Tensor with an Exact Sequence of Free Module Still Exact?

Question

Let $F\rightarrow G\rightarrow H$ be an exact sequence of free $R$-modules and $M$ be any $R$-module.

Is that true that $F\otimes_RM\rightarrow G\otimes_RM\rightarrow H\otimes_RM$ is exact?

If the answer is no, does that hold when all the modules above are finitely generated?

Answer 1

Let $R=\mathbb{Z}$. Then $$0\rightarrow\mathbb{Z}\xrightarrow{\times2}\mathbb{Z}$$ is an exact sequence of free $R$-modules, but the tensor product with $M=\mathbb{Z}/2\mathbb{Z}$ is not exact.