# Sheaf of Relative Differentials for Curves (Hartshorne)

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 $\Omega_{X,P}$.

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.

Thanks.

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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 $P$.
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 $\mathcal O_{X,P}$.
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