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.


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.


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