degree of polynomial satisfied by real part of complex algebraic number If a complex number $z=x+iy$ is algebraic over $\mathbb Q$ of degree $n$ then its real part $x$ will also be algebraic over $\mathbb Q$. What will be the degree of polynomial satisfied by $x$? Can it be greater than the degree of $z$ over $\mathbb Q$?
 A: Let $F = \mathbf{Q}(x_1,\ldots,x_n)$, and let $E = \mathbf{Q}(s_1,s_2,\ldots,s_n)$ where $s_i$ are the standard symmetric polynomials, e.g. $s_1 = x_1 + x_2 + \ldots$, $s_n = x_1 x_2 \ldots x_n$. Note that $S_n$ acts naturally on $F$ and $E = F^{S_n}$, and $F/E$ is Galois with Galois group $S_n$. Now let $K=E(x_1 + x_2)$. It is similarly easy to see that $K$ is the fixed field of $S_2 \times S_{n-2} \subset S_n$, from which we find that
$$[K:E] = \frac{n(n-1)}{2}.$$
Equivalently, the degree of $x_1 + x_2$ over $E$ is $n(n-1)/2$. By Hilbert Irreducibility, for generic specializations of the $s_i$ (so certainly for asymptotically 100% of them), it follows that the degree of the sum of any two distinct roots of a generic degree $n$ polynomial is $n(n-1)/2$. If $z = x+iy$ is a complex root of $f(z)$, then certainly $\overline{z}$ is also a root distinct from $z$, so (generically) $z + \overline{z}$ has degree $n(n-1)/2$. (Of course if $z$ is real then $\overline{z}$ is not a distinct root.)
Random examples:
If $z^4 - z - 1 = 0$, then $z + \overline{z}$ is a root of the irreducible  polynomial $t^6 + 4t^2 - 1$.
If $z^5 - z - 1 = 0$, then $z + \overline{z}$ is a root of the irreducible  polynomial $t^{10} + 3t^6 + 11t^5 - 4t^2 + 4t - 1$.
A: EDIT (see the comment of @ pisco)
Let $K:=\mathbf Q(z)$ of degree $n$, with $z=x+iy$ and $y\neq 0$. Obviously $K=\mathbf Q (z^{-1})= \mathbf Q(z, z^{-1})$. The problem is to "locate" $\mathbf Q(x)$ in the diagram of extensions and see what information we can derive
 on its degree over $\mathbf Q$.
1) Denote by $r$ the modulus of $z$, and assume first that $r=1$. Then  $x=\frac 12(z+z^{-1})$, and $K$ contains $K^+=\mathbf Q (x)$ . Since $K^+$ is the subfield of $K$ fixed by complex conjugation, $[K:K^+]=2$ (this implies that $n$ is even) and $[\mathbf Q (x):\mathbf Q]=n/2$.
2) In the general situation, introduce the real extension $k=\mathbf Q(r)$. If $k \neq \mathbf Q$, we can repeat the argument of 1) using $x/r \in K(r)/k$ instead of $x \in K/\mathbf Q$, and conclude that $[K(r):k(x/r)]=2$. Because $k$ contains $r$, this is equivalent to $[K(r):k(x)]=2$ $(*)$. Then everything can be taken back to $[\mathbf Q(r,x):\mathbf Q(r)]$ : (i) In the simplest case where $r \in K$, $[K:k(x)]=2$ because of $(*)$, and $n=[K:\mathbf Q]=2[k(x):\mathbf Q]$. Then, if $m$ denotes $[\mathbf Q(r,x):\mathbf Q(x)]$ (which depends on $y$), we have $[\mathbf Q(x):\mathbf Q]=n/2m$ ; (ii) If $r\notin K$, introduce  $F=\mathbf Q(x)\cap k$, so that $\mathbf Q(r)$ and $\mathbf Q(x)$ are linearly disjoint over $F$ and $[\mathbf Q(r): F]=[\mathbf Q(r,x):\mathbf Q(r)]$, so that finally $[\mathbf Q(r,x):\mathbf Q(r)].[F:\mathbf Q]$ = $[\mathbf Q(x):\mathbf Q]$. But this is not very satisfying.  
A: Let us assume $z\notin \mathbb{R}$, otherwise there is nothing to be said. Here is an elementary proof that $x$ has degree $\leq n(n-1)/2$ over $\mathbb{Q}$.
Consider the extension $\mathbb{Q}(z)\subseteq\mathbb{Q}(z,\overline{z})$. Let $f\in\mathbb{Q}[t]$ be the minimal polynomial of $z$ over $\mathbb{Q}$. Since $\overline{z}$ is a root of $f$, its minimal poynomial over $\mathbb{Q}(z)$ divides $f$, and so has degree $\leq n$. But, moreover, $f$ factorizes over $\mathbb{Q}(z)$ as \begin{equation}f=(t-z)g\end{equation} for some polynomial $g\in\mathbb{Q}(z)[t]$ of degree $n-1$. Then, since $z\neq\overline{z}$, we have $g(\overline{z})=0$ so actually $\overline{z}$ has degree $\leq n-1$ over $\mathbb{Q}(z)$.
We also have $[\mathbb{Q}(z,\overline{z}):\mathbb{Q}(x)]\geq 2$ simply because $z\notin\mathbb{R}$, so from
\begin{equation}
2[\mathbb{Q}(x):\mathbb{Q}]\leq [\mathbb{Q}(z,\overline{z}):\mathbb{Q}(x)][\mathbb{Q}(x):\mathbb{Q}]=[\mathbb{Q}(z,\overline{z}):\mathbb{Q}(z)][\mathbb{Q}(z):\mathbb{Q}]\leq (n-1)n
\end{equation}
it follows that $[\mathbb{Q}(x):\mathbb{Q}]\leq n(n-1)/2$.
