# Nonspliting short exact sequence

The short exact sequence $0\rightarrow \mathbb Z \stackrel{\alpha}{\longrightarrow} \mathbb Z \oplus \mathbb Q \stackrel{\beta} {\longrightarrow} \mathbb Q \rightarrow 0$ is splits because we have split map $\gamma: \mathbb Q\rightarrow \mathbb Z \oplus \mathbb Q$ defined by $\gamma(q) = (0,q)$ such that $\gamma\circ\beta$ is identity on $Q.$ My question here is if we have $0\rightarrow \mathbb Z \stackrel{\alpha}{\longrightarrow} G \stackrel{\beta} {\longrightarrow} \mathbb Q \rightarrow 0,$ then what is $G$? One of the possibility is $\mathbb Z \oplus \mathbb Q.$ The sequence is not necessarily split because we do not know exactly whether $G$ has a copy of $\mathbb Q$ or copy of $\mathbb Z.$ Is there any source where I can get more information about it? Thank you.

• arxiv.org/abs/1310.3488 . In a nutshell, there is myriad such extensions. Jun 12 '15 at 21:35
• @ arij grinberg Thank you. I will look.
– CAA
Jun 12 '15 at 22:06

## 1 Answer

The short exact sequences $0\to\mathbb{Z}\to G\to\mathbb{Q}\to0$ are classified by the group $\operatorname{Ext}(\mathbb{Q},\mathbb{Z})$. Two exact sequences define the same element in $\operatorname{Ext}(\mathbb{Q},\mathbb{Z})$ if and only if there exists a homomorphism $G\to G'$ making the diagram $$\require{AMScd} \begin{CD} 0 @>>> \mathbb{Z} @>>> G @>>> \mathbb{Q} @>>> 0 \\ @. @| @VVV @| \\ 0 @>>> \mathbb{Z} @>>> G' @>>> \mathbb{Q} @>>> 0 \end{CD}$$ commutative.

Now $\operatorname{Ext}(\mathbb{Q},\mathbb{Z})$ can be computed from an injective resolution of $\mathbb{Z}$, for instance $0\to\mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}\to0$, applying the $\operatorname{Hom}(\mathbb{Q},-)$ functor and obtaining the long exact sequence \begin{multline} 0\to\operatorname{Hom}(\mathbb{Q},\mathbb{Z}) \to\operatorname{Hom}(\mathbb{Q},\mathbb{Q}) \to\operatorname{Hom}(\mathbb{Q},\mathbb{Q}/\mathbb{Z})\\ \to\operatorname{Ext}(\mathbb{Q},\mathbb{Z}) \to\operatorname{Ext}(\mathbb{Q},\mathbb{Q}) \to\operatorname{Ext}(\mathbb{Q},\mathbb{Q}/\mathbb{Z}) \to0 \end{multline} Since $\operatorname{Hom}(\mathbb{Q},\mathbb{Z})=0$, $\operatorname{Hom}(\mathbb{Q},\mathbb{Q})\cong\mathbb{Q}$ and $\operatorname{Ext}(\mathbb{Q},\mathbb{Q})=0$ (because $\mathbb{Q}$ is injective) this boils down to the exact sequence $$0\to\mathbb{Q} \to\operatorname{Hom}(\mathbb{Q},\mathbb{Q}/\mathbb{Z}) \to\operatorname{Ext}(\mathbb{Q},\mathbb{Z})\to0$$ It's easy to see that each of these groups is in fact a vector space over $\mathbb{Q}$, but the middle group is huge.

A way for seeing this is to consider again the exact sequence above and applying to it the functor $\operatorname{Hom}(-,\mathbb{Q}/\mathbb{Z})$, getting the exact sequence $$0\to\operatorname{Hom}(\mathbb{Q}/\mathbb{Z},\mathbb{Q}/\mathbb{Z}) \to\operatorname{Hom}(\mathbb{Q},\mathbb{Q}/\mathbb{Z}) \to\operatorname{Hom}(\mathbb{Z},\mathbb{Q}/\mathbb{Z}) \to0$$ that can be rewritten as $$0\to\prod_p \mathbb{Z}_p \to\operatorname{Hom}(\mathbb{Q},\mathbb{Q}/\mathbb{Z}) \to\mathbb{Q}/\mathbb{Z}\to0$$ where the product runs over all prime numbers $p$ and $\mathbb{Z}_p$ is the ring of $p$-adic integers, which has the same cardinality as $\mathbb{R}$. So also $\operatorname{Ext}(\mathbb{Q},\mathbb{Z})$ has the same cardinality.

Any book on homological algebra will have the relevant information. Note that in general the long exact sequence does not stop and one has to consider $\text{Ext}^k$, but in the case of abelian groups the higher order Ext groups vanish.

• @ egreg Thank you so much. I got it. This is an awesome answer.
– CAA
Jun 12 '15 at 22:56