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A footnote to the multiplicative basis theorem
| Content Provider | Semantic Scholar |
|---|---|
| Author | Gustafson, William I. |
| Copyright Year | 1985 |
| Abstract | We characterize those perfect fields k such that for each integer n > 1, but there are but finitely many isomorphism types of k-algebras of dimension n that are of finite representation type. Some remarks on the imperfect case are also presented. A finite-dimensional algebra A over a field k is of finite representation type if it has only finitely many isomorphism types of indecomposable modules. A multiplicative basis for A is a k-basis B such that B U {0} is a semigroup under the multiplication in A. Recently, Bautista, Gabriel, Roiter and Salmeron [1] have shown that an algebraically closed field k has the property that each k-algebra of finite representation type has a multiplicative basis. Let us point out at once that this property characterizes algebraically closed fields. For, if a field k has a finite extension field F, then F is of finite representation type, and any multiplicative basis B for F would be a finite semigroup with cancellation, hence a group. Then, F would be isomorphic to the group algebra kB, but kB can never be simple if B is nontrivial. An important consequence of the theorem above is that when k is algebraically closed, there are but finitely many k-algebras of finite representation type of any fixed k-dimension ("finite representation type is finite"). Let us express this by saying that k has property (N). We wish here to discuss other fields with property (N). Recall from [3, III-29] that a field k is of type (F) if it is perfect and, for each n > 1, there are only finitely many k-isomorphism types of field extensions of degree n over k. Examples are finite fields, local fields with finite residue field and the fields F((T)) of quotients of power series over algebraically closed fields F of characteristic zero. THEOREM. A perfect field k has property (N) if and only if it is of type (F). PROOF. One implication is clear. If k is of type (F), let k be an algebraic closure of k, and let G be the Galois group of k/k. By [2], a finite-dimensional k-algebra A is of finite representation type if and only if k ?k A is. Hence, it suffices to show that a k-algebra A of the form Ik ?k A has only finitely many k-forms, up to isomorphism. Such forms are classified by the set H' (G, Autkaig (A)). This set is finite by [3, III-30] since Aut-kaig (A) is a linear algebraic group defined over k, and the proof is complete. Let us remark on the case of imperfect fields. The degree of imperfection of a field k of characteristic p is that integer r so that [k: kP] = pr (or infinity, if [k: kP] is infinite). Thus, k is perfect just when its degree of imperfection is zero. It is well Received by the editors October 8, 1984. 1980 Mathematics Subject Clasification. Primary 16A46. (?)1985 American Mathematical Society 0002-9939/85 $1.00 + $.25 per page |
| Starting Page | 7 |
| Ending Page | 8 |
| Page Count | 2 |
| File Format | PDF HTM / HTML |
| DOI | 10.1090/S0002-9939-1985-0796436-1 |
| Volume Number | 95 |
| Alternate Webpage(s) | http://www.ams.org/journals/proc/1985-095-01/S0002-9939-1985-0796436-1/S0002-9939-1985-0796436-1.pdf |
| Alternate Webpage(s) | https://doi.org/10.1090/S0002-9939-1985-0796436-1 |
| Language | English |
| Access Restriction | Open |
| Content Type | Text |
| Resource Type | Article |
