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Let $X$ be a scheme of finite type over a discrete valuation ring $R$ with fraction field $K$, such that the generic fibre $X_K$ is smooth over $K$. Let $Y$ be a closed subscheme of $X$ which contains no irreducible component of $X$.

Is it true - maybe under some additional assumptions on $X$ and/or $R$ - that $X(R) \setminus Y(R)$ is dense in $X(R)$ for the natural topology defined by $R$? If so, is there an easy way to see this or does it require heavy machinery?

Also, do there exist simple counterexamples when $X_K$ is not smooth over $K$?

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How is $X(R)$ defined? – Giovanni De Gaetano May 14 '12 at 16:34
$X(R)$ is just the set of $R$-morphisms $\text{Spec}(R) \to X$. – Evariste May 14 '12 at 16:41
up vote 3 down vote accepted

This is false without further assumption on $R$. Just consider a smooth projective curve $X$ over $R=\mathbb Z_{p\mathbb Z}$ with generic fiber $X_{\mathbb Q}$ of genus $>1$. Let $Y_\mathbb Q$ be the finite set of rational points of $X_{\mathbb Q}$ and $Y$ the Zariski closure of $Y_\mathbb Q$ in $X$. Then $X(\mathbb Z_{p\mathbb Z})=X_\mathbb Q(\mathbb Q)$ (by valuative criterion of properness), but the latter is equal to $Y(\mathbb Z_{p\mathbb Z})$.

Suppose now $R$ is complete (and $X_K$ is smooth), then $X(R)\setminus Y(R)$ is dense in $X(R)$. To see this, fix a section $S\in X(R)$ and let $x_0\in X(K)$ be the generic fiber $S$. As $X_K$ is smooth at $x_0$, Zariski locally around $x_0$ we have an étale quasi-finite morphism $X\to \mathbb A^d_K$ which maps $x_0$ to $(0,..., 0)$. The theorem of implicit functions implies that a small open (analytic) neighborhood of $x_0$ in $X(K)$ is a polydisk. As $Y(K)$ (Zariski locally) is contained in the zero locus of some non-zero (analytic) function $f$, no small polydisk centered at $x_0$ is entirely contained in $Y(K)$ (otherwise $f=0$ by induction on $d$). So any small polydisk centered at $x_0$ contains a point of $X(K)\setminus Y(K)$. To finish, it is easy to see that a small enough open (analytic) neighborhood $U$ of $x_0$ is contained in $X(R)$. This proves the claim (leaving some details to the readers...).

If $R$ is excellent and henselian, the result still hold by algebraic approximation (M. Artin), see "Néron models", § 3.6. But his requires heavy machinery.

Let's terminate with a counterexample when $X_K$ is not smooth. Let $R=\mathbb Z_p$ (complete), $$X=\mathrm{Spec} R[x,y]/(x^2-py^2).$$ Then $X(R)$ is reduced to one section $x=y=0$. Take $Y$ equal to this section.

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Dear QiL, what is this topology on $X(R)$ that Evariste and you allude to? – Georges Elencwajg May 14 '12 at 19:35
@GeorgesElencwajg, suppose for simplicity that $X$ is separated over $R$. Then $X(R)\to X_K(K)$ is injective, and the topology on $X(R)$ is induced by that of $X_K(K)$. – user18119 May 14 '12 at 19:44
Thank you, QiL. – Georges Elencwajg May 14 '12 at 19:51
Dear QiL, one more question. Say we take $R$ henselian (and excellent if necessary). Let $Z$ be another scheme of finite type over $R$ and let $f: Z \to X$ be an $R$-morphism (without any further assumptions). If $X(R) \setminus Y(R) \subseteq f(Z(R))$, do we also have $X(R) \subseteq f(Z(R))$? If so, why? – Evariste May 14 '12 at 21:03
@Evariste: no, consider the inclusion $Z=X\setminus Y\to X$ ! – user18119 May 15 '12 at 20:32

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