Exercise 8.2.2

Proof. $f(x)=x^3+x^2-2x-1\in \mathbb{Q}[x]$: ${\sigma }_1=-1,{\sigma }_2=-2,{\sigma }_3=1$.

Note that $f$ of degree 3 is irreducible over $\mathbb{Q}$, otherwise it would have a rational root $\alpha =p∕q\in \mathbb{Q},p\wedge q=1,q>0$.

Then $p^3+p^{2}q-2p{q}^2-q^3=0$, so $p\mid q^3,p\wedge q=1$, therefore $p\mid 1$, and similarly $q\mid 1$, thus $\alpha =\pm 1$, but neither 1 nor $-1$ is a root of $f$, so $f$ is irreducible.

By Exercise 8.2.1, the splitting field of $f$ over $\mathbb{Q}$ is

$\mathrm{discr}(f)=-4{\sigma }_2^3-27{\sigma }_3^2+{\sigma }_1^2{\sigma }_2^2-4{\sigma }_1^3{\sigma }_3+18{\sigma }_1{\sigma }_2{\sigma }_3=32-27+4+4+36=49$.

$\mathrm{discr}(f)=49$ is a square in $\mathbb{Q}$, so $\mathrm{Gal}(L∕\mathbb{Q})$ is the cyclic group $C_3\simeq \mathbb{Z}∕3\mathbb{Z}$ (Prop. 7.4.2), and so $[L:\mathbb{Q}]=3$.

Assume that the extension $\mathbb{Q}\subset L$ is radical.

As $[L:\mathbb{Q}]=3$, it doesn’t exists any strict sub-extension of $\mathbb{Q}\subset L$, so the definition of a radical extension imply the existence of $\gamma \in L$ and of an integer $m>1$ such that $L=\mathbb{Q}(\gamma )$ and ${\gamma }^m\in \mathbb{Q}$. As the extension is of degree 3, it it not a quadratic extension, so $m\ge 3$.

We conclude the reasoning (Example 8.2.3):

Let $p$ be the minimal polynomial of $\gamma$ over $\mathbb{Q}$. Then $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)=[L:\mathbb{Q}]=3$. As $\gamma$ is a root of $x^m-{\gamma }^m\in \mathbb{Q}[x]$, $p$ divides $x^m-{\gamma }^m$.

As the extension $\mathbb{Q}\subset L$ is a Galois extension, all the roots of $p$ are in $L$ so are real since $L\subset ℝ$. Thus the three real distinct roots of $p$ are among the roots of $x^m-{\gamma }^m$, that is

But this is impossible since at most two of these roots are real.

Conclusion: $\mathbb{Q}\subset L$ is not a radical extension (but $\mathbb{Q}\subset L\subset \mathbb{Q}({\zeta }_7)$, so $\mathbb{Q}\subset L$ is solvable). □