GENERIC KERNELS AND A QUESTION OF CARLSON, FRIEDLANDER AND SUSLIN

SHAWN BALAND

Abstract. Let E ≅ (ℤ/p)^{2} be an elementary abelian p-group of rank two, k and algebraically closed field of characteristic p, and let J = J(kE). We investigate finitely generated kE-modules M of constant Jordan type and their generic kernels 𝔎(M). In particular, we answer a question posed in [7] regarding whether or not the submodules J^{−i}𝔎(M) have constant Jordan type for all i ≥ 0. We show that this question has an affirmative answer whenever p = 3 or J^{2}𝔎(M) = 0. We also show that this question has a negative answer in general by constructing a kE-module M of constant Jordan type for p ≥ 5 such that J^{−1}𝔎(M) does not have constant Jordan type.

1. Introduction

Let p be a prime number and let k be an algebraically closed field of characteristic p. Throughout this paper we consider an elementary abelian p-group E = ⟨g_{1}, …, g_{r}⟩ ≅ (ℤ/p)^{r} of rank r, where the g_{i} are pairwise-commuting generators of order p. We shall focus mainly on the case in which r = 2, although we begin our exposition in a more general context. As usual, we define generators X_{i} = g_{i} − 1 for the Jacobson radical J = J(kE), and for a nonzero α = (λ_{1}, …, λ_{r}) ∈ 𝔸^{r}(k) we define the element X_{α} = λ_{1}X_{1} + ⋯ + λ_{r}X_{r}. Because X_{α}^{p} = 0, X_{α} acts on a finite dimensional kE-module M via a matrix whose Jordan canonical form contains blocks of length at most p. The Jordan type of X_{α} on M is the partition [p]^{a_{p}} … [1]^{a_{1}} of dim_{k} M, where X_{α} acts on M via a_{j} Jordan blocks of length j. We say that M has constant Jordan type if the Jordan type of X_{α} on M is independent of 0 ≠ α ∈ 𝔸^{r}(k).

Modules of constant Jordan type were introduced by Carlson, Friedlander and Pevtsova [6]. One of the major questions regarding such modules concerns which partitions of dim_{k} M are realised as the Jordan types of kE-modules of constant Jordan type. This question was investigated in the papers [1, 2, 3, 4, 5], although little progress has been made towards a full classification of realisable Jordan types.

Recently, Carlson, Friedlander and Suslin [7] have investigated the constant image property for kE-modules in any rank, and for r = 2 they have defined the generic kernel 𝔎(M) of a kE-module M, which can be characterised as the largest submodule of M having the constant image property. One of the striking features of the generic kernel is that, whenever M has constant rank, the kernel of the action of X_{α} on M is contained in 𝔎(M). Benson [3] used this property to prove that if a kE-module (for any r) has constant Jordan type containing no blocks of length one, then the total number of Jordan blocks is divisible by p. (This statement proves a special case of an unpublished conjecture of Rickard.)

In an attempt to obtain stronger connections between modules of constant Jordan type and their generic kernels, Carlson, Friedlander and Suslin have posed the following question.

Question 1.1 ([7, Question 8.11]). If a k(ℤ/p)^{2}-module M has constant Jordan type, is it true that J^{−i}𝔎(M) has constant Jordan type for all i ≥ 0?

Here J^{−i}𝔎(M) denotes the set of elements m ∈ M for which J^{i}m ⊆ 𝔎(M). We should point out that the original statement of this question was given in terms of the submodules X_{1}^{−i}𝔎(M), but as we shall see in Proposition 2.8, these two formulations are equivalent.

In addition to the above property having been true in all prior known examples, motivation for Question 1.1 came from the observation that if M has constant Jordan type, then there exists a filtration

0 = J^{p}𝔎(M) ⊆ J^{p−1}𝔎(M) ⊆ ⋯ ⊆ J^{−p+2}𝔎(M) ⊆ J^{−p+1}𝔎(M) = M

of M in which J^{i}𝔎(M) has constant Jordan type for all i ≥ 0 and for i = −p+1. For further details, see Theorem 8.10 of [7].

In this paper we present a thorough treatment of Question 1.1 and prove that it has an affirmative answer in the special cases where p = 3 or J^{2}𝔎(M) = 0. The latter case involves a duality formula and an inductive process that allows us to use the constant image property and the generic kernel of the dual M^{∗}. Finally, we show that Question 1.1 does not have an affirmative answer in general by constructing a k(ℤ/p)^{2}-module such that J^{−1}𝔎(M) does not have constant Jordan type in the first interesting case, namely that in which p ≥ 5, J^{2}𝔎(M) ≠ 0 and J^{3}𝔎(M) = 0.

2. Constant Jordan type and the generic kernel

In this section we introduce properties of modules of constant Jordan type and the generic kernel that will be needed in the sequel. The exposition roughly follows that of [7]. We begin by fixing some notation. If M is a kE-module, we denote by M^{∗} the k-linear dual of M. Throughout this paper we restrict our attention to the group theoretic kE-module structure on M^{∗} defined by the action

(g.f)(m) = f(g^{−1}m) for all f ∈ M^{∗}, g ∈ E, m ∈ M.

One should contrast this with the p-restricted Lie kE-module structure given by

((g − 1).f)(m) = f(−(g − 1)m) for all f ∈ M^{∗}, g ∈ E, m ∈ M.

We also consider the Heller shift Ω(M) of M, which is defined to be the kernel of the projective cover P_{M} → M. Given these structures, we now recall some important properties enjoyed by modules of constant Jordan type.

Theorem 2.1. (i) If M is a kE-module of constant Jordan type, then M^{∗} and Ω(M)

have constant Jordan type. (ii) If M ⊕ N has constant Jordan type, then M and N have constant Jordan type. (iii) If M and N have constant Jordan type, then M ⊗_{k} N has constant Jordan type.

Proof. These are Propositions 5.2 and 1.8, Theorem 3.7 and Corollary 4.3 of [6]. □

Definition 2.2. If M is a kE-module, we say that M has the constant image property if the image of X_{α} on M is independent of 0 ≠ α ∈ 𝔸^{r}(k).

It is clear from the definition that M has the constant image property if and only if X_{α}M = JM for all nonzero α ∈ 𝔸^{r}(k). The following proposition illustrates the connection between the constant Jordan type and constant image properties.

Proposition 2.3. (i) If M has the constant image property, then X_{α}^{i}M = J^{i}M for all 0 ≠ α ∈ 𝔸^{2}(k) and all i ≥ 0.

(ii) If M has the constant image property, then M has constant Jordan type.

(iii) The class of kE-modules with the constant image property is closed under finite sums, quotients and radicals.

Proof. For (i), we assume by induction that X_{α}^{i}M = J^{i}M. Then by the commutativity of kE and the constant image property we have

J^{i+1}M = JJ^{i}M = JX_{α}^{i}M = X_{α}^{i}JM = X_{α}^{i+1}M.

(ii) follows from (i) since M has constant Jordan type if and only if, for each i ≥ 0, the dimension of X_{α}^{i}M is independent of α. (iii) follows from the definition if the constant image property. □

Recall that a kE-module M has constant rank if the rank of X_{α} on M is independent of 0 ≠ α ∈ 𝔸^{r}(k). In particular, if M has constant Jordan type then M has constant rank. We now give a proposition that implies, in particular, that M has constant rank if and only if M^{∗} has constant rank.

Proposition 2.4. Let M be a kE-module. If α is a nonzero point in 𝔸^{r}(k) such that the rank of X_{α} on M is maximal amongst all points in 𝔸^{r}(k), then the rank of X_{α} on M^{∗} is equal to the rank of X_{α} on M.

Proof. This is a special case of the argument given in Proposition 5.2 of [6]. For completeness we recount that proof, making the necessary modifications. Let S : kE → kE be the k-linear map given by g ↦ g^{−1} for all g ∈ E, i.e., S is the group theoretic antipode of kE when viewed as a k-Hopf algebra. For each 1 ≤ i ≤ r we have

X_{i} + S(X_{i}) = (g_{i} − 1) + (g_{i}^{−1} − 1) = −(g_{i} − 1)(g_{i}^{−1} − 1) ∈ J^{2},

hence X_{α} + S(X_{α}) ∈ J^{2} for all 0 ≠ α ∈ 𝔸^{r}(k). If X_{α} achieves the maximal rank on M, then it follows from Theorem 1.9 of [8] that rank(X_{α}, M) = rank(S(X_{α}), M). The proof is now complete by noting that the representation affording M^{∗} is obtained from the representation affording M by precomposing with S and then composing with the transpose map. □

Throughout the rest of this paper (unless otherwise specified) we shall assume that r = 2, that is, E = ⟨g_{1}, g_{2}⟩ ≅ (ℤ/p)^{2} and the Jacobson radical J = J(kE) is generated by the elements X_{1} = g_{1} − 1 and X_{2} = g_{2} − 1.

Definition 2.5. Let M be a finitely generated kE-module. For any 0 ≠ α = (λ_{1}, λ_{2}) ∈ 𝔸^{2}(k), note that Ker(X_{α}, M) is uniquely determined by the class ᾱ ∈ ℙ^{1}(k). For a subset S ⊆ ℙ^{1}(k), we denote by _{S}M the submodule ∑_{ᾱ∈S} Ker(X_{α}, M) of M. We define the generic kernel of M to be the submodule

𝔎(M) = ⋂_{S ⊆ ℙ^{1}(k) cofinite}_{S}M.

Remark 2.6. A subset S ⊆ ℙ^{1}(k) is cofinite if its complement in ℙ^{1}(k) is a finite set. Since M is finite dimensional, there always exists a cofinite S ⊆ ℙ^{1}(k) for which 𝔎(M) = _{S}M.

The following theorem shows that the generic kernel 𝔎(M) is particularly well-behaved in the case where M has constant rank.

Theorem 2.7. (i) If M is a kE-module, then 𝔎(M) has the constant image property. Moreover, if N is any submodule of M having the constant image property, then N is contained in 𝔎(M).

(ii) M has constant rank if and only if 𝔎(M) = _{ℙ^{1}(k)}M. In this case, Ker(X_{α}, M) is contained in 𝔎(M) for each 0 ≠ α ∈ 𝔸^{2}(k).

Proof. This follows from Theorem 7.10 and Corollary 7.7 of [7], along with the definition of the generic kernel. □

We now use Theorem 2.7 to prove a proposition that allows us to replace “J” by “X_{1}” in the statement of Question 1.1; hence our formulation is equivalent to that given in [7]. The reader should compare this with Proposition 2.3 (i).

Proposition 2.8. If M is a kE-module of constant rank, then for all 0 ≠ α ∈ 𝔸^{2}(k) and all i ≥ 0 we have

X_{α}^{−i}𝔎(M) = J^{−i}𝔎(M).

Proof. We proceed by induction on i, beginning with the case i = 1. For this it suffices to show that if m ∈ M and X_{α}m ∈ 𝔎(M), then Jm ⊆ 𝔎(M). Note for any β ∈ 𝔸^{2}(k) that X_{β}X_{α}m ∈ J𝔎(M), hence by Theorem 2.7 (i) and the constant image property there exists m′ ∈ 𝔎(M) such that X_{α}m′ = X_{β}X_{α}m = X_{α}X_{β}m. Thus

m′ − X_{β}m ∈ Ker(X_{α}, M) ⊆ 𝔎(M)

by Theorem 2.7 (ii). It follows that X_{β}m also lies in 𝔎(M), showing that Jm ⊆ 𝔎(M).

Now let i > 1 and assume by induction that X_{α}^{−i}𝔎(M) = J^{−i}𝔎(M). We clearly have J^{−i−1}𝔎(M) ⊆ X_{α}^{−i−1}𝔎(M), so suppose m ∈ X_{α}^{−i−1}𝔎(M). Then X_{α}^{i}X_{α}m = X_{α}^{i+1}m ∈ 𝔎(M) so that X_{α}m ∈ X_{α}^{−i}𝔎(M) = J^{−i}𝔎(M). Hence X_{α}J^{i}m = J^{i}X_{α}m ⊆ 𝔎(M), that is, J^{i}m ⊆ X_{α}^{−1}𝔎(M) = J^{−1}𝔎(M) by the first step of the proof. It follows that J^{i+1}m = JJ^{i}m ⊆ 𝔎(M) as required. □

The following corollary will be used implicitly in Section 4 whenever we invoke the duality provided by Theorem 3.3.

Corollary 2.9. If M is a kE-module of constant rank, then M = J^{−p+1}𝔎(M)/J^{p}𝔎(M).

Proof. For all all 0 ≠ α ∈ 𝔸^{2}(k) we have X_{α}^{p}M = 0, hence X_{α}^{p−1}M ⊆ Ker(X_{α}, M) ⊆ 𝔎(M) by Theorem 2.7 (ii). It follows that M ⊆ X_{α}^{−p+1}𝔎(M) = J^{−p+1}𝔎(M) by Proposition 2.8. Finally, we have J^{p}𝔎(M) = 0 by part (i) of Proposition 2.3. □

3. Duality for generic kernels

In this section we introduce a duality formula that will be used to study Question 1.1 in terms of the generic kernels of both M and M^{∗}. This is a slightly different approach to duality than that given in [7], although the underlying machinery is essentially the same. We begin with a couple of lemmas. Recall that if M is a kE-module and N is a submodule of M, then the orthogonal space

N^{⊥} = {f ∈ M^{∗} | f(n) = 0 for all n ∈ N}

is naturally a kE-submodule of M^{∗}, and we have (N^{⊥})^{⊥} ≅ N.

Lemma 3.1. If M is a kE-module of constant rank, then 𝔎(M^{∗}) ≅ (M/J𝔎(M))^{∗}.

Proof. By Proposition 8.4 and Theorem 8.6 of [7] we have

𝔎(M^{∗}) = (X_{1}𝔎(M))^{⊥} = (J𝔎(M))^{⊥} ≅ (M/J𝔎(M))^{∗}

since 𝔎(M) has the constant image property. □

Lemma 3.2. If M is a kE-module in any rank r and N is a submodule of M, then (i) J^{−1}(N^{⊥}) = (JN)^{⊥} and (ii) J(N^{⊥}) = (J^{−1}N)^{⊥}.

Proof. First note for any 1 ≠ g ∈ E and any f ∈ M^{∗} that the element X = g − 1 ∈ kE acts on f via the relation

(X.f)(m) = f((g^{−1} − 1)m) = f(−g^{−1}Xm) for all m ∈ M.

Similarly, we have f(Xm) = (−g^{−1}X.f)(m) for all m ∈ M.

To prove (i), suppose f ∈ J^{−1}(N^{⊥}) and n ∈ JN. Writing n = X_{1}n_{1} + ⋯ + X_{r}n_{r} with n_{i} ∈ N, we have

f(n) = f(X_{1}n_{1}) + ⋯ + f(X_{r}n_{r}) = −(g_{1}^{−1}X_{1}.f)(n_{1}) − ⋯ − (g_{r}^{−1}X_{r}.f)(n_{r}) = 0

since X_{i}.f ∈ N^{⊥}, hence f ∈ (JN)^{⊥}. Conversely, if f ∈ (JN)^{⊥} and n ∈ N, then

X_{i}.f(n) = f(−g_{i}^{−1}X_{i}n) = 0 for all 1 ≤ i ≤ r

since −g_{i}^{−1}X_{i}n ∈ JN. It follows that X_{i}.f ∈ N^{⊥} for all i, that is, J.f ⊆ N^{⊥}.

To prove (ii), note that

J^{−1}(N) = J^{−1}((N^{⊥})^{⊥}) = (J(N^{⊥}))^{⊥}

by (i) so that (J^{−1}N)^{⊥} = J(N^{⊥}). □

We are now in a position to prove the main theorem of this section.

Theorem 3.3. If M is a kE-module of constant rank, then for all a, b ∈ ℤ with a ≤ b we have

J^{a}𝔎(M^{∗})/J^{b}𝔎(M^{∗}) ≅ (J^{−b+1}𝔎(M)/J^{−a+1}𝔎(M))^{∗}.

Proof. We fist claim for all a ≥ 0 that J^{−a}J𝔎(M) = J^{−a+1}𝔎(M). If m ∈ J^{−a+1}𝔎(M), then J^{a−1}m ⊆ 𝔎(M) so that J^{a}m ⊆ J𝔎(M) and m ∈ J^{−a}J𝔎(M). Conversely, suppose J^{a}m ⊆ J𝔎(M) and let Y ∈ J^{a−1}. Then X_{1}Y m ∈ J^{a}m ⊆ J𝔎(M) so that X_{1}Y m = X_{1}m′ for some m′ ∈ 𝔎(M) by the constant image property. It follow by Theorem 2.7 (ii) that Y m − m′ ∈ 𝔎(M). Hence Y m ∈ 𝔎(M), showing that J^{a−1}m ⊆ 𝔎(M).

Having established the claim, it follows by Lemmas 3.1 and 3.2 that

\frac{J^{a}𝔎(M^{∗})}{J^{b}𝔎(M^{∗})} ≅ \frac{J^{a}(J𝔎(M))^{⊥}}{J^{b}(J𝔎(M))^{⊥}} = \frac{(J^{−a+1}𝔎(M))^{⊥}}{(J^{−b+1}𝔎(M))^{⊥}} ≅ (\frac{J^{−b+1}𝔎(M)}{J^{−a+1}𝔎(M)})^{∗} .  □

4. Question 1.1 in the cases p = 3 and J^{2}𝔎(M) = 0

In this section we identify two special cases in which Question 1.1 has an affirmative answer. We begin with a preliminary lemma.

Lemma 4.1. If M is a kE-module of constant rank then

Ker(X_{α}, M) ∩ Im(X_{α}, M) = Ker(X_{α}, J^{−1}𝔎(M)) ∩ Im(X_{α}, J^{−1}𝔎(M)).

Proof. The leftward containment is clear since Ker(X_{α}, M) = Ker(X_{α}, J^{−1}𝔎(M)) by Theorem 2.7 (ii). So let m ∈ Ker(X_{α}, M) ∩ Im(X_{α}, M). Since Ker(X_{α}, M) ⊆ 𝔎(M), we have m ∈ 𝔎(M) ∩ Im(X_{α}, M). Hence there exists m′ ∈ M such that X_{α}m′ = m. By Proposition 2.8 this implies m′ ∈ J^{−1}𝔎(M) so that m ∈ Im(X_{α}, J^{−1}𝔎(M)). This establishes the reverse containment. □

Proposition 4.2. If M is a kE-module of constant rank, then for any 0 ≠ α ∈ 𝔸^{2}(k), the number of blocks of length one in the Jordan type of X_{α} on J^{−1}𝔎(M) is equal to the number of blocks of length one in the Jordan type of X_{α} on M.

Proof. By the lemma and its proof we have

\frac{Ker(X_{α}, M)}{Ker(X_{α}, M) ∩ Im(X_{α}, M)} = \frac{Ker(X_{α}, J^{−1}𝔎(M))}{Ker(X_{α}, J^{−1}𝔎(M)) ∩ Im(X_{α}, J^{−1}𝔎(M))}.

The number of Jordan blocks of length one in the action of X_{α} on M is the dimension of the left-hand term, and the number of Jordan blocks of length one in the action of X_{α} on J^{−1}𝔎(M) is the dimension of the right-hand term. □

Theorem 4.3. If p = 3 and M is a kE-module of constant Jordan type, then J^{−i}𝔎(M) has constant Jordan type for all i ≥ 0.

Proof. If M is a k(ℤ/3)^{2}-module of constant Jordan type, then M = J^{−2}𝔎(M) by Corollary 2.9. Hence we need only check that J^{−1}𝔎(M) has constant Jordan type. We fix 0 ≠ α ∈ 𝔸^{2}(k) and let [1]^{a}[2]^{b}[3]^{c} be the Jordan type of X_{α} on J^{−1}𝔎(M). Since the number of Jordan blocks in the action of X_{α} on J^{−1}𝔎(M) is equal to the number of Jordan blocks in the action of X_{α} on M by Theorem 2.7 (ii), we see that a + b + c is independent of α. We also have a + 2b + 3c = dim_{k} J^{−1}𝔎(M), which is clearly independent of α. Finally, a is equal to the number of Jordan blocks of length one in the action of X_{α} on M by Proposition 4.2. The fact that J^{−1}𝔎(M) has constant Jordan type now follows from the nonsingularity of the matrix

(\begin{matrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 0 & 0 \end{matrix}) .  □

In order to address the case in which J^{2}𝔎(M) = 0, we make the following observation. Let M be any kE-module and let n ≥ 0. For each nonzero α ∈ 𝔸^{2}(k) there is a natural short exact sequence

(1)  0 ⟶ L_{n,α} ⟶ Im(X_{α}^{n}, M/J^{j+1}𝔎(M)) ⟶ Im(X_{α}^{n}, M/J^{j}𝔎(M)) ⟶ 0

where

L_{n,α} = Im(X_{α}^{n}, M/J^{j+1}𝔎(M)) ∩ J^{j}𝔎(M)/J^{j+1}𝔎(M).

If M/J^{j+1}𝔎(M) has constant Jordan type, then the dimension of the middle term is independent of α. We now introduce a criterion under which the dimensions of the outer two terms would also be independent of α.

Proposition 4.4. Suppose M/J^{j+1}𝔎(M) has constant Jordan type and fix n ≥ 0. If there exists a submodule N of M/J^{j+1}𝔎(M) such that L_{n,α} = Im(X_{α}^{n}, N) for all 0 ≠ α ∈ 𝔸^{2}(k), then rank(X_{α}^{n}, M/J^{j}𝔎(M)) is independent of α.

Proof. Since M/J^{j+1}𝔎(M) has constant Jordan type, we see from the sequence (1) that

rank(X_{α}^{n}, N) + rank(X_{α}^{n}, M/J^{j}𝔎(M))

is constant as a function of α. By Lemma 1.2 of [8], there exists a dense open subset U ⊆ 𝔸^{2}(k) such that α ∈ U if and only if rank(X_{α}^{n}, M/J^{j}𝔎(M)) is maximal. Similarly, there exists a dense open subset V ⊆ 𝔸^{2}(k) such that α ∈ V if and only if rank(X_{α}^{n}, N) is maximal. Since 𝔸^{2}(k) is irreducible, U ∩ V is nonempty. Choosing α ∈ U ∩ V, we simultaneously have rank(X_{α}^{n}, M/J^{j}𝔎(M)) maximal and rank(X_{α}^{n}, N) maximal. This forces both summands to be constant in α. □

We now use the above theory to show that Question 1.1 is true in the special case where J^{2}𝔎(M) = 0.

Theorem 4.5. If M is a kE-module of constant Jordan type such that J^{2}𝔎(M) = 0, then J^{−i}𝔎(M) has constant Jordan type for all i ≥ 0.

Proof. Since J^{2}𝔎(M) = 0, we have M^{∗} = J^{−1}𝔎(M^{∗}) by Theorem 3.3. We first show that M^{∗}/J^{j}𝔎(M^{∗}) has constant Jordan type for all j ≥ 0 using downwards induction on j, beginning at M^{∗} = M^{∗}/J^{p}𝔎(M^{∗}). So assume M^{∗}/J^{j+1}𝔎(M^{∗}) has constant Jordan type. Note for 0 ≠ α ∈ 𝔸^{2}(k) that any Jordan block in the action of X_{α} on M^{∗}/J^{j}𝔎(M^{∗}) has length at most j + 1. If n ≤ j, let m ∈ M^{∗}/J^{j+1}𝔎(M^{∗}) such that X_{α}^{n}m ∈ J^{j}𝔎(M^{∗})/J^{j+1}𝔎(M^{∗}). Since 𝔎(M^{∗})/J^{j+1}𝔎(M^{∗}) has the constant image property, there exists m′ ∈ 𝔎(M^{∗})/J^{j+1}𝔎(M^{∗}) such that X_{α}^{n}m′ = X_{α}^{n}m. Applying the sequence (1) to M^{∗}, this yields

L_{n,α} = Im(X_{α}^{n}, 𝔎(M^{∗})/J^{j+1}𝔎(M^{∗})) for all nonzero α.

It follows from Proposition 4.4 that rank(X_{α}^{n}, M^{∗}/J^{j}𝔎(M^{∗})) is independent of α for all n ≤ j. Letting [1]^{a_{1}} … [j + 1]^{a_{j+1}} denote the Jordan type of X_{α} on 𝔎(M^{∗})/J^{j+1}𝔎(M^{∗}), we have

a_{1} + 2a_{2} + 3a_{3} + ⋯ + ja_{j} + (j + 1)a_{j+1} = dim_{k} M^{∗}/J^{j}𝔎(M^{∗})
a_{2} + 2a_{3} + ⋯ + (j − 1)a_{j} + ja_{j+1} = rank(X_{α}, M^{∗}/J^{j}𝔎(M^{∗}))
⋮
a_{j} + 2a_{j+1} = rank(X_{α}^{j−1}, M^{∗}/J^{j}𝔎(M^{∗}))
a_{j+1} = rank(X_{α}^{j}, M^{∗}/J^{j}𝔎(M^{∗})).

Because the right-hand terms are all independent of α, these equations have a solution for a_{1}, …, a_{j+1} that is also independent of α. This shows that M^{∗}/J^{j}𝔎(M^{∗}) has constant Jordan type.

Having proved the claim, setting j = −i + 1, it now follows by another application of Theorem 3.3 that J^{−i}𝔎(M) has constant Jordan type for all i ≥ 0 as required. □

5. A counterexample to Question 1.1

We now present an example of a kE-module M of constant Jordan type for which J^{−1}𝔎(M) does not have constant Jordan type. Let M be the 26-dimensional kE-module with basis a, b, c, …, x, y, z on which g_{1} acts via the matrix

( \begin{matrix}& a & b & c & d & e & f & g & h & i & j & k & l & m & n & o & p & q & r & s & t & u & v & w & x & y & z \\ a & 1 \\ b & & 1 \\ c & 1 & & 1 \\ d & & & & 1 \\ e & & & & & 1 \\ f & & & & & & 1 \\ g & & 1 & & & & & 1 \\ h & & & & & & & & 1 \\ i & & & & & & & & & 1 \\ j & 1 & & & & & & & & & 1 \\ k & & & 1 & & & & & & & & 1 \\ l & & & & 1 & & & & & & & & 1 \\ m & & & & & & & & & & & & & 1 \\ n & & & & & & & & & & & & & & 1 \\ o & & & & & & & & & & & & & & & 1 \\ p & & & & & 1 & & & & & & & & & & & 1 \\ q & & & & & & & & & & & & & & & & & 1 \\ r & & & & & & & & & & & & & & & & & & 1 \\ s & & & & & & & 1 & & & & & & & & & & & & 1 \\ t & & & & & & & & & 1 & & & & & & & & & & & 1 \\ u & & & & & & & & & & 1 & & & & & & & & & & & 1 \\ v & & & & & & & & & & & & & & & 1 & & & & & & & 1 \\ w & & & & & & & & & & & & & & & & 1 & & & & & & & 1 \\ x & & & & & & & & & & & & & & & & & & 1 & & & & & & 1 \\ y & & & & & & & & & & & & & & & & & & & & & 1 & & & & 1 \\ z & & & & & & & & & & & & & & & & & & & & & & & 1 & & & 1 \end{matrix} )

and g_{2} acts via the matrix

( \begin{matrix}& a & b & c & d & e & f & g & h & i & j & k & l & m & n & o & p & q & r & s & t & u & v & w & x & y & z \\ a & 1 \\ b & & 1 \\ c & & & 1 \\ d & 1 & & & 1 \\ e & & & & & 1 \\ f & & & & & & 1 \\ g & & & & & & & 1 \\ h & & 1 & & & & & & 1 \\ i & & & & & & & & & 1 \\ j & & & & & & & & & & 1 \\ k & & & & & & & & & & & 1 \\ l & & & 1 & & & & & & & & & 1 \\ m & & & & 1 & & & & & & & & & 1 \\ n & 1 & & & & & & & & & & & & & 1 \\ o & & & & & & & & & & & & & & & 1 \\ p & & & & & & & & & & & & & & & & 1 \\ q & & & & & 1 & & & & & & & & & & & & 1 \\ r & & & & & & & & & & & & & & & & & & 1 \\ s & & & & & & 1 & & & & & & & & & & & & & 1 \\ t & & & & & & & & 1 & & & & & & & & & & & & 1 \\ u & & & & & & & & & 1 & & & & & & & & & & & & 1 \\ v & & & & & & & & & & & & & & 1 & & & & & & & & 1 \\ w & & & & & & & & & & & & & & & 1 & & & & & & & & 1 \\ x & & & & & & & & & & & & & & & & & 1 & & & & & & & 1 \\ y & & & & & & & & & & & & & & & & & & & & 1 & & & & & 1 \\ z & & & & & & & & & & & & & & & & & & & & & & 1 & & & & 1 \end{matrix} ) .

Here we have omitted entries equal to zero. This module structure is illustrated in the following diagram, where X_{1} = g_{1}−1 acts via single edges downwards to the left, X_{2} = g_{2}−1 acts via double edges downwards to the right, and the dotted edges indicate that X_{1}.a = c+j and X_{2}.a = d + n.

[figure: a; b; c; d; e; f; g; h; i; j; k; l; m; n; o; p; q; r; s; t; u; v; w; x; y; z]

We first show that M has constant Jordan type by verifying that rank(X_{α}^{n}, M) is independent of 0 ≠ α ∈ 𝔸^{2}(k) for n = 1, 2, 3. By inspection we see that rank(X_{2}, M) = 13. For any λ ∈ k, the image of X_{1} + λX_{2} on M is spanned by the elements

c + j + λ(d + n), g + λh, k + λl, l + λm, p + λq, s, t, u, v, w, x, y, z,

which are readily confirmed to be linearly independent. Hence rank(X_{1} + λX_{2}, M) = 13 as well. Next observe that rank(X_{2}^{2}, M) = 5, and for any λ ∈ k, the image of (X_{1} + λX_{2})^{2} on M is spanned by the linearly independent elements

k + u + 2λl + λ^{2}(m + v), s + λ^{2}t, w + λ^{2}x, y, z

so that rank((X_{1} + λX_{2})^{2}, M) = 5. Finally, we have rank(X_{2}^{3}, M) = 2, and for any λ ∈ k, the image of (X_{1} + λX_{2})^{3} on M is the submodule spanned by λ^{3}y, y + λ^{3}z and z, which in any case has dimension two. Thus rank((X_{1} + λX_{2})^{3}, M) = 2 as required. This shows that M has constant Jordan type [4]^{2}[3][2]^{5}[1]^{5}.

To see that J^{−1}𝔎(M) does not have constant Jordan type, note that 𝔎(M) is the submodule of M generated by

f, g, h, i, j, k, l, m, n, o, p, q, r

since this is the largest submodule of M having the constant image property. It follows that J^{−1}𝔎(M) is the submodule of M generated by

f, b, i, j, c, d, n, o, e, r.

Observe that J^{−1}𝔎(M) has a direct summand N generated by the elements f, b, i, j. Clearly N cannot have constant Jordan type because, for example, X_{2}^{3}N = span_{k}(y), whereas X_{1}^{3}N = 0. By Theorem 2.1 (ii), this shows that J^{−1}𝔎(M) cannot have constant Jordan type.

References

1. S. Baland, Modules of constant Jordan type with two non-projective blocks, J. Algebra 346 (2011), 343–350.

2. D. J. Benson, Modules of constant Jordan type with one non-projective block, Algebras and Representation Theory 13 (2010), 315–318.

3. , Modules of constant Jordan type and a conjecture of Rickard, preprint, 2011.

4. , Modules of constant Jordan type with small non-projective part, to appear in Algebras and Representation Theory, 2011.

5. J. F. Carlson and E. M. Friedlander, Exact category of modules of constant Jordan type, Algebra, arithmetic and geometry: Manin Festschrift, Progr. in Math., vol. 269, Birkhäuser Verlag, Basel, 2009, pp. 259– 281.

6. J. F. Carlson, E. M. Friedlander, and J. Pevtsova, Modules of constant Jordan type, J. Reine & Angew. Math. 614 (2008), 191–234.

7. J. F. Carlson, E. M. Friedlander, and A. A. Suslin, Modules for ℤ/p × ℤ/p, Comment. Math. Helvetici 86 (2011), 609–657.

8. E. M. Friedlander, J. Pevtsova, and A. Suslin, Generic and maximal Jordan types, Invent. Math. 168 (2007), 485–522.

Institute of Mathematics, University of Aberdeen, Aberdeen AB24 3UE