Suppose a system of polynomial equations with rational coefficients has a real solution. Does necessarily there exists a real solution with algebraic coordinates? What about the simplest case of one polynomial equation in two variables? 



Real algebraic solution
The is an immediate application of the "real Nullstellensatz", in *exactly* the same way that one answers the analogous questions between $\overline{\mathbf{Q}}$ and $\mathbf{C}$ by using the usual Nullstellensatz (over $\overline{\mathbf{Q}}$!). I trust that the OP knows this latter application of the usual Nullstellensatz. It is just a proof by contradiction.


Namely, suppose $F \hookrightarrow F'$ is an inclusion between real closed fields (such as $F' = \mathbf{R}$ and $F$ the subfield of algebraic numbers inside $\mathbf{R}$) and let $f\_1, \dots, f\_n$ be a finite collection of polynomials in variables $X\_1, \dots, X\_m$ over $F$. Given that the $f\_i$'s have a common zero in $F'$ we want to claim the same in $F$. Suppose to the contrary that the $f\_i$'s have no common zero in $F$. Then the "real Nullstellensatz" (for real closed fields) says that $-1 = \sum g\_j f\_j + \sum h\_k^2$ for some $g\_j, h\_k \in F[X\_1,\dots,X\_m]$. But this identity persists over $F'$, where it obstructs the existence of a common zero of the $f\_j$'s over $F'$, a contradiction.


