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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 $I(D)|_U$?

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  • $\begingroup$ Yes. By $S$-flatness and Nakayama can reduce (2)--(4) to case $S$ is spec of field (even alg. closed). For (1), can assume $S$ noetherian, then local, then complete (by f. flatness reasons), say spec of $A$. The completed local ring on $X$ at special pt of section is abstractly $A[[t]]$ because local structure thm. for smooth maps gives etale map from $(X,x)$ to affine line (and etale maps induce formal isoms at residually trivial fibers). So enough that any $A$-alg. map $A[[t]] \rightarrow A$ has principal kernel. If $t \mapsto a$ then by completeness of $A$ can replace $t$ with $t-a$. QED $\endgroup$
    – BCnrd
    Commented Sep 15, 2010 at 5:01
  • $\begingroup$ Dear workitout: It looks like you're still reading the early parts of Katz-Mazur. I hope you have someone local with whom to discuss things in person. Why did you give yourself the pseudonym "workitout"? $\endgroup$
    – BCnrd
    Commented Sep 15, 2010 at 5:03
  • $\begingroup$ Dear Bcnrd, I'm not reading Katz-Mazur and I was not asking for any kind of reduction. $\endgroup$
    – Workitout
    Commented Sep 15, 2010 at 13:40
  • $\begingroup$ Dear workitout: I was giving you the key points of the arguments to prove the 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! $\endgroup$
    – BCnrd
    Commented Sep 15, 2010 at 15:22
  • $\begingroup$ Dear workitout: OK, now that I have K-M nearby I can tell you more: look at their Corollary 1.1.5.2. They do give a nice overview about divisors, sometimes using external references, which may or may not make it useful, depending on your background. For example, their Lemma 1.2.2 is exactly your (1). $\endgroup$
    – BCnrd
    Commented Sep 15, 2010 at 16:31

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