Proof that exactness is a local property

Question

I know that, if $M$ is a $R$-module, $M$ is $0$ if and only if $M_p$ is $0$ for every prime ideal $p\subseteq R$. So if I have an exact sequence of $R$-modules, $0\to A \xrightarrow f B \xrightarrow g C \to 0$ for example, it seems to me that $$(\text{Ker}g_p)/(\text {Im} f_p)=(\text{Ker}g)_p/(\text {Im} f)_p=(\text{Ker}g/\text {Im} f)_p.$$ So if the identities above are true, the argument written in bold makes obvious that exactness is local. However I'm not very sure about those identities, do they really hold?

Answer 1

$\newcommand{\p}{\mathfrak{p}}\DeclareMathOperator{\im}{im}\newcommand{\into}{\hookrightarrow}$You haven't mentioned this but you are implicitly using that localisation is an exact functor.

In other words, if the sequence $$0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$$ is exact, then so is $$0 \to A_{\p} \xrightarrow{f_{\p}} B_{\p} \xrightarrow{g_{\p}} C_{\p} \to 0$$ for every prime $\p$.
(In general, localising at any multiplicative subset preserves exactness, not necessarily at the complement of a prime.)

In particular, this shows you that $$B_{\p}/A_{\p} \cong C_{\p} = (B/A)_{\p}.$$

(In the above, I have naturally associated $A$ with a submodule of $B$ and $C$ with $B/A$. Similarly for the localisations.)

(Sidenote: By breaking a general exact sequence into short exact sequences of the form $0 \to \ast \to \ast \to \ast \to 0$, one sees that localisation preserves the exactness of any exact sequence.)

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Now suppose that you have a complex $$A \xrightarrow{f} B \xrightarrow{g} C.$$ (The above simply means that $g \circ f = 0$.)

This tells you that $\im(f) \subset \ker(g)$. By the previous discussion, we have $$\frac{(\ker(g))_{\p}}{(\im(f))_{\p}} \cong \left(\frac{\ker(g)}{\im(f)}\right)_{\p}.$$

However, there is one small check to be done. We need to show that $$\frac{((\ker(g))_{\p}}{\im(f))_{\p}} \cong \frac{\ker(g_{\p})}{\im(f_{\p})}.$$

To see this, note that exactness of $$A \xrightarrow{f} \im(f) \to 0$$ gives us exactness of $$A_{\p} \xrightarrow{f_{\p}} \im(f)_{\p} \to 0.$$ In other words, the map $f_{\p}$ is onto, i.e., $\im(f_{\p}) = \im(f)_{\p}$.
Similarly, considering $$0 \to \ker(g) \to B \xrightarrow{g} C$$ shows us that $(\ker(g_{\p})) = \ker(g)_{\p}$, as desired.