If $G_f\cong D_5$ for $f(x)=x^5+ax+b\in \mathbb{Q}[x]$, show that these criteria are true.

Question

As a followup to this post, I was wondering if the converse of this statement was true. That is, if $G_f$ (the galois group of $f(x)=x^5+ax+b$ for $a,b\in\mathbb{Q}$) is isomorphic to $D_5$ (the dihedral group of order 10), I want to show that:

(a) $f(x)$ is irreducible in $\mathbb{Q}[x]$

(b) The discriminant $D(f)$ is a square in $\mathbb{Q}$

(c) $f(x)=0$ is solvable by radicals.

What I've tried:

(a) I have no idea how to show that $f$ is irreducible.

(b) $G_f\cong D_5\leq A_5$, which is true if and only if $D(f)$ is a square.

(c) Consider $R:=\{1, r, r^2, r^3, r^4\}\subset D_5$. One can easily show that $R$ is a subgroup. Also, since $[D_5:R]=|D_5/R|=2$, we see that $R\unlhd D_5$ and $D_5/R\cong \mathbb{Z}/2\mathbb{Z}$, which is abelian. Since $R$ is cyclic, it is abelian, so it is solvable. Hence $D_5$ is solvable, but this is true if and only if $f(x)=0$ is solvable by radicals.

My questions:

1. How do I show (a)?
2. Are my proofs for (b) and (c) correct?

Any help is appreciated!

Answer 1

Your proofs for (b) and (c) look fine. For (a), consider the degree of the roots of $f$ and the order of the splitting field for $f$.

Suppose $f$ is not irreducible. Let $\alpha$ be a root of $f$. $\newcommand{\rat}{\mathbb{Q}}$ Then $[\rat(\alpha):\rat]\in\{1,2,3,4\}$. Let $K$ be the spiltting field for $f$. We know that $[K:\rat]=|D_5|=10$. Then we have $[\rat(\alpha):\rat]\in\{1,2\}$ by divisibility considerations. This shows that the irreducible factors of $f$ are either linear or quadratic.

Now, let $g$ be the product of all quadratic irreducible factors of $f$, and $E$ be the splitting field for $g$. Then $K=E$. However, $g$ has degree at most $4$, so $[E:\rat]\mid 4!$. This is impossible because $[E:\rat]=[K:\rat]=10$. Hence, $f$ must be irreducible.