# A nonsplit short exact sequence of abelian groups with $B \cong A \oplus C$

A homework problem asked to find a short exact sequence of abelian groups $$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$ such that $$B \cong A \oplus C$$ although the sequence does not split.

My solution: $$0 \rightarrow \mathbb{Z} \overset{i}{\rightarrow} \mathbb{Z} \oplus (\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}} \overset{p}{\rightarrow} (\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}} \rightarrow 0$$ with $i(x)=(2x,0,0,\dotsc)$ and $p(x,y_1,y_2,\dotsc)=(x+2\mathbb{Z},y_1,y_2,\dotsc)$.

My new questions:

1. Is there an example with finite/finitely generated abelian groups?
2. If the answer to $(1)$ is negative, will it help to pass to general $R$-modules for some ring $R$?
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FYI: A different example, still with infinite groups, is Example 6.35 of Rotman's Advanced Modern Algebra (2nd ed., AMS). – KCd Apr 22 '12 at 21:55

I'm sorry, but my answer uses Ext groups, which may not be in the scope of your course.

I don't know of a simple example with finitely generated abelian groups for the following reason: if $0 \rightarrow \mathbb Z \rightarrow E \rightarrow \mathbb Z / n \mathbb Z \rightarrow 0$ is an extension represented by $r+n \mathbb Z \in \operatorname{Ext}^1(\mathbb Z,\mathbb Z/n\mathbb Z)\cong \mathbb Z/n\mathbb Z$, then one can show that $E \cong \mathbb Z \oplus \mathbb Z/d\mathbb Z$ where $d$ is the highest common factor of $r$ and $n$. If you assume this sequence doesn't split, then $r$ is not a multiple of $n$ and $d<n$, so the middle term will never be isomorphic to the direct sum of the of the first and third.

However, you can get a simple example using modules in the following way. Let $G=\langle g\rangle$ be an infinite cyclic group and let $\mathbb Z G$ be the associated commutative group ring. Consider $0 \rightarrow \mathbb Z \rightarrow \mathbb Z \oplus \mathbb Z \rightarrow \mathbb Z \rightarrow 0$ a short exact sequence of $\mathbb Z G$-modules where the first map is inclusion into the first component and the second map is projection onto the second. Let $g$ act on $\mathbb Z \oplus \mathbb Z$ by $\bigl(\begin{smallmatrix} 1&1\\ 0&1 \end{smallmatrix} \bigr)$. Then $\operatorname{Ext}^1_{\mathbb Z G}(\mathbb Z, \mathbb Z) \cong \mathbb Z$ and this extension corresponds to $1$. If you let $g$ act by $\bigl(\begin{smallmatrix} 1&n\\ 0&1 \end{smallmatrix} \bigr)$ it corresponds to $n$. As long as $n>0$ this is a non-split short exact sequence where the direct sum of the outside terms are isomorphic to the middle one. However, this example is far from elementary, and I apologize for that.

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I should add that there is a 1-1 correspondence between elements of $Ext^1(C,A)$ and short exact sequences $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ and this sequence splits iff it corresponds to the $0$ element of the Ext group. – Parsa Apr 22 '12 at 22:50
This looks great, but I don't yet have the tools to understand it. When I do, I will verify it's correct and (probably) accept it. Thank you very much! – user3533 Apr 30 '12 at 20:34
In your SES of $\mathbb{Z}G$-modules, the middle term is isomorphic to the direct sum of the other two as groups, but not as $\mathbb{Z}G$-modules: $g$ acts trivially on the left and right copies of $\mathbb{Z}$, but not on the middle $\mathbb{Z}\oplus\mathbb{Z}$. – Julian Rosen Dec 8 '15 at 5:02
I don't think the last example is valid. You should study the split-ness of $0\to\mathbb Z\to\mathbb Z\oplus\mathbb Z\to\mathbb Z\to0$ in the Abelian category of Abelian groups, not in the category of $\mathbb ZG$-modules. – Frank Science Jul 2 at 3:58

There is no counterexample with $A,B,C$ finitely generated abelian groups. There is, more generally, no counterexample with $A,B,C$ finitely generated modules over any noetherian ring $R$.

To see this, consider the exact sequence

$$0\rightarrow Hom(C,A)\rightarrow Hom(B,A)\rightarrow Hom(A,A)$$

The original sequence splits if and only if this sequence is exact on the right. If $A$, $B$ and $C$ are of finite length as modules, this follows immediately just by counting lengths. Otherwise, it's enough to show exactness after localizing and completing at an arbitrary prime $P$, and for this it's enough to show exactness after tensoring with $R/P^n$, and for this you can assume the lengths are finite, which is the case we've already dealt with.

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Sorry, I'm quite unfamiliar with techniques on localization and completion you're referring to. Do you have any reference? – Frank Science Jul 1 at 13:48
@FrankScience: Any standard textbook on commutative algebra will do. My favorite is Atiyah-MacDonald. – WillO Jul 1 at 21:45
I didn't read last chapters of Atiyah-Macdonald (that involves things like Artin-Rees). By localization and completion, I mean using this technique to reduce general cases to Artinian cases, which I cannot remember that appears in Atiyah-Macdonald. – Frank Science Jul 2 at 3:49
And on this specific problem, we know that exactness is stalk-local, and I believe that under some finiteness condition (like Noetherian, finitely presented, etc), split-ness is also stalk-local (since we have some base-change property of $\operatorname{Ext}$, maybe one can find some elementary proof, say passage from stalks to a fundamental open set). But for completion, I know nothing, except that it's an exact functor (under some finiteness condition). – Frank Science Jul 2 at 3:52
@FrankScience : Once you've reduced to $R$ local, with maximal ideal $P$, the completion $\hat{R}$ is faithfully flat over $R$, which means that a sequence of $R$-modules is exact if and only if it becomes exact after tensoring with $\hat{R}$.. But $\hat{R}$ is an inverse limit of the $R/P^k$, and exactness behaves well with respect to inverse limits, so that exactness mod $P^k$ for each $k$ (which follows from counting lengths) implies exactness over the completion, as needed. – WillO Jul 2 at 16:41

I believe that the answer to Question 1 is 'yes'.

Consider the following sequence of finitely generated abelian groups: $$0 \longrightarrow \mathbb{Z} \stackrel{f}{\longrightarrow} \mathbb{Z} \oplus \mathbb{Z} \stackrel{g}{\longrightarrow} \mathbb{Z} \longrightarrow 0,$$ where $f(n) := (2n,3n)$ and $g(a,b) := 3a - 2b$ for all $a,b,n \in \mathbb{Z}$.

Firstly, observe that $f$ is injective. Secondly, as $\text{gcd}(2,3) = 1$, we see that $g$ is surjective. Lastly, notice that $$\ker(g) = \{ (a,b) \in \mathbb{Z}^{2} \,|\, 3a - 2b = 0 \} = \text{im}(f).$$ With these three facts established, we conclude that the sequence above is short exact. However, the sequence does not split.

Erratum: I just realized that the last sentence is totally wrong. We have the following splitting map $s$ from $\mathbb{Z} \oplus \mathbb{Z}$ to the first $\mathbb{Z}$: $$\forall (a,b) \in \mathbb{Z}^{2}: \quad s(a,b) := 2a - b.$$ A quick way of seeing that the sequence splits is to note that it ends in $\mathbb{Z}$, which is free. I thank the users for their comments.

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Since $\mathbb{Z}$ is free, doesn't every short exact sequence ending in $\mathbb{Z}$ split? – Jason DeVito Sep 3 '12 at 12:20
split by $k \mapsto (k,k)$ – m_t_ Sep 3 '12 at 12:38