2
$\begingroup$

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 $\leq d$?

$\endgroup$
1
  • 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". $\endgroup$ Apr 18, 2010 at 20:12

3 Answers 3

8
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 $t$.

$\endgroup$
1
$\begingroup$

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 $N' = (x,y)$.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Apr 18, 2010 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.