Let $f(x) \in \mathbb{Q}[x]$ be an irreducible polynomial of degree $4$ with exactly $2$ real roots.

Question

Show that the Galois group of $f$ over $\mathbb{Q}$ is either $S_4$ or the dihedral group of order $8$.

So I'm studying from quals remotely, in a state hundreds of miles away from my university that is also quarantining the state that my university is in do to the pandemic. Before the pandemic one of my friends gave me this solution.

Since $f$ is degree $4$ we know that the Galois group is either $Z_4, V_4, D_8, A_4$, or $S_4$. Since it has $2$ real roots it has a pair of complex conjugate roots. Since complex conjugate roots are equivalent to conjugation a transposition, and since the only transitive subgroups containing a transposition are $D_8$ and $S_4$ we are done.

So I understand why we have $2$ complex roots but I don't understand why they have to be complex conjugate to one another. I am also struggling to figure out an element in $D_8$ that is equivalent to a transposition.

Any help is appreciated.

Answer 1

As Theo says, the key for seeing that the complex roots come as conjugate pairs is that $f$ has real coefficients.

For finding transpositions in $D_8$, why not think of $D_8$ as the symmetry group of the square? By this analogy, the elements of $D_8$ are permuting the four corners of the square (whereas in your original problem, the elements of $D_8$ are permuting the four roots of $f(x)$).

Numbering the vertices of the square in clockwise order, the elements of $D_8$ are:

• The identity, $e$.
• The 90-degree rotations, $(1234)$ and $(1432)$.
• The 180-degree rotation, $(13)(24)$,
• The reflections across the horizontal and the vertical, $(14)(23)$ and $(12)(34)$
• The reflections across the diagonals, $(13)$ and $(24)$.

The reflections across the diagonals are your transpositions.