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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.


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

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 $P$? – Arturo Magidin Jun 19 '11 at 23:25
@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). – Arturo Magidin Jun 19 '11 at 23:58
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? – Olivier Bégassat Jun 19 '11 at 23:59
@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. – Arturo Magidin Jun 20 '11 at 0:33
@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. – Arturo Magidin Jun 20 '11 at 0:58

As stated the assertion is incorrect. (See below)

Take $R=\mathbb{Z}$, $S=\mathbb{Z}/6\mathbb{Z}$ (a torsion module).

Note that for a rational prime $p$ and $x\in S$, $p^nx = 0$ if and only if $6|p^nx$. In particular, we must either have $x=0$, or $x=3$ and $p=2$; or $x=2$ or $x=4$ and $p=3$. So we have: $$S_P = \left\{\begin{array}{ll} S & \text{if }P=\langle 0\rangle;\\ \{0,3\} &\text{if }P=\langle 2\rangle;\\ \{0,2,4\} & \text{if }P=\langle 3\rangle;\\ \{0\} &\text{otherwise.} \end{array}\right.$$

Now, $\mathrm{Hom}_R(S,S)$ has six elements (one each for the possible images of the generator of $S$). And $$|\mathrm{Hom}_R(S_P,S_P)| = \left\{\begin{array}{ll} 6 & \text{if }P=\langle 0\rangle;\\ 2 &\text{if }P=\langle 2\rangle;\\ 3 &\text{if }P=\langle 3\rangle;\\ 1 &\text{otherwise.} \end{array}\right.$$

So $$\left|\prod_{P\text{ prime}}\mathrm{Hom}_R(S_P,S_P)\right| = 6\times 2 \times 3 = 36\neq 6 = \left|\mathrm{Hom}_R(S,S)\right|,$$ and I still don't see how they can be isomorphic, even if $S$ is torsion and $R$ is a PID.

The OP notes he's taken it from Rotman's book. The problem is from the section on modules over PIDs, so it is clear that $R$ is meant to be a PID, not merely a domain. In addition, the problem takes the product over all $P$-primary components of $S$, and the definition of $P$-primary explicitly excludes the zero prime. So, what we really have is:

Let $R$ be a PID. Let $S$ be a torsion $R$-module, and for each nonzero prime $P=(p)$ of $R$, let $S_P$ be the $P$-primary component of $S$, $$S_P = \{x\in S\mid p^nx=0\text{ for some }n\in\mathbb{N}\}.$$ Then $$\mathrm{Hom}_R(S,S) = \prod_{P\text{ nonzero prime}} \mathrm{Hom}_R(S_P,S_P).$$

To see this, simply use the fact that since $S$ is torsion, then $S$ is the sum of its $P$-primary components, $$S = \bigoplus_{P\text{ nonzero prime}} S_P.$$Take $R=\mathbb{Z}$, $S=\mathbb{Z}/4\mathbb{Z}$. Then $\mathrm{Hom}_R(S,S)$ has four elements. If $P=(0)$, then $S_P = S$; if $P=(2)$, then $S_P=\{0,2\}$. If $P$ is neither of these, then $S_P=\{0\}$. $\mathrm{Hom}_R(S_2,S_2)$ has two elements. So the product on the right has $8$ elements in it, while $\mathrm{Hom}_R(S,S)$ has $4$, so the two cannot be isomorphic. So in fact, any module homomorphism from $S_P$ to $S$ will have image in $S_P$.

A map form a direct sum $\oplus M_i$ to a module $M$ is equivalent to a family of maps $f_i\colon M_i\to M$.

To see this, simply use the universal property of the direct sum: if $\iota_j\colon M_j\to \oplus M_i$ is the canonical embedding of $M_j$ into the direct sum, then for any family of homomorphisms $\f_j\colon M_j\to M$ there exists a unique homomorphism $f\colon \oplus M_i\to M$ such that $f_j = f\circ \iota_j$; this shows that for every element of $\prod\mathrm{Hom}_R(M_j,M)$ there is a unique corresponding element of $\mathrm{Hom}_R(\oplus M_i,M)$. Conversely, given any $f\colon\oplus M_i\to M$, we obtain a family of maps $f_j\colon M_j\to M$ by letting $f_j = f\circ\iota_j$ (equivalently, thinking of $M_j$ as a submodule of $\oplus M_i$, then $f_j$ is simply the restriction of $f$ to $M_j$). This defines a map $\mathrm{Hom}_R(\oplus M_i,M)$ to $\prod\mathrm{Hom}_R(M_j,M)$. It is now easy to see that the uniqueness clause of the universal property ensures that the two constructions are inverses of each other, proving the isomorphism. (This isomorphism is in fact categorical: for any category $\mathcal{C}$, if $A_i$ are objects and $\coprod A_i$ is their coproduct, then the elements of $\mathcal{C}(\coprod A_i,B)$ corresponds to families $\{f_i\}_{i\in I}$ with $f_i\in\mathcal{C}(A_i,B)$).

So $$\begin{align*} \mathrm{Hom}_R(S,S) &= \mathrm{Hom}_R(\oplus S_P, S)\\ &\cong \prod \mathrm{Hom}_R(S_P, S)\\ &=\prod\mathrm{Hom}_R(S_P,S_P), \end{align*}$$ since, as discussed, any map from $S_P$ to $S$ actually has image in $S_P$.

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sorry! I'm very sorry, I forgot to state that $S$ is a torsion $R$-module, this appears in Rotman's book (Advanced Modern algebra) page 663, ex 9.5 – new Jun 20 '11 at 0:25
although I'm not sure if $R$ being a domain suffices. Don't we need the fact that $R$ is a PID so that $M$ is the direct sum of $S_{P}$ where $P$ rans over all prime ideals $P$ to show it is surjective/injective? – new Jun 20 '11 at 0:28
thanks, asuming I don't know the fact that $Hom_{R}(\oplus S_{p},S) \cong \prod Hom_{R}(S_{P},S)$ can we construct the map as you did previously, taking the restrictions? – new Jun 20 '11 at 1:58
@new: I'll add the isomorphism to the body. – Arturo Magidin Jun 20 '11 at 2:56

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