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Let $\theta$ be a root of $f(x) := x^3-3x +1$. Now , we can easily check that $\mathbb{Q}(\theta)$ is a splitting field of $f(x)$ and that the Galois group is cyclic of order $3$.

Then, how do I determine the other roots explicitly in terms of $\theta$ ? ( That is , $a \theta^2 + b \theta +c$ for some $a, b, c \in \mathbb{Q}$)

[My attempt]

(1) Let the Galois group be $<\sigma>$ , order $3$ cyclic group. Then, $\sigma(\theta)$ is also root of $f(x)$.

Since $[\mathbb{Q}(\theta) : \mathbb{Q}] = 3 $ (dimension) , so, $\sigma(\theta) = a \theta^2 + b \theta +c$ for some $a, b, c \in \mathbb{Q}$.

Using $\theta^3 = 3 \theta -1$, I tried to find $\sigma(\theta)^3 -3\sigma(\theta)+1 =0$ but this is so complicate.

(2) Let $\theta , \alpha , -\theta -\alpha$ be roots of $f(x)$. By relation of roots and coefficients of $f(x)$, we know that $$\theta^2 +\theta \alpha + \alpha^2 = 3 \;\;\; \text{and} \;\;\; \theta \alpha (\theta + \alpha) = 1 $$

First, using $\theta^2 +\theta \alpha + \alpha^2 = 3$ , $$ \alpha = \frac{-\theta \pm \sqrt{12-3\theta^2} }{2} $$ So, let $ \alpha = a \theta^2 + b \theta +c$ for some $a, b, c \in \mathbb{Q}$. So $12 - 3 \theta^2 = \alpha^2 = (a \theta^2 + b \theta +c)^2$. This looks simple than (1) but also so complicate..

How can I solve this problem?

Community
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hew
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    Possible duplicate of https://math.stackexchange.com/questions/1767252/expressing-the-roots-of-a-cubic-as-polynomials-in-one-root – lhf Dec 09 '19 at 12:14

3 Answers3

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According to this answer, the other roots are $$ -\dfrac{\theta}{2} \pm \dfrac{1}{2}(2\theta^2+\theta-4) = \theta^2-2 \text{ and } -\theta^2-\theta+2 $$

lhf
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Hint: Divide your polynomial by $x-\theta$ the use the quadratic formula

How you know that the Galois group is cyclic of order 3 and you still did not know if $\mathbb{Q}(\theta)$ is the splitting field or not. And you still do not know the other two roots?

For example $x^3-2$ if you join $\sqrt[3]{3}$ the automorphism fixing $Q$ is just the identity even though $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$

IrbidMath
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There is a striking similarity between $f(2z)$ and $T_3(z)=4z^3-3z$, the third Chebyshev polynomial of the first kind. By setting $x=2z$ we have that the equation $f(x)=0$ is equivalent to $$ T_3(z)=-\frac{1}{2} $$ or, by letting $z=\cos\theta$, $$ \cos(3\theta)=-\frac{1}{2}.$$ Three distinct solutions are clearly given by $\theta\in\left\{\frac{2\pi}{9},\frac{4\pi}{9},\frac{8\pi}{9}\right\}$, so the roots of the original polynomial are

$$ \alpha=2\cos\frac{2\pi}{9},\quad\beta=2\cos\frac{4\pi}{9}=2T_2\left(\frac{\alpha}{2}\right),\quad \gamma=2\cos\frac{8\pi}{9}=2 T_4\left(\frac{\alpha}{2}\right).$$

Jack D'Aurizio
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