Direct summand of a free module

Let $M$, $L$, $N$ be $A$-modules and $M=N\oplus L$. If $M$ and $N$ are free, is $L$ necessarily free?

Solution 1:

Let $M$ be a module over a commutative ring. Recall three properties:

$M$ is projective if it is a direct summand of a free module.
$M$ is weakly stably free if there exists a free module $F$ such that $M \oplus F$ is free.
$M$ is stably free if there exists a finitely generated free module $F$ such that $M \oplus F$ is free.

Then free implies stably free implies weakly stably free implies projective.

But actually projective implies weakly stably free: this is called the Eilenberg swindle. Thus if you don't require $F$ to be finitely generated, then any nonfree projective module gives a negative answer to this question. There are "cheap" examples of this: e.g. let $R_1$ and $R_2$ be any two nonzero rings, let $R = R_1 \times R_2$, and consider $R_1$ as an $R$-module: it is projective but not free.

What about stably free modules which are not free? Here counterexamples lie much deeper. Plop has described what is in fact the most standard counterexample, which reduces the problem to a nontrivial fact in differential topology. A somewhat more comprehensive discussion of this class of examples can be found in $\S 6.4.3$ of my commutative algebra notes. Also, at the end of $\S 3.5$ I give the Eilenberg Swindle as an exercise, with a big hint.

Added: To the best of my recollection, the terminology "weakly stably free" is not standard but something that I made up while writing up my notes. But the point of course is that after this little discussion involving the Eilenberg Swindle, one sees that such terminology is not needed.

Solution 2:

No, for example the tangent vector bundle of the sphere $S^2$ is non-trivial, whereas the normal vector bundle is. More concretely, $A=\mathbb{R}[x,y,z]/(x^2+y^2+z^2-1)$, $L=\{(P_1,P_2,P_3)\ |\ xP_1+yP_2+zP_3=0\}$, $N=A \cdot (x,y,z)$, $M=A^3$.