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.
 A: 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 $P_2$.
