I am unable to see whether this statement is true or false, any help would be appreciated.
Let $E$ be the splitting field of a polynomial over $\Bbb{Q}$ which has both real and complex roots. Then the conjugation automorphism is an element of $Gal(E/\Bbb{Q})$ of order 2. Let $\Omega$={$ \alpha_1, …, \alpha_n, \beta_1, …, \beta_{2m} $} be the set of roots of the polynomial, where the $\alpha_i$ are the real roots are the $\beta_i$ the complex ones, with $\beta_{2i-1}$ and $\beta_{2i}$ conjugates.
Is it true that $\Bbb{Q}(\alpha_1, …, \alpha_n, \beta_1+\beta_2, …, \beta_{2m-1}+\beta_{2m})$ is always the field fixed by the conjugation automorphism? The $\subseteq$ inclusion is obvious, but I can't figure out $\supseteq$.
Best Answer
The fixed field of the complex conjugation is $M=E\cap\Bbb{R}$, and by Galois theory we always have $[E:M]=2$. The field $K$ generated by the real zeros and the real parts of the complex zeros (which is what you get when you use sums of complex conjugate pairs) is obviously contained in $M$, but may be smaller.
Daniel Fischer's example $f(X)=(X-1)(X^2+3)(X^2+5)$ is a simple example of this phenomenon. As a product of two roots $\sqrt{15}=\sqrt{-3}\sqrt{-5}$ is an element of $M$. But all the irrational zeros of $f(X)$ are purely imaginary so we have $K=\Bbb{Q}$.
I want to add examples of irreducible polynomials such that $K$ is a proper subfield of $M$. If we didn't have the extra requirement that $f(X)$ must also have real zeros, then $f(X)=X^4+4X^2+2$ would do well. It is irreducible by Eisenstein, $p=2$. The zeros of the quadratic $q(X)=X^2+4X+2$ are $-2\pm\sqrt2$, both negative, so as $f(X)=q(X^2)$, all the zeros of $f(X)$ are purely imaginary, again implying $K=\Bbb{Q}$.
This was really reusing Daniel's idea. A problem is that the above example polynomial does not meet the requirement that it should also have real zeros. Let's modify the situation marginally, and use the cubic $q(X)=X^3+4X^2-2$ instead of a quadratic. It is still irreducible by Eisenstein as is $f(X)=q(X^2)=X^6+4X^4-2$. A difference is that $q(X)$ has a single positive zero and two negative ones: $q(-1)>0>q(0)$ and $\lim_{x\to\pm\infty}q(x)=\pm\infty$. Therefore $f(X)$ has two real zeros and four purely imaginary zeros. If $\alpha_1$ is one of the real zeros, then $-\alpha_1$ is the other. Consequently $K=\Bbb{Q}(\alpha_1)$ is a degree six extension of $\Bbb{Q}$.
We observe that the discriminant of $q(X)$ is $d=404$, a non-square. Therefore the Galois group of $q(X)$ is $S_3$, and its splitting field, call it $L$, is a degree six extension. Let's denote the zeros of $q(X)$ by $\beta_1=\alpha_1^2>0$, $\beta_2$ and $\beta_3$. The Vieta relations tell us that $\beta_1\beta_2\beta_3=2$. As the zeros of $f(X)$ are $\pm\sqrt{\beta_i},i=1,2,3$, we can deduce that $\sqrt2\in M$. The discriminant gives the quadratic subfield $\Bbb{Q}(\sqrt{d})=\Bbb{Q}(\sqrt{101})$ of $L$. As $L\subset M$ we see that $\Bbb{Q}(\sqrt{101},\sqrt2)\subseteq M$, implying that $4\mid [M:\Bbb{Q}]$.
Consequently $K$ is a proper subfield of $M$ also in this case.
I didn't double check everything, but a study of the cycles structure of Frobenius elements in $G=Gal(E/\Bbb{Q})$, when viewed as a group of permutations of the six roots of $f(X)$ seems to imply that $G\simeq C_2\wr S_3$. In other words $[E:\Bbb{Q}]=48$, and consequently $[M:K]=4$.