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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3$\begingroup$ If $d=1$ then aren't you asking "is a submodule of a free module free"? And the answer is well-known to be "no". $\endgroup$– Kevin BuzzardCommented Apr 18, 2010 at 20:12
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3 Answers
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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.
answered Apr 18, 2010 at 18:23
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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 $t$.
answered Apr 18, 2010 at 18:26
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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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$\begingroup$ If $k$ is a field then the ring $R=k[x,y]$ has global dimension $2$ and all modules over $R$ have a projective resolution $$0\to P_2\to P_1\to P_0\to M\to 0.$$ By the Quillen-Suslin theorem we may take the $P_i$ to be free. $\endgroup$ Commented Apr 18, 2010 at 18:54