# Is a "separable" algebra over a field finite-dimensional?

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.

• What is the enveloping algebra? I only know this term in the context of Lie algebras.
– Echo
Feb 9, 2022 at 11:12
• $A^e$ is the tensor product algebra $A \otimes_k A^{op}$ of $A$ and the opposite algebra $A^{op}$. Feb 9, 2022 at 12:25
• 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. Feb 9, 2022 at 13:39
• If you take the product of all matrix algebras $\mathrm{Mat}_n(k)$. Would that be semisimple in your definition?
– Echo
Feb 9, 2022 at 14:08
• @ 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. Feb 9, 2022 at 14:13

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.

• 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. Feb 9, 2022 at 15:29
• @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. Feb 9, 2022 at 15:38
• @UriyaFirst, I found what I think is a correct proof now using your idea. Feb 9, 2022 at 16:02
• 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. Feb 9, 2022 at 16:38
• 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 $k$. Feb 9, 2022 at 17:43