MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\leq d$?

share|cite|improve this question
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". – Kevin Buzzard Apr 18 '10 at 20:12
up vote 8 down vote accepted

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.

share|cite|improve this answer

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 $t$.

share|cite|improve this answer

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 $N' = (x,y)$.

share|cite|improve this answer
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. – Robin Chapman Apr 18 '10 at 18:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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