= Coefficient normalization in an ideal of a tensor product
{title2=$0\ne z=1\otimes1+\sum_{i=2}^n c_i\otimes d_i$}
Let $A,B$ be unital <associative algebras> over a <field>, each a <simple ring>, and $J\ne0$ 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 $Aa_1A=A$, choose $\sum_j u_ja_1v_j=1$ and replace $x$ by $\sum_j(u_j\otimes1)x(v_j\otimes1)$. 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 $a\otimes1$ 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 $1,c_2,\ldots,c_n$ then forces $n=1$. Thus $1\otimes1\in J$, proving simplicity of the <tensor product> without a splitting-field argument.
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