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 $\leq d$?
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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. |
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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 $N' = (x,y)$. |
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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 $t$. |
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