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Let $X/S$ be a smooth, projective scheme of relative dimension one over a scheme $S$ (which we may assume is affine Noetherian, but need not be reduced nor irreducible nor even connected). For $s \in S$, the fibre $X_s = X\times_S \operatorname{Spec}(k(s))$ of $X$ is therefore a smooth, projective curve which we assume moreover to be geometrically connected.

Recall that a scheme is normal if all its local rings are normal (i.e. a domain which is integrally closed in its fraction field). The local rings of $X_s$ are regular of dimension one which is equivalent to being normal (Atiyah and MacDonald, Proposition 9.2). So $X/S$ has normal fibres, but:

Question: From this setup, how can we prove that $X/S$ is normal? If not, what (minimal set of) extra assumptions are required?

By Serre's Criterion (EGA IV.2, Section 5.8 or Liu, Chapter 8, Theorem 2.23) it suffices to prove that, for all $x \in X$,

  • if $\dim \mathscr{O}_{X,x} \leq 1$, then $\mathscr{O}_{X,x}$ is regular, and
  • if $\dim \mathscr{O}_{X,x} \geq 2$, then $\operatorname{depth}\mathscr{O}_{X,x} \geq 2$.

But I don't see immediately how to use the regularity of the local rings of the fibres to show these results. Ideas?

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Since $X/S$ is smooth, for every point $x$ of $X$, there is an open set $U$ around $x$ such that it factors as $U\rightarrow A^1_S \rightarrow S$, where the first map is étale. (See SGA1 exposé 2 for the proof of the equivalence of this definition with smoothness.) By SGA1 1.9.5, $X$ is normal at $x$ if and only if $A^1_S$ is normal at $f(x)$.

It isn't clear to me if there's an easy way to check $A^1_S$; it seems like there should be, but nothing jumps out at me and SGA1 doesn't seem to have a normality criteria for this.

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    @only: sorry I did'nt read correctly your answer.If $\mathbb A^1_S$ is normal, than $S$ is obviously normal. The converse can be checked with Serre's criterion or directly.2012-10-28