Flat Modules and Localization at prime ideal

Question

Suppose $M$ is a flat $A$ module and $\mathfrak{p}$ is a prime ideal in $A$, question is to prove that

$M_{\mathfrak{p}}$ is a flat $A_{\mathfrak{p}}$ module.

Let $S=R\setminus \mathfrak{p}$. Consider sequence of $S^{-1}A$ modules $$0\rightarrow S^{-1}N_1\rightarrow S^{-1}N_2\rightarrow S^{-1}N_3\rightarrow 0.$$ We have to prove that we have exact sequence $$0\rightarrow S^{-1}N_1\otimes_{S^{-1}A}S^{-1}M\rightarrow S^{-1}N_2\otimes_{S^{-1}A}S^{-1}M\rightarrow S^{-1}N_3\otimes_{S^{-1}A}S^{-1}M\rightarrow 0.$$ As $S^{-1}(N\otimes_AM)\cong S^{-1}N\otimes_{S^{-1}A}S^{-1}M$ it suffices to prove that we have an exact sequence $$0\rightarrow S^{-1}(N_1\otimes_AM)\rightarrow S^{-1}(N_2\otimes_AM)\rightarrow S^{-1}(N_3\otimes_AM)\rightarrow 0.$$

To prove this, it suffices to prove that we have following exact sequence $$0\rightarrow N_1\otimes_AM\rightarrow N_2\otimes_AM\rightarrow N_3\otimes_AM\rightarrow 0.$$

Then, as localization preserves exactness we would have the sequences that we have desired. This is definitely more than what we want.

Suppose we have $$0\rightarrow N_1\rightarrow N_2\rightarrow N_3\rightarrow 0$$ and not just $$0\rightarrow S^{-1}N_1\rightarrow S^{-1}N_2\rightarrow S^{-1}N_3\rightarrow 0.$$ This is definitely more than what we have.

As $M$ is a flat $A$ module $$0\rightarrow N_1\rightarrow N_2\rightarrow N_3\rightarrow 0$$ implies we have an exact sequence $$0\rightarrow N_1\otimes_AM\rightarrow N_2\otimes_AM\rightarrow N_3\otimes_AM\rightarrow 0$$ and we are done.

I am not able to solve this. Any suggestions are welcome.

Answer 1

Take it from this point of view: consider an exact sequence of $S^{-1}A$-modules: $$0\longrightarrow N_1\longrightarrow N_2\longrightarrow N_3\longrightarrow 0 $$ and tensor by $S^{-1}M$ over $S^{-1}A$: the sequence: $$0\longrightarrow N_1\otimes_{S^{-1}A}S^{-1}M\longrightarrow N_2\otimes_{S^{-1}A}S^{-1}M\longrightarrow N_3\otimes_{S^{-1}A}S^{-1}M\longrightarrow 0 $$remains exact because $$N_i\otimes_{S^{-1}A}S^{-1}M\simeq N_i\otimes_{S^{-1}A}(S^{-1}A\otimes_{A}M)\simeq (N_i\otimes_{S^{-1}A}S^{-1}A)\otimes_{A}M\simeq N_i\otimes_{A}M, $$ and $N$ is $A$-flat.