4
$\begingroup$

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 $K$?

$\endgroup$
2
  • $\begingroup$ How is $X(R)$ defined? $\endgroup$ – Giovanni De Gaetano May 14 '12 at 16:34
  • $\begingroup$ $X(R)$ is just the set of $R$-morphisms $\text{Spec}(R) \to X$. $\endgroup$ – Evariste May 14 '12 at 16:41
3
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ Dear QiL, what is this topology on $X(R)$ that Evariste and you allude to? $\endgroup$ – Georges Elencwajg May 14 '12 at 19:35
  • $\begingroup$ @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 $X_K(K)$. $\endgroup$ – user18119 May 14 '12 at 19:44
  • $\begingroup$ 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? $\endgroup$ – Evariste May 14 '12 at 21:03
  • $\begingroup$ @Evariste: no, consider the inclusion $Z=X\setminus Y\to X$ ! $\endgroup$ – user18119 May 15 '12 at 20:32
  • $\begingroup$ I'm confused, I'll think more. Thank you. $\endgroup$ – Evariste May 15 '12 at 23:51

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.