Exercise 13.2.13

Proof.

(a)

We use the formulas of Exercise 15 for $f=x^5+\mathit{ax}+b$: $$\begin{array}{llll}\hfill \mathrm{\Delta }(f)& =256a^5+3125b^4,& \hfill & \\ \hfill {𝜃}_f(y)& ={\left(y^3-20a{y}^2+240a^{2}y+320a^3\right)}^2-2^{10}\mathrm{\Delta }(f)y.& \hfill & \end{array}$$

With $a=0,b=-2$, we obtain

$$\begin{array}{llll}\hfill \mathrm{\Delta }(f)& =50000=2^4{5}^5,& \hfill & \\ \hfill {𝜃}_f(y)& =y^6-2^{14}{5}^{5}y.& \hfill & \end{array}$$

Let $L$ be the splitting field of $x^5-2$ over $\mathbb{Q}$. $\mathrm{\Delta }(f)$ is not a square in $\mathbb{Q}$, and ${𝜃}_f(y)$ has a root 0 in $\mathbb{Q}$. So, by Theorem 13.2.6 and Exercise 11, $\mathrm{Gal}(L∕\mathbb{Q})$ is isomorphic to $\mathrm{AGL}(1,{𝔽}_5)$. This result is already proved in Section 6.4.

(b)

We know that ${\zeta }_5=({\zeta }_5\sqrt[5]{2})∕\sqrt[5]{2}\in L$, and also $\sqrt{5}={\zeta }_5+{\zeta }_5^{-1}-{\zeta }_5^2-{\zeta }_5^{-2}\in L$ (see the quadratic Gauss sum page 249).

Since $\mathbb{Q}\subset \mathbb{Q}(\sqrt{5})$ is a quadratic extension, by the Galois correspondence, $\mathrm{Gal}(L∕\mathbb{Q}(\sqrt{5}))$ is a subgroup of index 2 in $\mathrm{Gal}(L∕\mathbb{Q})$ and the subgroup $H\subset S_5$ corresponding to $\mathrm{Gal}(L∕\mathbb{Q}(\sqrt{5}))$ has index 2 in $G\simeq \mathrm{AGL}(1,{𝔽}_5)$. Thus $|H|=10$, and since $5\mid |H|$, $H$ contains a 5-cycle and is a transitive subgroup of $S_5$. By Theorem 13.2.2, $H$ is conjugate to $⟨(12345)⟩$ or to $\mathrm{AGL}(1,{𝔽}_5)\cap A_5$. Since $(G:H)=2$, $H$ is conjugate to $\mathrm{AGL}(1,{𝔽}_5)\cap A_5$, so

$$\mathrm{Gal}(L∕\mathbb{Q}(\sqrt{5}))\simeq H\simeq \mathrm{AGL}(1,{𝔽}_5)\cap A_5.$$