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I'm calculating the $\text{Ext}^1_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Z}, \mathbb{Z})$. In particular, $\text{Ext}^1_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Z}, \mathbb{Z}) \cong \text{Hom}_{\mathbb{Z}}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})$ the Pontragin dual. However how should I see what this group is? I only know that for finite abelian groups on the first position, this dual is isomorphic to it self. But no idea for infinitely generated groups.

nekodesu
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2 Answers2

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Throughout $\Bbb Q_p$ and $\Bbb Z_p$ denote the $p$-adic numbers and $p$-adic integers.

Observe that $$\Bbb Q/\Bbb Z\cong\bigoplus_p \Bbb Q_p/\Bbb Z_p.$$ Thus $$\textrm{Hom}(\Bbb Q/\Bbb Z,\Bbb Q/\Bbb Z) \cong\prod_p \textrm{Hom}(\Bbb Q_p/\Bbb Z_p,\Bbb Q/\Bbb Z).$$ Any homomorphism from $\Bbb Q_p/\Bbb Z_p$ to $\Bbb Q/\Bbb Z$ has image within $\Bbb Q_p/\Bbb Z_p$ and so $$\textrm{Hom}(\Bbb Q/\Bbb Z,\Bbb Q/\Bbb Z) \cong\prod_p \textrm{Hom}(\Bbb Q_p/\Bbb Z_p,\Bbb Q_p/\Bbb Z_p).$$ Each endomorphism of $\Bbb Q_p/\Bbb Z_p$ is induced by multiplication by an element of $\Bbb Z_p$. Therefore $$\textrm{Hom}(\Bbb Q/\Bbb Z,\Bbb Q/\Bbb Z) \cong\prod_p \Bbb Z_p.$$ This group is often denoted as $\hat {\Bbb Z}$, the profinite completion of $\Bbb Z$. It's also the absolute Galois group of each finite field.

Angina Seng
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Slightly different argument: Note that $\mathrm{Hom}_\mathbb{Z}(\varinjlim_i A_i, B) \simeq \varprojlim_i\mathrm{Hom}_\mathbb{Z}(A_i, B)$ (this is just restating of the universal property for $\varinjlim:$ the left hand side consists of morphisms $\varinjlim_i A_i \rightarrow B$, the right hand side describes co-cones $A_i \rightarrow B, i \in I, $ of compatible systems of maps).

Then, as $\mathbb{Q}/\mathbb{Z} = \varinjlim_{n | m} \mathbb{Z}[\frac{1}{n}]/\mathbb{Z},$ we have

$$\mathrm{Hom}_\mathbb{Z}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})=\mathrm{Hom}_\mathbb{Z}(\varinjlim_{n | m} \mathbb{Z}[\frac{1}{n}]/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \simeq \varprojlim_{n | m}\mathrm{Hom}_\mathbb{Z}(\mathbb{Z}[\frac{1}{n}]/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})\simeq \varprojlim_{n | m} \mathbb{Z}/n\mathbb{Z}=\widehat{\mathbb{Z}}.$$

Pavel Čoupek
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  • The first isomorphism for topological groups needs some care, see the comments here : https://math.stackexchange.com/questions/1036377 – Watson Nov 26 '18 at 10:35
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    @Watson: The comment there is concerning the "true" Pontrjagin dual, i.e .$\mathrm{Hom}_{cont}$. This one should be, if I understand the OP's motivation correctly, "discrete" hom, i.e. the one showing up in long exact sequences for $\mathrm{Ext}$. So the first isomorphism is used for abstract groups. (And sure, you can endow the result with the profinite topology e.g. by the last isomorphism; but that does not affect the computations before.) – Pavel Čoupek Nov 28 '18 at 14:22