22
$\begingroup$

I expect this question has a very simple answer.

We all know from primary school that there are no non-trivial continuous homomorphisms from $\hat{\mathbb{Z}}$ to $\mathbb{Z}$. What if we forget continuity: can anybody give an explicit example of a homomorphism?

Note that $\hat{\mathbb{Z}}$ is torsion-free, and not divisible (since it's isomorphic to $\prod_p \mathbb{Z}_p$ and $\mathbb{Z}_p$ is not divisible by $p$). There is the canonical injection $\mathbb{Z} \to \hat{\mathbb{Z}}$; is there some abstract reason why it ought to have a left inverse, and if so can we write it down?

$\endgroup$
6
  • 4
    $\begingroup$ The equation $(x^2−13)(x^2−17)(x^2−221)=0$ has a solution in $\hat{\mathbb{Z}}$ but not in $\mathbb{Z}$. See Page 3 of "Number Theory" by Borevich and Shafarevich. $\endgroup$ May 24, 2011 at 13:16
  • 2
    $\begingroup$ $\hat{\mathbb Z}/\mathbb Z$ is a rational vector space (isomorphic to $\hat{\mathbb Z}\otimes\mathbb Q/\mathbb Z\otimes \mathbb Q$). $\endgroup$ May 24, 2011 at 13:16
  • 5
    $\begingroup$ Perhaps someone should make the category clear in which we talk about homomorphisms (groups? rings?). Because the comment by SJR obviously makes use of the ring structure, but I think that groups are meant. $\endgroup$ May 24, 2011 at 13:26
  • $\begingroup$ @Martin: Indeed, I was thinking of group homomorphisms. But rings are interesting too... $\endgroup$ May 24, 2011 at 13:33
  • 19
    $\begingroup$ I would have given the question +1, except for the "We all know from primary school" affectation. I'm sure it's supposed to be funny, but it's a ridiculous exaggeration: probably the majority of research mathematicians have never learned what $\widehat{\mathbb{Z}}$ is. I would be willing to bet that at least one person read that sentence, didn't understand it, and felt at least a little bit bad about themselves because of it. There's really no need for this sort of schoolboy humor. $\endgroup$ May 24, 2011 at 20:17

4 Answers 4

14
$\begingroup$

The answer is that there are no such homomorphisms. See the following preprint of Nik Nikolov https://arxiv.org/abs/0901.0244.

$\endgroup$
5
  • $\begingroup$ A perfect answer! Except that I was hoping to use $H^1(\mathbb{F}_p,\mathbb{Z})$ as an example to show why it's important to use continous cochains... $\endgroup$ May 24, 2011 at 12:38
  • 2
    $\begingroup$ I'm not sure an Annals paper, relying on the CFSG is quite necessary here. $\endgroup$
    – Junkie
    May 24, 2011 at 13:54
  • $\begingroup$ Junkie you are right as you can see from Neil Strickland's answer. But the paper is of course much more general. $\endgroup$ May 24, 2011 at 15:50
  • 1
    $\begingroup$ Although this is an excellent answer, Yiftach, this preprint of Nikolay and Dan Segal supercedes the above one arxiv.org/abs/1102.3037 $\endgroup$ May 24, 2011 at 22:06
  • $\begingroup$ In addition by an old result of R. Alperin, every homomorphism from any compact group to $\mathbf{Z}$ is continuous. Hence $\mathrm{Hom}(K,\mathbf{Z})=\{0\}$ for every compact group $K$. $\endgroup$
    – YCor
    Apr 22, 2018 at 9:08
60
$\begingroup$

Let $\phi:\hat{\mathbb{Z}}\to\mathbb{Z}$ be a nontrivial homomorphism. As every nontrivial subgroup of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$, we may suppose that $\phi$ is surjective, with kernel $K$ say. Now $\phi$ induces a surjective homomorphism $\phi_n:\hat{\mathbb{Z}}/n\hat{\mathbb{Z}}\to\mathbb{Z}/n\mathbb{Z}$, but it is standard that $\hat{\mathbb{Z}}/n\hat{\mathbb{Z}}$ has order $n$, so $\phi_n$ must be an isomorphism. This implies that $K\leq n\hat{\mathbb{Z}}$ for all $n$, but $\bigcap_n n\hat{\mathbb{Z}}=0$, so $\phi$ is injective, which is clearly impossible.

$\endgroup$
0
10
$\begingroup$

Let $\phi$ be such a homomorphism, on additive groups $\hat Z\rightarrow Z$. Write $(\vec x,\vec y)\in\hat Z$ for the element that is $x$ on primes that are 1 mod 3, and $y$ on primes that are 2 mod 3.

Then $\phi(\vec x,\vec 0)=0$ for all $x\in Z$, for $(\vec x,\vec 0)$ is $l$-divisible for any prime $l$ that is 2 mod 3. The symmetrical argument claims $\phi(\vec 0,\vec y)=0$ too.

Without loss of generality, we can assume that a preimage of $1$ is given by $(\vec 1,\vec 1)$.

Next, applying the group law and setting $\alpha=\phi(\vec 1,\vec{-1})$, we derive the system $$1+\alpha=\phi(\vec 1,\vec 1)+\phi(\vec 1,\vec{-1})=\phi(\vec 2,\vec 0)=0$$ $$1-\alpha=\phi(\vec 1,\vec 1)-\phi(\vec 1,\vec{-1})=\phi(\vec 0,\vec 2)=0$$ This is impossible, so $\phi$ does not exist.

$\endgroup$
1
  • 4
    $\begingroup$ One formalization of this argument: $\hat{\mathbf{Z}}=A\oplus B$ where $A=\mathbf{Z}_2$ and $B=\prod_{p\ge 3}\mathbf{Z}_p$. Then $A$ is 3-divisible and $B$ is 2-divisible. So the images of both $A$ and $B$ in $\mathbf{Z}$ have to be zero. $\endgroup$
    – YCor
    Apr 22, 2018 at 9:29
1
$\begingroup$

Here a slightly different proof. Decompose $\hat{\mathbb{Z}}$ into product of the pro-l Sylows. Now a vector in this decomposition having a zero coordinate at the prime p must be divisible by any power of p.

Any other element is the sum of two such elements, hence you are done (or alternatively fix only one prime, apply the above, now you have a map from the p-adics to the integers and it must be 0, as the p-adics are l-divisible for any l coprime to p).

$\endgroup$
1
  • $\begingroup$ Oops I did not see it was already given above $\endgroup$
    – Richard
    Jul 27 at 5:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.