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Let's consider definition of reduced modules and torsion modules:

A reduced module $C$ is defined by the property that $\mathrm{Hom}(A,C)=0$ for every divisible module $A$.

(see for example A Foundation of Torsion Theory for Modules Over Geeneral Rings by Hattori)

I can not find any example about reduced module in this sense.

Please help me find some. I really appreciate your ideas!

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All caps is considered to be "shouting"; please don't use it. –  Arturo Magidin Mar 1 '12 at 4:56
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3 Answers

Let $M$ be any module.

If $\{N_i\}_{i\in I}$ is a family of divisible submodules of $M$ (that is, each $N_i$ is a submodule of $M$, and each $N_i$ is divisible), then $N=\langle N_i\rangle$ is also a divisible submodule of $M$: it is a submodulee, and given $n_1+\cdots+n_k\in N$, with $n_j\in N_{i_j}$, and $r\in R$, there exist $m_j\in N_{i_j}$ such that $rm_j = n_j$, and then $r(m_1+\cdots + m_k) = n_1+\cdots + n_k$.

Therefore, if $M$ is any module, then it has a largest divisible submodule (namely, the submodule generated by all divisible submodules; note that the trivial submodule is divisible, so this collection is not empty). Call it $M_d$.

Claim 1. The only divisible submodule of $M/M_d$ is the trivial submodule. that is, $(M/M_d)_d = \{0\}$.

Proof. Let $K$ be a submodule of $M$ that contains $M_d$ such that $K/M_d$ is divisible; let $k\in K$, and let $r\in R$. Then there exists $u+M_d$ such that $r(u+M_d) = (k+M_d)$, so $k = ru+x$ for some $x\in M_d$, with $u\in K$. Since $M_d$ is divisible, there exists $v\in M_d$ (hence in $K$) such that $rv= x$. Thus, $k=r(u+v)$, and $u+v\in K$. Hence $K$ is a divisible submodule of $M$, so $K\subseteq M_d\subseteq K$. That is, $K/M_d$ is the trivial submodule.

Claim 2. $M/M_d$ is reduced.

Proof. Let $A$ be a divisible module, and let $\varphi\colon A\to M/M_d$ be a module homomorphism. Since $\varphi(A)$ is divisible (a quotient of a divisible module is divisible), then by Claim 1 $\varphi(A)=\{0+M_d\}$. Thus, $\varphi$ is the trivial map. Hence, $M/M_d$ is divisible.

This gives you lots of nontrivial reduced modules: just pick any module $M$ that is not divisible, and consider $M/M_d$

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How about letting the ring be $R=\mathbb Z$ and $C=\mathbb Z$? I claim $\mathbb Z$ is reduced. Suppose $0\neq\varphi\in Hom(A,C)$ where $A$ is divisible. If $\varphi(a)=k$, then $\varphi(-a)=-k$, so I can assume some positive value is achieved by $\varphi$. Let $n$ be the smallest positive integer in the image of $\varphi$. Then $\varphi(a)=n$, and also $2\varphi(a/2)=n$. Thus $0<\varphi(a/2)<n$, which is a contradiction.

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Jim and Arturo have given you everything you need to see this through. Let me take a slightly different tack. If there is a nonzero $f:A \rightarrow C$, with $A$ divisible, then the image of $f$, which is isomorphic to $A/ker(f)$, is a nonzero divisible subgroup of $C$ (as Arturo pointed out, homomorphic images of divisibles are divisible). This is not hard to show, and you probably already have. In any event, if $C$ is reduced, then $Hom(A,C)$ must be zero. To show the converse, prove that $C$ divisible implies $Hom(A,C)$ is not zero for some divisible module $A$. For this, take $A = C$ and $f$ the identity map on $C$. As Jim points out, $\mathbb{Z}$ is not divisible. The subgroups of $\mathbb{Z}$ are the cyclic subgroups generated by nonnegative integers $a$. Can such a subroup be divisible? If so, the equation $nx = a$ has a solution in $<a>$ for every nonzero integer $n$. But this cannot happen; try $a = 3$ and $n = 2$, or, in general, take $n$ relatively prime to $a$ so that the rational number $a/n$ is not an integer. Put everything together and you have that $Hom(A, \mathbb{Z}) = 0$ for every divisible group $A$. Arturo has given you a nice way to extend things to modules over a general ring $R$. In this context, the usual interpretation of the "divisibility equation" $nx = a$ is that $n$ is a regular element of the ring $R$, so that an $R$-module $A$ is divisible if and only if $nA = A$ for all regular $n$ in $R$.

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