Is $\mathbb R$ saturated? Consider $\mathbb R$ as a model of the field theory. Is it true that it is saturated?
$\mathbb C$ is saturated, since it is a model of ACF; $\mathbb Q$ is not, since it is not $\omega$-universal: $\mathbb Q(\pi)$ is a model of $Th(\mathbb Q,+,\cdot,0,1)$, but one cannot immerge it elementary in $\mathbb Q$; right?
What about $\mathbb R$? Since now I tried to repeat the proof that $\mathbb C$ is saturated, but it didn't work.
Thank you in advance.
 A: No. First, note that $x \leq y$ is definable in $\mathbb{R}$ by the formula $\exists z\,(z^2 = y-x)$. Now the partial type $\{n\leq x\mid n\in \mathbb{N}\}$ (over the empty set) is finitely satisfiable in $\mathbb{R}$ but not realized in $\mathbb{R}$.
For the case of $\mathbb{Q}$, the partial type $\{ax + b \neq 0\mid a,b\in \mathbb{N}\}\cup \{-ax + b\neq 0\mid a,b\in \mathbb{N}\}$ (over the empty set) is finitely satisfiable in $\mathbb{Q}$, but not realized in $\mathbb{Q}$. 
As I noted in the comments, your proposed argument does not work, because $\mathbb{Q}(\pi)$ is not elementarily equivalent to $\mathbb{Q}$. But any elementary extension of $\mathbb{Q}$ must contain an element realizing the partial type above, and hence can't be embedded elementarily in $\mathbb{Q}$.
A: There is an elementary way to see that $\mathbb{Q}(\pi)$ is not elementarily equivalent to $\mathbb{Q}$. Note that by Lagrange's four square theorem, every element $x\in \mathbb{Q}$ or $-x$ is a sum of four squares, whereas this is not the case in $\mathbb{Q}(\pi)$ ; neither $\pi$ nor $-\pi$ are the sum of four squares in $\mathbb{Q}(\pi)$.
