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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！

Y.X.
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    Why did you delete the original question, together with an answer? You lost also all upvotes. The hint by Will Jagy was:"the three roots are $$ 2 \cos \left( \frac{2\pi}{7} \right), ; ; 2 \cos \left( \frac{4\pi}{7} \right), ; ; 2 \cos \left( \frac{6\pi}{7} \right). $$ I think, it is better t edit the original question. For the person who gave an answer, it is also not fair to delete it. – Dietrich Burde Apr 16 '17 at 08:46
  • @Dietrich Brude The answer was deleted by the original poster. I did not delete it. And also I do need some explaination. Sorry for deleting the others comment, which I requested further explaination but have not get replied. – Y.X. Apr 16 '17 at 09:18

2 Answers2

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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?

Angina Seng
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  • Could you please avoid using Galois theory？I cannot understand it so far. – Y.X. Apr 16 '17 at 08:48
  • Thanks！I have not learnt Galois theory. Could you please give some explain how can I consider the map to $\Bbb R$ to conclude that there are 3 such maps？ – Y.X. Apr 16 '17 at 09:15
  • Is that because the fact that, as the homomorphism from $\Bbb Q(\alpha)$ to $\Bbb R$ is the identity map restrict to $\Bbb Q$, the homomorphism is uniquely determined by where we send $\alpha$ to. As $\phi(0)=\phi(\alpha^3+\alpha^2-2\alpha-1)=\phi(\alpha^3)+\phi(\alpha^2)-2\phi(\alpha)-\phi(1)=0$, there are 3 choices of $\phi(\alpha)$ because all the 3 roots are real, whereas as $x^3-a$ has a unique real solution, there is only one choice where we send the root to? – Y.X. Apr 16 '17 at 09:53
  • @PropositionX that's exactly it! – Angina Seng Apr 16 '17 at 10:36
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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.

Hagen von Eitzen
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  • See also https://math.stackexchange.com/questions/1767252/expressing-the-roots-of-a-cubic-as-polynomials-in-one-root – lhf Feb 01 '22 at 14:41