# What is $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}(p^{\infty}))$?

What is $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}(p^{\infty}))$?

I have a reference that says the group in question is $\mathbb{Q}_p,$ the additive group of the quotient field of the $p$-adic integers. Can anyone provide a reasonable derivation of this result?

• I'm guessing that $\mathbb Z(p^\infty)$ means $\mathbb Z[\tfrac1p]/\mathbb Z$. – André Henriques May 17 '18 at 21:18
• @Andre Henriques - Yes, that is correct. – Chris Leary May 18 '18 at 2:10

Note that this equality is as topological groups, where Hom is endowed with uniform convergence on compact subsets.

Every such homomorphism is zero on $p^n$ for some $n$. Hence the given homomorphism group $G=\mathrm{Hom}(\mathbf{Q},\mathbf{Z}(p^{\infty}))$ can be written as $\bigcup G_n$, where $G_n$ is the set of homomorphisms that vanish on $p^n$. Note that $G_n$ is open.

By definition, $$G_0=\mathrm{Hom}(\mathbf{Q}/\mathbf{Z},\mathbf{Z}(p^\infty))=\mathrm{Hom}(\bigoplus_\ell\mathbf{Z}(\ell^\infty),\mathbf{Z}(p^\infty))=\mathrm{End}(\mathbf{Z}(p^\infty)),$$ because for prime $\ell\neq p$ we have $\mathrm{Hom}(\mathbf{Z}(\ell^\infty),\mathbf{Z}(p^\infty))=0$. It is classical that $\mathrm{End}(\mathbf{Z}(p^\infty))$ is isomorphic to $\mathbf{Z}_p$ as a topological ring (details upon request).

Hence each $G_n$ is isomorphic to $\mathbf{Z}_p$. Multiplication by $p$ maps $G_n$ into $G_{n-1}$. Also, mapping $f$ to $x\mapsto f(p^{-1}x)$ maps $G_{n-1}$ into $G_n$ and is an inverse to multiplication by $p$. So indeed $pG_n=G_{n-1}$, and we deduce the result.

Edit: consider $\mathbf{Q}_p$ acting on itself as its own endomophism group (as topological group). Restricting to $\mathbf{Q}$ at the origin, and modding out by $\mathbf{Z}_p$ at the target yields a topological group homomorphism $\phi:\mathbf{Q}_p\to\mathrm{Hom}(\mathbf{Q}_p/\mathbf{Z}_p)$ (where the latter Hom is in topological groups). This homomorphism $\phi$ is injective: indeed, an element $f$ in the kernel of $\phi$ would, in restriction to $\mathbf{Q}$, satisfy $f(\mathbf{Q}_p)\subset\mathbf{Z}_p$. Since $\mathbf{Q}_p$ is divisible it has no nontrivial finite quotient and hence no nonzero homomorphism into any profinite group. So $f=0$.

Keeping in mind that there is a canonical isomorphism $\mathbf{Z}(p^\infty)\to\mathbf{Q}_p/\mathbf{Z}_p$, since we previously obtained that $\mathrm{Hom}(\mathbf{Q}_p/\mathbf{Z}_p)\simeq\mathbf{Q}_p$ as topological group, we deduce that $\phi$ is a isomorphism.

I should also add that it's important to phrase the result in terms of topological groups. Indeed, as an abstract group, $\mathbf{Q}_p$ is just isomorphic to $\mathbf{Q}^{(c)}$ (and also to $\mathbf{R}$).

• This is a nice answer. However, I don't see the connection with $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}(p^{\infty}))$. Am I being dense? – Chris Leary May 18 '18 at 2:15
• @ChrisLeary It seems to me that $G$ is the Hom group you're looking for. – Najib Idrissi May 18 '18 at 6:51
• @YCor - The edit helps. Thanks for your assistance. – Chris Leary May 18 '18 at 15:55

YCor's answer is good, but here's a slightly different way of seeing it.

Given $x \in \mathbb Q$, let $[f(x)]$ be any lift of $f(x)$ from $\mathbb Z[1/p]/\mathbb Z$ to $\mathbb Z[1/p]$. Then $p^n [ f(1/p^n)]$ is a $p$-adic Cauchy sequence because $p^n [f(1/p^n)] - p^m [ f(1/p^m)]$ is an integer multiple of $p^{\min (n,m)}$. Changing the lift also causes a difference by an integer multiple of $p^n$ and thus does not affect the limit. This defines for each $f$ an element of $\mathbb Q_p$.

To go the other way, given $\alpha \in \mathbb Q_p$, observe that multiplication by $\alpha$ defines a map $\mathbb Q$ to $\mathbb Q_p$. Compose this with the projection to $\mathbb Q_p / \mathbb Z_p \cong \mathbb Z[1/p] / \mathbb Z$ to obtain a map $\mathbb Q \to \mathbb Z[1/p]$.

It is straightforward to check that these are inverses. Starting with $\alpha$, $[f(1/p^n)]$ is an element of $\mathbb Z_p$ plus $\alpha /p^n$ so $p^n [f(1/p^n)]$ is an approximation of $\alpha$. Starting with $f$, the limit $\alpha$ differs from $p^n [f(1/p^n)]$ by an element of $p^n \mathbb Z_p$, so $f(1/p^n)$ differs from $\alpha p^{-n}$ by an element of $\mathbb Z_p$, and then multiply these by elements of $\mathbb Z$ inverting every prime other than $p$ to get all rational numbers.

• I wonder if $\mathbb R$ can be obtained in some similar way... – მამუკა ჯიბლაძე May 18 '18 at 8:22
• @მამუკაჯიბლაძე Hom between two discrete groups will always be totally disconnected. – YCor May 18 '18 at 12:35
• @მამუკაჯიბლაძე There is a construction as maps $\mathbb Z \to \mathbb Z$ that are homomorphisms up to bounded error, modulo bounded functions. – Will Sawin May 18 '18 at 12:59
• @Will Sawin - I like this answer very much as well. I appreciate your time and effort – Chris Leary May 18 '18 at 15:56