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Let $f(x) = x^3 - 3x + 1 \in \mathbb{Q}[x]$. It is irreducible over $\mathbb{Q}$ since it is irreducible over $\mathbb{F}_2$. The discriminant of $f$ is $81 = 9^2$, so the Galois group of $f$ is isomorphic to $A_3$ and the splitting field of $f$ is $\mathbb{Q}(\alpha)$, where $\alpha$ is a root of $f$.

Now I want to represent the elements of $G(\mathbb{Q}(\alpha)/\mathbb{Q})$ as matrices $3 \times 3$, but to do that I need a basis of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$. And the question is:

What basis of $\mathbb{Q}(\alpha)/\mathbb{Q}$ is better to choose and how to express the roots $\alpha, \beta$ and $\gamma$ in terms of that basis?

If I take ${1, \alpha, \alpha^2}$ as a basis, what is a good way of finding the coordinates of $\beta$ and $\gamma$?

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  • $\begingroup$ $1, \alpha, \alpha^2$ is a basis $\endgroup$ – lhf Nov 22 '16 at 0:03
  • $\begingroup$ @lhf, yes, I see it. But I do not understand what is $\alpha^2$ and what, for example, $\sigma : \alpha \mapsto \beta, \ \beta \mapsto \gamma$ do with $\alpha^2$. In other words, how do I express two other roots $\beta$ and $\gamma$ in terms of $1, \alpha$ and $\alpha^2$? $\endgroup$ – nightked Nov 22 '16 at 0:08
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    $\begingroup$ The best basis to use is obviously $\{\alpha, \beta, \gamma \}$ $\endgroup$ – mercio Nov 22 '16 at 20:45
  • $\begingroup$ @mercio hahaha, well stated. :) $\endgroup$ – Adam Hughes Nov 22 '16 at 20:50
  • $\begingroup$ (I didn't actually read the polynomial) it would have worked if the coefficient of $x^2$ was nonzero. $\endgroup$ – mercio Nov 22 '16 at 21:18
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Do synthetic division to see that

$$x^3-3x+1 = (x-\alpha)(x^2+\alpha x+\alpha^2-3)$$

Do synthetic division to see that

Then the other roots are just

$$\beta,\gamma = {-\alpha\pm \sqrt{12-3\alpha^2}\over 2}$$

Now, the discriminant here must be a square in the field, and as $\alpha$ is an algebraic integer, the other roots are also integers, and so we suspect that if

$$12-3\alpha^2 = (a\alpha^2+b\alpha+c)^2$$

then let's guess $b=1$ since that will cancel with the $-\alpha$, and further we suspect that $a, b$ are both even so that division by $2$ in the denominator will keep them in the $\Bbb Z$-module spanned by $1,\alpha,\alpha^2$.

Writing

$$12-3\alpha^2 = (2k\alpha^2+\alpha + 2j)^2$$ $$=4k^2\alpha^4+\alpha^2+4j^2+2(2k\alpha^3+4kj\alpha^2+2j\alpha)$$ $$=4k^2(3\alpha^2-\alpha)+\alpha^2+4j^2+2(2k(3\alpha-1)+4kj\alpha^2+2j\alpha)$$

$$=\alpha^2(12k^2+1+8kj)+\alpha(12k-4k^2+4j)+1(4j^2-4k)$$

We conclude

$$\begin{cases} j^2-k = 3 \\ 3k-k^2+j=0 \\ 3k^2+2kj = -1 \end{cases}$$

which seems onerous, but the last one gives us $k=\pm 1$ since $k$ is a divisor of a unit, and using that with the first shows $j^2=4$ i.e. $j=\pm 2$. Checking against the second one, we see $3k+j = 1$ so $k=1, j=-2$ works and we get one solution is $\beta = \alpha^2 -2$. And clearly the opposite will still square to this, so negating all the coefficients verifies $\gamma = -\alpha^2-\alpha+2$. The image of $\alpha^2$ is the image of $\alpha$, squared, i.e. is $\beta^2$, so we square this to get:

$$\beta^2 = \alpha^4-4\alpha^2+4 = 3\alpha^2-\alpha-4\alpha^2+4 = -\alpha^2-\alpha + 4$$

So our matrix for a generator of the Galois group which corresponds to the permutation $\alpha\mapsto\beta\mapsto\gamma$ is just

$$\sigma = \begin{pmatrix} 1 & -2 & 4 \\ 0 & 0 & -1 \\ 0 & 1 & -1 \end{pmatrix} $$

We can even confirm that $\sigma^3= I$, the $3\times 3$ identity matrix, verifying its order.

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The roots of $f(x)$ will be of the form $\alpha, \beta, -\alpha - \beta$, as the coefficient of $x^{2}$ is zero. Now a basis of $\mathbb{Q}(\alpha)/\mathbb{Q}$ will be $$\tag{basis} 1, \alpha, \beta. $$ You know that one of the generators $g$ of the Galois group will permute the roots cyclically $$ \alpha \mapsto \beta \mapsto - \alpha - \beta \mapsto \alpha. $$ Then, using (basis), a matrix for this generator will be $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1\\ 0 & 1 & -1\\ \end{bmatrix}. $$

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First, I think you meant $C_3$ and not $A_3$ since $K=\mathbb Q(\alpha)$ is Galois. That said, it means that the quotient of $x^3-3x+1$ by $x-\alpha$, which is $x^2+\alpha x+\alpha^2-3$, is reducible in $K$. Thus, its discriminant $12-3\alpha^2$ is a square in $K$. Now write $z=a+b\alpha+c\alpha^2$ and solve $z^2=12-3\alpha^2$. With some patience, you'll find $(a,b,c)=\pm (2,1,-4)$ and thus the other roots of $f$.

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    $\begingroup$ When I write $A_3$, I usually mean $A_3$. But you can write $C_3$ since they are isomorphic. Thank you for the idea. Previously I put $a + b\alpha + c\alpha^2$ directly into $x^3 - 3x + 1$. Is it the only possible strategy for such problems? $\endgroup$ – nightked Nov 22 '16 at 18:34

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