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I don't know why I'm having trouble with this, but I can't quite see whether the cyclotomic polynomials are considered solvable. Obvioulsy we can write the solution of the nth cyclotomic polynomial as the nth root of unity, which seems to be a perfectly good algebraic solution; but aren't there usually "better" solutions? I've done the fifth and seventh, and I think Gauss is credited for solving the eleventh if I'm not mistaken; the seventeenth falls out of the construction of the regular 17-gon...but I'm still not clear: are all of them, in theory, solvable "explicitly", which is to say beyond the level of just saying that the nth root of unity is a solution?

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Define "better". What was your better root of the 5th cyclotomic polynomial beside a 5th root of unity? What's "beyond the level of" saying that the $n$th root of unity is a solution? Define "beyond the level of". –  Barry Smith Dec 18 '11 at 19:09
    
I think "better" means e.g. in terms of real radicals (so the difference between saying that $\zeta_3 = \sqrt[3]{1}$ and saying that $\zeta_3 = - \frac{1}{2} + i \frac{ \sqrt{3} }{2}$). –  Qiaochu Yuan Dec 18 '11 at 19:15
    
Ah, solvable by radicals. I don't know that I'd consider $\frac{-1+\sqrt{5}}{4}+i \sqrt{\frac{5+\sqrt{5}}{8}}$ "better" than $e^{2\pi i/5}$, but I understand the question now. –  Barry Smith Dec 18 '11 at 19:53

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up vote 10 down vote accepted

The method of Lagrange resolvents succeeds to solve cyclotomic equations in radicals (in the useful sense, not the "nth root of 1" sense). This is discussed in fairly elementary terms, with examples, in chapter 19 of my algebra book/course-notes, at http://www.math.umn.edu/~garrett/m/algebra/

However, that essentially-elementary approach becomes burdensome quickly, certainly difficult to do or understand by hand calculation. But if we observe that the resolvents are _Gauss_sums_, then admixture of some ideas of Kummer and Eisenstein gives a much-improved, but less elementary approach to expression of roots of unity by radicals. This is discussed, with several examples, in http://www.math.umn.edu/~garrett/m/v/kummer_eis.pdf

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As I said in my comment to Qiaochu, I'm not capable of following your math, but I looked through your Chapter 19 from beginning to end, and if you haven't nailed it completely, then you're an awfully good faker. Answer accepted. –  Marty Green Dec 18 '11 at 22:59
    
@Marty: it's worth pointing out that the expression on p. 253 is not much more explicit than the expression $\zeta_7 + \zeta_7^{-1}$, seeing as it requires that we take the cube root of a complex number with nonzero real part (this despite the fact that $\zeta_7 + \zeta_7^{-1}$ is a root of a cubic polynomial with all real roots!). See en.wikipedia.org/wiki/Casus_irreducibilis . –  Qiaochu Yuan Dec 18 '11 at 23:10
    
@Marty: AFAICT Paul's nice notes do not contain a formula for $\zeta_7$ with real and imaginary parts expressed using real radicals, if that is what you are hoping to see. I don't know, if such a formula exists at all. –  Jyrki Lahtonen Dec 20 '11 at 6:34
    
No, I'm sorry, that was never a consideration for me. I should have recognized from Qiaochu's first comment that he was interpreting my question that way as well, because I think that was the source of our whole miscommunication. I see now that he's making exactly that point in this comment thread about the real part of the seventh root of unity: I never cared about separating the real and imaginary parts. I clarify this point in my follow up question, "Non-trivial Solutions for the Cyclotomic Polynomials". –  Marty Green Dec 20 '11 at 12:41

A polynomial is solvable iff its splitting field is solvable iff its Galois group is solvable. The Galois group of the cyclotomic polynomial $\Phi_n(x)$ is $(\mathbb{Z}/n\mathbb{Z})^{\ast}$, which is abelian, and all abelian groups are solvable.

An interesting question is how to find "nice" radical expressions for roots of unity. One way to do this is to go through the proof that the roots of a solvable polynomial are expressible in radicals by finding a composition series of the Galois group and constructing Kummer extensions.

This is perhaps easiest to describe by example, so take $n = 5$. Then $\Phi_5(x) = x^4 + x^3 + x^2 + x + 1$ has Galois group $(\mathbb{Z}/5\mathbb{Z})^{\ast} \cong C_4$, so it has a composition series with two factors of $C_2$. This implies that $\mathbb{Q}(\zeta_5)$ is a quadratic extension of a quadratic extension. Basic facts about Gauss sums imply that the second quadratic extension is $\mathbb{Q}(\sqrt{5})$, so $\zeta_5$ satisfies a quadratic polynomial with coefficients in $\mathbb{Q}(\sqrt{5})$. This polynomial must be $$x^2 - (\zeta_5 + \zeta_5^{-1}) x + 1.$$

Again, basic facts about Gauss sums imply that $\zeta_5 + \zeta_5^{-1} = \frac{-1 + \sqrt{5}}{2}$, so it follows by the quadratic formula that $$\zeta_5 = \frac{ 1 - \sqrt{5} + i \sqrt{10 + 2 \sqrt{5}} }{4}.$$

In general one may need to take more than just square roots, in which case things get complicated. There will be expressions one can write down generalizing Gauss sums that are guaranteed to land in a specific subfield of $\mathbb{Q}(\zeta_n)$ but I don't know a good way of actually figuring out what those expressions are by hand, and in writing down Kummer extensions one may need to adjoin smaller roots of unity (so I believe the most important case is when $n$ is prime).

A basic remark is that the problem reduces to the case that $n$ is a power of a prime, since for arbitrary $n$ it is possible to write $\zeta_n$ as a product of roots of the form $\zeta_{p^k}$ where $p^k \parallel n$.

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Is your $\zeta_5$ actually $\zeta_{10}$? –  Barry Smith Dec 18 '11 at 19:55
    
I was thinking of constructibility by ruler and compass in my now deleted answer. A temporary aberration! –  Geoff Robinson Dec 18 '11 at 19:56
    
Qiaochu, as usual your answer is beyond reproach; my complain is not with the quality of your answer but only perhaps because my knowledge is insufficient to make obvious the small point you have left out. The Galois group is abelian; hence the equation is solvable by radicals. Is not, however, the expression (1)^(1/n) also an expression in radicals? I know you have given the example of Kummer extensions, but I don't understand that well enough to see immediately that such techniques can be extended to polynomials of any degree. –  Marty Green Dec 18 '11 at 20:12
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$p=5$ is probably not the best example. The equation $x^{p-1}+x^{p-2}+...+x+1=0$ is reciprocal, thus the standard substitution $x+\frac{1}{x}$ reduces it to an equation of degree $\frac{p-1}{2}$. For $p=5$ you get a quadratic, so clearly solvable... –  N. S. Dec 18 '11 at 20:14
    
I might ask as well if such explicit solutions have been tabulated? I already speculated that Gauss was responsible for the explicit solution of the eleventh; wouldn't all the prime orders be of interest, and in that case what is the lowest prime order cyclotomic polynomial that hasn't yet been solved "explicitly"? –  Marty Green Dec 18 '11 at 20:14

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