 Solution

Let $J\ne 0$ be a two-sided ideal of $A{\otimes }_{k}B$. Choose a nonzero $x\in J$ of least possible two-factor tensor rank $n$, and take a shortest expressionBoth families $(a_i)$ and $(b_i)$ are linearly independent: a linear dependence in either family would combine terms and shorten the expression. In particular $a_1,b_1\ne 0$.

Because $A$ is a simple ring, the two-sided ideal generated by $a_1$ is $A$. Thus there are finitely many $u_j,v_j\in A$ with ${\sum }_j{u}_j{a}_1{v}_j=1$. FormSince the $b_i$ are linearly independent, applying a linear functional selecting $b_1$ shows $y\ne 0$. Its two-factor tensor rank is at most $n$ and hence, by the choice of $x$, exactly $n$. Consequently $c_1=1,c_2,\dots ,c_n$ are linearly independent.

Similarly, simplicity of $B$ gives finitely many $p_j,q_j\in B$ with ${\sum }_j{p}_j{b}_1{q}_j=1$. NowThe linear independence of the $c_i$ shows $z\ne 0$, and minimality again makes its two-factor tensor rank exactly $n$. We have obtainedBoth coefficient families in this expression are linearly independent. Retaining this additional property is the key to the next part. This is coefficient normalization in an ideal of a tensor product.