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?

  • 2
    $\begingroup$ For $-z$ to also be a root you would need $f$ to be biquadratic (even degree terms only). $\endgroup$ Jun 26 '18 at 15:52
  • $\begingroup$ I understand. I assumed there were none which is my mistake. $\endgroup$
    – xzeo
    Jun 26 '18 at 15:53
  • $\begingroup$ Any tips on how the solve the problem, then? $\endgroup$
    – xzeo
    Jun 26 '18 at 15:54
  • 3
    $\begingroup$ 1. show f is irreducible. 2. [Q(z) : Q] =4, [E: Q(z)] =3, let w be a root not in Q(z), the mininal polynomial g of w over Q(z) must be of degree 3, real polynomial of odd degree has a real root, say y, then Q(z,y) already equal E, and Q(z,y) is real. $\endgroup$
    – user119882
    Jun 26 '18 at 16:13
  • 1
    $\begingroup$ sorry for the smartphone-typing, so make that comment. $\endgroup$
    – user119882
    Jun 26 '18 at 16:16

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?
  • $\begingroup$ I think I understand everything now, apart from the second tip. Is this because the order of complex conjugation is 2? $\endgroup$
    – xzeo
    Jun 26 '18 at 16:04
  • 1
    $\begingroup$ Yes, @xzeo. The non-real roots come in complex conjugate pairs... $\endgroup$ Jun 26 '18 at 16:13
  • 1
    $\begingroup$ @xzeo Because complex conjugation is of order 2, you will break a set of size 4 into orbits of size 1 and 2. $\endgroup$
    – Aaron
    Jun 26 '18 at 16:14

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