Let $f:X\longrightarrow S$ be a morphism of preschemes which is smooth of pure relative dimension 1. Let $\sigma:S\longrightarrow X$ be a section of $X$. Let $D$ be the (positive) divisor associated to $\sigma$.

(1) Is this divisor automatically a relative(to $f$) one?

(2) If $x$ is a point of $D$, and $t$ a regular which generates $I(D)_x$. Does $t$ automatically has the property that $\mathcal O_x dt=(\Omega^1_{X/S})_x$?

(3) Does $t$ has automatically the property that $\mathcal{O}_s\longrightarrow \mathcal{O}_x/t\mathcal{O}_x$ is an isomorphism?

(4) Questions (2) and (3) with $t$ a regular section of $\mathcal O_X$ over an open $U$ containing $x$ which generates $I(D)|_U$?

completenessof $A$ can replace $t$ with $t-a$. QED – BCnrd Sep 15 '10 at 5:01provethe affirmative answers. (That is, for (2) through (4), if you use flatness and Nakayama Lemma arguments you can get the result from the field case, a case which you presumably know how to do. For (1) one can't easily reduce to that case, which is why I gave you a more detailed explanation as to how to exploit the smoothness over the base ring.) Chapter 1 of Katz-Mazur covers many many things along the lines of your (1)--(4) (they even prove (1) in there somewhere), so I highly recommend that you read it. Enjoy! – BCnrd Sep 15 '10 at 15:22