Exercise 6.17

Proof. Let $f(x)=x^3-2$. Then $f(\sqrt[3]{2})=0$. If $f(x)$ was not irreducible, then $f(x)=u(x)v(x)$, with $1\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(v)\le 2,\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)+\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(v)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=3$, so $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)=1,\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(v)=2$.

Then $f(x)=(\mathit{ax}+b)(c{x}^2+\mathit{dx}+e),a,b,c,d,e\in \mathbb{Q}$. Let $w=-b∕a$. Then $f(w)=w^3-2=0$ and $w\in \mathbb{Q}$, so there exist $p,q\in \mathbb{Z}$, such that $w=p∕q,p\wedge q=1$.

Thus $p^3=2q^3$, so $p^3$ is even, thefore $p$ is even : $p=2p^{\prime },p^{\prime }\in \mathbb{Z}$.

$8{p{}^{\prime }}^3=2q^3$, $4{p{}^{\prime }}^3=q^3$, so $q^3$ is even, which implies that $q$ is even. Then $2\mid p\wedge q=1$ : this is a contradiction.

So $f(\sqrt[3]{2})=0$, and $f$ is monic, irreducible: $f$ is the minimal polynomial of $\sqrt[3]{2}$ on $\mathbb{Q}$. □