3
$\begingroup$

A standard example of both a monomorphism and epimorphism that is not a isomorphism is the projection $f:\mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$ in the category $\mathsf{DivAb}$ of divisible abelian groups. The kernel $\ker(f)$ should be equal to $\mathbb{Z}$, but $\mathbb{Z} \notin \mathsf{DivAb}$, so it must be something smaller, but the only divisible subgroup of $\mathbb{Z}$ is $0$. Therefore, $f$ is mono.

In every category, we have a notion of subobject category. As $f$ is mono, I write $\mathbb{Q} \in \mathsf{Sub}(\mathbb{Q/Z})$. I want to know, if it's a maximal subobject, i.e. if the following claim is true or false:

For any $X\in \mathsf{Sub}(\mathbf{Q/Z})$ if $g:\mathbb{Q} \to X$ is a inclusion of subobjects and the composition $\mathbb{Q} \to X \to \mathbb{Q/Z}$ is equal to $f$, then $g$ is an isomorphism.

I think it's false, and there should be a zoo of examples, but I cannot find any. Put more profoundly, the question is about the preorder of epi-mono factorizations of $f$, as defined here. The examples of other standard "fake isomorphisms" are beautifully explained in terms of this preorder, but I don't see it in this situation.

I tried to use some facts about divisible groups, for example that $\mathbb{Q/Z} = \bigoplus_{p \text{ - prime}} \mathbb{Z}(p^\infty)$, where $\mathbb{Z}(p^\infty)$ is a Prüfer p-group.

$\endgroup$
1
  • 1
    $\begingroup$ As written this is trivially false since the map $X\to\mathbb{Q}/\mathbb{Z}$ could be an isomorphism instead. I imagine you want to exclude this, so that by "maximal subobject" you mean "maximal proper subobject". (Any object in any category has a unique maximal subobject, namely the identity map.) $\endgroup$ Sep 6, 2021 at 2:35

2 Answers 2

3
$\begingroup$

What about factorization of $f$ as $ \mathbb{Q} \to \mathbb{Q}/{2\mathbb{Z}} \to \mathbb{Q}/\mathbb{Z} $? The kernel of the second map in $\mathsf{Ab}$ is $\mathbb{Z}/2\mathbb{Z}$ whose only divisible subgroup is trivial as well. So this would be a nontrivial subobject of $\mathbb{Q}/\mathbb{Z} $ that is larger than the one given by $f$.

$\endgroup$
3
  • $\begingroup$ Thanks! Yeah, I thought that such a subobject is isomorphic to Q/Z. While Q/2Z is isomorphic to Q/Z, the projection isn't a isomorphism. Great! So now it looks like Q/pZ is maximal for each prime p? $\endgroup$
    – mz71
    Sep 6, 2021 at 9:17
  • $\begingroup$ I don't think the $Q/pZ\to Q/Z$ is maximal either -- you can factor it e.g. through the canonical map $Q/2pZ\to Q/Z$. $\endgroup$
    – Jonas Frey
    Sep 6, 2021 at 15:56
  • $\begingroup$ Oh, I thought it was the other way around. As $\mathbb{Q}$ is smaller than $\mathbb{Q}/n\mathbb{Z}$... Nevermind, thanks for your time. $\endgroup$
    – mz71
    Sep 6, 2021 at 21:18
3
$\begingroup$

No, $f$ is not a maximal proper subobject of $\mathbb{Q}/\mathbb{Z}$. For any $n\in\mathbb{Z}$, there is a quotient map $\mathbb{Q}/n\mathbb{Z}\to\mathbb{Q}/\mathbb{Z}$ which is also monic, and your $f$ factors through this quotient.

For some more exotic examples, note that for any $\alpha\in\hat{\mathbb{Z}}$ (the profinite completion of $\mathbb{Z}$), we can consider multiplication by $\alpha$ as a homomorphism $\mathbb{Q}/\mathbb{Z}\to\mathbb{Q}/\mathbb{Z}$ (since to multiply a fraction $\frac{m}{n}$ mod $\mathbb{Z}$ by $\alpha$, we only need to know what $\alpha$ is mod $n$). If $\alpha$ is not infinitely divisible by any prime, then this homomorphism is surjective, and its kernel contains no nontrivial divisible subgroup so it is monic in our category.

Now suppose additionally that $\alpha$ is divisible by infinitely many different primes (but still not infinitely divisible by any single prime); for instance, identifying $\hat{\mathbb{Z}}$ with $\prod_p\mathbb{Z}_p$, $\alpha$ could be the element whose coordinate in $\mathbb{Z}_p$ is $p$ for each prime. I claim then that in fact the map $h:\mathbb{Q}\oplus\mathbb{Q}/\mathbb{Z}\to \mathbb{Q}/\mathbb{Z}$ which is $f$ on the first coordinate and multiplication by $\alpha$ on the second coordinate is monic (and so is a subobject that nontrivially contains $f$ as a direct summand!).

To prove this, suppose $x\in\ker(h)$ is nonzero; we wish to show there is no divisible subgroup of $\ker(h)$ containing $x$. If the first coordinate of $x$ is $0$, the same must be true of any $y$ such that $ny=x$, so such a $y$ must have second coordinate that is in the kernel of multiplication by $\alpha$. So if $x$ were contained in a divisible subgroup of $\ker(h)$, restricting to the elements whose first coordinate is $0$ we would get a nontrivial divisible subgroup of the kernel of multiplication by $\alpha$. Since $\alpha$ is not infinitely divisible by any prime, no such subgroup exists.

So, we may assume the first coordinate of $x$ is nonzero. Multiplying $x$ by a nonzero integer, we may then assume the second coordinate of $x$ is $0$, and its first coordinate is some integer $a$. Now note that an element $y\in\ker(h)$ satisfies $ny=x$ iff the first coordinate of $y$ is $a/n$ and the second coordinate is $b/n$ mod $\mathbb{Z}$ for some $b\in\mathbb{Z}$, with $y\in\ker(h)$ saying that $\alpha b=-a$ mod $n$. In particular, if $\alpha$ is $0$ mod $n$, then $a$ must be $0$ mod $n$ as well, assuming such a $y$ exists. Since $\alpha$ is divisible by infinitely many primes, this forces $a=0$, contradicting our assumption that the first coordinate of $x$ was nonzero.

(If all you care about is the poset of subobjects, then the use of $\hat{\mathbb{Z}}$ here is perhaps a little misleading, since the subobject $\alpha:\mathbb{Q}/\mathbb{Z}\to\mathbb{Q}/\mathbb{Z}$ only depends on how many times $\alpha$ is divisible by each prime. If you prefer, you could say we have picked some $d_p\in\mathbb{N}$ for each prime $p$ (the number of times $\alpha$ is divisible by $p$) and are considering the map $\mathbb{Q}/\mathbb{Z}\to\mathbb{Q}/\mathbb{Z}$ which is given by multiplication by $p^{d_p}$ on each summand $\mathbb{Z}/(p^\infty)$.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .