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

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?

• Your assumption that it is free over $A$ is already wrong. For a counter example, take $A = M = \mathbb{Z}$ and $B = \mathbb{F}_2$. – Dirk Apr 25 at 12:00
• @Dirk Thanks, I changed $\rightarrow$ to $\subset$ – bart Apr 25 at 12:10
• That doesn't fix the problem actually : if $A=M$ the tensor product is $B$, which is not necessarily a free $A$-module. – Arnaud D. Apr 25 at 12:22

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.