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Show that an exact sequence of abelian groups $$0 \rightarrow \mathbb{Q} \rightarrow M \rightarrow N \rightarrow 0$$ always splits.

If $f: \mathbb{Q} \rightarrow M$ and $g: M \rightarrow N$, what needs to be found is a map $h: N \rightarrow M$ such that $gh=1_N$. My first question is simply this: isn't that the same as requiring that $g$ is an isomorphism?

Secondly, I have no idea how to even begin here. Could someone give a hint or two? Thanks.

Erik Vesterlund
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  • $g$ needn't be an isomorphism, but it must be a surjection for there to be such an $h$ – ant Sep 29 '13 at 07:33

1 Answers1

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This is no direct answer to your question, but some sidenotes to put it in a more general context. I'd like to write it as a comment, but this would be way too long. Maybe you are interested in understanding the background why your claim is true, so this could be intersting for you.

First note, that abelian groups are precisly the modules over the integers $\mathbb{Z}$.

There is the notion of injective modules, which plays a big role in e.g. homological algebra and is definitly worth to get used to it. The second equivalent defintion in the Wikipedia-article shows you, that you're question reduces to: Is $\mathbb{Q}$ an injective abelian group ($\mathbb{Z}$-module)?

With a little effort (essentially using a lemma called "Baers criterion". Be aware that you need Zorns Lemma, respectivly the axiom of choice to prove it) one can show, that an abelian group is injective, iff it's divisible. If you take a look at the definition of divisibility, $\mathbb{Q}$ is clearly divisible, so this shows your question.

Of course this way is maybe away from beeing as straight-forward as possible to answer your question, but I suggest you to try to understand this way for two reasons: Firstly, you will get to know concepts, which are very important in algebra (Of course, if you just want to get your exercices done, this doesn't play a role for you). Secondly if you will take a straight-forward way, you will essentially need most of the arguments, which you will need for the proofs in the more generel way I suggested (At least I think there is no way to prove your claim without lemma of zorn, etc.), so why don't take the more conceptual way?

archipelago
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  • I agree that it's a good idea to think more generally, but you could have possible gone half-way. Namely, it's not hard to show that every divisible group $Q$ satisfies the following "injectivity property": if $A\subseteq M$ (where $M$ is any other abelian group) then $A$ is a direct summand of $M$ (i.e. there exists $K\subseteq M$ such that $M=Q\oplus K$). This can be proven easily for divisible groups by a Zorn's lemma argument. You can then easily dedue the OPs problem from this. Namely, since $\mathbb{Q}$ is divisible, and $f(\mathbb{Q})\cong \mathbb{Q}$ (where $f:\mathbb{Q}\to M$) – Alex Youcis Sep 29 '13 at 08:41
  • we have that $f(\mathbb{Q})$ is divisible and so injective. Thus, since $f(\mathbb{Q})\leqslant M$, there exists some $K\leqslant M$ such that $M=f(\mathbb{Q})\oplus K$. Then, the map $M\to \mathbb{Q}$ defined by $(f(q),x)\mapsto q$, is clearly a splitting map for the sequence. – Alex Youcis Sep 29 '13 at 08:43
  • @archipelago Although I haven't reached the section on injective modules yet I know the notion and thought that could provide the solution. However I didn't read very carefully, so I missed it. Your answer is complete, thank you :) – Erik Vesterlund Sep 30 '13 at 21:03