If $A$ is an integral domain and $k \subset K$ is an algebraic field extension, is $A \otimes_k K$ a free $A$-module?

Question

I am curious if the question in the title is true. For reference I was thinking about exercise 11.1.G in Vakil's FOAG. He claims such a result is true. I can neither think of how to prove it nor think of a counterexample. However, in the context of the question, all we need is that $A \otimes_k K$ is flat and hence torsion free over $A$, which is a weaker condition.

Answer 1

Yes, it's true if $A$ is a $k$-algebra: since $K$ is free over $k$, hence $K\simeq k^{(I)}$ for some set $I$, and as tensor product commutes with direct sums $$A\otimes_k K\simeq A\otimes_k k^{(I)}\simeq(A\otimes_k k)^{(I)}\simeq A^{(I)}.$$

Answer 2

I assume that $A$ is a $k$-algebra, since otherwise the tensor product does not make sense to me. Now $K$ is free over $k$, hence $K\cong k^{\oplus I}$ as $k$-modules for some index set $I$. Right exactness of the tensor product then yields $A\otimes_k K \cong A^{\oplus I}$ as $A$- modules, whence the claim follows.