I'm doing this Galois Theory question:
$f = x^3 + x + 3$ is known to be irreducible and has just one real root, call it $\theta$.
What is the minimal polynomial for $\theta^2$? Is it just $f$?
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4There is a standard trick here. The polynomial $f(x)f(-x)$ is even. Therefore we can write $$g(x^2)=-f(x)f(-x)$$ for some monic polynomial $g(x)\in\Bbb{Z}[x]$ of degree three. It follows that $g(\theta^2)=0$. This is also a duplicate. Gimme a minute. – Jyrki Lahtonen Nov 05 '15 at 10:59
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2If $\theta$ is real, then so is $\theta^2$. But $f$ has only one real root. Can the required minimal polynomial be $f$? – Git Gud Nov 05 '15 at 10:59
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Good point @Git Gud – John.P Nov 05 '15 at 11:04
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The method is described with a quartic example here. I have seen more recent incarnations of it here, but my search-fu is weak today :-( – Jyrki Lahtonen Nov 05 '15 at 11:16
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The trick plays a role in Graeffe's method. – Jyrki Lahtonen Nov 05 '15 at 11:20
2 Answers
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The minimal polynomial of $\theta^2$ must have degree at most $3$.
So, we need to look for a linear equation for $1, \theta^2, \theta^4, \theta^6$.
$\theta^3 = -\theta-3$ implies $\theta^4= -\theta^2-3\theta$ and $\theta^6= \theta^2+6\theta+9$. So $\theta^6+2\theta^4=-\theta^2+9$.
A systematic way of refining this relation is to solve a linear system $Ax=0$, where the columns of $A$ are the coordinates of $1, \theta^2, \theta^4, \theta^6$ in the basis $1, \theta, \theta^2$.
lhf
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I was trying something like this but it just wouldn't work! Thanks a lot for that, I like your method. – John.P Nov 05 '15 at 11:05
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There’s another systematic way of doing this, involving the Norm. Let $R=\Bbb Q[X]$, and look at $S=R[\theta]/(\theta^3+\theta+3)$, and the Norm from $S$ down to $R$. You can calculate the Norm of an element $z$ by looking at the $R$-linear map $w\mapsto zw$, and calculating its determinant. It’s just a $3\times3$ with coefficients in $R$, not bad at all to do.
But the Norm of $X-\theta^2$ is clearly a cubic polynomial in $R=\Bbb Q[X]$ that has $\theta^2$ for a root. In this case, it is irreducible, and the computation gives $-9 + X + 2X^2 + X^3$.
Lubin
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