“Real part” of a number field Let ${\mathbb K} \subseteq {\mathbb C}$ be a finite extension of $\mathbb Q$, and let $n=[{\mathbb K} : {\mathbb Q}]$. Let $X_{\mathbb K}$ denote the set of all “components” (i.e., real and imaginary parts) of elements in ${\mathbb K}$. Let ${\mathbb L}=\mathbb Q(X_{\mathbb K})$ ; this is a subfield of $\mathbb R$. 
It is easy enough to see that $[{\mathbb L}:{\mathbb Q}]$ is finite, and in fact, it is  $\leq 2n^2$ (see proof below). Denote by $f(n)$ the best possible bound ; I have computed that $f(2)=2$ and $f(3)=6$, and I ask : what is $f(n)$ in general ? 
 Proof of the upper bound  
Let ${\mathbb M}_1$ denote the image of $\mathbb K$ by complex conjugation, ${\mathbb M}_2$ the smallest subfield of $\mathbb C$ containing both $\mathbb K$ and ${\mathbb M}_1$ (the so-called “compositum”), and finally ${\mathbb M}_3={\mathbb M}_2[i]$.
By construction, ${\mathbb M}_3$ contains $i$ and is invariant by complex conjugation, so it is also invariant with respect to taking real and imaginary parts. Since it contains $\mathbb K$, we see that ${\mathbb L} \subseteq {\mathbb M}_3$.
Now $[{\mathbb M}_2:{\mathbb Q}]$ is smaller than the product $[{\mathbb K}:{\mathbb Q}][{\mathbb M}_1:{\mathbb Q}]=n^2$, and in turn $[{\mathbb M}_3:{\mathbb Q}]$ is smaller than $2[{\mathbb M}_2:{\mathbb Q}]
$, qed.
 Update at 17:39  (in answer to Qiaochu’s comment) : it is not true that $f(n)=2n$ for every $n \geq 3$. Indeed, I can show that $f(4) \geq 12$. To check this, let $P$ be any rational polynomial of degree $4$ with no real roots, no purely imaginary root, and Galois group $S_4$ (for example, $P=X^4 - 6X^3 + 15X^2 - 19X + 13$ will do). 
   Then take ${\mathbb K}={\mathbb Q}(\lambda)$ where $\lambda$ is any root of $P$. Using Galois theory, it is easy to see that in this case $[{\mathbb L}:{\mathbb Q}]$ is 12.
Another way to put it : under those hypotheses, there is an automorphism $\sigma$ of $\mathbb C$  fixing $i$ and $\lambda$ but not $\bar{\lambda}$. This $\sigma$ acts as the identity on $\mathbb L$, but acts non-trivially on the real part of ${\mathbb K}[i]$.
 A: Pick a primitive element $x \in \mathbb K$. Then $\mathbb M_2 = \mathbb K[\overline{x}]$. But $\overline{x}$ has the same minimal polynomial as $x$ over $\mathbb Q$, which factors as $(X-x)Q$ with $Q$ of degree $n-1$ in $\mathbb K [X]$. Thus the extension $\mathbb K \subset \mathbb M_2$ is at most of degree $n-1$.
This gives you a better upper bound, $n(n-1)$, for $[\mathbb M_2 : \mathbb Q]$ (which is attained if and only if the action of the Galois group on the conjugates of $x$ is $2$-transitive).
Next, you have $\mathbb L[i] = \mathbb M_2[i]$, and $[\mathbb L[i] : \mathbb L] = 2$ because $\mathbb L$ is real. This implies that $[\mathbb L : \mathbb Q] = [\mathbb M_2 : \mathbb Q]$ or $[\mathbb M_2 : \mathbb Q]/2$ (depending if $i \in \mathbb M_2$ or not).
So $[\mathbb L : \mathbb Q] \le n(n-1)$.
The last thing to do is to show that the bound can be obtained. 
Suppose $n \ge 4$, that the Galois group of the Galois closure of $\mathbb K$ over $\mathbb Q$ is $S_n$, and that $i \in \mathbb Q[x_1,x_2]$ for a pair of distinct conjugates of $x$.
Since $\mathbb Q \subset \mathbb Q[i]$ is a Galois extension and $S_n$ is $2$-transitive, we must have $i \in \mathbb Q[x_i,x_j]$ for any pair of distinct conjugates of $x$. Since the permutations with two or more fixpoints generate $S_n$, we must have $i \in \mathbb Q$, which is false. Thus $i$ can't be in $\mathbb Q[x_1,x_2]$.
Finally, if $x$ has a nonreal conjugate, then we have that $i \notin \mathbb Q[x,\overline{x}]$, and so $[\mathbb L : \mathbb Q] = n(n-1)$
