# $X_K$ normal imply $X$ normal

In Algebraic Geometry and Arithmetic Curves of Qing Liu, I have two problems with the lemma 4.1.18 (page 119). The lemma is so: let $\mathcal{O}_K$ a DVR (uniformizing parameter $t$) with residue field $k$, $X$ an $\mathcal{O}_K$-scheme such that $\mathcal{O}_X(U)$ is flat over $\mathcal{O}_K$ for all open $U\subseteq X$. If $X_K$ is normal and $X_k$ reduced then $X$ is normal.

My first problem is that in the proof we're beginning to suppose that $X$ is affine. I understand the idea: if it is made for affine scheme and we take $X$ not affine then for all open affine $U\subseteq X$ one has $U$ normal. My problem is that it is not enough to conclude the normality of $X$ because I don't have the irreductility (Qing Liu has the definition of normality with normal at all points and irreducible). So my question is: how deduce the irreductibility of $X$ from my hypothesis?

My second problem: at a point in the proof we are reasonning in $A\otimes_{\mathcal{O}_K}K=A_t$ (with $A$ flat over $\mathcal{O}_K$ so that $A\subseteq A_t$) and we have $\alpha=\frac{a}{t^k}\in A_t$. Then QiL says that we can suppose that $a\notin tA$: why? I think it is equivalent to saying that $\cap_{k\in\mathbb{Z}}t^k A=(0)$, but I don't see why it should be true.

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1. Assuming $X=\textrm{Spec }A$, we have $X_K=\textrm{Spec }(A\otimes_{\mathcal O_K}K)$, with $A\otimes_{\mathcal O_K}K$ integral because $X_K$ is normal. Since the canonical homomorphism $A\to A\otimes_{\mathcal O_K}K$ is injective, $A$ has to be integral as well. Hence $X=\textrm{Spec }A$ is irreducible.

2. I think the assumption $a\notin tA$ is because of this: note that if for some $s\geq 0$ we are able to write $\alpha=t^s\beta$, with $\beta\in A$, then $\alpha \in A$ and we win. Now, if $a$ were of the form $ta_1$, with $a_1\in A$, then $t^{-r}a=\alpha=t^{-r+1}a_1$. Yes, $-r+1$ might still be negative, but we can repeat the process with $a_1$ instead of $a$, and we get $\alpha=t^{-r+2}a_2$. After $r$ steps, we fall into $A$.

Edit. $X$ is integral (i.e. reduced and irreducible) even if it is not affine.

$X$ is reduced because it is covered by affine integral schemes. So assume $X$ is reducible, covered by two irreducible components $Z_1,Z_2\subset X$. Either they intersect, or they do not. If there is a point $x\in Z_1\cap Z_2$, then $\mathcal O_{X,x}$ has at least two minimal primes. But for every affine (hence integral) open subset $U\subset X$ containing $x$, $\mathcal O_{U,x}$ has only one minimal prime. Contradiction, since $\mathcal O_{X,x}\cong \mathcal O_{U,x}$.

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For 1: why can we assume that $X$ is affine? It seems to me that for proving that $X$ is integral I need to proove that $\mathcal{O}_X(U)$ is integral for all open $U\subseteq X$ and not only for all affine open $U\subseteq X$. – Macadam Feb 13 '13 at 8:27
I think this assumption is part of the general strategy of that proof, as you wrote in your question. And in particular $A$ is integral. – Brenin Feb 13 '13 at 9:31
I don't understand your explaination. I'm totally ok that $A$ is integral. My problem is how to pass to the general case with $X$ not affine? (Because it'is not enough to be integral on all affine open $U\subseteq X$) – Macadam Feb 13 '13 at 10:03
For 2: I'm ok, thanks! – Macadam Feb 13 '13 at 10:05
@GabrielSoranzo: the proof of Prop. 2.4.17 in the book shows that it is enough to check integrality with affine open subsets. Of course it would be better to have this property in the statement. – user18119 Feb 13 '13 at 18:41