Representation Type under Excellent Extensions

In this paper, we prove that representation type of finite dimensional algebras over an algebraically closed field is an invariant under excellent extensions. Mathematics Subject Classification: 16K20, 16G60


Introduction
In studying the algebraic structure of group rings, Passman in [9] introduced the notion of the excellent extensions of rings( the name comes from [1]).Such extensions of rings are vital since they include two important classes of extensions of rings: finite matrix rings and skew group rings AG, where the finite group G satisfies the condition |G| −1 ∈ A. Many authors have studied the invariant properties of rings under excellent extensions ( [1,5,6,9,10,11,12]).
It is well known that determining the representation type of algebras is fundamental and important in representation theory of finite dimensional algebras.Let k be an algebraically closed field.Recall that a finite dimensional k-algebra A is said to be of finite representation type or a representation finite algebra, if there exist only finitely many isomorphism classes of finitely generated indecomposable A-modules.Recall that the a finite dimensional k-algebra A is said to be of tame type or a tame algebra if A is not of finite representation type, whereas for any dimension d > 0, there are a finite number of k[T ] − A−bimodules {M i : i = 1, 2, • • • , n}, which are free as left k[T ]−modules such that all but a finite number of indecomposable A-modules of dimension d are isomorphic to k[T ]/(T − λ) ⊗ k[T ] M i for λ ∈ k.And, a finite dimensional k-algebra A is said to be of wild type or a wild algebra if there is a finitely generated k < X, Y > −A-bimodule N, which is free as a left k < X, Y >-module such that the functor − ⊗ k<X,Y > N from mod k < X, Y > to modA preserves indecomposability and reflects isomorphisms.The famous tame-and-wild theorem of Drozd's in [3,4] states that a finite dimensional k-algebra, which is not of finite representation type, is either of tame type or of wild type, and not both.Therefore, it gives the classification of finite dimensional algebras over an algebraically closed field .
Let B ≥ A be an excellent extension of a finite dimensional k-module A. B is of finite representation type if and only if so is A by [6].On the other hand, AG (G is a finite group such that |G| −1 ∈ A, where A is a finite dimensional k-algebra) and A have the same representation type by [8].And S. Kasjan in [7] showed that A is a tame algebra if and only if so is A ⊗ k F , where F is a separable field over k.Based on these facts, it is natural to ask the following question: Question: Let B ≥ A be an excellent extension of an finite dimensional k-algebra, do A and B have the same representation type?
In this paper, we will study the invariance of the representation type of finite dimensional k-algebras under excellent extensions and give a correct answer to this question.This paper is organized as follows.
In Section 2, we give some notations in our terminology and some preliminary results which are often used in this paper.In Section 3, we prove the following Theorem 1.1 Let B be an excellent extension of a finite dimensional kalgebra A, then A and B have the same representation type.

Preliminaries
Throughout this paper, All modules are finitely generated right modules unless stated otherwise.

We begin with the definition of excellent extensions of finite dimensional k-algebras. Definition 2.1. Let A be a subalgebra of a finite dimensional k-algebra B, such that A and B have the same identity. Then B is called an algebraic extension of A, and denoted by A ≤ B. An algebraic extension
(2) B is a free normalizing extension of A with a basis that includes 1; that is, there is a finite set Let A be a finite dimensional k-algebra.A category is called generic category, denoted by GC(A), if its objects are all k[T ]-A-bimodules which are free as left k[T ]-modules, and its morphisms are all k[T ]-A-morphisms.It is closely related with tame algebras and thus with generic modules [2].The following lemma are taken from [8], which plays an important role in this paper.

Representation type
By the definition of excellent extensions, it is easy to prove the following and B is free with basis {b 1 , b 2 , • • • , b n } ∈ B as both a right and left A-module.Lemma 2.2.(See [[9], p.275, Lemma 2.3].)Let A be a finite dimensional k-algebra, and let F be a finite separable field extension of k.Then A ⊗ k F is an excellent extension of A. Lemma 2.3.(See[[6],Theorem 3.8].)Let B be an excellent extension of a finite dimensional k-algebra A. If A is of finite representation type if and only if so is B. Lemma 2.4.( See [[11], Lemma 1.1].)Let A ≤ B be an algebraic extension such that B is right A-projective.Then we have M B |(M ⊗ A B) B for any Bmodule.

Lemma 3 . 1 .Theorem 3 . 3
Let B ≥ A be an excellent extension.If A is a finite dimensional k-algebra, then so is B.Proposition 3.2 Let B be an excellent extension of a finite dimensionλ) ∼ = k as algebras.Since the set {N i j : j = 1, • • • , n; i = 1, • • • , ; m j }are finite, and so A is a tame algebra.According to the Drozd's tame-and-wild theorem and Lemma 2.3 and Proposition 3.1, we obtain our main result.Let B be an excellent extension of A, then A and B have the same representation type.