0
$\begingroup$

Suppose $B$ and $C$ are $F$-algebras. Assume that $B$ is central simple and $C$ is simple. Prove that $B\otimes C$ is simple.


I am stuck with how to use the fact that $B$ is central simple. I have come up with a "proof" that is most likely wrong since I only used the fact that $B$ is simple. However I don't know which step is wrong. Thanks for any help!

My attempt:

Let $B'\otimes C'$ be a nonzero two sided ideal of $B\otimes C$. Let $b_i'\otimes c_i'\in B'\otimes C'$. For all $(b\otimes c)\in B\otimes C$, $$(b\otimes c)(b_i'\otimes c_i')=(bb_i'\otimes cc_i')\in B'\otimes C'$$

$$(b_i'\otimes c_i')(b\otimes c)=(b_i'b\otimes c_i'c)\in B'\otimes C'$$

Thus $B'$ is an ideal of $B$. This implies $B'=B$. Similarly $C'=C$.

Thus $B\otimes C$ is simple.

Note that I haven't used the fact that $Z(B)=F$, thus definitely missing something here.

$\endgroup$
2
$\begingroup$

You seem to have wrong ideas about tensor products ; you use them like direct products. An ideal of $A\otimes B$ needs not be of the form $I\otimes I'$, and an element of $A\otimes B$ is certainly not of the form $a\otimes b$ (it is a sum of such elements).

Note that you do need the fact that $B$ is central : for instance, $B=C=K$ where $K$ is a quadratic extension of $F$ gives $B\otimes C \simeq K^2$ which is not simple.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.