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In Eisenbud's Commutative algebra with a view..., he shows that if an $A$-module $M$ has a finite length, then the sum of localization maps at maximal ideals is an isomorphism: $$\phi:M\overset{\sim}{\rightarrow}\oplus_{\mathfrak{m}}M_{\mathfrak{m}}.$$ But isn't this true for any module (i.e., not necessarily finite length)? My argument is that $\phi$ is an isomorphism because $\phi_{\mathfrak{m}}$ is an isomorphism for every maximal ideal $\mathfrak{m}$ since

$$\mbox{Supp}M_{\mathfrak{m}}=\mbox{Supp}M\cap\mbox{Spec}A_{\mathfrak{m}}.$$

Am I missing something here? I think only difference that $\ell(M)<\infty$ makes is maybe the sum is finite...

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We certainly know that $\phi:M\to \bigoplus_{m} M_{m}$ is an injective $A$-module homomorphism. How do you propose to prove that $\phi$ is a surjective $A$-module homomorphism? –  Amitesh Datta Aug 28 '11 at 23:33
    
I think I see where I went wrong. I answered my own question below. –  ashpool Aug 29 '11 at 0:03

2 Answers 2

up vote 2 down vote accepted

Let us discuss this in a number of steps:

Exercise 1: Let $A$ be a commutative ring and let $M$ be an $A$-module. Let $\text{MaxSpec}(A)$ be the set of all maximal ideals of $A$. Prove that the natural $A$-module homomorphism $\phi:M\to \bigoplus_{m\in\text{MaxSpec}(A)} M_m$ is injective. (Hint: if $x\in M$, choose a maximal ideal $m$ of $A$ such that $\text{Ann}(x)=\{a\in A:ax=0\}\subseteq m$. Consider the image of $x$ in $M_m$.)

Exercise 2: Let us consider the $\mathbb{Z}$-module $\mathbb{Z}$ and let $\phi:\mathbb{Z}\to \bigoplus_{m\in\text{MaxSpec}(\mathbb{Z})} \mathbb{Z}_m$ be the natural $\mathbb{Z}$-module homomorphism. Prove that $\phi$ is not surjective. (Hint: let us order the positive prime numbers in the manner $p_1,p_2,\dots$. If $n\in \mathbb{Z}$, let us remark that the image of $n$ in $\mathbb{Z}_{p_i}$ is $\frac{n}{1}$. Prove that the element $(1,0,0,\dots)\in \bigoplus_{i=1}^{\infty} \mathbb{Z}_{p_i}$ has no preimage in $\mathbb{Z}$ under $\phi$. If you are stuck, think about the kernel of the localization homomorphism $\mathbb{Z}\to \mathbb{Z}_{p_i}$ for $i\in\mathbb{N}$.)

Problem 1: Let $A$ be an integral domain. Prove that the natural $A$-module homomorphism $\phi:A\to \bigoplus_{m\in\text{MaxSpec}(A)} A_m$ is not surjective unless $A$ is a local domain. Can you generalize to other commutative rings?

Problem 2: Let $A$ be an integral domain. Prove that $A=\bigcap_{m\in\text{MaxSpec}(A)} A_m$.

I hope this helps!

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Thanks! Very interesting examples. As for Problem 2, I don't immediately see a connection to the current problem. Is there a connection? –  ashpool Aug 29 '11 at 0:25
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@ashpool Problem 2 should be easy if you understand the answer to your question very well. Indeed, the technique used to solve Problem 2 is not at all dissimilar to the one used to solve Exercise 1. Also, the result of Problem 2 is somewhat important. –  Amitesh Datta Aug 29 '11 at 0:29
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Dear Amitesh, Problem 1 doesn't seem to be correct: think of a local domain. –  Georges Elencwajg Aug 29 '11 at 0:36
    
@Amitesh Datta I was aware of the result in Problem 2, but dismissed it thinking that it is not related to the current problem. –  ashpool Aug 29 '11 at 0:47
    
@Georges Dear Georges, thanks! A careless mistake on my part. I will edit my answer. Regards, –  Amitesh Datta Aug 29 '11 at 0:50

I realized an error in my argument. The equation should be more properly read $$\mbox{Supp}_{A_\mathfrak{m}}M_{\mathfrak{m}}=\mbox{Supp}_AM\cap\mbox{Spec}A_{\mathfrak{m}}.$$ $\phi$ is an $A$-module homomorphism, and $M_{\mathfrak{m}}$ here should be viewed as an $A$-module, not an $A_{\mathfrak{m}}$-module. So $\mbox{Supp}_AM_{\mathfrak{m}}$ may be actaully larger than $\mbox{Supp}_{A_{\mathfrak{m}}}M_{\mathfrak{m}}$. So, even if $\mathfrak{m}\neq\mathfrak{n}$ are two distinct maximal ideals, $(M_{\mathfrak{m}})_{\mathfrak{n}}$ may not be zero. I don't have a definite example, but at least this makes my previous argument invalid.

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