$ \ker(g\otimes I ) =\ker(g) \otimes M, $ where $M$ is faithfully flat.

Question

Let $M$ be faithfully flat $R$ module and $g: N \to N'$ be a $R$-linear map. Then want to show that kernel of the tensor map $g\otimes I : N \otimes M \to N' \otimes M $ is equal to $\ker(g) \otimes M.$

Clearly $\ker(g)\otimes M \subseteq \ker (g \otimes I)$ but I cannot prove the converse. Even I can't use that $M$ is faithfully flat. I need some help. Thanks.

Answer 1

You only need flatness, you can forget about the faithful part. Every left exact functor commutes with kernels, as the following simple argument shows:

You have an exact sequence $$0 \to \operatorname{ker}(g) \to N \to N'$$ and after tensoring this stays exact:

$$0 \to \operatorname{ker}(g) \otimes M \to N \otimes M \to N' \otimes M$$

Hence there is a natural isomorphism $\operatorname{ker}(g) \otimes M \cong \operatorname{ker}(N \otimes M \to N' \otimes M)$.

Of course you can take any left exact functor instead of $- \otimes M$.