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From Hatcher, page 196, before corollary 3.3.

We are first given these three properties of the Ext functor:

$\text{Ext}(H \oplus H', G) \cong \text{Ext}(H, G)\oplus\text{Ext}(H′, G)$.

$\text{Ext}(H, G) = 0$ if $H$ is free.

$\text{Ext}(Z_n, G) \cong G/nG.$

Then, Hatcher writes "these three properties imply that $\text{Ext}(H, \mathbb{Z})$ is isomorphic to the torsion subgroup of $H$ if $H$ is finitely generated".

I'm trying to understand why that is.

I know a torsion subgroup is the subgroup composed of all elements of finite order. Also, a finitely generated abelian group $H$ is the direct sum of a free abelian group (which I'll call $H'$) and its torsion subgroup (which I'll call $T$).

So $$\text{Ext}(H)=\text{Ext}(H'\oplus T)$$ From the first property, this equals $$\text{Ext}(H')\oplus\text{Ext}(T).$$ As $H'$ is free, this equals (from the second property) $$0\oplus \text{Ext(T)}=\text{Ext(T)}.$$

But why does $\text{Ext}(T)=T$ hold?

Eric Wofsey
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man_in_green_shirt
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    Hint: Look at the classification of finitely generated abelian groups. – user60589 Nov 28 '15 at 01:03
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    The isomorphism is not canonical, so writing "$\text{Ext}(T) = T$" is a bit dangerous. It's more accurate to say that $\text{Ext}^1(H, \mathbb{Z})$ is canonically isomorphic to the Pontryagin dual of the torsion subgroup of $H$ (for example, by looking at the long exact sequence induced by the short exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to S^1 \to 0$, and using the fact that $\mathbb{R}$ is injective). – Qiaochu Yuan Nov 28 '15 at 01:07
  • @user60589 A finitely generated abelian group can be written as $\mathbb{Z}^n\oplus\mathbb{Z}{q_1}\oplus ...\oplus\mathbb{Z}{q_t}$. What does that tell me? – man_in_green_shirt Nov 28 '15 at 01:10
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    @man_in_green_shirt Apply it to $H$, then use property 1), 2) and 3). – user60589 Nov 28 '15 at 01:11
  • @QiaochuYuan I don't know what the Pontryagin dual is...I'll look it up and see if I can understand something – man_in_green_shirt Nov 28 '15 at 01:12

2 Answers2

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The third property (with $G=\mathbb{Z}$) tells you that if $C$ is a finite cyclic group, then $\operatorname{Ext}(C,\mathbb{Z})\cong C$. Furthermore, any finitely generated torsion abelian group is a direct sum of cyclic groups. So $T$ is a direct sum of cyclic groups, so by the first and third properties, $\operatorname{Ext}(T,\mathbb{Z})\cong T$. Note, however, that the isomorphism in the third property is not canonical, so while $\operatorname{Ext}(H,\mathbb{Z})$ is isomorphic to $T$, it is not canonically isomorphic to $T$.

Eric Wofsey
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    Ok, so $T$ is isomorphic to $Z_{q_1}\oplus...\oplus Z_{q_t}$, where we ignore the term $Z^n$ because if it were present, then there would be terms with infinite order. So then $\text{Ext}(Z_{q_1}\oplus...\oplus Z_{q_t})\cong\text{Ext}(Z_{q_1})\oplus...\oplus \text{Ext}(Z_{q_t})\cong Z_{q_1}\oplus...\oplus Z_{q_t}\cong T$.

    Thanks again, really appreciate it

     – man_in_green_shirt Nov 28 '15 at 01:25
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To elaborate on my comment, applying $\text{Ext}^{\bullet}(A, -)$ to the short exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to S^1 \to 0$ produces the long exact sequence

$$0 \to \text{Hom}(A, \mathbb{Z}) \to \text{Hom}(A, \mathbb{R}) \to \text{Hom}(A, S^1) \to \text{Ext}^1(A, \mathbb{Z}) \to 0$$

where we can ignore the rest of the sequence because $\text{Ext}^1(A, \mathbb{R})$ vanishes, thanks to the fact that $\mathbb{R}$ is divisible and hence injective. (So is $S^1$; in fact what we've written down above is an injective resolution of $\mathbb{Z}$.)

If $A$ is torsion, then $\text{Hom}(A, \mathbb{R}) = 0$, and the long exact sequence above produces a natural isomorphism

$$\text{Hom}(A, S^1) \cong \text{Ext}^1(A, \mathbb{Z}).$$

The group $\text{Hom}(A, S^1)$ is known as the Pontryagin dual $\widehat{A}$ of $A$. If $A$ is finite it is noncanonically isomorphic to $A$. In general it naturally has a topology making it a profinite abelian group, and every profinite abelian group arises in this way.

Qiaochu Yuan
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  • So from the SES $0\to \mathbb{Z}\to\mathbb{R}\to S^1$ we get the LES $0\to\text{Hom}(A,\mathbb{R})\to\text{Hom}(A,S^1)\to\text{Ext}^1(A,\mathbb{Z}) $ $\to\text{Ext}^1(A,\mathbb{R})\to\text{Ext}^1(A,S^1)\to \text{Ext}^2(A,\mathbb{Z}) \to \text{Ext}^2(A,\mathbb{R}) \to \text{Ext}^2(A,S^1)\to ...$

    An abelian group is divisible iff it is an injective object in the category of abelian groups. As $\mathbb{R}$ is divisible, it's injective. For any injective object $B$, $\text{Ext}^i(A,B)=0$ for $i=0$, hence ignore the part of the sequence from $\text{Ext}^1(A,\mathbb{R})$ onwards. As $A$ is torsion

     – man_in_green_shirt Nov 28 '15 at 11:40
  • (note: in the previous comment, I meant $i\neq 0$, and the SES was supposed to have $\to 0$ at the end)

    and $\mathbb{R}$ is torsion free, $\text{Hom}(A,\mathbb{R})=0$, from which the canonical isomorphism follows. Thanks for expanding on your comment, as when I tried reading up on the Pontryagin dual yesterday I didn't understand much.

     – man_in_green_shirt Nov 28 '15 at 11:47
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    @man_in_green_shirt: yes, that's right. – Qiaochu Yuan Nov 28 '15 at 19:58
  • Also, everything I've said above goes through if we replace $\mathbb{R}$ with $\mathbb{Q}$ and $S^1$ with $\mathbb{Q}/\mathbb{Z}$, and in particular $\text{Hom}(A, \mathbb{Q}/\mathbb{Z})$ (which is sometimes also called the Pontryagin dual) is still the Pontryagin dual of $A$ if $A$ is torsion. – Qiaochu Yuan Nov 29 '15 at 08:20