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Let $S$ be an $R$-module where $R$ is an integral domain and let $P = \langle p \rangle $ be a prime ideal.

Define:

$S_{P}=\{s \in S: p^{n} s=0 \ \textrm{for some natural }n\}$.

As usual, denote $\mathrm{Hom}_{R}(S,S)$ the set of all homomorphisms (of $R$-modules) from $S$ to $S$.

Let $S$ be a torsion $R$-module where $R$ is a domain. Can someone please explain why is the following isomorphism true?

$$\mathrm{Hom}_{R}(S,S) \cong \prod_{P\text{ prime}} \mathrm{Hom}_{R}(S_{P},S_{P})$$

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    The product over *what*? Are you taking the product over all prime ideals $P$?2011-06-19
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    @Arturo Magidin: yes, I'm sorry.2011-06-19
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    @new: *All* prime ideals? How do you define $S_P$ when $P$ is not principal? (you are only assuming $R$ is a domain, so you don't know whether primes are principal or not).2011-06-19
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    Just a question what you mean by '$P=\langle p \rangle$ a prime ideal' : are you only considering the principal prime ideals? Also, $\lbrace 0\rbrace$ is such an ideal, do you consider it too? If not, what do you make of the case where $R$ is a field?2011-06-19
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    @new: You still don't say what to do with nonprincipal ideals. Since you are quoting it from a book, please (i) quote it **correctly**; and (ii) give the source in the body of the post.2011-06-20
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    @new: Since Section 9.1 in Rotman's book is called "Modules over PIDs", I think it is fairly clear that this is supposed to be a PID, not an arbitrary domain. Also, the definition of $p$-primary **explicitly** excludes the $0$-ideal, something you did not exclude in your definition.2011-06-20

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