# Are smooth relative curves over an arbitrary base normal?

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.
• 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. – Hamish Oct 28 '12 at 13:35
• 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 $X$. – Hamish Oct 28 '12 at 13:42
• @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. – user18119 Oct 28 '12 at 16:26
• @QiL: Oops, thanks for correcting me. Why is $A^1_S$ being normal equivalent to $S$ being normal though? – only Oct 28 '12 at 16:35
• @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. – user18119 Oct 28 '12 at 16:48