Exercise 13.1.14

Proof.

(a)

$f=x^4+4x+2$.

$\mathrm{\Delta }(f)=-2^8\cdot 19$ is not a square in $\mathbb{Q}$, and ${𝜃}_f(y)=y^3-8y-16$ is irreducible over $\mathbb{Q}$, so ${\mathrm{Gal}}_{\mathbb{Q}}(f)\simeq S_4$ (part (a) of Theorem 13.1.11).

(b)

$f=x^4+8x+12$.

$\mathrm{\Delta }(f)=2^{12}\cdot 3^4$ is a square in $\mathbb{Q}$, and ${𝜃}_f(y)=y^3-48y-64$ is irreducible over $\mathbb{Q}$, so ${\mathrm{Gal}}_{\mathbb{Q}}(f)\simeq A_4$ (part (a) of Theorem 13.1.11).

(c)

$f=x^4+1$.

$\mathrm{\Delta }(f)=2^8$ is a square in $\mathbb{Q}$ and ${𝜃}_f(y)=y(y-2)(y+2)$ splits completely over $\mathbb{Q}$, so ${\mathrm{Gal}}_{\mathbb{Q}}(f)\simeq \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ (part (b) of Theorem 13.1.11).

(d)

$f=x^4+x^3+x^2+x+1$.

$\mathrm{\Delta }(f)=5^3$ is not a square, and ${𝜃}_f(y)=(y-2)(y^2+y+1)$ has a unique root in $\mathbb{Q}$, so part (c) of Theorem 13.1.1 applies. Let $\zeta$ a root of $f$. Then

$$f=(x-\zeta )(x-{\zeta }^2)(x-{\zeta }^3)(x-{\zeta }^4)$$

splits completely over $\mathbb{Q}(\zeta )$. By Exercise 13,

$$G\simeq \mathbb{Z}∕4\mathbb{Z}.$$

(we know already this result, since $f={\Phi }_5$.)

(e)

$f=x^4-2$.

By Exercise 13, Example 2, $\mathrm{\Delta }(f)=-2^{11}$ is not a square, and ${𝜃}_f(y)=y(y^2+8)$ has a unique root in $\mathbb{Q}$. Moreover if $a=\sqrt[4]{2}$,

$$f=(x-a)(x+a)(x^2+a^2)$$

doesn’t splits completely over $\mathbb{Q}$, so

$$G\simeq D_8.$$