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In another question I asked how I would find the miminal polynomial of a primitive nth root of unity over $\mathbb Q$, which was very well answered and easy to follow.

  • Taking the same example, let $ζ_1$ be a primitive 21st root of 1. Find the minimal polynomial of $ζ_1$ over $\mathbb Q$.

$\Phi_{21}(x) = \dfrac{x^{21}-1}{\Phi_1(x) \Phi_3(x) \Phi_7(x)} = x^{12}-x^{11}+x^9-x^8+x^6-x^4+x^3-x+1$.

My questions are

  • what is the most effective way of dividing $x^{21}-1$ by $\Phi_1(x) \Phi_3(x) \Phi_7(x)$? All the ways I have been through are very long and arduous, so I was wondering if there is a more efficient way?

  • What is the best approach to finding the splitting field for this polynomial over $\mathbb Q$? I'm quite stuck on this, should I factorise then try to find the roots?

Any help would be amazing!

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Polynmial division yields the result, if you know already $\Phi_1(x)=x-1$, $\Phi_3(x)=x^2+x+1$ and $\Phi_7(x)=x^6+\cdots +x+1$. First, start dividing by $x-1$, this is easy. Then divide by the next, and so on.
Otherwise, a very effective way is to use a factorisation algorithm for integer polynomials, yielding $$ x^{21}-1=(x^{12} - x^{11} + x^9 - x^8 + x^6 - x^4 + x^3 - x + 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)(x^2 + x + 1)(x - 1). $$ The splitting field is, by definition, the cyclotomic field $\mathbb{Q}(\zeta_{21})$, of degree $12$ over $\mathbb{Q}$.

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  • $\begingroup$ I see, thank you! Would you be able to describe how to find the splitting field of this polynomial? $\endgroup$ – user484410 Nov 6 '17 at 19:32
  • $\begingroup$ See my answer.. $\endgroup$ – Dietrich Burde Nov 6 '17 at 19:38
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By the inclusion/exclusion principle or the Moebius inversion formula we have

$$ \Phi_{21}(x)=\frac{(x^{21}-1)(x-1)}{(x^3-1)(x^7-1)} = (1-x-x^{21}+x^{22})\sum_{n\geq 0}x^{3n}\sum_{n\geq 0}x^{7n} $$ and we know in advance that $\deg\Phi_{21}=\varphi(21)=12$, hence we may compute the coefficients of $\Phi_{21}(x)$ by expanding $(1-x)\sum_{m=0}^{12}r(m) x^m$ (where $r(m)$ is the number of ways for writing $m$ as the sum between a multiple of $3$ and a multiple of $7$) and truncating at the $x^{12}$ term. $$(1-x)(1+x^3+x^6+x^7+x^9+x^{10}+x^{12})=\underbrace{\color{green}{1 - x + x^3 - x^4 + x^6 - x^8 + x^9 - x^{11} + x^{12}}}_{\Phi_{21}(x)}+\ldots.$$

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