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Let $f:S→B$ be an elliptic fibration from an integral surface $S$ to integral curve $B$

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Here I use following definitions:

A surface (resp. curve) is a $2$ -dim (resp. $1$-dim) proper k scheme over fixed field $k$.

Fibration has two properties: 1. $O_B = f_*O_S$ 2. all fibers of f are geometrically connected

Futhermore a fibration is elliptic if the generic fiber $S_{\eta}=f^{-1}(\eta)$ is an elliptic curve (over $k(\eta)$.

Denote by $i_S: S_{\eta} \to S$ the canonical immersion. Here I'm ot sure to 100% but I guess that for the structure sheaf holds $O_{S_{\eta}}= O_S \otimes_k k(\eta)$.

Now the QUESTION:

Since $S_{\eta}$ is elliptic curve and therefore smooth the restriction of the Kähler differentials $\omega_{S/B} \vert _{S_{\eta}}$ is invertible.

My question is how to see that there exist open neighbourhood $U \subset S$ of $S_{\eta}$ such that the restriction $\omega_{S/B} \vert _U$ is still invertible?

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    $\begingroup$ $\Omega ^2_{S/B}$ is zero. $\endgroup$
    – abx
    Commented Jun 30, 2019 at 4:45
  • $\begingroup$ I suspect that the OP meant $\omega_{S/B}$... $\endgroup$
    – Sándor Kovács
    Commented Jun 30, 2019 at 23:05
  • $\begingroup$ @Karl_Peter: try to use Nakayama's lemma. $\endgroup$
    – Sándor Kovács
    Commented Jun 30, 2019 at 23:07
  • $\begingroup$ If the generic fiber is smooth, there are only finitely many singular fibers $F_i$. $\omega _{S/B}$ is invertible in $S\smallsetminus \cup F_i$. $\endgroup$
    – abx
    Commented Jul 1, 2019 at 14:35
  • $\begingroup$ ...and this is based on a little Nakayama argument as Sándor states: Concretely if $x \in S_\eta$ then since $f^{\#}m_{\eta} \subset m_x \subset O_{S,x}$ then since $(\omega_{S/B} \vert _{S_{\eta}})_x= \omega_{S/B} \otimes O_{S,x}/m_{\eta}$ is invertible then also $\omega_{S/B} \otimes O_{S,x}$ (by Nakayama) and then there exist an open neighbourhood around $x$ where $\omega_{S/B} $ is also open. $\endgroup$
    – user267839
    Commented Jul 1, 2019 at 14:56

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