= Solution
Let $J\ne0$ be a <two-sided ideal> of $A\otimes_kB$. Choose a nonzero $x\in J$ of least possible <two-factor tensor rank> $n$, and take a shortest expression
$$
x=\sum_{i=1}^n a_i\otimes b_i.
$$
Both families $(a_i)$ and $(b_i)$ are <linearly independent>: a <linear dependence> in either family would combine terms and shorten the expression. In particular $a_1,b_1\ne0$.
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 $\sum_j u_ja_1v_j=1$. Form
$$
y=\sum_j(u_j\otimes1)x(v_j\otimes1)
=\sum_{i=1}^n c_i\otimes b_i\in J,\qquad
c_i=\sum_j u_ja_iv_j,\qquad c_1=1.
$$
Since the $b_i$ are <linearly independent>, applying a <linear functional> selecting $b_1$ shows $y\ne0$. Its <two-factor tensor rank> is at most $n$ and hence, by the choice of $x$, exactly $n$. Consequently $c_1=1,c_2,\ldots,c_n$ are <linearly independent>.
Similarly, simplicity of $B$ gives finitely many $p_j,q_j\in B$ with $\sum_jp_jb_1q_j=1$. Now
$$
z=\sum_j(1\otimes p_j)y(1\otimes q_j)
=\sum_{i=1}^n c_i\otimes d_i\in J,\qquad
d_i=\sum_jp_jb_iq_j,\qquad d_1=1.
$$
The <linear independence> of the $c_i$ shows $z\ne0$, and minimality again makes its <two-factor tensor rank> exactly $n$. We have obtained
$$
\boxed{z=1\otimes1+\sum_{i=2}^n c_i\otimes d_i\in J\setminus\{0\}.}
$$
Both 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>.
Back to article page
