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In the book Skew Fields, by P.K.Draxl, at page 60, states there as lemma 2:

Let $A,B,C$ be finite-dimensional $K$-algebras such that $|C:K|\leq|A:K||B:K|$ and let $f: A\rightarrow C$, and $g: B\rightarrow C$ be $K$-algebra homomorphisms. Then $A \otimes B \cong C$ provided $A,B$ are central simple $K$-algebras.

There it is alluded to theorem 2 in section 5, where one states that if, notations as above, f and g commute, then there will be an R-algebra homomorphism from $A \otimes B$ to C. But, thus far as I observe, there appears no such condition in the statements of the lemma, so I could not draw conclusions from the theorem, could I?
Any help would be greatly appreciated.
Thanks for any attention.

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up vote 2 down vote accepted

The "obvious" candidate is $f\otimes g:A\otimes B\rightarrow C$. If it turned out to be a ring homomorphism, then by simplicity of $A$ and $B$, $A\otimes B$ is also simple, and so the map would be injective. By the dimensionality condition given, the map would have to be surjective. Thus, $f\otimes g$ would be the candidate isomorphism.

You are right though that it is necessary that $f(a)g(b)=g(b)f(a)$ for all $a\in A$, $b\in B$.

Consider $C=M_2(F)\otimes M_3(F)$, and $A=B=M_3(F)$. Let $f,g$ both be the inclusion maps of $A$ and $B$. The dimension condition is satisfied: $9*9\geq 9*4$. Obviously though, $A\otimes B\ncong C$.

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Thanks for clarifying the matter. –  awllower May 29 '12 at 4:25
    
@awllower The version of that book in googlebooks looks pretty horribly typeset, so maybe the condition was accidentally omitted? –  rschwieb May 29 '12 at 10:54
    
From my own experiences, it appears that the book is within my standard about the typeset. But it might happen that it typed there wrongly... –  awllower May 30 '12 at 2:49
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