= Solution
Use the left $A$-<module> structure $c\cdot(a\otimes b)=ca\otimes b$ on the <tensor product of associative algebras>. Expanding each $b\in B$ in the <basis> $b_1,\ldots,b_m$ shows that every <tensor> can be written
$$
x=\sum_{j=1}^m a_j\otimes b_j
=\sum_{j=1}^m a_j\cdot(1\otimes b_j).
$$
Thus the displayed family generates the <module>.
For uniqueness, let $\beta_j:B\to k$ be the corresponding <dual basis>, with $\beta_j(b_i)=\delta_{ij}$. Applying the <linear map> $\operatorname{id}_A\otimes\beta_j$ to a <linear relation> $\sum_i a_i\otimes b_i=0$ gives $a_j=0$. Hence the family is <linearly independent in a module>, and
$$
\boxed{A\otimes_kB\cong A^{\oplus m}\quad\text{as left }A\text{-modules}.}
$$
In particular $1\otimes b_1,\ldots,1\otimes b_m$ is a <basis of a module>, with coefficients uniquely in $A$.
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