I was trying to do the exercise 4.17 in Bresar "Introduction to noncommutative algebra" which states the following
Let $A,B$ be simple unital $F$-algebras. Show that the algebra $A\otimes B$ is semiprime if only if $Z(A\otimes B)$ is semiprime.
Here my attempt:
First I will assume that $Z(A\otimes B)$ is semiprime. It is well know that $Z(A\otimes B)=Z(A)\otimes Z(B)$. Let $I\neq 0$ be a nilpotent ideal of $A\otimes B$ and consider $z=\sum_{i=1}^{d}a_{i}\otimes b_{i}$ a non-zero element in $I$ with $d$ minimal. (This was the hint given in the book.) Since $d$ is minimal, both the sets $\{a_{1},\ldots,a_{d}\}$ and $\{b_{1},\ldots, b_{d}\}$ must be linearly independent. Otherwise, we could, using the linear dependence of some elements, reduce the expression of $z$ and this contradicts the minimality of $d$. Now, since $A$ is simple, we have $Aa_{1}A=A$ and therefore there exists $r_{i},s_{i}\in A$ such that $\sum_{j}r_{j}a_{1}s_{j}=1$. In this way, for each $j$ we can multiply $z$ on the left by $(r_{j}\otimes 1)$ and on the right by $(s_{j}\otimes 1)$ to obtain an element in $I$ of the form
$r_{j}a_{1}s_{j}\otimes b_{1}+\ast...$
So, adding these elements we arive at an element in $I$ of the form
$u=1\otimes b_{1}+a_{2}'\otimes b_{2}+\cdots+a_{d}'b_{d}$.
And since $d$ is minimaly, once again $\{a_{2}',\ldots,a_{d}'\}$ must be linearly independent. Now given any $a\in A$ the element
$(a\otimes 1)u-(a\otimes 1)u=\sum_{i=2}^{d}(aa_{i}'-a_{i}'a)\otimes b_{i}$
is in $I$ and by minimality of $d$ it is $0$. Moreover taking into account that $\{b_{1},\ldots,b_{d}\}$ is linearly independent we have
$a_{2}',\ldots,a_{d}'\in Z(A)$.
On the other hand, analogously using the fact that $B$ is simple, from $u$ we can get an element in $I$ of the form
$v=1\otimes 1+a_{2}'\otimes b_{2}'+\cdots+a_{d}'\otimes b_{d}'$
with $\{b_{2}',\ldots,b_{d}'\}$ linearly independent and $b_{2}',\ldots,b_{d}'\in Z(B)$.
Now consider the ideal $L:=Z(A\otimes B)v Z(A\otimes B)$ of $Z(A\otimes B)$. Since $I$ is being assumed to be nilpotent, in particular $v$ is nilpotent. As a consequence, $L$ is nilpotent and since $Z(A\otimes B)$ is semiprime we conclude that $v=0$ and by minimality of $d$ follows that $I=0$, and that implies that $A\otimes B$ is semiprime.
Conversely, assume that $A\otimes B$ is semiprime and let $a\in Z(A\otimes B)$ such that $a(Z(A\otimes B))a=0$. In particular, $a^{2}=0$ (since $1\otimes 1\in Z(A\otimes B)$). In this way, for each $r\in R$ we have
$ara= (aa)r=a^{2}r=0$
and since $A\otimes B$ is semiprime we must have $a=0$. As a consequence, $Z(A\otimes B)$ is semiprime.
Remark: Here I'm using two characterizations of semiprime rings: A ring $R$ is semiprime $\Leftrightarrow$ $R$ has not non-zero nilpotent ideals $\Leftrightarrow$ For each $a\in R$ with $aRa=0$, we must have $a=0$.
I would like to know if my attempted solution is correct. Sorry by my English maybe I made some grammatical mistakes.