# The isomorphism class of $\mathrm{Ext}^1_\mathbb{Z}(\mathbb{R}/\mathbb{Z},\mathbb{Z})$

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.

• 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 $R$.
– YCor
Oct 18, 2017 at 20:31
• If we are trying to compute Ext with $\mathbb{Q}$ in the first argument, shouldn’t we take a <i>projective</i> resolution of $\mathbb{Q}$? Oct 18, 2017 at 20:59
• 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.
– YCor
Oct 18, 2017 at 21:14
• 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. Oct 18, 2017 at 22:37
• Ah sorry it's a typo: I indeed of course meant an injective resolution of $Z$.
– YCor
Oct 18, 2017 at 22:55

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

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

• 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. Oct 20, 2017 at 17:37
• It is already quite involved, but could you also briefly explain why does the $\lim^1$-term vanish? Oct 11, 2020 at 5:50
• Each of the individual groups in the $\lim^1$ is zero; nothing hard needed here. Oct 11, 2020 at 6:27
• 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 Oct 11, 2020 at 11:52
• I am saying something stronger: the input to the derived functor is zero. Even derived functors are still additive! Oct 11, 2020 at 17:53

$Ext(\mathbb Q/\mathbb Z,\mathbb Z)\cong \hat{\mathbb Z}$.

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 $\hat{\mathbb Z_p}$.

• 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? Oct 19, 2017 at 5:57

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.