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Let P be an R-module.

1.If P is a quotient of the R-module M then P is isomorphic to a direct summand of M.

2.Every short exact sequence 0→L→M→P→0 splits.

How do I show equivalence between statements 1 and 2?

Sharmi C
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1 Answers1

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In homological algebra, we have a device that is called the splitting lemma. It states that a short exact sequence $0 \rightarrow L \rightarrow M \rightarrow P$ splits if and only if it is isomorphic (in the category of exact sequences) to the short exact sequence $0 \rightarrow L \rightarrow L \oplus P \rightarrow P \rightarrow 0$.

Hence, if 2. holds, then if $P$ is the quotient of $M$, we get a s.e.s. $0 \rightarrow L \rightarrow M \rightarrow P \rightarrow 0$ where $L = \ker (M \rightarrow P)$, and the splitting lemma yields that $P$ is isomorphic to a direct summand of $M$.

Vice versa, if 1. holds, note that $P$ is a quotient of the free module $F \langle P \rangle$ via the map that sends $p \mapsto p$. Thus, $P$ is the direct summand of a free module, hence projective (ie. 2. holds).

Cloudscape
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  • If you don't know about projective modules and the equivalent characterisations, you might want to study the Wikipedia page. It will give you everything you need. (Wikipedia also has a page on the splitting lemma.) – Cloudscape Aug 21 '18 at 11:18
  • Thank you. In fact I have just started the basic course of module theory and I came across this statement in the Algebra book by Dummit and Foote(page 389). – Sharmi C Aug 21 '18 at 13:30