Showing that localization is an exact functor

I'm again in this awfully familiar situation where I'm struggling to prove simple statements mostly because I have no idea how a template of a proof should look like in this specified context.

I'm trying to prove these two statements:

1. Given a multiplicative subset of a ring $S \subset R$. The localization functor from $R-Mod$ to $S^{-1}R-Mod$ defined by: $$M \mapsto S^{-1}M ,\phi \mapsto S^{-1}\phi=(\frac{m}{s} \mapsto \frac{\phi(m)}{s}) $$ Preserves homology of all complexes.
2. Let $A \to B \to C$ be a complex in $R-Mod$. If all localization by prime ideals $\mathfrak p \subset R$ make $A_{\mathfrak p} \to B_{\mathfrak p} \to C_{\mathfrak p}$ exact in $R_{\mathfrak p} -Mod$ then $A \to B \to C$ is itself exact.

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My actual problem with both questions is that i keep getting confused by torsion elements of the modules and if annhilators intersect the multiplicative set i'm stuck altogether...

I think it's a partly language barrier so I would be really thankful if whoever answers this question will give a detailed answer that would demonstrate a correct and elegant use of the relevant language and tools in this context.

Solution 1:

It is enough to prove it preserves short exact sequences: $\;0\to M\to N\to P\to 0$. As the tensor product is right-exact, and $S^{-1}M\simeq M\otimes_A S^{-1}A$, it is even enough to prove it preserves injectivity.

So consider an injective morphism $\varphi\colon M\to N$ and suppose $\;(S^{-1}\varphi)\Bigl(\dfrac ms\Bigr)=0$ in $S^{-1}N$. This means there exists $t\in S$ such that $\;t\mkern1mu\varphi(m)=\varphi(tm)=0$. But then $$\frac ms=\frac{tm}{ts}=\frac0{ts}=0,$$ which shows $\;S^{-1}\varphi\;$ is injective.