 Solution

Write $x\in Z(A{\otimes }_{k}B)$ uniquely as $x={\sum }_j{a}_j\otimes b_j$ using part (i). For each $a\in A$, the equality $(a\otimes 1)x=x(a\otimes 1)$ becomesThe coefficient uniqueness from part (i) gives $a{a}_j=a_{j}a$ for every $a$, so each $a_j$ belongs to the center of an associative algebra $Z(A)$. Consequently $x\in Z(A){\otimes }_{k}B$.

Choose a basis $z_1,\dots ,z_t$ of the vector space $Z(A)$ and rewrite $x={\sum }_{i=1}^t{z}_i\otimes d_i$. For $b\in B$, commuting with $1\otimes b$ givesExtend the linearly independent family $z_i$ to a basis of $A$ and apply its dual basis to the first factor. It follows that $b{d}_i=d_{i}b$ for every $b$, hence $d_i\in Z(B)$. This proves $Z(A{\otimes }_{k}B)\subseteq Z(A){\otimes }_{k}Z(B)$.

Conversely, if $z\in Z(A)$ and $w\in Z(B)$, thenThe pure tensors $a\otimes b$ span the tensor product, so $z\otimes w$ is central, as is every sum of such pure tensors. Therefore the center of a tensor product of associative algebras isThese tensor products are identified with their natural vector subspaces of $A{\otimes }_{k}B$; extending bases justifies these embeddings over the field $k$.