Is $B \otimes_A M$ a free $B$ module if $M$ is a free $A$ module?

Question

Let $A\subset B$ be commutative rings with identity. Let $M$ be a free $A$ module. Then $B \otimes_A M$ is a $B$ module. It is also a free $A$ module . But is it a free $B$ module?

Answer 1

Yes, it is a free $B$-module, and this is true for any ring homomorphism $A\to B$. Indeed, the tensor product with $B$ defines a functor $A-\mathbf{Mod}\to B-\mathbf{Mod}$, which is left adjoint to the restriction of scalars functor $B-\mathbf{Mod}\to A-\mathbf{Mod}$. Since restriction of scalars commute with the forgetful functors to $\mathbf{Set}$, their left adjoints must commute as well, which means that the functor $B\otimes_A\_$ takes the free $A$-module over a set to the free $B$-module over the same set.

Another way to prove it is to see that a free $A$-module is isomorphic to a direct sum $\bigoplus_{i\in I}A$, and that tensor product preserves direct sums, so that $$B\otimes_A M\cong B\otimes_A\left(\bigoplus_{i\in I}A\right)\cong \bigoplus_{i\in I}(B\otimes_A A)\cong \bigoplus_{i\in I}B,$$ which is a free $B$-module.