Let $P(x, y) \in \mathbb{Q}[x, y]$. We consider the equation $P(x, y)=0$. If $a, b \in \mathbb{Q}$ such that $P(a, b)=0$ then $(a, b) \in \mathbb{Q}^2$, is called a rational solution.
If $K$ a field, $\mathbb{Q} \leq K$, then if $(a, b) \in K \times K$ and $P(a, b)=0$ then $(a, b)$ is called a $K-$rational solution.
When $(a, b)$ is a $K-$rational solution does this mean that $(a, b) \in \mathbb{Q}$ ??
For example, $x^2-2$, is $\sqrt{2}$ a $\mathbb{R}-$solution ??
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I have also an other question...
Is $\mathbb{Q}$ a subfield of $\mathbb{Q}_p, \forall p \in \mathbb{P}$ ??
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EDIT:
If $(a, b) \in \mathbb{Q} \times \mathbb{Q}$ a rational solution of $P(x, y)=0 (1)$, where $P(x, y) \in \mathbb{Q}[x, y]$, then the equation $(1)$ has a $\mathbb{Q}_p-$rational solution $\forall p \leq \infty$.
The (rational) $x \in \mathbb{Q}$ is a perfect square of a rational $\Leftrightarrow$ when $x$ is a perfect square in $\mathbb{Q}_p, \forall p \leq \infty$.
So, something stands in $\mathbb{Q}$ iff it stands in $\mathbb{Q}_p, \forall p \in \mathbb{P}$ ??
Yes, $\mathbb{Q}$ is a subfield of its p-adic completion $\mathbb{Q}_p$. As for any completion of a field $K$, the field itself is embedded into its completion $\widehat{K}$. If we define the completion as the space of all Cauchy sequences modulo zero sequences, we have the embedding $K\hookrightarrow \widehat{K}$, given by $a\mapsto (a,a,\ldots)$, where $a$ is mapped to the constant Cauchy sequence. As a consequence, the prime field of $\mathbb{Q}_p$ is $\mathbb{Q}$, so that the characteristic of $\mathbb{Q}_p$ is zero.
For the first question, see K-rational points on algebraic varieties. It does not necessarily mean that the coordinates are in $\mathbb{Q}$.