# Monomorphism $\mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$ is a maximal subobject of $\mathbb{Q}/\mathbb{Z}$?

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.

• 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.) Sep 6, 2021 at 2:35

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 $$f$$.

• 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?
– mz71
Sep 6, 2021 at 9:17
• I don't think the $Q/pZ\to Q/Z$ is maximal either -- you can factor it e.g. through the canonical map $Q/2pZ\to Q/Z$. Sep 6, 2021 at 15:56
• 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.
– mz71
Sep 6, 2021 at 21:18

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 $$\mathbb{Z}/(p^\infty)$$.)