= Solution
Keep the nonzero $z$ of minimal <two-factor tensor rank> constructed in part (i). For every $a\in A$, its <commutator> with $a\otimes1$ lies in $J$ and equals
$$
(a\otimes1)z-z(a\otimes1)
=\sum_{i=2}^n(ac_i-c_ia)\otimes d_i.
$$
If this <tensor> were nonzero, its <two-factor tensor rank> would be at most $n-1$, contradicting minimality. Thus it is zero. The <linear independence> of $d_2,\ldots,d_n$ forces $ac_i=c_ia$ for every $a$ and $i\geq2$. Since $Z(A)=k$, each $c_i$ is a <scalar multiple> of $1$.
But $1,c_2,\ldots,c_n$ are <linearly independent>. Hence $n=1$, so $z=1\otimes1\in J$. A <two-sided ideal> containing the multiplicative identity is the entire <ring>; therefore $J=A\otimes_kB$.
We have proved the stronger result that $A\otimes_kB$ is a <simple ring>. It has finite <dimension> $(\dim_kA)(\dim_kB)$, so part (b) proves
$$
\boxed{A\otimes_kB\text{ is a semisimple ring}.}
$$
Part (c)(ii) also gives $Z(A\otimes_kB)=k$, so this proves that a <tensor product of central simple algebras> is again a <central simple algebra>.
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