From splitting lemma, we know in $R$-Mod Category, short exact sequence $0 \to A \stackrel{f}{\rightarrow} B \stackrel{g}{\rightarrow} C \to0$ splits if it satisfies one of the following equivalent conditions:

$(1)\ \exists f_1\in\text{Hom}(B,A)\text{ s.t. } f_1\circ f=\text{Id}_A$. $(2)\ \exists g_1\in\text{Hom}(C,B)\text{ s.t. } g\circ g_1=\text{Id}_C$.

$(3)\ \text{Im }f=\text{Ker }g$ is direct summand of $B$.

$(4)\ \exists \text{ isomorphism } h:B\to A \oplus C \text{ s.t. } $

$h \circ f \text{ is natural injection and }g \circ h^{-1} \text{ is natural projection.}$

And $0 \to A \stackrel{f}{\rightarrow} B \stackrel{g}{\rightarrow} C \to0$ splits $\implies B\cong A \oplus C$.

If we only have $B\cong A \oplus C$, is there any example for $0 \to A \to B \to C \to0$ doesn't split?

Related questions:

$(1)$ Example for infinitely generated modules.

$(2)$ Example for abelian groups.

As proved in answer below, it's ture for modules on commutative ring with finite length.

Special thanks for jgon, for his knowledge, time, patience and friendliness.

  • $\begingroup$ You mean, there exists an isomorphism, but it just doesn't satisfy the injection/projection conditions? $\endgroup$
    – rschwieb
    Commented Dec 13, 2018 at 14:40
  • $\begingroup$ Possible duplicate of A nonsplit short exact sequence of abelian groups with $B \cong A \oplus C$ $\endgroup$ Commented Dec 13, 2018 at 14:42
  • $\begingroup$ What do you mean by "drop the condition"? $\endgroup$ Commented Dec 13, 2018 at 14:57
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    $\begingroup$ I'm confused. Are you looking for another answer? How is this not answered by your linked question (2) (after all, abelian groups are $\Bbb{Z}$-modules)? Your answer doesn't prove that short exact sequences of finitely generated modules split when the middle module is isomorphic to the direct sum. It proves that they split when the modules have finite length. Finite generation is equivalent to finite length only when $R$ is Artinian. Also the answer to the linked question gives an example of a sequence of f.g. modules over a noncommutative ring that doesn't split. $\endgroup$
    – jgon
    Commented Mar 27, 2019 at 16:25
  • $\begingroup$ @jgon Thank you for your edit and kindly explaination, and sorry for my mistake. The linked questions completely answered my question. The answer below surely proves they split when the modules have finite length. I could't open that link when I edit hours ago so I made a big mistake. Thank you for your time and patience. $\endgroup$
    – Andrews
    Commented Mar 27, 2019 at 20:39

1 Answer 1


Let $R$ be a commutative ring and let $$\begin{equation}\label{eq:ses}0\to A\to B\to C\to 0\tag{*}\end{equation}$$ be a short exact sequence of modules of finite length over $R$. We show that if $B\cong A\oplus C$, then $\eqref{eq:ses}$ splits.

Since $B\cong A\oplus C$, we have that $$\newcommand\Hom{\operatorname{Hom}}\Hom_R(C,B)\cong\Hom_R(C,A)\oplus\Hom_R(C,C).$$ Hence $$\ell(\Hom_R(C,B))=\ell(\Hom_R(C,A))+\ell(\Hom_R(C,C))$$ where $\ell$ denotes length.

Write $f:B\to C$ for the map in $\eqref{eq:ses}$. Apply $\Hom_R(C,-)$ to the sequence $\eqref{eq:ses}$.

From the left exactness of $\Hom_R(C,-)$, we obtain the exact sequence $$0\to \Hom_R(C,A)\to \Hom_R(C,B)\to \Hom_R(C,C),$$ where the map $f_* : \Hom_R(C,B)\to \Hom_R(C,C)$ is given by $g\mapsto f\circ g$.

Let $N$ denote the cokernel of $f_*$, which evidently has finite length. It follows that $$0\to \Hom_R(C,A)\to \Hom_R(C,B)\to \Hom_R(C,C)\to N \to 0$$ is exact, and hence that $\ell(N)=0$.

Hence $f_*$ is surjective, which shows that the sequence $\eqref{eq:ses}$ splits.

Here ring $R$ is commutative to make Hom$_R(C, - )$ a $R$-module.

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    $\begingroup$ The proof only needs that the hom spaces have finite length, so the result holds in any hom-finite abelian category. In particular, this includes coherent sheaves on a projective $k$-scheme, for a field $k$. $\endgroup$ Commented Mar 27, 2019 at 21:42

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