Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I am trying to understand the sheaf of relative differentials for the case of nonsingular curves. Let's use Hartshorne as a reference, thus a curve is an integral scheme of dimension 1, proper over $k$, all of whose local rings are regular.

Based on the definition of the curve, and Theorem 8.15 at p. 177, i see that if $X$ is a curve, then $\Omega_{X/k}=\Omega_X$ is a locally free $O_{X}$-module of rank $1$. Now in page 300, it is mentioned that if $u$ is a local parameter at $P \in X$, then $du$ is a generator of the free $O_P$-module $\Omega_{X,P}$.

Could somebody please explain:

1) How does it follow from the fact that $\Omega_{X}$ is a locally free $O_X$-module of rank $1$, that the stalk $\Omega_{X,P}$ is a free $O_{X,P}$ module of rank $1$? 2) I understand that $d$ is some universal derivation; is it the universal $O_{X,P}$ derivation corresponding to the $O_{X,P}$-module $\Omega_{X,P}$? 3) Why is $\Omega_{X,P}$ generated by $du$?

As it might be obvious, i am completely missing the picture here.


share|cite|improve this question
up vote 4 down vote accepted
  1. If $\mathcal F$ is a locally free $\mathcal O_X$-module of rank $n$, then $\mathcal F_P$ is a free $\mathcal O_{X,P}$ module of rank $n$ for every $P \in X$. Indeed, take an open neighborhood $U$ of $P$ which trivializes $\mathcal F$, and choose an isomorphism $\mathcal F|_U \cong \mathcal O_U^n$; thus the stalk of $\mathcal F$ at $P$ is isomorphic to the stalk of $\mathcal O_U^n$ at $P$.
  2. Yes, because formation of differentials commutes with localization.
  3. The local ring $\mathcal O_{X,P}$ embeds into its completion, which is the power series ring $k[u]$. From this it can be seen that $\Omega_{X,P}$ is generated by $du$ over $\mathcal O_{X,P}$.
share|cite|improve this answer
Thanks. 1) What do you mean by "$U$ trivializes $F$? 3) I can't really understand this argument. Is there any alternative, without a completion argument? – Manos Nov 17 '12 at 2:06
You're welcome! By "$U$ trivializes $\mathcal F$", I just mean that $U$ is small enough so that there exists an isomorphism $\mathcal F|_U \cong \mathcal O_U^n$. Restricting the sheaf to $U$ does not change the local ring at $P$. What don't you understand about (3)? – Bruno Joyal Nov 17 '12 at 2:14
Thanks a lot. First, i don't know about completions. Second even if i accept that $O_{X,P}$ is embedded in $k[u]$, i can't see what this has to do with $\Omega_{X,P}$ :) – Manos Nov 17 '12 at 3:58
Coming back to the issue of why $\Omega_{X,P}$ is generated by $du$, regarding your argument, i think the universal property of $\Omega_{X,P}$ comes into place, but i can not see it clearly... – Manos Nov 18 '12 at 20:08

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.