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Refer to exercises 9, 10 of chapter 3 in Lang's algebra, page 167.

In particular, let $A$ be a commutative ring, $p$ a prime ideal and $M, N$ $A$-modules. Then $M_p, N_p$ are the localized $A_p$ modules. Let $f: M_p \rightarrow N_p$ be a homomorphism of $A_p$ modules. How can we lift this homomorphism to an $A$-homomorphism $M \rightarrow N$ of the initial modules?

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    Dear Manos, if you refer to an exercise in Lang, you should copy it in your question and not force users to try to get hold of that book.2011-10-31

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You are starting with a sequence of $A$-modules 0 \to M' \stackrel f\to M \stackrel g\to M'' \to 0. You want to show that if for each prime $\mathfrak{p}$ of $A$ the induced sequence of $A_\mathfrak{p}$-modules 0 \to M'_\mathfrak{p} \stackrel{f_\mathfrak{p}}\longrightarrow M_\mathfrak{p} \stackrel{g_\mathfrak{p}}\longrightarrow M''_\mathfrak{p} \to 0 is exact, then the original sequence is exact. Here the map $f_\mathfrak{p}$, for example, just sends $x/s$ to $f(x)/s$. You don't need to perform any sort of "lifting" on $f_\mathfrak{p}$, because you already have $f$. One way of solving the problem is to find ways to apply the fact that if $N_\mathfrak{p} = 0$ for all primes $\mathfrak{p}$, then $N = 0$. [You could try this with $N = \operatorname{Ker} f$. What is the relationship between $\operatorname{Ker} f$ and $\operatorname{Ker} f_\mathfrak{p}$?]

I should mention that if M' is finitely presented then there is a relationship between $A$-linear M' \to M and $S^{-1}A$-linear S^{-1}M' \to S^{-1}M. Proposition 2.10 of Eisenbud's book shows that in this situation we have a natural isomorphism \operatorname{Hom}_{S^{-1}A}(S^{-1}M', S^{-1}M) \approx S^{-1}\operatorname{Hom}_A(M', M). So you might suggestively call Georges' example "the identity map divided by $2$".

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    Since no one has mentioned it, perhaps it is worth noting that the functor $M\rightarrow M_p$ is the tensor product $\otimes_A A_p$. So we know certain things about it on more-general principles...2011-10-31
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In general, given an $A_p$-linear morphism $f:M_p\to N_p$ it is impossible to find a morphism of $A$-modules $g: M\to N$ lifting $f$ in the sense that $g_p=f.$

For example, let $A,M$ and $N$ all be equal to $\mathbb Z$. Choose for $p$ the zero ideal $p=(0)$.
Then $A_p=M_p=N_p=\mathbb Q$ and you can choose for $f:\mathbb Q \to \mathbb Q$ the morphism $f(q)=\frac{q}{2}$.
The only $\mathbb Z$-linear maps $g:\mathbb Z \to \mathbb Z$ are of the form $g(n)=an$ for some $a\in \mathbb Z$ and clearly none of them lifts $f$.

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    @Dylan: Precisely!2011-10-31