Does every short exact sequence $0\to \mathbb Z\to A \to \mathbb R \to 0$ split in the category of Abelian groups?
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3$\begingroup$ This is the same as asking whether $Ext^1(R,Z)=0$, and since $R=Q^{(c)}$ and on the left Ext^1 commutes with direct sums, this is the same as asking whether $Ext^1(Q,Z)=0$, i.e., the same question with $R$ replaced with $Q$. Unless you only consider continuous exact sequences only, in which cases it's an easy yes. $\endgroup$– YCorCommented Aug 25, 2020 at 14:49
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2$\begingroup$ The answer is then no: actually this $Ext^1$ is uncountable if I'm correct. Basically, the argument is that in $Q^2/Z^2$, there are continuum many copies of $Q/Z$, but only countably many, when pulled back to $Q^2$, correspond to a direct decomposition. $\endgroup$– YCorCommented Aug 25, 2020 at 14:51
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$\begingroup$ @YCor I though that Ext on the left takes direct sums to direct products. Nevertheless, the question still reduces to whether $\text{Ext}^1(\mathbb Q,\mathbb Z)=0$. $\endgroup$– Andreas BlassCommented Aug 25, 2020 at 14:53
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6$\begingroup$ Another way to see that $Ext^1(Q,Z)$ is big is to look at the long exact sequence arising from $0\to Z\to Q\to Q/Z\to 0$. It includes $\dots\to Hom(Q,Q)\to Hom(Q,Q/Z)\to Ext(Q,Z)\to Ext(Q,Q)\to\dots$. Here $Hom(Q,Q)\cong Q$ is countable, $Hom(Q,Q/Z)$ has the cardinality of the continuum, and $Ext(Q,Q)=0$. So $Ext(Q,Z)$ has the cardinality of the continuum. $\endgroup$– Andreas BlassCommented Aug 25, 2020 at 14:57
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2$\begingroup$ Semiconcretely, let $A$ be the ring of adeles; this is the subring of $\prod_p \mathbb{Q}_p$ consisting of $(\alpha_p)$ such that, for all but finitely many $p$, we have $\alpha_p \in \mathbb{Z}_p$. Then $A/\prod_p \mathbb{Z}_p \cong Q/Z$. Given any $\alpha \in \mathbb{A}$, the map $q \mapsto \alpha \cdot q + \prod_p \mathbb{Z}_p$ is a homorphism $Q \to Z/\prod_p \mathbb{Z}_p \cong Q/Z$. Now I need to remember how the boundary map from Hom to Ext works. $\endgroup$– David E SpeyerCommented Aug 25, 2020 at 15:14
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The calculation of $\text{Ext}^1(\mathbb{Q}, \mathbb{Z})$ can be found in this MO answer; in terms of just its isomorphism type the conclusion is that it's an uncountable-dimensional vector space over $\mathbb{Q}$, abstractly isomorphic to $\mathbb{R}$. It can also be written as a quotient $\mathbb{A}_{\mathbb{Q}}/\mathbb{Q}$ where $\mathbb{A}_{\mathbb{Q}} \cong \hat{\mathbb{Z}} \otimes \mathbb{Q}$ is the finite rational adeles.
answered Aug 25, 2020 at 20:35
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$\begingroup$ I believe but don't know how to prove that the relevant copy of $\mathbb{Q}$ sitting inside $\mathbb{A}_{\mathbb{Q}}$ is the obvious one (spanned by the identity), but fortunately it doesn't matter here. $\endgroup$ Commented Aug 25, 2020 at 21:51