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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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    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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    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