Projective dimension Is it true that if a module has a free resolution of length $d$ then any of its submodule has a free resolution of length $\leq d$?
 A: No. A submodule of a free module need not have finite projective dimension.
As a simple example let $R=\mathbb{Z}/p^2\mathbb{Z}$. The free module $R$
has a submodule $p\mathbb{Z}/p^2\mathbb{Z}\cong\mathbb{Z}/p\mathbb{Z}$
which has no finite projective resolution.
A: No, a ring will always be free viewed as a module over itself, but its ideals certainly don't have to be free.
For example, consider the ring $R = k[t]/t^2$ and consider the submodule $I = (t),$ the ideal generated by $t$.  Then $R \to I$ by multiplication by $t$ and has kernel $I$.  It's then easy to see that $\ldots \to R \to R \to I \to 0$ is an infinite free resolution of $I$ where each map is multiplication by $t$.
A: Counterexamples can even be found in a domain, by taking rings of higher dimension or singular rings—once you're no longer over a PID, ideals will suffice.  Take $R = k[x,y]$, the module $M = R$ itself, and the submodule $M' = (x,y)$.  Or, take the ring $S = k[x,y]/(x^3-y^2)$, the module $N = S$ itself, and the submodule $N' = (x,y)$.
