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.

*

*$A$ is projective as an $A^e$-module.

*For any field extension $K$ of $k$, the algebra $A \otimes_k K$ is semsimple.

Also consider the following condition.


*$A$ 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.
 A: 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.
