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.



Deceptively short proof of Regev's $A \otimes B$ theorem 
I spoke about this with Louis Rowen (who, among other, wrote a few books on the subject) and here is what I got from this conversation:


1. There is no circular dependency in this argument.
2. The Razmyslov-Kemer-Brown theorem is more difficult and complicated result than Regev's theorem, so there is no much point to infer the latter from the former.


