Tensor of s.e.s. of projective modules

Question

Let $f:R \rightarrow S$ be a ring homomorphism. All modules are right R-modules.Let $\otimes_R S:\mathbb{P}(R) \rightarrow \mathbb{P}(S)$ be the functor sending a projective $R$-module $P$ to the projective $S$-module $P\otimes_R S$. Is this functor left exact?

Answer 1

If by left exact you mean "preserves finite limits", the answer is no : consider $R= \mathbb Z, S= \mathbb Z/2$, and the following equalizer diagramm $0\to \mathbb Z \rightrightarrows\mathbb Z$ where the top map is $2$ and the bottom map is $0$.

Then when you tensor it, it's no longer an equalizer, as the two maps become $0 : \mathbb{Z/2\to Z/2}$, and the equalizer of that is $\mathbb Z/2$, not $0$.

If by left exact you mean "sends short exact sequences $0\to A\to B\to C\to 0$ to exact sequences $0\to F(A)\to F(B)\to F(C)$", then it's not very relevant (because $\mathbb P(R)$ is not abelian), but the answer is yes : indeed any short exact sequence of projectives splits, so the splitting is preserved by $\otimes_R S$.