2}f_{1}^{n}. Since f_{1}, f_{2} is a regular sequence in A so also is f_{1}^{n}, f_{2}^{m}. This implies that a_{2} = af_{2}^{m} for some a ∈ A. Therefore h = af_{2}^{m}/f_{2}^{m} = a lies in A. □

Lemma 5.2: Suppose that A is a subalgebra of R and I is an ideal of R. If I ⊆ A and f ∈ \sqrt{I} ∩ A then A_{f} = R_{f}.

Proof: There exists m ∈ Z such that f^{m} ∈ I. Take h ∈ R_{f} and write h = r/f^{k} with r ∈ R. Then rf^{m} = hf^{k+m} ∈ I ⊆ A and h = rf^{m}/f^{k+m} ∈ A_{f}. □

Corollary 5.3: Suppose A is a subalgebra of k[V]^{G} and that A contains the image of the transfer, Im Tr^{G}. If there exist f_{1}, f_{2} ∈ \sqrt{Im Tr^{G}} ∩A such that f_{1}, f_{2} is a regular sequence on A, then A = k[V]^{G}.

Remark 5.4: Suppose that A is a subalgebra of k[V]^{G} containing the image of the transfer. Further suppose that f_{1} and f_{2} are non-associate primes of A lying in \sqrt{Im Tr^{G}}. A relatively routine calculation shows that f_{1}, f_{2} is a regular sequence in A and so A = k[V]^{G}.

Remark 5.5: Let \overline{k} denote the algebraic closure of k and define \overline{V} := \overline{k}⊗V. By [27, Theorem 2.1], we have that \sqrt{Im Tr^{G}} consists of those invariant polynomials in k[V] which vanish on the subvariety V of \overline{V} defined by V = ∪_{σ∈Σ}\overline{V}^{σ} where Σ consists of all the elements σ of G of order p. Thus given an element f ∈ k[V]^{G} we may check that f ∈ \sqrt{Im Tr^{G}} by verifying that f vanishes on \overline{V}^{σ} for every element σ ∈ G of order p.

Suppose that f_{1} and f_{2} are elements of an algebra A. The syzygy module, syz_{A}(f_{1}, −f_{2}), is the kernel of the map from A^{2} to A which takes (c_{1}, c_{2}) to c_{1}f_{1} − c_{2}f_{2}.

Lemma 5.6: Suppose that A is a subalgebra of k[V], {f_{1}, f_{2}} ⊂ A, and f_{1}, f_{2} is a regular sequence in k[V]. Then f_{1}, f_{2} is regular on A if and only if syz_{A}(f_{1}, −f_{2}) is a principal A-module.

Proof: Suppose that f_{1}, f_{2} is regular on A. For an arbitrary (a, b) ∈ syz_{A}(f_{1}, −f_{2}), we have af_{1} − bf_{2} = 0. Thus bf_{2} = af_{1}. Since f_{1}, f_{2} is regular on A, there exists c ∈ A such that b = cf_{1}. Thus f_{1}(a − cf_{2}) = af_{1} − cf_{1}f_{2} = 0. Since f_{1} is not a zero divisor, we conclude a = cf_{2}. Therefore (a, b) = c(f_{2}, f_{1}) and syz_{A}(f_{1}, −f_{2}) is the principal A-module generated by (f_{2}, f_{1}).

Suppose that syz_{A}(f_{1}, −f_{2}) is a principal A-module. Clearly (f_{2}, f_{1}) ∈ syz_{A}(f_{1}, −f_{2}). Furthermore, since f_{1}, f_{2} is a regular sequence in k[V], f_{1} and f_{2} have no positive degree common factors. Thus syz_{A}(f_{1}, −f_{2}) is the principal A-module generated by (f_{2}, f_{1}). By hypothesis, f_{1} is not a zero divisor in A. Thus to show that f_{1}, f_{2} is regular on A, it is sufficient to show that f_{2} is not a zero divisor on A/(f_{1})A. For an arbitrary a, b ∈ A with bf_{2} = af_{1}, we have (a, b) ∈ syz_{A}(f_{1}, −f_{2}). Thus (a, b) = c(f_{2}, f_{1}) for some element c ∈ A. Therefore b = cf_{1} as required. □

Algorithm 5.7: Suppose that V is a triangular representation of G and the height of the image of the transfer is at least 2. Then a generating set for k[V]^{G} can be constructed as follows.

Step 1: Use the homogeneous system of parameters {x_{1}, N(x_{2}), N(x_{3}), …, N(x_{n})}. Note that LM(N(x_{i})) = x_{i}^{d_{i}} where d_{i} is the index of the isotropy subgroup, G_{x_{i}}, in G. Thus the monomials dividing x_{2}^{d_{2}−1}x_{3}^{d_{3}−1} ⋯ x_{n}^{d_{n}−1} are a basis for k[V] over k[x_{1}, N(x_{2}), …, N(x_{n})] and

T := {Tr^{G}(β) | β divides x_{2}^{d_{2}−1} ⋯ x_{n}^{d_{n}−1}}

is a generating set for Im Tr^{G} as a module over R := k[x_{1}, N(x_{2}), …, N(x_{n})].

Step 2: Choose f_{1}, f_{2}, a partial homogeneous system of parameters for S with \{f_{1}, f_{2}\} ⊂ \sqrt{Im Tr^{G}} and take C := \{f_{1}, f_{2}, x_{1}, N(x_{2}), …, N(x_{n})\} ∪ T .

Step 3 : Take A to be the subalgebra of k[V]^{G} generated by C.

Step 4 (optional): If one of the generators for A divides another, perform the division and add the quotient to C. Remove redundant generators from C.

Step 5: Compute a generating set for the syzygy module syz_{A}(f_{1}, −f_{2}). This module is the kernel of the map from A^{2} to A which takes (c_{1}, c_{2}) to c_{1}f_{1} − c_{2}f_{2}. The syzygy module computation involves the construction of a SAGBI basis for A. It follows from Lemma 3.1 that A has a finite SAGBI basis. The details of the syzygy module computation, for algebras with a finite SAGBI basis, can be found in [20, Section 5].

Step 6: If syz_{A}(f_{1}, −f_{2}) is a principal A-module, stop. From Lemma 5.6, the sequence f_{1}, f_{2} is regular on A if and only if syz_{A}(f_{1}, −f_{2}) is a principal module. Therefore, using Corollary 5.3, if syz_{A}(f_{1}, −f_{2}) is principal then A = k[V]^{G}.

Step 7: By construction, f_{1}, f_{2} is a regular sequence in k[V]^{G} and therefore syz_{k[V]^{G}}(f_{1}, −f_{2}) is a principal k[V]^{G}-module with generator (f_{2}, f_{1}). Hence for each generator, (h_{1}, h_{2}) ∈ syz_{A}(f_{1}, −f_{2}) ⊆ syz_{k[V]^{G}}(f_{1}, −f_{2}), we have (h_{1}, h_{2}) = c(f_{2}, f_{1}) for c = h_{1}/f_{2} ∈ k[V]^{G}. For each generator, add the corresponding c to C. Go to Step 3.

Proof: The algorithm generates an increasing sequence of R-submodules of the Noetherian R-module k[V]^{G}. Therefore the algorithm terminates. □

Remark 5.8: In practice, it is probably best to combine Algorithm 5.7 with a certain amount of “preprocessing”. We can start Step 5 with any subalgebra of k[V]^{G} containing the set C. One might as well include any known invariants before computing the generators of the syzygy module.

Remark 5.9: In Algorithm 5.7, we need the hypothesis that the height of the image of the transfer is at least 2 in order to guarantee the existence of a suitable f_{1} and f_{2}. We can rephrase this restriction. The reduced variety corresponding to the image of the transfer is the set V described in Remark 5.5. The height of the ideal Im Tr^{G} is the codimension of the variety, V. The codimension of a union of subspaces is the minimum of the codimensions of the subspaces. We wish to exclude height 1. This means that \overline{V}^{σ}, or equivalently V^{σ}, must have codimension at least 2 for every σ ∈ G of order p. The subspace V^{σ} is a codimension 1 subspace of V if and only if σ is a (pseudo) reflection of order p (i.e., a transvection). Thus Algorithm 5.7 applies as long as the representation V is triangular and G contains no transvections.

Remark 5.10: There is a variation on Algorithm 5.7 in which the finitely generated algebra A is identified with a quotient k[W]/I. The syzygy module calculation can then be performed in k[W]. This means that SAGBI basis is not required and the representation need not be triangular. However, when using this approach it is necessary to construct a generating set for the ideal I. This can be quite difficult.

Example 5.11: Suppose k has characteristic p > 2 and consider k[V_{2} ⊕ V_{3}]^{Z/p}. Let {y_{1}, x_{1}} be a triangular basis of V_{2}^{∗} where y_{1} is a distinguished variable. Let {z_{2}, y_{2}, x_{2}} be a triangular basis of V_{3}^{∗} where z_{2} is a distinguished variable. A relatively simple calculation gives Tr^{Z/p}(y_{i}^{p−1}) = −x_{i}^{p−1}. Thus x_{1} and x_{2} lie in the radical of the image of the transfer and we may apply Algorithm 5.7 with f_{1} := x_{1} and f_{2} := x_{2}. For the initial iteration, A is generated by the homogeneous system of parameters and the image of the transfer. However, using [28, Section 5], we know that k[V_{2} ⊕ V_{3}]^{Z/p} contains three “rational invariants”, u := y_{1}x_{2} −y_{2}x_{1}, d := y_{2}^{2} −2x_{2}z_{2} −x_{2}y_{2} and w := y_{1}^{2}x_{2} +x_{1}y_{1}x_{2} −2x_{1}y_{1}y_{2} +x_{1}^{2}z_{2}, which, at least for general p, are not in A (this can be easily verified for small p using MAGMA [4]). Therefore x_{1}, x_{2} is not regular in A, syz_{A}(x_{1}, −x_{2}) is not a principal A-module and, referring to Remark 5.4, x_{1}, x_{2} are not distinct primes in A. To see this last fact directly note that Tr^{Z/p}(uy_{1}^{p−1}) = −ux_{1}^{p−1}, Tr^{Z/p}(wy_{2}^{p−1}) = −wx_{2}^{p−1} and Tr^{Z/p}(uwy_{2}^{p−1}) = −uwx_{2}^{p−1} are all in A. Thus (ux_{1}^{p−1})(wx_{2}^{p−1}) = (uwx_{2}^{p−1})(x_{1}^{p−2})x_{1} ∈ (x_{1})A. However neither ux_{1}^{p−1} nor wx_{2}^{p−1} lie in (x_{1})A. Therefore (x_{1})A is not a prime ideal. To see the algorithm in action, start with the fact that Tr^{Z/p}(uy_{1}^{i}y_{2}^{p−1−i}) = −ux_{1}^{i}x_{2}^{p−1−i} ∈ A. Thus (ux_{1}^{p−2}x_{2}, ux_{1}^{p−1}) lies in syz_{A}(x_{1}, −x_{2}) and the first iteration of Step 7 would add ux_{1}^{p−2} to A. However, judicious use of Step 4 would have produced a generating set and Step 5 would then have produced a principal module. A finite SAGBI basis for k[V_{2} ⊕ V_{3}]^{Z/p} is given in [28, Section 5].

See Remark 8.1 for a second example.

Remark 5.12: At the beginning of this section we assumed that R contains no zero-divisors. This was done to (slightly) simplify the discussion and because it is true for all our applications. However, this assumption is not really necessary; all that is required is that the elements f_{1} and f_{2} with respect to which we localized must not be zero-divisors.

6. Cohomology

To apply the method of ladders described in Section 7 to a modular representation will require group cohomology computations for the cyclic group Z/p. In this section we develop the group cohomology results we will need.

For a Z/p-module M, the first cohomology group of Z/p with coefficients in M is given by

H^{1}(Z/p, M) = \frac{kernel(Tr |_{M})}{image( Δ|_{M} )}.

A Z/p-module decomposition of M gives a vector space decomposition of H^{1}(Z/p, M). Using the fact that Tr = Δ^{p−1}, we see that H^{1}(Z/p, V_{p}) = 0 and, for n < p, any element v with Δ^{n−1}v ≠ 0 represents a non-zero class in the one dimensional vector space H^{1}(Z/p, V_{n}). One way to identify such an element is to chose a non-zero element u ∈ V_{n}^{Z/p} and then to find v such that Δ^{n−1}v = u. A detailed discussion of group cohomology may be found in [10].

Consider a Z/p-module W and decompose W into a direct sum of indecomposable Z/p-summands: W = ⨁_{i=1}^{t} W_{i} where W_{i} ≅ V_{dim(W_{i})} for all i. As usual we choose a distinguished variable z_{i} ∈ W_{i}^{∗} for each i and use the corresponding triangular basis for W_{i}^{∗}. Also as usual we let N_{i} denote the norm of z_{i}. Let B := k[N_{1}, …, N_{t}].

Suppose f ∈ k[W] and consider the norm decomposition f = f_{1}N_{1} + f_{2}N_{2} + ⋯ + f_{t}N_{t} + r. Since N_{i} ∈ k[W]^{Z/p}, Tr(f) = Tr(f_{1})N_{1} + ⋯ + Tr(f_{t})N_{t} + Tr(r) and Δ(f) = Δ(f_{1})N_{1} + ⋯ + Δ(f_{t})N_{t} + Δ(r). Thus f represents an element of H^{1}(Z/p, k[W]) if and only if f_{i} for i = 1, …, t and r represent elements of H^{1}(Z/p, k[W]). Similarly, [f] = 0 if and only if [f_{i}] = 0 for i = 1, …, t and [r] = 0.

Lemma 6.1: N_{i} acts injectively on H^{1}(Z/p, k[W]).

Proof: Suppose f represents an element of H^{1}(Z/p, k[W]) and N_{i}[f] = 0. Then fN_{i} = Δh for some h ∈ k[W]. Divide h by N_{i} to get h = qN_{i} +r with deg_{z_{i}}(r) < p. Thus Δ(h) = Δ(q)N_{i} + Δ(r) and, since Δ does not increase the z_{i}-degree, we have deg_{z_{i}}(Δ(r)) < p. However, we also have Δ(h) = fN_{i}. Therefore, using the uniqueness of the division algorithm, Δ(r) = 0 and Δ(q) = f. Thus [f] = 0. □

Note that if U is a vector space over k, then B ⊗_{k} U is a free B-module of rank dim(U).

Proposition 6.2: H^{1}(Z/p, k[W]) is isomorphic to the free B-module B ⊗_{k} H^{1}(Z/p, k[W]^{♭}).

Proof: Note that k[W] is isomorphic to the free B-module B ⊗_{k} k[W]^{♭}. Also k[W]^{♭} is a Z/p-submodule of k[W] and B ⊆ k[W]^{Z/p}. Thus

H^{1}(Z/p, k[W]) ≅ H^{1}(Z/p, B ⊗_{k} k[W]^{♭}) ≅ B ⊗_{k} H^{1}(Z/p, k[W]^{♭})

as required. □

The multi-grading of k[W] described near the end of Section 2 is inherited by H^{1}(Z/p, k[W]).

Corollary 6.3: The free B-module H^{1}(Z/p, k[W]) is generated in multi-degrees (d_{1}, …, d_{t}) with d_{i} ≤ p − dim(W_{i}).

Proof: It follows from Proposition 6.2 that a vector space basis for H^{1}(Z/p, k[W]^{♭}) gives a generating set for the free B-module H^{1}(Z/p, k[W]). Furthermore one can choose a basis for H^{1}(Z/p, k[W]^{♭}) with one basis element for each non-free Z/p-module summand of k[W]^{♭}. In fact the basis element can be represented by an element of the summand and has the same multi-degree as the summand. It is easy to see that

k[W]_{(d_{1},...,d_{t})}^{♭} ≅ k[W_{1}]_{d_{1}}^{♭} ⊗ k[W_{2}]_{d_{2}}^{♭} ⊗ ⋯ ⊗ k[W_{t}]_{d_{t}}^{♭}.

From [1] (see also [15, Lemma 2.10]), k[V_{n}]_{d}^{♭} is a free Z/p-module for all d ≥ p − n + 1. Since the tensor product of any finite dimensional Z/p-module with a free Z/p-module is free (see, for example, [2, II §7 Lemma 4]), we see that k[W]_{(d_{1},...,d_{t})}^{♭} is free if any d_{i} ≥ p − dim(W_{i}) + 1. Therefore the B-module generators lie in multi-degrees with d_{i} < p − dim(W_{i}) + 1. □

In Section 8 we will need to understand H^{1}(Z/p, k[3V_{2}]) as a module over k[3V_{2}]^{Z/p}. We use the notation introduced in Section 4. Using Proposition 6.2 and Corollary 6.3, it is sufficient to consider k[3V_{2}] in multi-degrees (d_{1}, d_{2}, d_{3}) with d_{i} ≤ p − 2. If p = 2, we are left with a single generator in degree zero and H^{1}(Z/2, k[3V_{2}]) is isomorphic to k[N(y_{1}), N(y_{2}), N(y_{3})].

Assume p > 2. Let M denote the subspace of k[V_{2}] given by M = ∑_{d=0}^{p−2} k[V_{2}]_{d}. As a graded Z/p-module, M is isomorphic to V_{1} ⊕ V_{2} ⊕ ⋯ ⊕ V_{p−1}. Furthermore H^{1}(Z/p, k[3V_{2}]^{♭}) ≅ H^{1}(Z/p, M^{⊗3}). Thus it is sufficient to consider (V_{1} ⊕ V_{2} ⊕ ⋯ ⊕ V_{p−1})^{⊗3}. In degree 1 this gives 3V_{2}. A basis for the invariants is given by {x_{1}, x_{2}, x_{3}} and a basis for the cohomology is given by {[y_{1}], [y_{2}], [y_{3}]}. In multidegree (1, 1, 0) we have V_{2} ⊗V_{2} ≅ V_{3} ⊕V_{1} (see [2, p. 50]). Since Δ^{2}(y_{1}y_{2}) = 2x_{1}x_{2} we can choose y_{1}y_{2} as a generator for V_{3} and u_{12} = y_{1}x_{2} − x_{1}y_{2} as a generator for V_{1}. The analogous results hold for the multi-degrees (1, 0, 1) and (0, 1, 1). In multi-degree (ℓ, 0, 0) with ℓ < p − 1 we have V_{ℓ+1} generated by y_{1}^{ℓ}.

In the next lemma we consider how the invariants x_{i} and x_{i}^{2} act on the cohomology classes represented by certain simple monomials.

Lemma 6.4: Suppose ℓ < p − 1 and i ≠ j. Then x_{i}[y_{i}^{ℓ}] = 0 and, if ℓ > 0, x_{i}^{2}[y_{j}y_{i}^{ℓ−1}] = 0, but x_{i}[y_{j}^{ℓ}] ≠ 0.

Proof: We first show that x_{i}[y_{i}^{ℓ}] = 0. The proof is by induction on ℓ. For ℓ = 0 we have Δ(y_{i}) = x_{i} and thus x_{i}[1] = 0. Suppose 0 < ℓ < p − 1. Using the definition of Δ we have

Δ(y_{i}^{ℓ+1}) = (ℓ + 1)x_{i}y_{i}^{ℓ+1} + ∑_{t=2}^{ℓ} \binom{ℓ + 1}{t}x_{i}^{t}y_{i}^{ℓ+1−t}.

By induction, if t > 1 then x_{i}[y_{i}^{ℓ+1−t}] = 0. Since ℓ < p − 1, ℓ + 1 is a unit in k and x_{i}[y_{i}^{ℓ}] = 0.

Next we show that x_{i}^{2}[y_{j}y_{i}^{ℓ−1}] = 0 if ℓ > 0. First suppose i < j. Using our first result we have [u_{ij}x_{i}y_{i}^{ℓ−1}] = u_{ij}x_{i}[y_{i}^{ℓ−1}] = 0. Using the definition of u_{ij} gives x_{j}x_{i}[y_{i}^{ℓ}] − x_{i}^{2}[y_{j}y_{i}^{ℓ−1}] = 0. Thus x_{i}^{2}[y_{j}y_{i}^{ℓ−1}] = x_{j}x_{i}[y_{i}^{ℓ}] = 0. For i > j, repeat the argument using u_{ji}.

Finally we show x_{i}[y_{j}^{ℓ}] ≠ 0. Suppose first that j = 1 and i = 2. The multidegree component (ℓ, 1, 0) of M^{⊗3} is isomorphic to V_{ℓ+1} ⊗ V_{2} ≅ V_{ℓ+2} ⊕ V_{ℓ}. Note that for any f ∈ k[V], either Δ(f) = 0 or deg_{x}(Δ(f)) > deg_{x}(f). Therefore y_{1}^{ℓ}y_{2} generates V_{ℓ+2} and the corresponding invariant, Δ^{ℓ+1}y_{1}^{ℓ}y_{2} is a non-zero multiple of x_{1}^{ℓ}x_{2}. The second invariant in this bidegree is x_{1}^{ℓ−1}u_{12}. Thus the generator, h, of the summand isomorphic to V_{ℓ} lying in multi-degree (ℓ, 1, 0) must satisfy deg_{y}(h) ≥ ℓ. Therefore we may take this generator to be a linear combination of x_{1}y_{1}^{ℓ−1}y_{2} and y_{1}^{ℓ}x_{2}. Using the fact that Δ is a twisted derivation, Δ(y_{1}^{ℓ}y_{2}−y_{1}^{ℓ}x_{2}) = y_{2}Δ(y_{1}^{ℓ}) + x_{2}y_{1}^{ℓ} and x_{2}[y_{1}^{ℓ}] = −[y_{2}Δ(y_{1}^{ℓ})]. However

y_{2}Δ(y_{1}^{ℓ}) = y_{2} ∑_{t=1}^{ℓ} \binom{ℓ}{t}x_{1}^{t}y_{1}^{ℓ−t}

= ℓy_{2}x_{1}y_{1}^{ℓ−1} + x_{1}^{2} ∑_{t=2}^{ℓ} \binom{ℓ}{t}x_{1}^{t−2}y_{2}y_{1}^{ℓ−t}.

Thus using the fact that x_{1}^{2}[y_{2}y_{1}^{ℓ−t}] = 0 we have [y_{2}Δ(y_{1}^{ℓ})] = ℓx_{1}[y_{2}y_{1}^{ℓ−1}]. Combining this with our earlier calculation gives x_{2}[y_{1}^{ℓ}] = −ℓx_{1}[y_{2}y_{1}^{ℓ}]. Thus x_{2}[y_{1}^{ℓ}] + ℓx_{1}[y_{2}y_{1}^{ℓ}] = 0 but any other non-zero linear combination may be chosen to generate V_{ℓ−1}. Hence x_{2}[y_{1}^{ℓ}] ≠ 0. The analogous argument works for all choices of i and j. □

In multi-degree (1, 1, 1) we have V_{2} ⊗ V_{2} ⊗ V_{2} ≅ (V_{3} ⊕ V_{1}) ⊗ V_{2}. For p = 3 this is isomorphic to 2V_{3} ⊕ V_{2} and {[u_{12}y_{3}]} is a basis for the cohomology. For p > 3 we have V_{3} ⊗ V_{2} ≅ V_{4} ⊕ V_{2} and so the module is isomorphic to V_{4} ⊕ 2V_{2}. A simple computation gives Δ^{3}(y_{1}y_{2}y_{3}) = 6x_{1}x_{2}x_{3}. Also, note that y_{1}u_{23}−y_{2}u_{13}+y_{3}u_{12} = 0 and x_{1}u_{23} − x_{2}u_{13} + x_{3}u_{12} = 0. In this multi-degree a basis for the invariants is given by {x_{1}x_{2}x_{3}, x_{1}u_{23}, x_{3}u_{12}} and {[y_{1}y_{2}y_{3}], [y_{1}u_{23}], [u_{12}y_{3}]} is a basis for the cohomology.

Theorem 6.5: Take p = 3. Then H^{1}(Z/3, k[3V_{2}]) is the free k[N(y_{1}), N(y_{2}), N(y_{3})]- module generated by {[1], [y_{1}], [y_{2}], [y_{3}], [u_{12}], [u_{13}], [u_{23}], [u_{12}y_{3}]}. As a module over k[3V_{2}]^{Z/3}, H^{1}(Z/3, k[3V_{2}]) is generated by {[1], [y_{1}], [y_{2}], [y_{3}]} The action of k[3V_{2}]^{Z/3} is determined by x_{i}[y_{i}] = 0, x_{1}[y_{2}] = −x_{2}[y_{1}] = u_{12}[1], x_{1}[y_{3}] = −x_{3}[y_{1}] = u_{13}[1] x_{2}[y_{3}] = −x_{3}[y_{2}] = u_{23}[1], and u_{23}[y_{1}] = −u_{13}[y_{2}] = u_{12}[y_{3}].

Proof: For p = 3, M = V_{1} ⊕ V_{2} and each non-zero multi-degree was discussed in the paragraph preceding the statement of the theorem. It remains to show that the action of H^{1}(Z/3, k[V]) is as described. Note that Δ(y_{i}y_{j}−x_{i}y_{j}) = x_{i}y_{j}+x_{j}y_{i}. Thus, if i = j we have [x_{i}y_{i}] = 0 and if i < j we have [u_{ij}] = [x_{j}y_{i} − x_{i}y_{j}] = 2[x_{j}y_{i}] = −2[x_{i}y_{j}] and x_{i}[y_{j}] = −x_{j}[y_{i}] = u_{ij}[1]. A straight forward computation gives Δ(y_{1}y_{2}y_{3} − u_{12}y_{3} + u_{23}y_{1} − x_{1}x_{3}y_{2}) = y_{1}u_{23} − y_{3}u_{12}. Thus u_{23}[y_{1}] = u_{12}[y_{3}]. Since y_{1}u_{23} − y_{2}u_{13} + y_{3}u_{12} = 0, we have u_{13}[y_{2}] = u_{23}[y_{1}] + u_{12}[y_{3}] = −u_{12}[y_{3}]. □

7. Ladders

Suppose that p is the characteristic of k and G is a p-group. Then there exists a normal subgroup, N, with G/N isomorphic to the cyclic group of order p, Z/p. The ring of invariants is given by

k[V]^{G} = (k[V]^{N})^{G/N} = (k[V]^{N})^{Z/p} .

Suppose we have computed k[V]^{N}. More precisely, suppose we have a short exact sequence of Z/p-modules 0 → J ⟶^{i} A := k[a_{1}, …, a_{k}] ⟶^{ρ} k[V]^{N} → 0. This gives rise to a long exact sequence in group cohomology

0 → J^{Z/p} → A^{Z/p} → k[V]^{G}→H^{1}(Z/p, J) →^{i^{1}} H^{1}(Z/p, A)→ ….

All of the maps in this long exact sequence are A^{Z/p}-module maps. Furthermore k[V]^{G} is generated by ρ(A^{Z/p}) and the pre-image of the A^{Z/p}-module generators of the kernel of i^{1}.

We may choose the ring A so that A ≅ k[W] for some graded Z/p-module W. One way to do this is take W to be ⊕_{j=1}^{m}k[V]_{j}^{N} for a sufficiently large m. In practice one should choose W so as to minimize dim(W).

As in Section 6 we can decompose W, choose distinguished variables, and construct norms. Let B = k[N_{1}, …, N_{t}]. From Proposition 6.2, H^{1}(Z/p, A) is a finitely generated, free B-module.

Proposition 7.1: Im Tr^{G} ⊆ ρ(A^{Z/p}).

Proof: The action of G/N ≅ Z/p on A induces an action of the group ring of Z/p on A. Thus Tr acts on A and Tr(A) ⊆ A^{Z/p}. Since ρ is a map of Z/p-modules, we have ρ ∘ Tr = Tr ∘ρ. Furthermore, ρ(A) = k[V]^{N} implies Im Tr^{N} ⊆ ρ(A). Either by interpreting Tr as the relative transfer, Tr_{N}^{G} (see [27] for details), or by direct observation, we see that Tr^{G} = Tr ∘ Tr^{N}. Thus Im Tr^{G} = Tr(Im Tr^{N}) ⊆ Tr(ρ(A)) = ρ(Tr(A)) ⊆ ρ(A^{Z/p}). □

Remark 7.2: As a consequence of the proposition, as long as G/N, in its action on W, is not generated by a transvection, we can use ρ(A^{Z/p}) as input to Step 5 of Algorithm 5.7 (see Remark 5.8) to compute k[V]^{G}. Thus replacing the cohomology calculation with a syzygy module calculation.

For a p-group G there is a composition series {e} = G_{0}◁G_{1}◁G_{2}◁⋯◁G_{m+1} = G with G_{i+1}/G_{i} ≅ Z/p. Using the above method we may first compute k[V]^{G_{2}} = (k[V]^{G_{1}})^{G_{2}/G_{1}}. Then having computed k[V]^{G_{2}} we may again use the method to compute k[V]^{G_{3}} = (k[V]^{G_{2}})^{G_{3}/G_{2}}. Continuing in this manner we finally may compute k[V]^{G} = (k[V]^{G_{m}})^{G/G_{m}}. This iterated process is the ladder algorithm for computing k[V]^{G}. The strength of the ladder algorithm is that at each rung we are computing the invariants and group cohomology with respect to the relatively simple group G_{i+1}/G_{i} ≅ Z/p.

8. An example: k[2V_{3}]^{U_{3}}

Here we illustrate the ladder algorithm by using it to compute the invariants of an interesting representation of a non-abelian group of order p^{3}.

Let k be a field of characteristic p and let V_{3} be a 3 dimensional vector space over k. Choose a basis, {x, y, z} for V_{3}^{∗} and define

G = U_{3} := \{[ \begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix} ]| a, b, c ∈ F_{p}\} ,

with the action on V_{3}^{∗} given by

x ↔ [ \begin{matrix}1 \\ 0 \\ 0 \end{matrix} ] , y ↔ [ \begin{matrix}0 \\ 1 \\ 0 \end{matrix} ] , z ↔ [ \begin{matrix}0 \\ 0 \\ 1 \end{matrix} ] .

Let β denote the element of U_{3} formed by taking a = 0, b = 1, and c = 0. Let α denote the element formed by taking a = 1, b = 0, and c = 0. Let γ denote the element formed by taking a = 0, b = 0, and c = 1. The (pseudo) reflections α, β and γ generate U_{3}. The invariants x, N(y) and N(z) form a homogeneous system of parameters for k[V_{3}]^{U_{3}} such that the product of the degrees equals the order of the group. Therefore k[V_{3}]^{U_{3}} = k[x, N(y), N(z)] (see, for example, [29, Proposition 5.5.5]).

Consider the representation of U_{3} afforded by W = V_{3} ⊕ V_{3} = 2V_{3}. Since U_{3} is generated by elements that act on W as bi-reflections, the representation satisfies Kemper’s criteria [17, Corollary 3.7] and the invariants could be Cohen-Macaulay. In fact a simple MAGMA [4] calculation shows that for p = 2 the invariants are Cohen-Macaulay. Using [19, Theorem A] we know that the depth of the invariants of the isotropy subgroups give an upper bound on the depth of the invariants. However, for the representation W, all proper isotropy subgroups have Cohen–Macaulay rings of invariants so this result imposes no restriction on depth(k[W]^{U_{3}}). Thus there appears to be no simple method for deciding whether of not k[W]^{U_{3}} is Cohen–Macaulay for p > 2.

Using the ladder algorithm we will compute a complete list of generators for k[W]^{U_{3}} for the prime 3. The limitations of our computing resources prevent us from obtaining a complete list of generators for k[W]^{U_{3}} for primes greater than 3. However, the method does provide enough information for us to prove that k[W]^{U_{3}} is not Cohen–Macaulay for all primes p ≥ 3.

Remark 8.1: Using Remark 5.5 one sees that xN(y) lies in the radical of the image of the transfer in k[V_{3}]^{G}. Therefore, in principle, one can construct a generating set for k[2V_{3}]^{U_{3}} using Algorithm 5.7 with f_{1} := x_{1}N(y_{1}) and f_{2} := x_{2}N(y_{2}). We were able to do this calculation, using MAGMA [4], for p = 3. However the p = 5 calculation was beyond the capabilities of the computers and algorithms at our disposal.

Take G_{1} to be the subgroup generated by β and let G_{2} be the subgroup generated by α and β. We have a two rung ladder G_{1} ◁ G_{2} ◁ G.

The action of β on W is the action of Z/p on 2V_{1} ⊕ 2V_{2}. Using [5], we see that k[W]^{β} is generated by x_{i}, y_{i}, N_{β}(z_{i}) := z_{i}^{p} − z_{i}x_{i}^{p−1} and u_{β} := z_{1}x_{2} − z_{2}x_{1}. Take A_{1} := k[x_{1}, y_{1}, N_{1}, x_{2}, y_{2}, N_{2}, U_{β}] with α(N_{i}) = N_{i} and α(U_{β}) = U_{β}. Define ρ_{1} : A_{1} → k[W]^{β} by ρ_{1}(N_{i}) = N_{β}(z_{i}) and ρ_{1}(U_{β}) = u_{β}. Then ρ_{1} is an αequivariant algebra epimorphism and the kernel of ρ_{1}, J_{1}, is the principal ideal generated by r := U_{β}^{p} − x_{2}^{p}N_{1} + x_{1}^{p}N_{2} − (x_{1}x_{2})^{p−1}U_{β}.

Lemma 8.2: The inclusion of J_{1} into A_{1} induces a monomorphism from H^{1}(⟨α⟩, J_{1}) to H^{1}(⟨α⟩, A_{1}) where ⟨α⟩ ≅ Z/p.

Proof: An element in H^{1}(⟨α⟩, J_{1}) which maps to zero in H^{1}(⟨α⟩, A_{1}) is represented by rf with f ∈ A_{1} and rf = Δh for some h ∈ A_{1}. View h and r as polynomials in U. Since r is monic in U we can divide h by r to get h = qr + ℓ with deg_{U}(ℓ) < deg_{U}(r) = p. Apply Δ and use the fact that r is α-invariant to get rf = Δh = (Δq)r + Δℓ. Thus Δℓ = (f − Δq)r. The operator Δ does not increase the U-degree of a polynomial. Therefore deg_{U}(Δℓ) < p. However deg_{U}(r) = p. Thus f − Δq = 0. Therefore Δ(rq) = rf and rf represents zero in H^{1}(⟨α⟩, J_{1}). □

As a consequence of the lemma, ρ_{1} induces an epimorphism from A_{1}^{α} to (k[W]^{β})^{α} = k[W]^{G_{2}}. The action of α on the generators of A_{1} is the action of Z/p on 3V_{1}⊕2V_{2}. Again using [5], we see that k[W]^{G_{2}} is generated by x_{i}, N(y_{i}) = y_{i}^{p}− y_{i}x_{i}^{p−1}, u_{α} := y_{1}x_{2}−y_{2}x_{1}, N_{β}(z_{i}) and u_{β}. Take A_{2} := k[x_{1}, H_{1}, N_{1}, x_{2}, H_{2}, N_{2}, U_{α}, U_{β}] with γ(N_{i}) = N_{i} + H_{i}, γ(U_{β}) = U_{β} + U_{α} and the other generators invariant. Define ρ_{2} : A_{2} → k[W]^{G_{2}} by ρ_{2}(H_{i}) = N(y_{i}), ρ_{2}(N_{i}) = N_{β}(z_{i}), ρ_{2}(U_{α}) = u_{α} and ρ_{2}(U_{β}) = u_{β}. Then ρ_{2} is a γ-equivariant algebra epimorphism. Let J_{2} denote the kernel of ρ_{2}. A straight-forward Gröbner basis calculation can be used to show that the kernel of the map from A_{2} to k[W], given by ρ_{2} followed by inclusion, is generated by r := U_{β}^{p} − x_{2}^{p}N_{1} + x_{1}^{p}N_{2} − (x_{1}x_{2})^{p−1}U_{β} and s := Δr = U_{α}^{p} − x_{2}^{p}H_{1} + x_{1}^{p}H_{2} − (x_{1}x_{2})^{p−1}U_{α}. Thus we conclude that J_{2} is generated by r and s.

The action of γ on A_{2} is the action of Z/p on 2V_{1} ⊕ 3V_{2}. Again using [5], we see that A_{2}^{γ} is generated by x_{i}, H_{i}, N_{γ}(N_{i}), U_{α}, N_{γ}(U_{β}), U_{12} := N_{1}H_{2} − N_{2}H_{1}, U_{13} := N_{1}U_{α} − H_{1}U_{β}, U_{23} := N_{2}U_{α} − H_{2}U_{β} and Tr(N_{1}^{e_{1}}N_{2}^{e_{2}}U_{β}^{e_{3}}) for e_{j} ≤ p − 1. Using Corollary 4.4, we need only include transfers with e_{1} + e_{2} + e_{3} > 2(p − 1). Applying ρ_{2} to this set gives

A := {x_{i}, N(y_{i}), N(z_{i}) | i = 1, 2} ∪ {u_{α}, N_{γ}(u_{β}), u_{12}, u_{13}, u_{23}}

∪ {Tr_{G_{2}}^{G}(N_{β}(z_{1})^{e_{1}}N_{β}(z_{2})^{e_{2}}u_{β}^{e_{3}}) | e_{j} ≤ p − 1 and e_{1} + e_{2} + e_{3} > 2(p − 1)} ,

where u_{12} := ρ_{2}(U_{12}) = N_{β}(z_{1})N(y_{2})−N_{β}(z_{2})N(y_{1}), u_{13} := ρ_{2}(U_{13}) = N_{β}(z_{1})u_{α}− u_{β}N(y_{1}) and u_{23} := ρ_{2}(U_{23}) = N_{β}(z_{2})u_{α} − u_{β}N(y_{2}).

The ring k[W]^{G} is generated by the union of A and the preimage of a set of A_{2}^{γ}-module generators for the kernel of the map from H^{1}(⟨γ⟩, J_{2}) to H^{1}(⟨γ⟩, A_{2}).

We now turn our attention to the kernel of i^{1} : H^{1}(⟨γ⟩, J_{2}) → H^{1}(⟨γ⟩, A_{2}). Define s′ := s − U_{α}^{p} = x_{1}^{p}H_{2} − x_{2}^{p}H_{1} − (x_{1}x_{2})^{p−1}U_{α} and let K denote the A_{2}^{γ}- submodule of H^{1}(⟨γ⟩, A_{2}) consisting of the classes annihilated by s′

Lemma 8.3: There is an epimorphism, say Φ, of A_{2}^{γ}-modules from K to kernel(i^{1}) which takes a class represented by h ∈ A_{2} to an element represented by sh in kernel(i^{1}). If ℓ ∈ A_{2} with Δℓ = sh then the connecting homomorphism takes ρ_{2}(ℓ) to [sh].

Proof: Since s ∈ A_{2}^{γ}, if h = h′ + Δm then sh = sh′ + Δ(sm) and, thus, sh and sh′ represent the same class in H^{1}(Z/p, J_{2}). Therefore multiplication by s gives a well defined map from H^{1}(Z/p, A_{2}) to H^{1}(Z/p, J_{2}).

Elements in A_{2}^{γ} contained in the image of the transfer annihilate all elements of H^{1}(Z/p, A_{2}) (see, for example, [18, Corollary 2.4]). Furthermore, Tr(U_{β}^{p−1}U_{α}) = U_{α}^{p}. Thus s and s′ = s−U_{α}^{p} annihilate the same submodule of H^{1}(Z/p, A_{2}). Hence if we restrict to K then [sh] ∈ kernel(i^{1}) and we have a well defined map from K to kernel(i^{1}). Since A_{2}^{γ} is a commutative ring, this is a map of A_{2}^{γ}-modules.

Suppose μ ∈ kernel(i^{1}). Then μ = [Δf] for some f ∈ A_{2}. View f and r as polynomials in U = U_{β} and divide f by r to get f = qr + ℓ with deg_{U}(ℓ) < p. Thus ℓ = f − qr and Δℓ = Δf − Δ(qr). Since qr ∈ J_{2}, Δ(qr) represents zero in H^{1}(Z/p, J_{2}). Thus Δℓ and Δf represent the same element in H^{1}(Z/p, J_{2}) and μ = [Δℓ]. Furthermore, Δ does not increase the U-degree. Therefore deg_{U}(Δℓ) < p and, since deg_{U}(r) = p, Δℓ lies in the principal ideal generated by s. Thus μ = [Δℓ] = [sh] for some h ∈ A_{2}. However μ ∈ kernel(i^{1}) implies [h] ∈ K. Hence the map is surjective.

Suppose that ℓ ∈ A_{2} with Δℓ = sh. Note that Δ gives the first differential in the cochain complex used to compute H^{∗}(Z/p, A_{2}). As a consequence, using basic homological algebra, the connecting homomorphism takes ρ_{2}(ℓ) to [sh]. □

Suppose that h ∈ A_{2}^{γ} . Then Δ(hr) = hΔ(r) = hs and, if Tr(h) = 0, [h] ∈ K and Φ([h]) = 0. Thus Φ determines an epimorphism of A_{2}^{γ}-modules from K_{γ} to kernel(i^{1}) where K_{γ} is the quotient of K by the submodule of cohomology classes represented by elements of A_{2}^{γ}.

Theorem 8.4: The A_{2}^{γ}-module kernel(i^{1}) is generated in degrees greater than 3p.

Proof: Since Φ increases degree by 2p, it is sufficient to show that K_{γ} is zero in degrees less than or equal to p. From Proposition 6.2, H^{1}(Z/p, A_{2}) is a free module over k[x_{1}, x_{2}, N_{γ}(N_{1}), N_{γ}(N_{2})]. Furthermore, any basis for H^{1}(Z/p, A_{2}^{♭}) gives a set of module generators. Recall that deg(N_{i}) = deg(H_{i}) = p and deg(U_{β}) = deg(U_{α}) = 2. Therefore, if we restrict to degrees less than or equal to p we can take U_{α}^{i}[1], [N_{1}], [N_{2}], and U_{α}^{j}[U_{β}^{ℓ}] with i < (p − 1)/2 and j + ℓ < (p − 1)/2 as the generating set. Since we are interested in K_{γ} we may omit U_{α}^{i}[1]. Using Lemma 6.4, U_{α}[U_{β}^{ℓ}] = 0 and so we can take j = 0. Let f := c_{1}N_{1}+c_{2}N_{2}+c_{3}U_{β}^{ℓ} where c_{1}, c_{2} ∈ k and c_{3} ∈ k[x_{1}, x_{2}] with deg(c_{3}) ≤ p−2ℓ. Suppose s′[f] = 0. Using Lemma 6.4, [H_{1}N_{1}] = [H_{2}N_{2}] = [U_{α}U_{β}^{ℓ}] = 0. Thus s′[f] = x_{1}^{p}c_{1}[N_{1}H_{2}]+x_{1}^{p}c_{3}[H_{2}U_{β}^{ℓ}]−x_{2}^{p}c_{2}[H_{1}N_{2}]−x_{2}^{p}c_{3}[H_{1}U_{β}^{ℓ}]−(x_{1}x_{2})^{p−1}[U_{α}(c_{1}N_{1}+c_{2}N_{2})]. We may assume that f is homogeneous. If deg(f) < p then c_{1} = c_{2} = 0 and s′[f] = c_{3}(x_{1}^{p}[H_{2}U_{β}^{ℓ}] − x_{2}^{p}[H_{1}U_{β}^{ℓ}]). From Lemma 6.4 and Proposition 6.2, (x_{1}^{p}[H_{2}U_{β}^{ℓ}] − x_{2}^{p}[H_{1}U_{β}^{ℓ}]) ≠ 0. Thus c_{3} = 0 and f = 0. If deg(f) = p then deg(c_{3}) = p − 2ℓ > 0 and {x_{1}^{p}, x_{2}^{p}, c_{3}x_{1}^{p}, c_{3}x_{2}^{p}, (x_{1}x_{2})^{p−1}} is a linearly independent subset of k[x_{1}, x_{2}]. Therefore c_{1}[H_{1}N_{2}] = c_{2}[H_{2}N_{1}] = c_{3}[H_{1}U_{β}^{ℓ}] = 0. Again using Lemma 6.4, [H_{1}N_{2}] ≠ 0, [H_{2}N_{1}] ≠ 0 and [H_{1}U_{β}^{ℓ}] ≠ 0. Thus c_{1} = c_{2} = c_{3} = 0 and f = 0. □

Theorem 8.5: For p = 3, k[2V_{3}]^{U_{3}} is generated by A and one additional generator:

κ := (u_{α}^{p−1} − (x_{1}x_{2})^{p−1}) u_{12} (u_{β}^{2} − u_{β}u_{α})
−x_{2}^{p}u_{23} (N_{β} (z_{1}^{2}) − N(y_{1})N_{β} (z_{1})) − x_{1}^{p}u_{13} (N_{β} (z_{2}^{2}) − N(y_{2})N_{β} (z_{2}))

− (x_{1}^{p}N(y_{2}) + x_{2}^{p}N(y_{1})) (N_{β}(z_{1}z_{2})u_{β} − u_{12}u_{β} + u_{23}N_{β}(z_{1}) − N(y_{1})u_{α}N_{β}(z_{2})) .

Proof: We use the description of H^{1}(Z/3, 3V_{2}) given in Theorem 6.5. Since we are interested in identifying K_{γ} we may omit cohomology classes represented by invariants. Thus it is sufficient to consider the module generated by [N_{1}], [N_{2}], [U_{β}], and U_{12}[U_{β}]. Since U_{α}U_{12}[U_{β}] = 0, H_{1}U_{12}[U_{β}] = 0 and H_{2}U_{12}[U_{β}] = 0, we see that U_{12}[U_{β}] is in K_{γ}. To describe the corresponding invariant we need to find f ∈ A_{2} such that Δ(f) = sU_{12}U_{β}. Let h := N_{1}N_{2}U_{β} − U_{12}U_{β} + U_{23}N_{1} − H_{1}U_{α}N_{2}. Referring to the proof of Theorem 6.5, we see that

U_{12}Δ(U_{β}^{2} − U_{α}U_{β}) = 2U_{α}U_{12}U_{β}
H_{1}Δ(h) + U_{23}Δ (N_{1}^{2} − N_{1}H_{1}) = 2H_{1}U_{12}U_{β}
H_{2}Δ(h) + U_{13}Δ (N_{2}^{2} − N_{2}H_{2}) = H_{2}U_{12}U_{β}

Let f_{1} := U_{12}(U_{β}^{2}−U_{α}U_{β}), f_{2} := H_{1}h+U_{23} (N_{1}^{2} − N_{1}H_{1})), f_{3} := H_{2}h+U_{13} (N_{2}^{2} − N_{2}H_{2})) and f := f_{1}(U_{α}^{p−1} − (x_{1}x_{2})^{p−1}) − f_{2}x_{2}^{p} − f_{3}x_{1}^{p}. Then Δ(f) = s(2U_{12}U_{β}) and ρ_{2}(f) = κ is the corresponding invariant.

We still need to examine the contribution to K_{γ} from the module generated by [N_{1}], [N_{2}] and [U_{β}]. Multiplication by s′ induces a map from this module to the module generated by [U_{12}], [U_{13}] and [U_{23}]. Using Theorem 6.5 we see that the matrix representing s′ is

( \begin{matrix}−x_{1}^{3} & −x_{2}^{3} & 0 \\ (x_{1}x_{2})^{2} & 0 & −x_{2}^{p} \\ 0 & (x_{1}x_{2})^{2} & x_{1}^{3} \end{matrix} ) .

A generator for the kernel is given by (x_{2}^{3}, −x_{1}^{3}, (x_{1}x_{2})^{2}). In other words we get one more generator for K_{γ}, x_{2}^{3}N_{1} − x_{1}^{3}N_{2} + (x_{1}x_{2})^{2}U_{β} = U_{β}^{3} − r. However

Δ (r (r − s + N_{γ}(U_{β})) + sU_{β}U_{α} (U_{α} − U_{β})) = s(U_{β}^{3} − r).

Thus the corresponding invariant is zero. □

Remark 8.6: For p = 2, k[2V_{3}]^{U_{3}} can be computed rather quickly using the invariant theory packages in MAGMA [4]. When we first considered the problem, the p = 3 computation was beyond the capabilities of our computing facilities. Our original calculation for p = 3 was based, essentially, on Algorithm 5.7.

The ladder calculations evolved out of an attempt to better understand the result and a, so far unsuccessful attempt, to extend the calculation to p = 5. Recently, Gregor Kemper, using a computer with 4 GB of RAM, has been able to construct a generating set for p = 3 using the invariant theory packages in MAGMA. His calculations agree with ours. As far as we know, no one has been able to construct a generating set for p = 5. The most effective approach to the p = 5 problem may well be the hybrid of Algorithm 5.7 and the ladder technique consisting of using the algebra generated by A as the input to Step 5 of Algorithm 5.7. For p = 2, k[2V_{3}]^{U_{3}} is Cohen–Macaulay. As we prove below, for p > 2, the invariants are not Cohen-Macaulay. This at least partly explains the dramatic increase in computational complexity in passing from p = 2 to p = 3.

Theorem 8.7: If p > 2, then k[2V_{3}]^{U_{3}} is not Cohen–Macaulay.

Proof: We will show that the partial homogeneous system of parameters {x_{1}, N(y_{1}), N(y_{2})} is not a regular sequence in k[2V_{3}]^{U_{3}}. Clearly {x_{1}, N(y_{1})} is regular. Thus it is sufficient to show that N(y_{2}) is a zero divisor in k[2V_{3}]^{U_{3}}/(x_{1}, N(y_{1}))k[2V_{3}]^{U_{3}}. We break the argument into two steps.

Step 1: Show that u_{13}^{2}u_{α}^{p−2}N(y_{2})−(u_{12}N_{γ}(u_{β})+u_{13}u_{23}u_{α}^{p−2})N(y_{1}) ∈ (x_{1})k[2V_{3}]^{U_{3}}.

Step 2: Show that u_{13}^{2}u_{α}^{p−2} ∉ (x_{1}, N(y_{1}))k[2V_{3}]^{U_{3}}.

Proof of Step 1: Observe that u_{12}u_{α} − u_{13}N(y_{2}) + u_{23}N(y_{1}) = 0. Thus

u_{13}^{2}u_{α}^{p−2}N(y_{2}) − u_{13}u_{23}u_{α}^{p−2}N(y_{1}) = u_{13}u_{α}^{p−2} (u_{13}N(y_{2}) − u_{23}N(y_{1}))
= u_{13}u_{α}^{p−2}u_{12}u_{α}

and u_{13}^{2}u_{α}^{p−2}N(y_{2})−(u_{12}N_{γ}(u_{β})+u_{13}u_{23}u_{α}^{p−2})N(y_{1}) = u_{12}(u_{13}u_{α}^{p−1}−N_{γ}(u_{β})N(y_{1})). A simple calculation using the definitions of the appropriate invariants gives u_{13}u_{α}^{p−1} − N_{γ}(u_{β})N(y_{1})) ≡ 0 (mod (x_{1})k[2V_{3}]). Therefore there exists f ∈ k[2V_{3}] such that u_{13}u_{α}^{p−1} − N_{γ}(u_{β})N(y_{1}) = x_{1}f. Since x_{1}f and x_{1} are both elements of k[2V_{3}]^{U_{3}} we have f ∈ k[2V_{3}]^{U_{3}} concluding the proof of Step 1.

Proof of Step 2: We use the graded reverse lexicographic order with x_{1} < y_{1} < z_{1} < x_{2} < y_{2} < z_{2}. Suppose u_{13}^{2}u_{α}^{p−2} = fN(y_{1}) + hx_{1} for f, h ∈ k[2V_{3}]^{U_{3}}. Since LM(u_{13}^{2}u_{α}^{p−2}) = x_{2}^{p}y_{1}^{p}z_{1}^{2p} > LM(hx_{1}) we have LM(u_{13}^{2}u_{α}^{p−2}) = LM(fN(y_{1})). Therefore LM(f) = x_{2}^{p}z_{1}^{2p}. We will show that there is no element of k[2V_{3}]^{U_{3}} with this lead monomial. Using Theorem 8.4 we see that we need only consider elements in the algebra generated by A. Elements from A of the form Tr_{G_{2}}^{G}(N_{β}(z_{1})^{e_{1}}N_{β}(z_{2})^{e_{2}}u_{β}^{e_{3}}) have degree p(e_{1}+e_{2})+2e_{3} and must satisfy e_{i} ≤ p−1 and e_{1} + e_{2} + e_{3} > 2(p − 1). The smallest possible degree for such an element is 2(p − 1) + p^{2} and comes from taking e_{3} = p − 1 and e_{1} + e_{2} = p. As long as p > 2, p^{2} + 2(p − 1) > 3p and so we can restrict to the subalgebra generated by

{x_{i}, N(y_{i}), N(z_{i}) | i = 1, 2} ∪ {u_{α}, N_{γ}(u_{β}), u_{12}, u_{13}, u_{23}} .

The corresponding lead monomials are

{x_{i}, y_{i}^{p}, z_{i}^{p^{2}} | i = 1, 2} ∪ {y_{1}x_{2}, (z_{1}x_{2})^{p}, (z_{1}y_{2})^{p}, z_{1}^{p}y_{1}x_{2}, z_{2}^{p}y_{1}x_{2}} .

Clearly x_{2}^{p}z_{1}^{2p} is not a product of monomials from this list. Therefore f comes from a tête-a-tête. All of our generators are homogeneous with respect to the bidegree so the tête-a-tête must occur in bidegree (2p, p). The monomials of bidegree (2p, p) which greater than x_{2}^{p}z_{1}^{2p} are of the form z_{1}^{2p}m were m has bidegree (0, p) and m > x_{2}^{p}. Since there are no tête-a-tête’s generated by monomials of this form, there is no suitable f. □

Acknowledgement. We thank Gregor Kemper for reading an earlier draft of this paper and suggesting some improvements and corrections.

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