# Exactness of sequences of modules is a local property

It's well known, that passing to modules of fractions is exact, i.e. if $$M'\xrightarrow{f} M\xrightarrow{g} M''$$ is an exact sequence of $$A$$-modules ($$A$$ being a commutative ring with unity), then for every multiplicative subset $$S\subset A$$, the induced sequence $$S^{-1}M'\to S^{-1}M\to S^{-1}M''$$ is exact.

But none of the books on commutative algebra I know treats whether for $$M'\to M\to M''$$ to be exact it suffices that $$M'_\mathfrak{p}\to M_\mathfrak{p} \to M''_\mathfrak{p}$$ is exact for each prime ideal $$\mathfrak{p}\subset A$$. So I was looking for a proof, even if I didn't expected it to be true (at least without some finiteness assumptions), and surprisingly it seems to work. But it's too easy to be right, though I can't find the error — so where am I mistaken? These are my thought:

Let $$M'_\mathfrak{p}\to M_\mathfrak{p} \to M''_\mathfrak{p}$$ be exact for each prime ideal $$\mathfrak{p}\subset A$$ (one could take maximal ideals as well). Then (only for the case this wasn't assumed anyway) we have $$\operatorname{im}(f)\subset \ker(g)$$, since for each prime $$\mathfrak{p}\subset A$$ we have $$0=g_\mathfrak{p}\circ f_\mathfrak{p}(x) = (g\circ f)_\mathfrak{p}(x) = g\circ f (x)$$ in $$M''_\mathfrak{p}$$ for all $$x\in M'$$, hence there exists $$s\in A\setminus\mathfrak{p}$$ such that $$s\; (g\circ f(x))=0$$. But with a standard argument (take $$\mathfrak{p}\supset\operatorname{ann}(g\circ f(x))$$ to produce a contradiction to the contrary) we get $$g\circ f(x) = 0$$ and $$\operatorname{im}(f)\subset \ker(g)$$.

Now we know, that the inclusion map $$i\colon\operatorname{im}(f)\hookrightarrow \ker(g)$$ is defined. But for each prime $$\mathfrak{p}\subset A$$ we assumed $$i_\mathfrak{p}\colon \operatorname{im}(f_\mathfrak{p})\to\ker(g_\mathfrak{p})$$ to be bijective and this is a local property, so $$i\colon\operatorname{im}(f)\to \ker(g)$$ is too and finally $$\operatorname{im}(f)=\ker(g)$$. Of cause this requires that localization commutes with $$\ker$$ and $$\operatorname{im}$$, but I'm sure this is true.

I'm getting more and more convinced it's right the more often I look at it but still, something bothers me. Any Suggestions or even counter examples?

• Seeing this old question of mine, I think it's worth mentioning the term faithfully flat module and that one should have a look at Bourbaki's Algèbre commutative Ch. 1 and 2. As a matter of fact, that exactness is a local property is a formal consequence of exactness of localisations and the fact that $M_{\mathfrak{m}}=0$ for all maximal ideals $\mathfrak{m}\subset A$ if and only if $M = 0$. In other words, $\bigoplus_{\mathfrak{m}}A_m$ is a faithfully flat $A$-module.
– Ben
Oct 12, 2016 at 17:19

Yes, exactness is indeed local, and localisation commutes with $\ker$ and $\operatorname{im}$ (since localisation is exact). In fact, exactness is so local that you just need to check it at the maximal ideals. Here is a sketch:

1. $M = 0$ if and only if $M_\mathfrak{m} = 0$ for all maximal ideals $\mathfrak{m}$.

2. A homomorphism $M \to N$ is a monomorphism/epimorphism/isomorphism if and only if $M_\mathfrak{m} \to N_\mathfrak{m}$ is a monomorphism/epimorphism/isomorphism for all maximal ideals $\mathfrak{m}$. [Use (1).]

3. Suppose we have a sequence of modules and homomorphisms: $$0 \longrightarrow M'' \longrightarrow M \longrightarrow M' \longrightarrow 0$$ Suppose also that this sequence is exact after localising at $\mathfrak{m}$, for all maximal ideals $\mathfrak{m}$. Then, by (1), the sequence is a chain complex, and by (2), the sequence is exact at $M''$ and $M'$. Since we have a chain complex, there is an induced homomorphism $\ker (M \to M') \to \operatorname{coker} (M'' \to M)$; but this is an isomorphism after localising at each $\mathfrak{m}$, so the homomorphism is already an isomorphism, and thus the sequence is exact at $M$ as well.

• Thank you, I like it more than mine. But I still wonder why this is 'missing' in Atiyah-Macdonald, Eisenbud, Matusumra,... Where is yours from?
– Ben
Mar 17, 2012 at 10:10
• It's in Eisenbud: Lemma 2.8 is (1), Corollary 2.9 is (2). I'm not sure where (3) is, but it follows easily enough. Mar 17, 2012 at 10:35
• I would have excepted 1. and 2. as well known, this isn't the problem. But especially Atiyah-Macdonald give a big list of local properties and isn't exactness important enough to be mentioned as local? At least as one of the tons of Exercises..
– Ben
Mar 17, 2012 at 11:09
• Well, it is present (in a weaker, disguised form) in [Hartshorne, Ch. II] as Exercise 1.2(c), plus Proposition 5.1. Mar 17, 2012 at 11:15
• Hi, just searched for this now as I'm working through the Atiyah-Macdonald book and I think its assumed implicitly in their proof/exercise of the Chinese Remainder Theorem (see page 99 Ex 9.9). Aug 24, 2012 at 14:49

As this is about one of the only things I can comment on I thought I'd write an answer! Hopefully it will be of help to someone. It essentially the same as what has been written but a bit more condensed.

$$(1)$$ $$M = 0$$ if and only if $$M_\mathfrak{m} = 0$$ for all maximal ideals $$\mathfrak{m}$$.

Keep notation and hypothesis of the original post i.e. we are considering a sequence $$E$$: $$M'\xrightarrow{f} M\xrightarrow{g} M''$$ that is exact when localized at every maximal ideal.

Since $$((g\circ f)M')_{\mathfrak{m}}=(g_\mathfrak{m}\circ f_\mathfrak{m})M_{\mathfrak{m}}'=0$$, we have by $$(1)$$ that $$E$$ is a complex, and hence, the quotient module $$\ker(g)/\operatorname{im}(f)$$ is well-defined. Whence, $$(\ker(g)/\operatorname{im}(f))_\mathfrak{m}=\ker(g_{\mathfrak{m}})/\operatorname{im}(f_{\mathfrak{m}})=0$$, and the result follows.