Let $R$ be a unital ring. Suppose that $A$, $B$, and $C$ are unitary left $R$-modules such that there exists a non-split exact sequence $$0\to A \overset{\alpha}{\longrightarrow}B\overset{\beta}{\longrightarrow}C\to 0.$$ If $A$ and $C$ are indecomposable $R$-modules, does it follow that $B$ is also indecomposable? (What if $R$ is non-unital or the modules are not necessarily unitary $R$-modules?)
Edit: As Jeremy Rickard shows, my work below is faulty due to a bad assumption (italicized in the text below). Because of this, the statement is not true even when $A$ and $C$ are modules of finite length.
Wrong Attempt
I know the answer if $A$ and $C$ have finite length. In that case, $B$ has finite length. Since $\operatorname{im}\alpha\cong A$ is an indecomposable submodule of $B$, there exists by the Krull-Schmidt theorem a direct sum decomposition $$B=B_1\oplus B_2\oplus\ldots \oplus B_n$$ of $B$, where each $B_i$ is indecomposable, such that $\operatorname{im}\alpha\subseteq B_1$. Then, $$C\cong (B_1/\operatorname{im}\alpha)\oplus B_2\oplus \ldots\oplus B_n.$$ By indecomposability of $C$, either $n=2$ and $B_1=\operatorname{im}\alpha$ which gives $C\cong B_2$, or $n=1$ and $B$ is indecomposable. However, in the case $n=2$ and $B_1=\operatorname{im}\alpha$, it follows that the exact sequence splits since we have a retraction from $B$ to $A$ (given by $B=B_1\oplus B_2\overset{\operatorname{proj}_1}{\ -\!\!\!-\!\!\!\twoheadrightarrow\ } B_1\overset{\alpha^{-1}}{\longrightarrow} A$). I am struggling to see whether this result extends to the infinite length cases.
Even for finite length modules, it's not true.
For example, ($R=\mathbb{Z}$) there is a non-split exact sequence $$0\to\mathbb{Z}/4\mathbb{Z}\stackrel{\alpha}{\to} \mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}/4\mathbb{Z}\to 0,$$ where $\alpha(n)=(2n,n)$. Your claim that $\text{im}(\alpha)$ must be contained in an indecomposable summand of $B$ isn't necessarily true.