A homework problem asked to find a short exact sequence of abelian groups
$$0 \rightarrow A \longrightarrow B \longrightarrow C \rightarrow 0$$ such that $B \cong A \oplus C$ although the sequence does not split. My solution to this is the sequence
$$0 \rightarrow \mathbb{Z} \overset{i}{\longrightarrow}
\mathbb{Z} \oplus (\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}} \overset{p}{\longrightarrow}
(\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}} \rightarrow 0$$
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:
- Is there an example with finite/finitely generated abelian groups?
- If the answer to (1) is negative, will it help to pass to general $R$-modules for some ring $R$?
Best Answer
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
$$0\rightarrow\mathrm{Hom}(C,A)\rightarrow\mathrm{Hom}(B,A)\rightarrow \mathrm{Hom}(A,A).$$
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.