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In what follows there is an error. The problem is that I can't find it. Let $$0\longrightarrow L\overset{f}{\longrightarrow} M\overset{g}{\longrightarrow}N\longrightarrow 0$$ be a short exact sequence of $R$-modules and $R$-module homomorphisms. Then we have

$$M\cong\operatorname{Ker}g\oplus M/\operatorname{Ker}g$$ $$=\operatorname{Im}f\oplus M/\operatorname{Ker} g$$ $$\cong L\oplus g(M)$$ $$=L\oplus N$$

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    $\begingroup$ Does that sequence split? $\endgroup$
    – Xam
    Oct 14, 2016 at 15:32
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    $\begingroup$ See here. By the way, are you talking about vector spaces over a field $R$ as in the title (in which case the conclusion is correct) or are you talking about $R$-modules over a ring $R$ as the question itself suggests? $\endgroup$
    – user228113
    Oct 14, 2016 at 15:33
  • $\begingroup$ @G.Sassatelli Sorry, the title is wrong, I am going to edit it $\endgroup$
    – bateman
    Oct 14, 2016 at 15:37

2 Answers 2

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Your first line doesn't necessarily hold. Consider the short exact sequence of $\mathbb{Z}$-modules (abelian groups)

$$0\to\mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}/4\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\to 0$$

where the maps are the obvious ones. The issue is that not every submodule of $M$ necessarily has a complement.

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Direct sums, really? Think of $R$ as the integers, $M$ a cyclic group of order $4$, and $L, N$ cyclic of order $2$.

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