 Coefficient normalization in an ideal of a tensor product

Let $A,B$ be unital associative algebras over a field, each a simple ring, and $J\ne 0$ a two-sided ideal of their tensor product. Choose $0\ne x={\sum }_{i=1}^n{a}_i\otimes b_i\in J$ of minimal two-factor tensor rank. Both factor families are linearly independent. Since $A{a}_{1}A=A$, choose ${\sum }_j{u}_j{a}_1{v}_j=1$ and replace $x$ by ${\sum }_j(u_j\otimes 1)x(v_j\otimes 1)$. The result has first coefficient $1$, is nonzero by the linear independence of the $b_i$, and retains minimal two-factor tensor rank; its first factor family is therefore still linearly independent. Now apply the same construction in $B$ to make the first second-factor coefficient $1$. The result is again nonzero of minimal two-factor tensor rank, and both its coefficient families are linearly independent.

If $Z(A)=k$, commutation with $a\otimes 1$ gives a tensor in $J$ expressible with at most $n-1$ pure tensors, hence zero by minimality. Independence forces every $c_i$ to be central, hence a scalar multiple of $1$. Independence of $1,c_2,\dots ,c_n$ then forces $n=1$. Thus $1\otimes 1\in J$, proving simplicity of the tensor product without a splitting-field argument.