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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?2012-05-08
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    @AlexM I don't understand your notation for the ring $R$.....What do you mean by $R = (r_1, \ldots, r_k)$?2012-05-08
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    @AlexM Please explain your notation for $R$.2012-05-08
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    @AlexM It is hard for people to help you if you don't explain the notation.2012-05-08
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    It means they generate $R$ as an ideal2012-05-08

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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!2012-05-08
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    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.2012-05-08
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    @rschwieb Oh, sorry, just the $N$-th [ideal product](http://en.wikipedia.org/wiki/Ideal_(ring_theory)#Ideal_operations). I only wanted to say why the $\{r_i^{n_i}\}$ will still generate the unit ideal.2012-05-08
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    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.)2012-05-08
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    @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.2012-05-08
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    If this setup looks familiar to you, then that's probably what was meant. It just looks a little weird to me. For example you could put $(1,x)=R$ for any $x$ at all in $R$, and this would still be true? If so it amazes me a little :)2012-05-08
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    @rschwieb True, but this is a surprisingly practical condition: in the Zariski topology on $\operatorname{Spec} R$ this is the condition for the "distinguished opens" $R_{r_i}$ to cover the whole space, and these are a basis for the topology so there's usually no loss in assuming that your cover is of this form. This also shows that spectra are quasi-compact. So I think there's real content to the statement.2012-05-08
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    @DylanMoreland: that was exactly my idea too, but the problem that I had (and still have) is how can I deduce in the lemma that $x=0$ from the fact that $r_i^{n_i}x=0$ and $R=(r_1,\ldots,r_k)^N$ for any N. Could you explain it to me?2012-05-08
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    @AlexM By the bi/multinomial theorem or whatever you like, as well as the definition of an ideal, you'll show that $(r_1^{n_1}, \ldots, r_k^{n_k}) = R$. In particular, $1$ is a linear combination of these elements. An element is zero in $L$ if and only if it is annihilated by $1$.2012-05-08
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    @DylanMoreland: I'm sorry Dylan but I really cannot understand why $R=(r_1^{n_1},\ldots,r_k^{n_k})$. Can you write a little proof, please?2012-05-08
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    Let $N$ be the sum of all the $n_i$. If you have an expression for $1$ as a linear combination $\sum a_ir_i$ of the $r_i$, then raise that to the $N$-th power. Argue that in each term of the resulting sum, some $r_i$ occurs to a power $\geq n_i$. This is sort of like proving that the nilpotent elements form an ideal.2012-05-08