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Problem of the Week #40 - March 4th, 2013

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Chris L T521

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Jan 26, 2012
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Here's this week's problem.

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Problem: Let $\theta$ be a root of $x^3-3x+1$. Prove that the splitting field of this polynomial is $\mathbb{Q}(\theta)$ and that the Galois group is cyclic of order $3$. In particular the other roots of this polynomial can be written in the form $a+b\theta+c\theta^2$ for some $a,b,c\in\mathbb{Q}$. Determine the other roots explicitly in terms of $\theta$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's question was partially answered correctly by jakncoke (you got the first part, but I don't agree with your root values). You can find our fused solution below.

Proof: $x^3-3x+1$ is irreducible over $\mathbb{Q}$ with the mod 2 test (coefficient mod 2) have no zeros in $\mathbb{Z}_2$.The splitting field for this polynomial would be of the form $Q(\theta,\sqrt{D})$ where D is the discriminant, and $\alpha$ is a root.

Since D is 81, $\sqrt{81} = \pm 9$, so $Q(\theta,\sqrt{81}) = Q(\theta)$. Thus the degree of the Galois group is deg($x^3-3x+1) = 3$ which is cyclic since it is of degree 3 (prime order is cyclic).

Let us now find the other roots in terms of $\theta$. We know that $x^3-3x+1 = (x-\theta)(x^2+\theta x+\theta^2-3)$, so we're interested in finding out whether $x^2+\theta x+\theta^2-3$ has roots in $\mathbb{Q}(\theta)$. By the quadratic formula, the other roots are \[x=\frac{-\theta\pm\sqrt{\theta^2-4(\theta^2-3)}}{2}=\frac{-\theta\pm\sqrt{12-3\theta^2}}{2}.\]
However, for these solutions to be in $\mathbb{Q}(\theta)$, $12-3\theta^2$ must be a square. Since the roots are of the form $a+b\theta+c\theta^2$ for some $a,b,c\in\mathbb{Q}$, we hope to find $a,b,c$ such that
\[12-3\theta^2 = \left(a+b\theta+c\theta^2\right)^2.\qquad\qquad(1)\]
Since $\theta$ is a root of $x^3-3x+1$, this can help us define higher powers of $\theta$ in terms that have degree no more than $2$. With little effort, one finds that $\theta^3=3\theta-1$ and $\theta^4=3\theta^2-\theta$. Thus, after some algebraic manipulations, one finds that (1) can be rewritten as
\[a^2-2bc+(2ab+6bc-c^2)\theta+(2ac+b^2+3c^2)\theta^2=12-3\theta^2.\]
Therefore, we need to solve the system of equations
\[\left\{\begin{aligned}2ac+b^2+3c^2 &= -3\\ 2ab+6bc-c^2 &= 0\\ a^2-2bc &= 12\end{aligned}\right.\]
which (by sheer luck or computer) has the solutions $a=-4$, $b=1$, and $c=2$. Therefore $\left(2\theta^2+\theta-4\right)^2=12-3\theta^2$ and it follows that the other two roots are
\[x=\frac{-\theta+2\theta^2+\theta-4}{2}=\theta^2-2\]
and
\[x=\frac{-\theta-2\theta^2-\theta+4}{2} = 2-\theta-\theta^2.\]
(Note that you can also find the third root once you have the second, since $(\theta^2-2)^2-2=2-\theta-\theta^2$.)$\hspace{1in}\clubsuit$
 
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