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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.

Clara
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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 00:59
  • Sure, I thought of the same one. – Clara May 16 '14 at 01:01
  • Would you clarify the notation $\mathbb{Q}_p$? The additive group of $p$-adic numbers? – Kevin Carlson May 16 '14 at 01:02
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    @KevinCarlson I'm using $\mathbb{Q}_p$ to denote the field of $p$-adic numbers. – Clara May 16 '14 at 01:05
  • Well, it can't be the field, has to be the additive group of the field. :) @Clara – Thomas Andrews May 16 '14 at 01:12
  • Whoops, that's what I meant, sorry! – Clara May 16 '14 at 01:13
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    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 01:58
  • @TylerLawson Is that group isomorphic to $\mathbb{Q}\oplus (\prod_p\hat{\mathbb{Z}}_p)/\mathbb{Z}$? Because that's the group I got while working through exact sequences. – Clara May 16 '14 at 04:07
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    @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 04:49
  • @TylerLawson Great, thank you! I'm not too familiar with these groups so the isomorphisms aren't obvious to me yet. – Clara May 16 '14 at 04:53
  • @Clara: for your middle isomorphism, see exercise 2.25(ii) in Joseph J. Rotman, An introduction to homological algebra, Springer (2008). – Watson Aug 29 '16 at 21:07

2 Answers2

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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 $\mathbf R^3$.

Bruno Joyal
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From this answer, we know that $$\hom(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \cong \hat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p$$

We can therefore compute the long exact sequence

$$ 0 \to \hom(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \to \hom(\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \to \hom(\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \to \operatorname{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \to \ldots $$ to be the short exact sequence $$ 0 \to \hat{\mathbb{Z}} \to \hom(\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \to \mathbb{Q} / \mathbb{Z} \to 0$$

The first map is given by $$x \mapsto \left(\frac{m}{n} \mapsto \frac{x'm}{n} \right) \qquad \qquad (x' \equiv x \bmod n)$$ and the second is $$ f \mapsto f(1) $$

Furthermore, $\hom(\mathbb{Q}, \mathbb{Q}/\mathbb{Z})$ is a $\mathbb{Q}$-vector space; the action of $q$ on the function $f$ gives the function $(q \cdot f)(x) = f(qx)$, so the tensor product with $\mathbb{Q}$ gives an exact sequence

$$ 0 \to \mathbb{Q} \otimes \hat{\mathbb{Z}} \to \mathbb{Q} \otimes \hom(\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \to \mathbb{Q} \otimes \mathbb{Q}/\mathbb{Z} \to 0$$

which simplifies to

$$ \mathbb{Q} \otimes \hat{\mathbb{Z}} \cong \hom(\mathbb{Q}, \mathbb{Q}/\mathbb{Z})$$

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