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Let $k$ be a field and $A$ a unital associative (possibly non-commutative) $k$-algebra, and let $A^e$ denote the enveloping algebra of $A$, namely, $A^e = A \otimes_k A^{op}$.

It seems that there are two typical definitions for separable $k$-algebras. Consider the following two conditions.

  1. $A$ is projective as an $A^e$-module.
  2. For any field extension $K$ of $k$, the algebra $A \otimes_k K$ is semisimple.

Also consider the following condition.

  1. $A$ is finite-dimensional over $k$ and the condition 2 is satisfied.

In some literature (e.g. Corollary 10.6 in Pierce's Associative Algebra), it is shown that 1 is equivalent to 3.

I wonder whether 2 is actually equivalent to 1 and 3, that is,

Question: Suppose that the condition 2 is satisfied. Then is $A$ finite-dimensional over $k$?

I found some people say that this question is true, and even adopt the condition 2 as the definition of separable algebras: e.g. nLab's article on separable algebra, and Lemma 3.3 in Reyes-Rogalski's Graded twisted Calabi-Yau algebras are generalized Artin-Schelter regular. I looked up some cited references, but it seems that in the proofs, $A$ is assumed to be finite-dimensional.

I'm not sure whether the question is true even when $A$ is commutative, or $A$ is a filed.

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  • $\begingroup$ What is the enveloping algebra? I only know this term in the context of Lie algebras. $\endgroup$
    – user473423
    Commented Feb 9, 2022 at 11:12
  • $\begingroup$ $A^e$ is the tensor product algebra $A \otimes_k A^{op}$ of $A$ and the opposite algebra $A^{op}$. $\endgroup$
    – H. E.
    Commented Feb 9, 2022 at 12:25
  • $\begingroup$ What definition of semisimple are you using? I know a few different ones, but it's possible that they're only equivalent when you already assume finite-dimensionality. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 9, 2022 at 13:39
  • $\begingroup$ If you take the product of all matrix algebras $\mathrm{Mat}_n(k)$. Would that be semisimple in your definition? $\endgroup$
    – user473423
    Commented Feb 9, 2022 at 14:08
  • $\begingroup$ @ R. van Dobben de Bruyn A ring is semisimple if it is a semisimple module as a left (or equivalently right) A-module, and a semisimple algebra is just a $k$-algebra which is semisimple as a ring. $\endgroup$
    – H. E.
    Commented Feb 9, 2022 at 14:13

2 Answers 2

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I claim that no infinite dimensional algebra $A$ over $k$ satisfies 2. I am indebted to the comment of @UriyaFirst for the idea of the main step in the proof.

First I claim we may assume that $k$ is algebraically closed. Indeed, if $[A:k]=\infty$ and satisfies 2, then for an algebraic closure $\overline k$ of $k$ we have that $A'=A\otimes_k \overline k$ is infinite dimensional over $\overline k$ and satisfies 2 over $\overline k$ by transitivity of extension of scalars.

So assume $k$ is algebraically closed and $[A:k]=\infty$. We show that $A$ does not satisfy $2$. If $A$ is not semisimple, then it fails 2, so we may assume that $A$ is a direct product of matrix algebras over division algebras and at least one of these division algebras $D$ is infinite dimensional over $k$. Since extension of scalars commutes with direct product and matrix algebras, it suffices to show that $D\otimes_k K$ is not semisimple for some field extension $K/k$.

The key observation is that if $F/k$ is an infinite extension, then $F\otimes_k F$ is either not Artinian or not Noetherian. If $F/k$ is not finitely generated, then $F\otimes_k F$ is not Noetherian by Theorem 11 of P. Vámos, On the minimal prime ideals of a tensor product of two fields, Mathematical Proceedings of the Cambridge Philosophical Society, 84 (1978), pp. 25-35. If $F/k$ is finitely generated but infinite dimensional then it has positive transcendence degree. The Krull dimension of $F\otimes_k F$ is the transcendence degree of $F/k$, by Rodney Y. Sharp, The Dimension of the Tensor Product of Two Field Extensions, Bulletin of the London Mathematical Society 9 Issue 1 (1977) pp 42–48. Thus $F\otimes_k F$ is not Artinian in this case.

There are two cases. If $[Z(D):k]=\infty$, then $D\otimes_k Z(D)$ has center $Z(D)\otimes_k Z(D)$ by general facts on centers of tensor products over a field, and so the center of $D\otimes_k Z(D)$ is not a finite direct product of fields by the above observation (since it fails either the Artinian or Noetherian condition). But the center of a semisimple ring is a direct product of fields by Wedderburn-Artin, contradiction.

So assume $[Z(D):k]<\infty$. Then $Z(D)=k$ because $k$ is algebraically closed. Let $\alpha\in D\setminus k$. Then $\alpha$ is transcendental over $Z(D)=k$ and so $k\leq K=k(\alpha)$ with $K/k$ a purely transcendental extension. Now using @UriyaFirst comment (but with a descending chain argument), we have that $K\otimes_k K$ is not Artinian by the above and so has an strictly descending chain of ideals $I_1\supsetneq I_2\supsetneq \cdots$. Now $D\otimes_k K$ is a free right $K\otimes_k K$-module (since $D$ is a free $K$-module) and so the $(D\otimes_k K) I_k$ form a strictly descending chain of left ideals in $D\otimes_k K$. Thus $D\otimes_k K$ is not left Artinian and hence not semisimple.

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    $\begingroup$ In a previous version of your answer you said that if $D$ is infinite-dimensional over $Z(D)$, then there is a subfield $F$ of $D$ of infinite dimension over $k$. The ring $D$ is a free right $F$-module. Consequently, $D\otimes_kF$ is a free right $F\otimes_kF$-module. The latter ring is not noetherian, so there is a chain of ideals $I_1\subsetneq I_2 \subsetneq I_3\subsetneq\dots$ in $F\otimes_kF$. Since $D\otimes_kF$ is free over $F\otimes_kF$, we get a strictly increasing chain of left ideals $(D\otimes_k F)I_1\subsetneq(D\otimes_k F)I_2\subsetneq\dots$, so $D\otimes_k F$ is not noetherian. $\endgroup$
    – Uriya First
    Commented Feb 9, 2022 at 15:29
  • $\begingroup$ @Uriya First, The previous version was slightly wrong. What I really get are subfields of unbounded dimension. I guess they could still all be finite. It is an open question I believe whether a division algebra all of whose elements are algebraic over its center is necessarily finite dimensional over the center. $\endgroup$
    – Benjamin Steinberg
    Commented Feb 9, 2022 at 15:38
  • $\begingroup$ @UriyaFirst, I found what I think is a correct proof now using your idea. $\endgroup$
    – Benjamin Steinberg
    Commented Feb 9, 2022 at 16:02
  • $\begingroup$ For the commutative case there is an easier proof. If $A$ is commutative and satisfies 2., then $A$ is a direct product of fields and so $A\otimes_k A$ will be semisimple since tensor products commute with direct product. Thus $A$ is a projective $A\otimes_k A$-module and so $A$ satisfies 1 and hence 3. $\endgroup$
    – Benjamin Steinberg
    Commented Feb 9, 2022 at 16:38
  • $\begingroup$ The proof is now correct as far as I can tell. Going to the algebraic closure at first indeed bypasses the complications arising if $D$ is algebraic over $k$. $\endgroup$
    – Uriya First
    Commented Feb 9, 2022 at 17:43
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That $A$ is finite-dimensional by condition 2 is proved in Algebra: Chapter VIII by Bourbaki, p.232 Theorem 1, where separable algebra in sense of condition 2 is called "absolutely semisimple algebra".

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