Q: Can we have $N\oplus P=N' \oplus P$ for $N' \subsetneq N$? 
Let $M$ be a $R$-module. Is it possible that we have two submodules $N,P \subset M$ such that $M=N\oplus P$ and also $M=N'\oplus P$ for some $N' \subsetneq N$? 

If $R$ is a field and $M$ a finite dimensional $R$-vectorspace then this obviously can't happen because of dimension reasons. But if there aren't any constraints on $R$, is this still impossible?

A: When you say $M = N \oplus P$, what you are saying is two things: first, that $N \cap P = \{0\}$, and second, that every $m \in M$ can be expressed as $n + p$ for some $n \in N$ and $p \in P$.
Now suppose $N' \subsetneq N$, with $M = N \oplus P$ and $M = N' \oplus P$.  Let $n \in N$.  Then there exist $n' \in N'$, and $p \in P$, such that $n = n' + p$.  It follows that $n - n' = p \in N \cap P = \{0\}$, and so $n = n'$.  In particular, every element of $N$ lies in $N'$, contradiction.  
