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?

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


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