Consider a commutative Ring $R$ and a module $M$ over $R$. Now let $N \subset M$ be a submodule. Then we have a canonical short exact sequence:
$$0 \rightarrow N \xrightarrow{i} M \xrightarrow{p}M/N \rightarrow 0$$
The $i$ is the inclusion and $p$ is the projection onto the quotient module. Up until now, I was under the impression that this sequence always splits by the splitting Lemma. The way I thought about it was, that the map
$$ r: M/N \rightarrow M\; ; \; m+N \mapsto m$$
defines a right-inverse to $p$, i.e. $p \circ r = id_{M/N}$. Since there are multiple $m$, mapping to the same element in $M/N$, I would choose a member of each class, sucht that the map becomes a homomorphism.
But I recently read that it only splits, if $N$ is a direct summand of $M$, which means that there is another submodule $N'$ such that $N \oplus N' = M$. But the existence of $N'$ somehow makes the splitting "obsolete", since this is exactly the definition of a split sequence. In other words, this is an if and only if statement. So $N$ is a direct summand iff the above SES splits. But this doesn't give me any more information on when this sequence splits… So my questions are the following:
- Am I allowed to define $r$ in this way, and if yes, why is the map $r$ not a right-inverse to $p$?
- Are there ways to check, if $N$ is a direct summand, without trying to split the above SES?
- Can someone provide a counterexample of $N$ and $M$ such that the sequence does indeed not split?
Edit: I guess I have to clarify my question:
In the left map, I really want the canonical inclusion of $N$ into $M$ as a submodule. For example, I won't allow the sequence $0 \rightarrow \mathbb{Z} \xrightarrow{\cdot 2} \mathbb{Z} \rightarrow \mathbb{Z}_2 \rightarrow 0$. In this case, the left map would be $id$ and the right one $0$.
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