# The closure of the set of rational points in the Adeles

Let $X$ be a smooth geometrically integral projective variety over $\mathbb{Q}$. Then we may consider the closure $\overline{X(\mathbb{Q})}$ of $X(\mathbb{Q})$ inside the adelic points $X(\mathbb{A})=\prod_v X(\mathbb{Q}_v)$ of $X$. However, we may also take the closure $\overline{X(\mathbb{Q})}^v$ of $X(\mathbb{Q})$ inside $X(\mathbb{Q}_v)$ for any place $v$ of $\mathbb{Q}$. Obviously we have $$\overline{X(\mathbb{Q})} \subset \prod_v \overline{X(\mathbb{Q})}^v \subset X(\mathbb{A}).$$

My question whether this first inequality is actually an equality?

My motivation is that I am trying to understand better $\overline{X(\mathbb{Q})}$ and what it looks like. I will simply note that the answer to my question is yes in the easy cases where $X$ satisfies weak approximation and when $X(\mathbb{Q})$ is empty.

Edit: To make sure there are not simple counter-examples like the one David pointed out below, I am assuming that $X(\mathbb{Q})$ is Zariski dense. I should also note that I am particularly interested in the case where $X$ is a fano variety.

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If $X(\mathbb{Q})$ is finite of order $> 1$, as for an elliptic curve of rank 0 with nontrivial torsion, then $\overline{X(\mathbb{Q})}$ is finite, but $\prod_v \overline{X(\mathbb{Q})}^v = X(\mathbb{Q})^{\aleph_0}$ is uncountably infinite, isn't it? –  David Loeffler May 22 '11 at 8:47
@David: Yes of course thanks! I over looked this simple case as the case I have in mind $X$ is fano and $X(\mathbb{Q})$ is Zariski dense. I will edit my question. –  Daniel Loughran May 22 '11 at 9:03
I'm puzzled by the question: isn't true that, because $\mathbb{Q}$ is discrete in $\mathbb{A}$, the set $X(\mathbb{Q})$ is discrete (hence closed) in $X(\mathbb{A})$? (this is certainly the case for affine varieties). –  Alain Valette May 22 '11 at 10:24
@Daniel: WHAT??? –  Alain Valette May 22 '11 at 11:27
@Alain: the projective situation is completely different to the affine situation. $\mathbf{Z}$ is a long way from $\mathbf{Q}$, so $\mathbf{A}^1(\mathbf{Z})$ is a long way from $\mathbf{A}^1(\mathbf{Q})$, but $\mathbf{P}^1(\mathbf{Q})=\mathbf{P}^1(\mathbf{Z)}$ by the valuative criterion for properness. –  Kevin Buzzard May 22 '11 at 18:38

Here's an example where $X(\mathbf{Q})$ is Zariski-dense but the first inequality is not an equality.

Let $X$ be an elliptic curve over the rationals, such that the group $X(\mathbf{Q})$ is isomorphic to $\mathbf{Z}$. Let me first remind you what $X(\mathbf{Q}_p)$ looks like, for $p$ a prime where the curve has good reduction. There's a natural reduction map $X(\mathbf{Q}_p)\to\overline{X}(\mathbf{F}_p)$, with $\overline{X}$ the reduction of $X$, and this reduction map is surjective onto a finite target. Hence $X(\mathbf{Q}_p)$ naturally breaks up as a finite disjoint union of cosets of the kernel of this map, and the kernel is (and hence all the cosets are) clopen.

Now let $P$ be a generator of $X(\mathbf{Q})$, put the curve into minimal form, choose a random large prime $\ell$ and consider $Q:=\ell.P$. It will be easy to find an example where $Q=(x,y)$ and there are two primes $p_1$ and $p_2$ in the denominator of $x$ but not in the denominator of the coordinates of $P$, and such that $X$ has good reduction at these primes. In short it should be easy to find two primes $p_1$ and $p_2$ such that $P$ has order exactly $\ell$ modulo both $p_1$ and $p_2$.

But now the first inclusion must be strict, because even the closure of $X(\mathbf{Q})$ in $X(\mathbf{Q}_{p_1})\times X(\mathbf{Q}_{p_2})$ is easily seen to be strictly smaller than the product of the closures, for the "same reason" that if $P_1$ has order $\ell$ in the finite abelian group $G$ and $P_2$ has order $\ell$ in the finite abelian group $H$, then the group generated by $(P_1,P_2)$ in $G\times H$ is strictly smaller than the product of the subgroups generated by $P_1$ and $P_2$.

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You must mean $l=14$, as $P=(0,0)$ is a generator of 37A, and $7P=(−{5\over 9},{8\over 27})$. –  Junkie May 23 '11 at 3:09
For example if $X$ is the rank 1 curve of conductor 37 and $P$ is a generator then one can take $\ell=17$ and $p_1=11$ and $p_2=139$. –  Kevin Buzzard May 23 '11 at 7:02
[Junkie: you're right---thanks. I was using $2P$; I just deleted the comment and replaced it with a corrected one.] –  Kevin Buzzard May 23 '11 at 7:03
Great answer thanks! –  Daniel Loughran May 23 '11 at 9:32