Show that every root of a polynomial is real (using Galois theory)

Question

The question

Let $f$ be a rational polynomial (so $f \in \mathbb{Q}[x]$) of degree $4$. Let $E$ be the splitting field of $f$ such that $[E:\mathbb{Q}] = 12$ and assume that $f$ has one real root $z$. Show that all roots of $f$ are real.

The problem

So I tried looking at this in multiple ways. The degree of the polynomial is $4$, so I know that $-z$ is also a root (which seems right to me). Now I'm trying to figure out what the other roots are by using Galois Theory. The things I already know now are:

• The Galois group of this extension has degree $12$, because this is a Galois extension
• Now the issue I'm having is the fact that the degree is $12$. This seems somewhat 'unreal' to me: look at the field $\mathbb{Q}(z)$. This field lies between $E$ and $\mathbb{Q}$ because it is an extension of degree $4$. Now the two other roots (call them $x$ and $-x$) should be roots of a third degree polynomial to make the degree $12$, but there are only two roots left and their minimal polynomial is $f$ divided by the two factors with $z$ and $-z$. So in my eyes the degree is $8$ here...

What am I seeing wrong? And how could I solve this question then?

Answer 1

Extended hints:

Remember that we can think of the Galois group $G$ as a group of permutations of the roots, in this case $G\le S_4$.

1. Show that $A_4$ is the only subgroup of order $12$.
2. Complex conjugation is of order two. We know that its counterpart in $G$ has at least one fixed point, namely $z$. Actually, it must have an even number of fixed points, why?
3. Show that the only permutation of $A_4$ that has an even positive number of fixed points is the identity. Why does this settle your question?