# Deceptively short proof of Regev's $A \otimes B$ theorem

The following theorem, due to Regev, is one of the cornerstones of the theory of PI algebras (i.e., associative algebras satisfying a nontrivial polynomial identity):

Let $A$, $B$ be two PI algebras over a field $K$. Then their tensor product $A \otimes_K B$ is PI.

Consider the following "proof" of this theorem. Since $A$ and $B$ are PI, their Jacobson radicals $J(A)$ and $J(B)$ are nilpotent, and $A/J(A)$ and $B/J(B)$ are semisimple PI algebras which are known to be embedded into matrix algebras over a commutative ring, say $M_n(C)$ and $M_m(D)$. Now, $J(A) \otimes B + A \otimes J(B)$ is a nilpotent ideal of $A \otimes B$, quotient by which is isomorphic to $A/J(A) \otimes B/J(B)$ and hence is embedded into $M_n(C) \otimes M_m(D)$, which, in its turn, is embedded into $M_{nm} (C \otimes D)$. Therefore, $A \otimes B$ contains a nilpotent ideal quotient by which is embedded into a matrix algebra over a commutative ring, and hence is PI.

Regev's theorem is a relatively difficult result, first conjectured by Jacobson, and having resisted attempts by a few mathematicians. Thus it hardly admits such a short simple proof. Where is the catch?

The only weak spot a can think of, is that nilpotence of the Jacobson radical of a PI algebra is a relatively new (at least proved long after Regev's theorem in 1970) complicated result whose proof probably involves appeal to Regev's theorem. Is it true? Or am I missing something else?

Edit May 26, 28 2011: As was pointed out by Bugs Bunny, we should require that $A$ and $B$ are finitely generated, as this is the hypothesis of the Razmyslov-Kemer-Brown theorem about nilpotence of the Jacobson radical of a PI algebra (and the theorem does not hold for infinitely-generated algebras). But, the general statement of Regev's theorem obviously reduces to this case.

Your proof depends on $A$ and $B$ being finitely generated: $C[[x]]$ is a PI-algebra whose Jacobson radical is not nilpotent. Regev's theorem works, in general, if I remember correctly...
• Claim 1. An algebra is PI if and only if all its finitely generated subalgebras are PI. Claim 2. $A \otimes B$ is finitely generated if and only if $A$ and $B$ are finitely generated. – Pasha Zusmanovich May 27 '11 at 5:40
• Claim 1 seems to be rotten. $M_n (C)$ is a subalgebra in $M_{n^2} (C)$. Take their union $A$ as $n=2,4,16,32...$. I claim $A$ is not PI (as it contains bigger and bigger matrix subalgebras) but every finitely generated subalgebra is PI as it sits in one of $M_n$-s... – Bugs Bunny May 27 '11 at 13:57