FIELDS OF DEFINITION FOR REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS

DAVE BENSON AND ZINOVY REICHSTEIN

Abstract. We examine situations, where representations of a finite-dimensional F - algebra A defined over a separable extension field K/F, have a unique minimal field of definition. Here the base field F is assumed to be a C_{1}-field. In particular, F could be a finite field or k(t) or k((t)), where k is algebraically closed.

We show that a unique minimal field of definition exists if (a) K/F is an algebraic extension or (b) A is of finite representation type. Moreover, in these situations the minimal field of definition is a finite extension of F. This is not the case if A is of infinite representation type or F fails to be C_{1}. As a consequence, we compute the essential dimension of the functor of representations of a finite group, generalizing a theorem of N. Karpenko, J. Pevtsova and the second author.

1. Introduction

Notational conventions. Throughout this paper F will denote a base field and A a finite-dimensional associative algebra over F. If K/F is a field extension (not necessarily algebraic), we will denote the tensor product K ⊗_{F} A by A_{K}. Let M be an A_{K}-module. Unless otherwise specified, we will always assume that M is finitely generated (or equivalently, finite-dimensional as a K-vector space). If L/K is a field extension, we will write M_{L} for L ⊗_{K} M.

An intermediate field F ⊂ K_{0} ⊂ K is called a field of definition for M if there exists a K_{0}-module M_{0} such that M ≅ (M_{0})_{K}. In this case we will also say that M descends to K_{0}.

Minimal fields of definition. A field of definition K_{0} of M is said to be minimal if whenever M descends to a field L with F ⊂ L ⊂ K, we have K_{0} ⊂ L.

Minimal fields of definition do not always exist. For example, let F = ℚ and A be the quaternion algebra

A = ℚ{i, j, k}/(i^{2} = j^{2} = k^{2} = ijk = −1).

Date: Wednesday 8^{th} February, 2017 at 10:04. 2010 Mathematics Subject Classification. 16G10, 16G60, 20C05. Key words and phrases. Modular representation, field of definition, finite representation type, essential

dimension.

The research of the first author was supported by the Collaborative Research Group in Geometric and Cohomological Methods in Algebra at the Pacific Institute for the Mathematical Sciences, Vancouver, Canada (2016).

The second author was partially supported by NSERC Discovery Grant 250217-2012.

Then A_{K} has a two dimensional module M given by

i ↦ (\begin{matrix}a & b \\ b & −a \end{matrix}) ,  j ↦ (\begin{matrix}b & −a \\ −a & −b \end{matrix}) ,  k ↦ (\begin{matrix}0 & 1 \\ −1 & 0 \end{matrix})

over any field K of characteristic 0 having two elements a and b such that a^{2} + b^{2} = −1. Examples of such fields include ℂ, ℚ(\sqrt{−1}) or ℚ(\sqrt{−}5). If we take K to be “the generic field” of this type, i.e., the field of fractions of ℚ[a, b]/(a^{2} +b^{2} +1), then M has no minimal field of definition; see Proposition 6.3(b).

C_{1}-fields. Such examples arise because of the existence of noncommutative division rings of finite dimension over F. So, it makes sense to develop a theory over fields for which these do not exist. We say that F is a C_{1}-field if any homogeneous polynomial f_{1}(x_{1}, …, x_{n}) of degree d < n with coefficients in F has a non-trivial solution in F^{n}. Examples of C_{1}- fields include finite fields, k(t), and k((t)), where k is algebraically closed. An algebraic extension of a C_{1}-field is again C_{1}. Over a C_{1}-field every every central division algebra is commutative. For a detailed discussion of this class of fields, including proofs of the above assertions, we refer the reader to [GS, Section 6.2]. Our first main result is as follows.

Theorem 1.1. Let F be a C_{1}-field, A be a finite-dimensional F -algebra, K/F be a separable algebraic field extension and M be an an A_{K}-module. Then M has a minimal field of definition F ⊂ K_{0} ⊂ K such that [K_{0} : F] < ∞.

To illustrate Theorem 1.1, let us consider a simple case, where char(F) = 0, A := F G is the group algebra of a finite group G, and M is absolutely irreducible KG-module. Denote the character of G associated to M by χ: G → K. We claim that in this case the minimal field of definition is F (χ), the field generated over F by the character values χ(g), as g ranges over G. Indeed, it is clear that F (χ) has to be contained in any field of definition F ⊂ K_{0} ⊂ K of M. Thus to prove the above assertion, we only need to show that M descends to F (χ). The minimal degree of a finite field extension L/F (χ), such that M is defined over L (i.e., there exists an LG-module with character χ), is the Schur index s_{M}; cf. [CR, Definition 41.4]. Thus it suffices to show that s_{M} = 1. By [CR, Theorem (70.15)], s_{M} is the index of the endomorphism algebra End_{A}(M) of M, which is a central simple algebra over F (χ). Since F is a C_{1}-field, and F (χ) is a finite extension of F, F (χ) is also a C_{1}-field. Hence, the index of every central simple algebra over F (χ) is 1. In particular, s_{M} = 1, and M descends to F (χ), as claimed.

Algebras of finite representation type. A finite-dimensional F -algebra A is said to be of finite representation type if there are only finitely many indecomposable finitely generated A-modules (up to isomorphism).

Our next result shows that for algebras of finite representation type Theorem 1.1 remains valid even if the field extension K/F is not assumed to be algebraic.

Theorem 1.2. Let F be a C_{1}-field, A be a finite-dimensional F -algebra of finite representation type, K/F be a field extension, and M be an A_{K}-module. Assume further that F is perfectly closed in K. Then M has a minimal field of definition F ⊂ K_{0} ⊂ K such that [K_{0} : F] < ∞.

Essential dimension. Given the A_{K}-module M, the essential dimension ed(M) of M over F is defined as the minimal value of the transcendence degree trdeg(K_{0}/F), where the minimum is taken over all fields of definition F ⊂ K_{0} ⊂ K. The integer ed(M) may be viewed as a measure of the complexity of M. Note that ed(M) is well-defined, irrespective of whether M has a minimal field of definition or not. We also remark that this number implicitly depends on the base field F, which is assumed to be fixed throughout. As a consequence of Theorem 1.2, we will deduce the following.

Theorem 1.3. Let F be a C_{1}-field, A be finite-dimensional F -algebra of finite representation type, K/F be a field extension, and M be an A_{K}-module. Then ed(M) = 0.

Both Theorem 1.2 and 1.3 fail if we do not require F to be a C_{1}-field; see Section 6.

The essential dimension of the functor of A-modules. We will also be interested in the essential dimension ed(Mod_{A}) of the functor Mod_{A} from the category of field extensions of F to the category of sets, which associates to a field K, the set of isomorphism classes of A_{K}-modules. By definition,

ed(Mod_{A}) := sup ed(M) ,

where the supremum is taken over all field extensions K/F and all finitely generated A_{K}- modules M. The value of ed(Mod_{A}) may be viewed as a measure the complexity of the representation theory of A. For generalities on the notion of essential dimension we refer the reader to [BF, Re_{1}, Re_{2}, Me_{1}, Me_{2}]. Essential dimensions of representations of finite groups and finite-dimensional algebras are studied in [KRP] and [BDH, Section 3].

Note that while ed(M) < ∞, for any given A_{K}-module M (see Lemma 2.1), ed(Mod_{A}) may be infinite. In particular, in the case, where A = F G is the group algebra of a finite group G over a field F, it is shown in [KRP, Theorem 14.1] that ed(Mod_{A}) = ∞, provided that F is a field of characteristic p > 0 and G has a subgroup isomorphic to (ℤ/pℤ)^{2}. Our final main result is the following amplification of [KRP, Theorem 14.1].

Theorem 1.4. Let G be a finite group and F be a field of characteristic p. Then the following conditions are equivalent:

(1) The p-Sylow subgroup of G is cyclic,

(2) ed(Mod_{F G}) = 0,

(3) ed(Mod_{F G}) < ∞.

2. Preliminaries on fields of definition

Lemma 2.1. Let A be a finite-dimensional F -algebra, K/F be a field extension and M be an A_{K}-module. Then M descends to an intermediate subfied F ⊂ E ⊂ K, where E/F is finitely generated.

Proof. Suppose a_{1}, …, a_{r} generate A as an F -algebra. Choose an F -vector space basis for M. Then the A-module structure of M is completely determined by the matrices representing multiplication by a_{1}, …, a_{r} in this basis. Each of these matrices has n^{2} entries in K, where n = dim_{F} (M). Let E ⊂ K be the field extension of F obtained by adjoining these these rn^{2} entries to F. Then M descends to E. □

Next we recall the classical theorem of Noether and Deuring. For a proof, see [CR, (29.7)] or [BP, Lemma 5.1].

Theorem 2.2. (Noether-Deuring Theorem) Let K/E be a field extension, A be a finitedimensional E-algebra, and M_{1}, M_{2} and M be A-modules. If K ⊗_{E} M_{1} and K ⊗_{E} M_{2} are isomorphic as A_{K}-modules, then M_{1} and M_{2} are isomorphic as A-modules. □

Lemma 2.3. Let F be a field, A be a finite-dimensional F -algebra, F ⊂ E ⊂ K be a field extension, N be A_{E}-module, and M = N_{K} and F ⊂ E_{0} ⊂ E be an intermediate field. Then

(a) M descends to E_{0} if and only if N descends to E_{0}.

(b) If F ⊂ E_{min} ⊂ K is a minimal field of definition for M, then E_{min} is a minimal field of definition for N.

Proof. (a) If N descends to E_{0}, then clearly so does M. Conversely, suppose M descends to E_{0}. That is, there exists a E_{0}-module N_{0} such that K ⊗_{E_{0}} N_{0} ≃ M as an A_{K}-module. Consider the A_{E}-modules N_{1} := E ⊗_{E_{0}} N_{0} and N_{2} := N. Both become isomorphic to M_{K} over K. By Theorem 2.2, N_{1} ≃ N_{2} as A_{E}-modules. In other words, N descends to E_{0}, as desired.

(b) Since E is a field of definition for M, we have E_{min} ⊂ E. By part (a), E_{min} is a field of definition for N, and part (b) follows. □

We finally come to the main result of this section.

Proposition 2.4. Suppose F is a C_{1}-field, A is a finite-dimensional F -algebra, K/F is a field extension, M is a finitely generated A_{K}-module, and F ⊂ K_{0} ⊂ K is an intermediate field, such that [K_{0} : F] < ∞.

If M^{n} is defined over K_{0} for some positive integer n, then so is M.

Proof. Set End_{A_{K}}^{ss}(M) to be the quotient of End_{A_{K}}(M) by its Jacobson radical. By our assumption M^{n} ≃ K ⊗_{K_{0}} N for some A_{K_{0}}-module N. By Fitting’s Lemma,

End_{A_{K}}^{ss}(M^{n}) ≃ M_{n}(D),

where D is a finite-dimensional division algebra over some finite field extension K′ of K. On the other hand,

(2.5)  M_{n}(D) ≃ End_{A_{K}}^{ss}(M^{n}) ≃ End_{A_{K}}^{ss}(K ⊗_{K_{0}} N) ≃ K ⊗_{K_{0}} End_{A_{K_{0}}}^{ss}(N) .

We conclude that End_{A_{K_{0}}}^{ss}(N) is a simple algebra over K_{0}, i.e.,

(2.6)  End_{A_{K_{0}}}^{ss}(N) ≃ M_{m}(D_{0})

over K_{0}, for some integer m ⩾ 0 and some finite-dimensional central division algebra D_{0} over a field K′_{0} such that K′_{0} is a finite extension of K_{0}. Now recall that we are assuming that F is a C_{1}-field and

F ⊂ K_{0} ⊂ K′_{0}

are finite field extensions. Hence, K′_{0} is also a C_{1}-field, and thus every finite-dimensional division algebra over K′_{0} is commutative. In particular, D_{0} = K′_{0}, is a field, and

M_{n}(D) ≃ K ⊗_{K_{0}} End_{A_{K_{0}}}^{ss}(N) ≃ K ⊗_{K_{0}} M_{m}(K′_{0}) .

Since M_{n}(D) is a simple algebra, we conclude that K ⊗_{K_{0}} K′_{0} is a field. Moreover, the index of M_{m}(K ⊗_{K_{0}} K′_{0}) is 1; hence, D = K′ is commutative, K ⊗_{K_{0}} K′_{0} = K′, and m = n.

Now (2.6) tells us that N ≃ M_{0}^{n} as a A_{K_{0}}-module, for some indecomposable A_{K_{0}}- module M_{0}. Since K ⊗_{K_{0}} N ≃ M^{n}, by the Krull-Schmidt theorem K ⊗_{K_{0}} M_{0} ≃ M. Thus M descends to K_{0}, as claimed. □

3. Proof of Theorem 1.1

We begin with a simple criterion for the existence of a minimal field of definition.

Lemma 3.1. Let A be a finite-dimensional F -algebra, and K/F be a field extension, and M be an A_{K}-module, satisfying conditions (a) and (b) below. Then M has a minimal field of definition.

(a) Suppose M descends to an intermediate field F ⊂ L ⊂ K, i.e., M ≃ K ⊗_{L} N for some A_{L}-module N. Then N further descends to a subfield F ⊂ E ⊂ L, where [E : F] < ∞.

(b) Suppose M descends to an intermediate field F ⊂ E ⊂ K such that [E : F] < ∞. That is, M ≃ K ⊗_{E} N for some A_{E}-module N. Then N has a minimal field of definition E_{min} ⊂ E.

Proof. Condition (a) implies that M is defined over some F ⊂ E ⊂ K with [E : F] < ∞. Let the A_{E}-module N and the field E_{min} ⊂ E be as in (b).

We claim that E_{min} is independent of the choice of E. That is, suppose F ⊂ E′ ⊂ K is another field of definition of M with [E′ : F] < ∞, M := K ⊗_{E′} N′ for some A_{E′}-module N′. Let E′_{min} ⊂ E′ be the minimal field of definition of N′, so that N′ := E′ ⊗_{E′_{min}} N′_{min}. Then our claim asserts that E_{min} = E′_{min}. If we can prove this claim, then clearly E_{min} is the minimal field of definition for M. Our proof of the claim will proceed in two steps.

First assume E ⊂ E′. By Lemma 2.3(b), E′_{min} is a minimal field of definition for N. By uniqueness of the minimal field of definition for N, E_{min} = E′_{min}.

Now suppose F ⊂ E ⊂ K and F ⊂ E′ ⊂ K are fields of definition for M such that [E : F] < ∞ and [E′ : F] < ∞. Let E′′ be the composite of E and E′ in K and E′′_{min} be the minimal field of definition of N_{E′′} ≃ N′_{E′′}. (Note that N_{E′′} and N′_{E′′} become isomorphic over K; hence, by Theorem 2.2, they are isomorphic over E′′.) Then, [E′′ : F] < ∞, and E, E′ ⊂ E′′. As we just showed, E_{min} = E′′_{min} and E′_{min} = E′′_{min}. Thus E_{min} = E′_{min}, as desired. □

We now proceed with the proof of Theorem 1.1.

Reduction 3.2. For the purpose of proving Theorem 1.1, we may assume without loss of generality that

(a) K is a finite extension of F.

(b) K is a Galois extension of F.

Proof. (a) follows from Lemma 3.1. Indeed, we are assuming that Theorem 1.1 holds whenever K is a finite extension of F. That is, condition (b) of Lemma 3.1 holds. On the other hand, condition (a) of Lemma 3.1 follows from Lemma 2.1.

(b) By part (a), we may assume that K/F is finite. Let L be the normal closure of K over F. Then L/F is finite Galois. Lemma 2.3(b) now tells us that if M_{L} := L ⊗_{K} M has a minimal field of definition then so does M. □

Lemma 3.3. Let F be a C_{1}-field, A be a finite-dimensional F -algebra, K/F be a finite Galois extension, and M be an A_{K}-module. The Galois group G := Gal(K/F) acts on the set of isomorphism classes of A_{K}-modules via

g: N → ^{g}N := K ⊗_{g} N .

Let G_{M} be the stabilizer of M under this action. Then the fixed field K^{G_{M}} of G_{M} is the minimal field of definition for M.

Proof. Suppose M is defined over K_{0}, where F ⊂ K_{0} ⊂ K. Then clearly ^{g}M ≃ M for every g ∈ Gal(K/K_{0}). Hence, Gal(K/K_{0}) ⊂ G_{M} and consequently, K^{G_{M}} ⊂ K_{0}. This shows that K^{G_{M}} is contained in every field of definition of M.

It remans to show that M descends to K_{0} := K^{G_{M}}. Write M = M_{1}^{d_{1}} ⊕⋯⊕M_{r}^{d_{r}}, where M_{1}, …, M_{r} are distinct indecomposables. The condition that ^{g}M ≃ M for any g ∈ G_{M} is equivalent to the following: if M_{j} ≃ ^{g}M_{i} for some g ∈ Gal(K/K_{0}), then d_{i} = d_{j}. Grouping G_{M}-conjugate indecomposables together, we see that M ≃ S_{1} ⊕⋯⊕S_{m}, where each S_{1}, …, S_{m} is the G_{M}-orbit sum of one of the indecomposable modules M_{i}. (Here the orbit sums S_{1}, …, S_{m} may not be distinct.) It thus suffices to show that each orbit sum is defined over K_{0}.

Consider a typical G_{M}-orbit sum S := M_{1} ⊕ ⋯ ⊕ M_{s}, where we renumber the indecomposable factors of M so that M_{1}, …, M_{s} are the G_{M}-translates of M_{1}. Let H be the stabilizer of M_{1} in G_{M}. That is,

H := {h ∈ G_{M} | ^{h}M_{1} ≃ M_{1}} .

Let K_{1} := K^{H}. Then

K ⊗_{K_{1}} (M_{1})_{↓K_{1}} = ⨁_{h∈H} ^{h}M_{1} = M_{1}^{|H|} .

In particular, this tells us that M_{1}^{|H|} descends to K_{1}. By Proposition 2.4, so does M_{1}. In other words, M_{1} ≃ K ⊗_{K_{1}} N_{1} for some K_{1}-module N. We claim that

(3.4)  K ⊗_{K_{0}} (N_{1})_{↓K_{0}} ≃ S.

If we can prove this claim, then S descends to K_{0}, and we are done.

To prove the claim, note that on the one hand,

(3.5)  K ⊗_{K_{0}} (M_{1})_{↓K_{0}} = ∏_{g∈G_{M}} ^{g}M_{1} = S^{|H|} .

On the other hand, since M_{1} ≃ K ⊗_{K_{1}} N_{1}, we have

(M_{1})_{↓K_{0}} ≃ ((M_{1})_{↓K_{1}})_{↓K_{0}} ≃ (N_{1}^{|H|})_{↓K_{0}} ,

and thus

(3.6)  K ⊗_{K_{0}} (M_{1})_{↓K_{0}} = (K ⊗_{K_{0}} ((N_{1})_{↓K_{0}})^{|H|}) ≃ (K ⊗_{K_{0}} (N_{1})_{↓K_{0}}))^{|H|} .

Comparing (3.5) and (3.6), we obtain

(K ⊗_{K_{0}} (N_{1})_{↓K_{0}})^{|H|} ≃ S^{|H|} .

The desired isomorphism (3.4) follows from this by the Krull-Schmidt theorem. □

4. Algebras of finite representation type

A finite-dimensional F -algebra A is said to be of finite representation type if there are only finitely many indecomposable finitely generated A-modules (up to isomorphism).

Theorem 4.1. Let F be a C_{1}-field, A be finite-dimensional F -algebra of finite representation type, and K/F be a field extension (not necessarily algebraic) such that F is perfectly closed in K. (That is, for every subextension F ⊂ E ⊂ K with [E : F] < ∞, E is separable over F.) Suppose M is an indecomposable A_{K}-module. Then

(a) M descends to an intermediate subfield F ⊂ E ⊂ K such that [E : F] < ∞.

(b) M is a direct summand of K ⊗_{F} N for some indecomposable A_{F} -module N.

Proof. (a) Consider the A-module M_{↓ F}. Generally speaking this module is not finitely generated over A. Nevertheless, since A has finite representation type, thanks to a theorem of Tachikawa [Ta, Corollary 9.5], M_{↓ F} can be written as a direct sum of finitely generated indecomposable A-modules. Denote one of these modules by N. That is,

(4.2)  M_{↓ F} ≃ N ⊕ N′ ,

for some A-module N′ (not necessarily finitely generated).

Let us now take a closer look at N. By Fitting’s lemma, E := End_{A}^{ss}(N) is a finitedimensional division algebra over F. Since F is a C_{1}-field, E is a field extension of F. Now set F′ := E ∩ K and m = [F′ : F]. Since F is perfectly closed in K, F′ is finite and separable over F. Thus

End_{A}^{ss}(F′ ⊗_{F} N) ≃ F′ ⊗_{F} End_{A}^{ss}(N) ≃ E × ⋯ × E .

This tells us that over F′, N decomposes into a direct sum of m indecomposables,

(4.3)  F′ ⊗_{F} N = N_{1} ⊕ ⋯ ⊕ N_{m}.

By the definition of F′, K ⊗_{F′} E is a field. Hence, each indecomposable A_{F′}-module N_{i} remains indecomposable over K.

Tensoring both sides of (4.2) with K, we obtain an isomorphism of A_{K}-modules

K ⊗ M_{↓ F} ≃ (K ⊗_{F} N) ⊕ (K ⊗_{F} N′)

= (⨁_{i=1}^{m} K ⊗_{F′} N_{i}) ⊕ (K ⊗_{F} N′)

= (K ⊗_{F} N_{1}) ⊕ N′′ ,

where N′′ := (⨁_{i=2}^{m} K ⊗_{F′} N_{i}) ⊕ (K ⊗_{F} N′). Note that

K ⊗ M_{↓ F′} ≃ ⨁_{B} M ,

where B is a basis of K as an F′-vector space. As we mentioned above, K ⊗_{F′} N_{1} is an indecomposable A_{K}-module. Since K ⊗_{F′} N_{1} is finitely generated and is contained in ⨁_{B} M, it lies in the direct sum of finitely many copies of M, say, in M^{r} := M ⊕ ⋯ ⊕ M (r copies). Thus we have maps

K ⊗_{F} N_{1} ↪ M^{r} ↪ ⨁_{B} M ↠ K ⊗_{F} N_{1}

whose composite is the identity, and so K ⊗_{F} N_{1} is isomorphic to a direct summand of M^{r}. By the Krull-Schmidt Theorem, K ⊗_{F′} N_{1} ≃ M. In particular, M descends to F′, as claimed.

(b) By (4.3), N is an indecomposable A-module, and N_{1} is a direct summand of F′⊗_{F} N. Hence, M ≃ K ⊗_{F′} N_{1} is a direct summand of K ⊗_{F} N, as desired. □

Corollary 4.4. Let F be a C_{1}-field, A be finite-dimensional F -algebra of finite representation type, and K/F be a field extension such that F is perfectly closed in K. Then A_{K} is also of finite representation type.

Proof. By our assumption A has finitely many indecomposable modules N^{(1)}, …, N^{(d)}. By Theorem 4.1(b) every indecomposable A_{K}-module is isomorphic to a direct summand of K ⊗_{F} N^{(i)} for some i. By the Krull-Schmidt Theorem, each K ⊗_{F} N^{(i)} has finitely many direct summands (up to isomorphism), and the corollary follows. □

5. Proof of Theorems 1.2 and 1.3

We will deduce Theorem 1.2 from Lemma 3.1. M satisfies condition (b) of Lemma 3.1 by Theorem 1.1. It thus remains to show that M satisfies condition (a) of Lemma 3.1. For notational simplicity, we may assume that K = L and M = N. That is, we want to show that M descends to some intermediate field F ⊂ E ⊂ K with [E : F] < ∞. Note that in the case, where M is indecomposable, this is precisely the content of Theorem 4.1(a).

In general, write M = M_{1} ⊕ ⋯ ⊕ M_{r} as a direct product of (not necessarily distinct) indecomposables. By Theorem 4.1(a), each M_{i} descends to an intermediate field F ⊂ K_{i} ⊂ K such that [K_{i} : F] < ∞. Let E be the compositum of K_{1}, …, K_{r} inside K. Then [E : F] < ∞, and M descends to E. This completes the proof of Theorem 1.2. □

We now proceed with the proof of Theorem 1.3. Denote the perfect closure of F in K by F^{pf}. By Theorem 1.2, M descends to an intermediate field F^{pf} ⊂ K_{0} ⊂ K such that [K_{0} : F^{pf}] < ∞. Hence, K_{0} is algebraic over F, and consequently, ed(M) ⩽ trdeg_{F} (K_{0}) = 0, as desired. □

6. An example

In this section we will show by example that both Theorem 1.2 and 1.3 fail if we do not require F to be a C_{1}-field. Let F = ℚ and A be the quaternion algebra

A = ℚ{x, y}/(x^{2} = y^{2} = −1, xy = −yx).

and K/F be any field having two elements a and b satisfying a^{2} + b^{2} = −1. Then A has a two dimensional A_{K}-module M given by

(6.1)  x ↦ (\begin{matrix}a & b \\ b & −a \end{matrix}) ,  y ↦ (\begin{matrix}b & −a \\ −a & −b \end{matrix}) .

Lemma 6.2. The following conditions on an intermediate field ℚ ⊂ E ⊂ K are equivalent:

(a) φ descends to E,

(b) A splits over E,

(c) there exist elements a_{0}, b_{0} ∈ E such that a_{0}^{2} + b_{0}^{2} = −1.

Proof. (a) ⟹ (b). Suppose M descends to an A_{E}-module N. Since A_{E} := E ⊗_{ℚ} A is a central simple 4-dimensional algebra over E, the homomorphism of algebras given by

A_{E} → End_{E}(N) ≃ M_{2}(E)

is an isomorphism. In other words, E splits A.

(b) ⟹ (a). Conversely, suppose E splits A. Then the representation of A → End_{K}(M) factors as follows:

A → E ⊗_{ℚ} A ≃ M_{2}(E) → M_{2}(K) .

This shows that φ descends to E.

The equivalence of (b) and (c) a special case of Hilbert’s criterion for the splitting of a quaternion algebra; see the equivalence of conditions (1) and (7) in [Lam, Theorem III.2.7] as well as Remark (B) on [Lam, p. 59]. □

Proposition 6.3. Let a and b be independent variables over ℚ, E be the field of fractions of ℚ[a, b]/(a^{2} + b^{2} + 1), and M be the 2-dimensional A_{E}-module given by (6.1). Then

(a) ed(M) = 1,

(b) M does not have a minimal field of definition.

Proof. (a) The assertion of part (a), follows from [KRP, Example 6.1]. For the sake of completeness, we will give an independent proof.

Suppose M descends to an intermediate subfield ℚ ⊂ E_{0} ⊂ E. Since trdeg_{ℚ}(E) = 1, trdeg_{ℚ}(E_{0}) = 0 or 1. Our goal is to show that trdeg_{ℚ}(E_{0}) ≠ 0. Assume the contrary, i.e., E_{0} is algebraic over ℚ.

Note that E is the function field of the conic curve a^{2}+b^{2}+c^{2} = 0 in ℙ^{2}. Since this curve is absolutely irreducible, ℚ is algebraically closed in E. Thus the only possibility for E_{0} is E_{0} = ℚ, On the other hand, M does not descend to ℚ by Lemma 6.2, a contradiction.

(b) Suppose M descends to E_{1} ⊂ E. Our goal is to show that M descends to a proper subfield E_{3} ⊂ E_{1}. By Lemma 6.2(c) there exist a_{1} and b_{1} in E_{1} such that a_{1}^{3} + b_{1}^{3} = −1. If ℚ(a_{1}, b_{1}) is properly contained in E_{1}, then we are done. Thus we may assume without loss of generality that E_{1} = ℚ(a_{1}, b_{1}). Set E_{3} := ℚ(a_{3}, b_{3}). where a_{3} := a_{1}^{3} − 3a_{1}b_{1}^{2} and b_{3} = 3a_{1}^{2}b_{1} − b_{1}^{3}. We claim that (i) A splits over E_{3}, and (ii) E_{3} ⊊ E_{1}.

In order to establish (i) and (ii), let us consider the following diagram.

[figure: E_{1}(i); E_{3}(i); E_{1}; E_{3}]

Here as usual, i is a primitive 4th root of 1. It is easy to see that E_{1}(i) = ℚ(i)(a_{1}, b_{1}) =

ℚ(i)(z) is a purely transcendental extension of ℚ(i), where z = a_{1}+b_{1}i and \frac{1}{z} = −a_{1}+b_{1}i.

Similarly E_{3}(i) = ℚ(i)(z^{3}), where z^{3} = a_{3} + b_{3}i and \frac{1}{z^{3}} = −a_{3} + b_{3}i. In particular, this

shows a_{3}^{2} + b_{3}^{2} = −1, thus proving (i). Moreover, since z is transcendental over ℚ(i), we have [E_{1}(i) : E_{3}(i)] = [ℚ(i)(z) : ℚ(i)(z^{3})] = 3 and thus

[E_{1} : E_{3}] = \frac{[E_{3}(i) : E_{3}] · [E_{1}(i) : E_{3}(i)]}{[E_{1}(i) : E_{1}]} = \frac{2 · 3}{2} = 3 .

This proved (ii). □

Remark 6.4. Write z^{n} = a_{n} + b_{n}i for suitable a_{n}, b_{n} ∈ E_{1} and set [E_{1} : E_{n}] = n. We showed above that [E_{1} : E_{3}] = 3 and thus E_{3} ⊊ E_{1}. The same argument yields [E_{1} : E_{n}] = n for any positive integer n.

7. Proof of Theorem 1.4

We shall actually prove a stronger, more natural theorem, about blocks of finite group algebras. Theorem 1.4 will follow from the fact that p-Sylow p of a finite group G are cyclic if and only if every block over a field F of characteristic p has cyclic defect.

Theorem 7.1. Let B be a block of a finite group algebra F G, where F is a field of characteristic p. Then the following are equivalent:

(1) B has cyclic defect,

(2) ed(Mod_{B}) = 0,

(3) ed(Mod_{B}) < ∞.

The implication (1) ⟹ (2) is a direct consequence of Theorem 1.3. The implication (2) ⟹ (3) is obvious.

The remainder of this section will be devoted to proving that (3) ⟹ (1). We shall show that if B has non-cyclic defect, then ed(Mod_{B}) = ∞. Let K be an extension field of F, let e be the block idempotent of B, let D be a defect group of B, and let N = Φ(D), the Frattini subgroup of D. If D is not cyclic, D/N is elementary abelian of rank r ≥ 2, with basis the images of elements g_{1}, …, g_{r} ∈ D. Since D is a defect group of B, any KD-module M is a summand of Res_{G,D}(e. Ind_{D,G}(M)).

Now let n > 0, and let K = F (t_{1,1}, …, t_{n,r}) be a function field in nr indeterminates, and let M_{i} (1 ≤ i ≤ n) be the two dimensional KD-module

g_{j} ↦ (\begin{matrix}1 & t_{i,j} \\ 0 & 1 \end{matrix}) .

Then J^{2}(KD) is in the kernel of M_{i}, so M_{i} is really a module for KD/J^{2}(KD), which has a basis 1, (g_{1} − 1), …, (g_{r} − 1). The last r elements of this list form a basis for J(KD)/J^{2}(KD), and we form a vector space V with basis (g_{1} − 1), …, (g_{r} − 1). The kernel of M_{i} as a module for KD/J^{2}(KD) is the codimension one subspace H_{i} of

J(KD)/J^{2}(KD) ≅ V

given by

(7.2)  H_{i} := {λ_{j}(g_{j} − 1) | ∑_{j} t_{i,j}λ_{j} = 0}.

By the Mackey decomposition theorem, the module M′_{i} = Res_{G,D}(e. Ind_{D,G}(M_{i})) is a direct sum of at least one copy of M_{i}, some conjugates of M_{i} by elements of N_{G}(D), and some modules of the form Ind_{D∩^{g}D,D} Res_{^{g}D,D∩^{g}D} ^{g}M. It follows that the Jordan canonical form of elements of V on M′_{i} is constant, except on a set S_{i}, which is a finite union of hyperplanes N_{G}(D)-conjugates of H_{i} and linear subspaces of smaller dimension.

Now let M := ⨁_{i} M_{i}. Our goal is to show that

ed(e. Ind_{D,G}(M)) ⩾ n(r − 1) .

This will imply that ed(Mod_{B}) ≥ n(r − 1) for every n > 0 and thus ed(Mod_{B}) = ∞, as desired.

Note that e. Ind_{D,G}(M) is a module whose restriction to D is ⨁_{i} M′_{i}. If e. Ind_{D,G}(M) descends to an intermediate subfield F ⊂ K_{0} ⊂ K, then so does the set ⋃_{i} S_{i} ⊂ V and its natural image in ℙ(V) = ℙ^{r−1}, which we will denote by S. To complete the proof of Theorem 7.1, it remains to show that if S descends to K_{0}, then

(7.3)  trdeg_{F} (K_{0}) ⩾ n(r − 1) .

Lemma 7.4. Let S ⊂ ℙ^{r−1} be a projective variety defined over a field K. Assume that a hyperplane H given by a_{1}x_{1} + a_{2}x_{2} + ⋯ + a_{r}x_{r} = 0 is an irreducible component of S for some a_{1}, …, a_{r} ∈ K (not all zero). Suppose S descends to a subfield K_{0} ⊂ K. Then each ratio a_{j}/a_{l} is algebraic over K_{0}, as long as a_{l} ≠ 0.

To deduce the inequality (7.3) from Lemma 7.4, recall that in our case S is the union of the hyperplanes H_{1}, …, H_{n}, a finite number of other hyperplanes (translates of H_{1}, …, H_{n} by elements of N_{G}(D)) and lower-dimensional linear subspaces of ℙ(V) = ℙ^{r−1}. In the basis (g_{1} − 1), …, (g_{r} − 1) of V, H_{i} is given by t_{i,1}x_{1} + t_{i,2}x_{2} + ⋯ + t_{i,r}x_{r} = 0; see (7.2). Thus by Lemma 7.4 the elements t_{i,j}/t_{i,1} are algebraic over K_{0} for every i = 1, …, n and every j = 2, …, r. In other words, if K_{1} is the algebraic cosure of K_{0} in K, then each t_{i,j}/t_{i,1} ∈ K_{1}, and thus trdeg_{F} (K_{0}) = trdeg_{F} (K_{1}) ⩾ n(r − 1), as desired.

Proof of Lemma 7.4. We may assume without loss of generality that K_{0} is algebraically closed. To reduce to this case, we replace K_{0} by its algebraic closure \overline{K_{0}} and K by a compositum of K and \overline{K_{0}}. If we know that each a_{i,j} is algebraic over \overline{K_{0}} (or equivalently, is contained in \overline{K_{0}}), then a_{i,j} is algebraic over K_{0}.

Now assume that K_{0} is algebraically closed. Since S is defined over K_{0}, every irreducible component of S is defined over K_{0}. In particular, H is defined over K_{0}. That is, the point (a_{1} : ⋯: a_{r}) of the dual projective space ℙ̌^{r−1} is defined over K_{0}. Equivalently, a_{i}/a_{j} ∈ K_{0} whenever a_{l} ≠ 0. This completes the proof of the claim and thus of Lemma 7.4 and Theorem 7.1. □

Acknowledgements

The second author would like to thank Julia Pevtsova for helpful discussions.

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E-mail address: d.j.benson@abdn.ac.uk

Department of Mathematics, University of British Columbia, Vancouver, CANADA

E-mail address: reichst@math.ubc.ca