Exercise 7.4.2

Proof.

(a)

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

$f$ is irreducible by the Schönemann-Eisenstein Criterion with $p=2$.

$$\mathrm{\Delta }(f)=-4p^3-27q^2=-4{(-4)}^3-27{(2)}^2=256-108=148=2^2\times 37.$$

As $\mathrm{\Delta }(f)\ne 0$, $f$ is separable, so Proposition 7.4.2 applies to $f$.

Recall that an integer $k\in \mathbb{Z}$ is a square in $\mathbb{Q}$ if and only if it is a square in $\mathbb{Z}$. As $37$ is not a square, $\mathrm{\Delta }(f)$ is not a square in $\mathbb{Q}$, so

$${\mathrm{Gal}}_{\mathbb{Q}}(x^3-4x+2)=S_3.$$

(b)

$f=x^3-4x+2$ has discriminant $\mathrm{\Delta }(f)=148={(2\sqrt{37})}^2$, which is a square in $\mathbb{Q}(\sqrt{37})$, thus $${\mathrm{Gal}}_{\mathbb{Q}(\sqrt{37})}(x^3-4x+2)=A_3.$$

(c)

$f=x^3-3x+1$.

If $\alpha =p∕q,p\wedge q=1$ is a root of $f$ in $\mathbb{Q}$, then $p^3-3p{q}^2+q^3=0$, thus $p\mid q,q\mid p$ with $p\wedge q=1$, therefore $\alpha =\pm 1$, but neither 1 nor $-1$ is a root of $f$, thus $f$ has no rational root. As $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=3$, $f$ is irreducible over $\mathbb{Q}$. $\mathrm{\Delta }(f)=-4{(-3)}^3-27=81=9^2$, thus $\mathrm{\Delta }(f)\ne 0$ and so $f$ is separable. Moreover $\mathrm{\Delta }(f)=9^2$ is a square in $\mathbb{Q}$. By Proposition 7.4.2,

$${\mathrm{Gal}}_{\mathbb{Q}}(x^3-3x+1)=A_3.$$

(d)

Let $u$ a root of $f=x^3-t\in ℂ(t)$ in a splitting field of $f$ over $ℂ(t)$. Then $$f=(x-u)(x-\mathit{\omega u})(x-{\omega }^{2}u).$$

We have proved in Exercise 4.2.9 that $f$ has no root in $ℂ(t)$, and that $f$ is irreducible over $ℂ(t)$ (Proposition 4.2.6). Moreover $f$ is separable.

$\mathrm{\Delta }(f)=-27t^2={(i\sqrt{27}t)}^2$ is a square in $ℂ(t)$, thus

$${\mathrm{Gal}}_{ℂ(t)}(x^3-t)=A_3.$$

(e)

If $\mathrm{\Delta }(f)=-27t^2$ was the square of an element $\alpha =p(t)∕q(t)$ in $\mathbb{Q}(t)$, then $$-27={\left(\frac{p(t)}{\mathit{tq}(t)}\right)}^2,p,q\in \mathbb{Q}[t].$$

Applying the evaluation homomorphism defined by $t\mapsto t_0$, were $t_0\in \mathbb{Q},t_0\ne 0$ and $t_0$ is not a root of $q(t)$, we obtain that $-27$ is a square in $\mathbb{Q}$: this is false, thus $\mathrm{\Delta }(f)$ is not the square of an element in $\mathbb{Q}(t)$. Therefore

$${\mathrm{Gal}}_{\mathbb{Q}(t)}(x^3-t)=S_3.$$