Prove that $\mathbb Q[x]/\langle x^3+x^2-2x-1\rangle$ cannot be obtained by adjoining any cube root of rational number

Question

I am trying to find an extension of degree $3$ of $\Bbb Q$ which is not isomorphic to one of the form $\Bbb Q(\sqrt[3]{a})$. To show that $\mathbb Q[x]/\langle x^3+x^2-2x-1\rangle$ is such an example. I need to prove that it cannot be obtained by adjoining a cubic root of any rationals. And I am stuck now. Could someone please help？Thanks！

Edit：Could someone give a proof without any Galois theory？ Thanks！

Answer 1

Isn't $f(X)=X^3+X^2-2X-1$ the minimum polynomial of $u=2\cos(2\pi/7)$? If so then $K=\mathbb{Q}(u)$ is a degree $3$ Galois extension of $\mathbb{Q}$.

But a pure cubic field $L=\mathbb{Q}(a^{1/3})$ cannot be Galois, as its Galois closure contains all cubic roots of unity. These are quadratic over $\mathbb {Q}$, so $L$ is not Galois over $\mathbb{Q}$.

What's wrong with Galois theory! Anyway, the field $\mathbb{Q}[X]/\langle f(X)\rangle $ has three maps to $\mathbb R$ taking the image of $X$ to $2\cos(2\pi/7)$, $2\cos(4\pi/7)$ and $2\cos(6\pi/7)$ in the three cases. But $\mathbb{Q}[X]/\langle X^3-a\rangle $ only has one homomorphism to $\mathbb R$ taking $X$ to the unique real cube root of $a$.

How's that?

Answer 2

The field $\Bbb Q[X]/\langle X^3+X^2-2X-1\rangle$ has the interesting property that the map $X\mapsto X^2-2$ induces an automorphism of order $3$, something $\Bbb Q[\sqrt[3]a]$ does not have.