TITLE: Problem in understanding module complements
QUESTION [5 upvotes]: Let $A$ be a $\mathbb{K}$-algebra and $M$ be an $A$-module, $N \subset M$ be a submodule.
A module complement of $N$ is defined as a submodule $N'\subset M$ so that $M = N \oplus N'$.
Now in our lecture we defined a semisimple $A$-module as an $A$-module so that for every submodule exists a module complement.

But isn't that a trivial property that is true for all $A$-modules, because of this:
$M \cong N \oplus M / N$, where $M / N$ is an $A$-module by the operation $a.(m+N)=a.m+N$ for $a \in A$, $m+N \in M/N$, so $M/N$ is always a module complement of $N$?

REPLY [5 votes]: The module $M/N$ is not a submodule of $M$, and therefore cannot be a complement of $N$.
It is not necessarily true that $M \cong N \oplus (M/N)$.
Take for example $A = k[x]$, $M = A$ and $N = Ax = (x)$.
Then
$$
            N \oplus (M/N)
     =      (x) \oplus (k[x]/(x))
     \ncong k[x]
     =      M
   $$
since $k[x]$ is torsion-free, but $k[x]/(x)$ is not.
In the above example the submodule $N$ has no direct complement:
Otherwise they would exist some ideal $(f) \subseteq k[x]$ with $k[x] = (x) \oplus (f)$. Then $xf \in (x) \cap (f) = 0$ and therefore $f = 0$, which would result in $k[x] = (x)$.