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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:

  1. What is the isomorphism class of $\mathrm{Ext}^1_\mathbb{Z}(\mathbb{Q},\mathbb{Z})$?
  2. 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.

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    $\begingroup$ Using an injective resolution of $Q$, one gets that $Ext^1(Q,Z)$ is the cokernel of $Q=Hom(Q,Q)\to Hom(Q,Q/Z)$. I guess that $Hom(Q,Q/Z)$ is isomorphic to the adeles (restricted product of $Q_p$ for all $p$). So this $Ext^1(Q,Z)$ would be the quotient of the pontryagin dual of $Q$ (a complicated compact connected group) by a dense subgroup isomorphic to the reals $R$. $\endgroup$
    – YCor
    Oct 18, 2017 at 20:31
  • $\begingroup$ If we are trying to compute Ext with $\mathbb{Q}$ in the first argument, shouldn’t we take a <i>projective</i> resolution of $\mathbb{Q}$? $\endgroup$
    – anonymous
    Oct 18, 2017 at 20:59
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    $\begingroup$ For $Ext(M,N)$, use a projective resolution of $M$ and apply $Hom(-,N)$ or an injective resolution of $N$ and apply $Hom(M,-)$. I made the second choice. $\endgroup$
    – YCor
    Oct 18, 2017 at 21:14
  • $\begingroup$ But if that were the case, you would be taking an injective resolution of the second argument—$\mathbb{Z}$—and then applying the covariant hom functor $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Q},-)$ to the result. Or am I missing something? Edit: em dashes. $\endgroup$
    – anonymous
    Oct 18, 2017 at 22:37
  • $\begingroup$ Ah sorry it's a typo: I indeed of course meant an injective resolution of $Z$. $\endgroup$
    – YCor
    Oct 18, 2017 at 22:55

3 Answers 3

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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. 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

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

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  • $\begingroup$ It may be a bit surprising that $\widehat{\mathbb{Z}}/\mathbb{Z}$ is a $\mathbb{Q}$-vector space. Here is the idea, as far as I can tell: every element is $n$-divisible because it maps to some element in $\mathbb{Z}/n\mathbb{Z}$ which you can then lift to $\mathbb{Z}$ and subtract, and every element is uniquely divisible because $\mathbb{Z}$ is torsion-free. $\endgroup$ Oct 20, 2017 at 17:37
  • $\begingroup$ It is already quite involved, but could you also briefly explain why does the $\lim^1$-term vanish? $\endgroup$ Oct 11, 2020 at 5:50
  • $\begingroup$ Each of the individual groups in the $\lim^1$ is zero; nothing hard needed here. $\endgroup$ Oct 11, 2020 at 6:27
  • $\begingroup$ Sorry for being overcautious but in principle it may happen that the value of a functor on an object is zero but the value of its derived functor on that object is nonzero $\endgroup$ Oct 11, 2020 at 11:52
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    $\begingroup$ I am saying something stronger: the input to the derived functor is zero. Even derived functors are still additive! $\endgroup$ Oct 11, 2020 at 17:53
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$Ext(\mathbb Q/\mathbb Z,\mathbb Z)\cong \hat{\mathbb Z}$.

The exact sequence $0\to \mathbb Z\to \mathbb Q\to \mathbb Q/\mathbb Z\to 0$ is an injective resolution of $\mathbb Z$, so the answer is the cokernel of $Hom(\mathbb Q/\mathbb Z,\mathbb Q)\to Hom(\mathbb Q/\mathbb Z,\mathbb Q/\mathbb Z)$, i.e. the cokernel of $0\to \hat{\mathbb Z}$.

$Ext(\mathbb Q,\mathbb Z)\cong\hat{\mathbb Z}/\mathbb Z$, by the exact sequence $$ 0\to Hom(\mathbb Z,\mathbb Z)\to Ext(\mathbb Q/\mathbb Z,\mathbb Z)\to Ext(\mathbb Q,\mathbb Z)\to 0. $$ (Zeroes at the ends are $Hom(\mathbb Q,\mathbb Z)$ and $Ext(\mathbb Z,\mathbb Z)$.)

Replace $\mathbb Q$ throughout by the pre-image of the $p$-torsion part of $\mathbb Q/\mathbb Z$, and you get the answer to the third question. The endomorphism ring of the $p$-part of $\mathbb Q/\mathbb Z$ is $\hat{\mathbb Z_p}$.

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    $\begingroup$ It's a little ambiguous what "it's" means here since the OP asks a few questions; it looks to me like this is a computation of $\text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z})$, is that right? $\endgroup$ Oct 19, 2017 at 5:57
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For context: the group $LA=\text{Ext}(\mathbb{Z}/p^\infty,A)$ is called the derived $p$-completion of $A$. It has a natural map to the ordinary completion $CA=\lim_n A/p^nA$ which is often an isomorphism. In particular it is an isomorphism if $A$ is a free abelian group, or if it is finitely generated, or if there exists $n$ such that $\text{ann}(p^n,A)=\text{ann}(p^{n+1},A)$. In the cases where $CA\neq LA$ it typically works out that $LA$ has better behaviour and is more relevant for applications, especially in algebraic topology and homological algebra. One reference is the book "Homotopy Limits, Completion and Localization" by Bousfield and Kan.

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