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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.

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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...

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That's a good point, the Razmyslov-Kemer-Brown theorem about nilpotency of the Jacobson radical of a PI algebra definitely requires that the algebra will be finitely generated. But in Regev's theorem, the condition of being finitely generated seems to be not restrictive, the general case obviously reduces to the finitely-generated one (though this is not how it was proved originally by Regev, and, as far as I can tell, in subsequent textbooks - they consider countably-generated relatively free algebras). –  Pasha Zusmanovich May 26 '11 at 7:16
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Surely you can reduce all PI-question to countably generated algebras. But how would you reduce this question to finitely generated subalgebras? The PI-degrees of finitely generated subalgebras can grow to infinity, so you will need some common bound on PI-degrees... –  Bugs Bunny May 26 '11 at 9:00
    
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
    
You seem to be right, thanks. Still, there is probaly a way to reduce it to a finitely-generated situation, should think about it more. But even in the finitely-generated case Regev's theorem seems to be a highly nontrivial claim. –  Pasha Zusmanovich May 28 '11 at 5:25
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up vote 1 down vote accepted

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.
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