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I'm trying to compute $\hom_\mathbb{Z}(\mathbb{Z}_{p^\infty},\mathbb{Q})$. I believe it is zero, simply because $\mathbb{Z}_{p^\infty}$ is torsion (it is a $p$-group) and $\mathbb{Q}$ is torsion-free.

Is that argument correct? I'm suspicious of it for the following reason. The computation of $\hom_\mathbb{Z}(\mathbb{Z}_{p^\infty},\mathbb{Q})$ is an exercise in Anderson & Fuller (exercise 1, section 4, chapter 1), in which there is a hint that points to a series of three exercises. Those three exercises are more sophisticated that the argument that I sketched above, so I'm led to doubt of my argument.

I append below the three exercises cited in the hint.

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user46225
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  • It seems that the thrid one in 14 solves the problem. Isn't it? – Mikasa May 01 '13 at 16:07
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    @BabakS.: Yes, I agree wholeheartedly. But why cite the three exercises? The exercise in Anderson & Fuller also asked to compute $\hom_{\mathbb{Z}}(\mathbb{Z}{p^\infty},\mathbb{Z})$ and $\hom{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})$, but I don't see that exercises 15 or 16 are necessary for those, either. – user46225 May 01 '13 at 16:10

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Observe that if $\tau$ is for torsion and $f \in Hom(A,B)$ then $f(\tau (A)) \subseteq \tau(B)$, for $0=f(0) = f(na) = nf(a) = 0$

the rest is what you have.

I.e. $\forall f \in Hom(\mathbb{Z}_{p^{\infty}} , \mathbb{Q})$, $f(\mathbb{Z}_{p^{\infty}})= f(\tau(\mathbb{Z}_{p^{\infty}})) \subseteq \tau(\mathbb{Q}) = \lbrace 0 \rbrace $

therefore $f=0$

Sajid
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