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?


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.

  • $\begingroup$ Thanks for your comments. Where abouts in SGA1, Exposé 2, do you get the factorisation $X \to A_S^1 \to S$? I didn't see it. I presume $A_S^1 = \mathbb{A}_S^1$ is the affine line over $S$. I admit that I find it surprising that there would be an étale morphism from $X$ to $\mathbb{A}_S^1$ since $X$ is projective, but my intuition might not be serving me well here. $\endgroup$ – Hamish Oct 28 '12 at 13:35
  • $\begingroup$ Oh, I see that the defintion (1.1) of smooth in SGA1, Exposé 2, is precisely that there is a factorisation $X \to \mathbb{A}_S^n \to S$ for some $n$, and I guess $n$ must be determined by the dimension of $X$, which is 1. Then we reduce to SGA1, Theorem 1.9.5, by taking an open affine cover of $X$. $\endgroup$ – Hamish Oct 28 '12 at 13:42
  • $\begingroup$ @only: the factorization exists only locally on $X$. As $X$ is étale over $\mathbb A^1_S$, it is normal if and only if $\mathbb A^1_S$ (hence $S$ as you noticed) is normal. This should be somewhere is SGA 1. $\endgroup$ – user18119 Oct 28 '12 at 16:26
  • $\begingroup$ @QiL: Oops, thanks for correcting me. Why is $A^1_S$ being normal equivalent to $S$ being normal though? $\endgroup$ – only Oct 28 '12 at 16:35
  • $\begingroup$ @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. $\endgroup$ – user18119 Oct 28 '12 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.