In a recent conversation with a colleague, the following question arose:



> 
> 1. What is the isomorphism class of $\mathrm{Ext}^1\_\mathbb{Z}(\mathbb{R}/\mathbb{Z},\mathbb{Z})$? That is to say, what is $\mathrm{Ext}^1\_\mathbb{Z}(\mathbb{R}/\mathbb{Z},\mathbb{Z})$ as an Abelian group?
> 
> 
> 


I did not have any immediate response other than “That’s interesting; I’ll have to think about it.” Afterwards, I spent some time recalling some standard information about $\mathbb{R}/\mathbb{Z}$ and the Ext functor, and have arrived at the following:
$$\mathrm{Ext}^1\_\mathbb{Z}(\mathbb{R}/\mathbb{Z},\mathbb{Z})\simeq {\left(\prod\_{x\in\mathbb{R}}{\mathrm{Ext}^1\_\mathbb{Z}{\left(\mathbb{Q}\_x,\mathbb{Z}\right)}}\right)}\times{\left(\prod\_{p\in\mathbb{P}}{\mathrm{Ext}^1\_\mathbb{Z}{\left(\mathbb{Z}{\left(p^\infty\right)},\mathbb{Z}\right)}}\right)}\ .$$
Here, $\mathbb{Q}\_x=\mathbb{Q}$ for all $x\in\mathbb{R}$, $\mathbb{P}$ denotes the set of prime numbers, and for a prime $p$, $\mathbb{Z}{\left(p^\infty\right)}$ is the Prüfer $p$-group. This decomposition reduces the original problem to the following two questions:



> 
> 2. What is the isomorphism class of $\mathrm{Ext}^1\_\mathbb{Z}(\mathbb{Q},\mathbb{Z})$?
> 3. Given a prime number $p$, what is the isomorphism class of $\mathrm{Ext}^1\_\mathbb{Z}(\mathbb{Z}{\left(p^\infty\right)},\mathbb{Z})$?
> 
> 
> 


I was curious to know if anyone happens to know the answer to any of questions (1), (2), or (3), or can point me to such in the literature.



The isomorphism class of $\mathrm{Ext}^1_\mathbb{Z}(\mathbb{R}/\mathbb{Z},\mathbb{Z})$
Writing $\mathbb{R}/\mathbb{Z} \cong \mathbb{Q}/\mathbb{Z} \oplus \bigoplus\_I \mathbb{Q}$ where $I$ indexes a Hamel basis for $\mathbb{R}$ minus one element, we have


$$\text{Ext}^1(\mathbb{R}/\mathbb{Z}, \mathbb{Z}) \cong \text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z}) \times \prod\_I \text{Ext}^1(\mathbb{Q}, \mathbb{Z}).$$


At this point we can write $\mathbb{Q}/\mathbb{Z}$ as the direct product of its Sylow subgroups, but we can also observe the following: there is a short exact sequence $0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0$ giving rise to a long exact sequence many of whose terms vanish, namely


$$0 \to \text{Hom}(\mathbb{Z}, \mathbb{Z}) \to \text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z}) \to \text{Ext}^1(\mathbb{Q}, \mathbb{Z}) \to 0 \to \dots$$


establishing that $\text{Ext}^1(\mathbb{Q}, \mathbb{Z})$ is the quotient of $\text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z})$ by a copy of $\mathbb{Z}$, so to compute the former it suffices to compute the latter (and figure out what copy of $\mathbb{Z}$ comes into play).


Now we can compute $\text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z})$ by writing $\mathbb{Q}/\mathbb{Z}$ as a filtered colimit of its subgroups $\frac{1}{n} \mathbb{Z}/\mathbb{Z}$ and using that $\text{Ext}^1$ sends filtered colimits in the first argument to cofiltered limits. (**Edit, 10/10/20:** Excuse me, this is false. There is a $\lim^1$ term involving $\lim^1 \text{Hom}(\frac 1 n \mathbb{Z}/\mathbb{Z}, \mathbb{Z})$ which vanishes.) This gives that


$$\text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z}) \cong \widehat{\mathbb{Z}} \cong \prod\_p \mathbb{Z}\_p$$


is the profinite integers. Now it's very tempting to conjecture that the copy of $\mathbb{Z}$ we need is the obvious one, giving


$$\text{Ext}^1(\mathbb{Q}, \mathbb{Z}) \cong \widehat{\mathbb{Z}}/\mathbb{Z}.$$


But this description is somewhat unsatisfying, for the following reason: by functoriality $\text{Ext}^1(\mathbb{Q}, -)$ always takes values in $\mathbb{Q}$-modules, which is to say $\mathbb{Q}$-vector spaces, but we've written this $\mathbb{Q}$-vector space as a quotient of two things which are not $\mathbb{Q}$-vector spaces. We can rectify this a bit by tensoring the above by $\mathbb{Q}$, which fixes it, giving


$$\text{Ext}^1(\mathbb{Q}, \mathbb{Z}) \cong \left( \widehat{\mathbb{Z}} \otimes \mathbb{Q} \right) / \mathbb{Q}.$$


$\widehat{\mathbb{Z}} \otimes \mathbb{Q}$ has another name: it is the ring of finite rational adeles $\mathbb{A}\_{\mathbb{Q}}$. This is the form in which the answer is stated in [these notes](http://math.jhu.edu/~jmb/note/torext.pdf). This is a $\mathbb{Q}$-vector space of dimension the reals and so another amusing way to state the answer is that


$$\text{Ext}^1(\mathbb{Q}, \mathbb{Z}) \cong \mathbb{R}.$$


At this point we don't really need to know what copy of $\mathbb{Z}$ we need to quotient $\widehat{\mathbb{Z}}$ by.


Altogether we get, abstractly, that $\text{Ext}^1(\mathbb{R}/\mathbb{Z}, \mathbb{Z})$ is the product of $\widehat{\mathbb{Z}}$ and a $\mathbb{Q}$-vector space $\prod\_I \mathbb{R}$ of some very large dimension ($2^{2^{\aleph\_0}}$?).


Incidentally, although it's not needed for this computation, the same filtered colimit argument as above gives that


$$\text{Ext}^1(\mathbb{Z}(p^{\infty}), \mathbb{Z}) \cong \mathbb{Z}\_p.$$


