# Is $\operatorname{Hom}_\mathbb{Z}(\mathbb{Q},\mathbb{Q}/\mathbb{Z})\cong\bigoplus_p\mathbb{Q}_p$?

Is $\operatorname{Hom}_\mathbb{Z}(\mathbb{Q},\mathbb{Q}/\mathbb{Z})\cong\bigoplus_p\mathbb{Q}_p$? Or maybe $\prod_p\mathbb{Q}_p$?

I know $\mathbb{Q}/\mathbb{Z}\cong\bigoplus_p \mathbb{Z}_{p^\infty}$, and also that $\operatorname{Hom}(\mathbb{Q},\mathbb{Z}_{p^\infty})\cong\mathbb{Q}_p$. So I want to say $$\operatorname{Hom}_\mathbb{Z}(\mathbb{Q},\mathbb{Q}/\mathbb{Z})\cong\operatorname{Hom}_\mathbb{Z}(\mathbb{Q},\bigoplus_p\mathbb{Z}_{p^\infty})\cong\bigoplus_p\operatorname{hom}(\mathbb{Q},\mathbb{Z}_{p^\infty})\cong\bigoplus_p\mathbb{Q}_p$$

However, I'm not sure about the middle isomorphism. I only know of rules which allow one to pull a coproduct in the first term out into the front of Hom and change it to a product, or you can pull a product in the second term out into the front of Hom as a product.

Some digging here seems to imply that one cannot generally pull a direct sum in the second term out in front as a product, ans this is an isomorphism if the first term is finitely generated, but $\mathbb{Q}$ is certainly not a f.g. $\mathbb{Z}$-module.

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Thanks for re-opening, I was ass-backwards. $G\to G/H$ always has a homomorphism, and there are lots more. $\mathrm{Hom}(Q/Z,Z)\cong 0$. –  Thomas Andrews May 16 '14 at 0:59
Sure, I thought of the same one. –  Clara May 16 '14 at 1:01
@KevinCarlson I'm using $\mathbb{Q}_p$ to denote the field of $p$-adic numbers. –  Clara May 16 '14 at 1:05
This Hom group is $\Bbb Q \otimes \widehat{\Bbb Z}$, which is almost the additive group of adeles of $\Bbb Z$ (except that it is missing the factor of $\Bbb R$). –  Tyler Lawson May 16 '14 at 1:58
@Clara If, by that, you mean the quotient of $\Bbb Q \oplus \prod \hat{\Bbb Z}_p$ by the diagonal subgroup $\Bbb Z$, then yes, absolutely. –  Tyler Lawson May 16 '14 at 4:49

Not quite. The Pontryagin dual of $\mathbf Q$ is a weird object, called the solenoid. Since $\mathbf Q = \varinjlim \frac{1}{n}\mathbf Z$, where the limit is taken over the integers ordered by divisibility, it follows that

$$\widehat{\mathbf Q/\mathbf Z} = \varprojlim \widehat{\frac{1}{n}\mathbf Z}$$

We can replace $\frac{1}{n}\mathbf Z$ by $\mathbf Z$ in the limit, if we also replace the morphisms in the limit with the appropriate multiplication maps. From Fourier theory, the dual of $\mathbf Z$ is the circle group $S^1$, and therefore

$$\widehat{\mathbf Q/\mathbf Z} = \varprojlim S_n,$$

where $S_n=S^1$ for each $n$ and for any $n,m$, the map $S_{nm} \to S_m$ is multiplication by $n$.

(A word of warning: Here, I am using the "full" Pontryagin dual, namely the maps into $S^1$, rather than into $\mathbf Q/\mathbf Z$. Otherwise, the dual of $\mathbf Z$ is not $S^1$ but the torsion $\mathbf Q/\mathbf Z\subseteq S^1$. The solenoid is actually bigger than the group you asked about: your group consists of the topologically nilpotent elements of the solenoid, i.e. those $x$ such that $x^{n!}$ converges to $1$. Remark however that the solenoid has no torsion elements.)

The solenoid is a bizarre topological space, one of the simplest examples of an indecomposable continuum.

Objects like the solenoid appear naturally in the fourier theory of number fields. If $\mathbf A$ denotes the adele ring of $\mathbf Q$, then one has an exact sequence

$$0 \to \mathbf Q \to \mathbf A \to S^1 \to 1$$

where the last map is given by the adelic exponential. Taking duals, and using the self-duality of adeles, we get a corresponding exact sequence

$$0 \to \mathbf Z \to \mathbf A \to \widehat{\mathbf Q} \to 1$$

Thus, the group of additive adeles appears naturally as a (non-split) extension of the solenoid by $\mathbf Z$.

Interestingly, the solenoid admits an embedding as a compact subspace of $\mathbf R^3$.

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That was unexpected, thanks. –  Clara May 16 '14 at 2:05
@TylerLawson Yes, you are right. I have fixed it! –  Bruno Joyal May 16 '14 at 2:11
@Clara You are welcome, glad you liked it. Please see the changes I just made! –  Bruno Joyal May 16 '14 at 2:12