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Let $R$ be a commutative ring with identity. Suppose $R=(r_1,\ldots,r_k)$. Take an homomorphism of $R$-modules: $f:M\rightarrow N$. Suppose that the function $\frac{f}{1}:M_{r_i}\rightarrow N_{r_i}$ is an isomorphism for every $i=1,\ldots,k$; how can I prove that then $f$ is an isomorphism?

$M_{r_i}$ denotes the localization of $M$ at the multiplicatively closed set $\{r_i^n:n\geq0\}$.

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I take it that $(r_1,\dots,r_k)=R$ means they generate $R$ as a ring? –  rschwieb May 8 '12 at 13:47
@AlexM I don't understand your notation for the ring $R$.....What do you mean by $R = (r_1, \ldots, r_k)$? –  user38268 May 8 '12 at 14:06
@AlexM Please explain your notation for $R$. –  user38268 May 8 '12 at 14:14
@AlexM It is hard for people to help you if you don't explain the notation. –  user38268 May 8 '12 at 14:25
It means they generate $R$ as an ideal –  Alex M May 8 '12 at 14:47
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First, a lemma.

Let $L$ be an $R$ module. $L = 0$ $\Leftrightarrow$ $L_{r_i} = 0$ for each $i$.

$\Rightarrow$. Clear.

$\Leftarrow$. Let $x \in L$. $x$ goes to zero in $L_{r_i}$ if and only if $r_i^{n_i}x = 0$ for some $n_i$. Now, if $(r_1, \ldots, r_k) = R$ then one can write $(r_1, \ldots, r_k)^N = R$ for any $N$. Do you see how to finish this off?

Now, $f\colon M \to N$ is injective if and only if $\ker f = 0$. In any case, we have an exact sequence \[ 0 \to \ker f \to M \to N \] and localization is flat. Now show that $(\ker f)_{r_i} \approx \ker f_{r_i}$. You can also write down an exact sequence that expresses surjectivity and make the same argument.

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This is really close to being the proof that "being an (epi/mono/iso)morphism is local at the primes". You'll see $\{r_1, \ldots, r_k)$ called a "partition of unity", and for good reason! –  Dylan Moreland May 8 '12 at 13:52
Hi, I have a question. What do you mean by $(r_1, \ldots, r_k)^N$? So far I could only come up with "all N-fold products of elements in $(r_1, \ldots, r_k)$, but my interpretation doesn't make sense here. –  rschwieb May 8 '12 at 13:57
@rschwieb Oh, sorry, just the $N$-th ideal product. I only wanted to say why the $\{r_i^{n_i}\}$ will still generate the unit ideal. –  Dylan Moreland May 8 '12 at 13:59
I've been having some trouble with what was meant: is it safe to say the OP meant that those $k$ elements generate $R$ as an ideal, and not as a ring? (I should do some problems of the OP's type... I have very poor intuition on the topic.) –  rschwieb May 8 '12 at 14:04
@rschwieb I've only ever seen $(a, b, \ldots)$ to mean "the ideal generated by $a$, $b$, ...". If it is as a ring, then I don't have any ideas. –  Dylan Moreland May 8 '12 at 14:05
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