Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $R$?
share|cite|improve this question
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.

share|cite|improve this answer
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.

share|cite|improve this answer
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: \begin{equation} 0 \longrightarrow \mathbb{Z} \stackrel{f}{\longrightarrow} \mathbb{Z} \oplus \mathbb{Z} \stackrel{g}{\longrightarrow} \mathbb{Z} \longrightarrow 0, \end{equation} 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 \begin{equation} \ker(g) = \{ (a,b) \in \mathbb{Z}^{2} \,|\, 3a - 2b = 0 \} = \text{im}(f). \end{equation} 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} $: \begin{equation} \forall (a,b) \in \mathbb{Z}^{2}: \quad s(a,b) := 2a - b. \end{equation} 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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.