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I know the polynomial is irreducible over $\Bbb Q$ and the roots are real irrational. A $\Bbb Q$-automorphism $\sigma:\Bbb Q(\alpha)\to \Bbb Q(\alpha)$ is determined by $\sigma(\alpha)$, which is also a root of the polynomial. My problem is finding out if the other roots are also in $\Bbb Q(\alpha)$. If the other roots are $\beta, \gamma$, then I set the system $3=\alpha+\beta+\gamma, -3=\alpha\beta\gamma$. For example, clearing $\beta$, I obtain

$\beta=\frac{3-\alpha}{2}\pm\sqrt{\alpha^2-6\alpha+9+\frac{12}{\alpha}}$

(A solution can be with the positive square root, then the negative one is $\gamma$)

I don't know how to conclude if this $\beta$ is in $\Bbb Q(\alpha)$ or not.

I tried to search similar problems but they apply results I don't know, like using the discriminant or splitting stuff. With the current theory I have in lessons I only have very basics tools.

Aarón David Arroyo Torres
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1 Answers1

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This is pure cheating: I browsed the referenced links and transformed their answers to fit this problem:

$$ P(x) = x^3 - 3x^2+3 $$

$$ P(\alpha) = 0 $$

$$ \beta = R(\alpha) = -\alpha^2 + \alpha + 3 $$

$$ P(\beta) = P (R(\alpha) = P ( -\alpha^2 + \alpha + 3) $$ $$ = - \alpha^{6} + 3 \alpha^{5} + 3 \alpha^{4} - 11 \alpha^{3} - 3 \alpha^{2} + 9 \alpha + 3 = $$ $$ = \left( \alpha^{3} - 3 \alpha^{2} + 3 \right) \left( -\alpha^{3} + 3 \alpha + 1 \right) = P(\alpha) \left( -\alpha^{3} + 3 \alpha + 1 \right) = 0 $$

Which means that $\beta$ is a root of $P$ and $\beta \in \mathbb Q(\alpha)$

blamocur
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  • How is the polynomial $R(\alpha)$ obtained? – Aarón David Arroyo Torres May 02 '22 at 18:15
  • @AarónDavidArroyoTorres I derived it from other posts, which use discriminants. The goal is to find $R$ such that $P(R(x))$ is divisible by $P(x)$. I wonder if it is possible to solve the resulting equations for the coefficients of $R$, and thus avoid using discriminants? – blamocur May 02 '22 at 18:20
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    Yes, you could suppose $\beta = A \alpha^2 + B \alpha + C$, then multiply out $P(\beta)=0$, reduce the $\alpha^4$ and $\alpha^3$ terms using $\alpha^3 = 3 \alpha^2-3$, and finally solve a system of three equations in $A, B, C$. (Also, I wouldn't call learning techniques from other examples then applying them to a similar problem cheating!) – aschepler May 02 '22 at 18:28
  • @aschepler The equations I get for A,B,C are quite hairy though :( – blamocur May 02 '22 at 19:00
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    Quite. Which is why the Galois theorems are nice to have... – aschepler May 02 '22 at 19:04