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?
Cyclotomic polynomials explicitly solvable??
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2Define "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". – 2011-12-18
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0I think "better" means e.g. in terms of real radicals (so the difference between saying that $\zeta_3 = \sqrt[3]{1}$ and saying that $\zeta_3 = - \frac{1}{2} + i \frac{ \sqrt{3} }{2}$). – 2011-12-18
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0Ah, 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. – 2011-12-18
2 Answers
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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0As 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. – 2011-12-18
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0@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 http://en.wikipedia.org/wiki/Casus_irreducibilis . – 2011-12-18
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0@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. – 2011-12-20
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0No, 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". – 2011-12-20
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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0Is your $\zeta_5$ actually $\zeta_{10}$? – 2011-12-18
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0I was thinking of constructibility by ruler and compass in my now deleted answer. A temporary aberration! – 2011-12-18
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0Qiaochu, 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. – 2011-12-18
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1$p=5$ 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... – 2011-12-18
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0I 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"? – 2011-12-18
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0@Barry: I had a sign error. It should be correct now. – 2011-12-18
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0@Marty: yes, but it doesn't really tell you anything you didn't already know. The interesting question to my mind is how efficiently you can write $\zeta_n$ in terms of radicals of rational expressions together with possibly roots of unity $\zeta_m$ with $m < n$. I don't really have a good answer to this question. – 2011-12-18
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0I'm still having trouble making myself clear: just because the nth order cyclotomic polynomial is solvable, does that mean it must have "explicit" solutions other than "the nth roots of unity"? I think it does: I actually don't think the expressions for the nth roots of unity qualifies in an essential way as an "algebraic solution of the equation". But I'm just speculating here. – 2011-12-18
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0@Marty: you're not being very clear about what you mean by "explicit," but as I said in my answer, you can write down a "more explicit" expression by considering a factor series of the Galois group and doing various Kummer extensions. I don't know how good the answer you get from this procedure is in general, and I don't know what you mean by "in an essential way." – 2011-12-18
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0Qiaochu, I hope you will look at what Paul Garrett has written. I looked at his reference and it's way beyond me but it seems to answer my question exactly. So I'm hoping you'll agree that I ought to check his answer off as my accepted answer. – 2011-12-18
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0Qiaochu, I'm sorry we cross-posted: my comment went up before I read yours. The reason I say Paul answered my question explicitly is that he says Lagrangen resolvents solves the polynomial "in a useful" way, whereas your statement that an abelian group was solvable seemed to leave the possibility open that the polynomial might only be solvable in the trivial way (nth roots of unity). Appreciate your contribution and hope this clarifies my confusion. – 2011-12-18
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0@Marty: the first sentence of your question asks whether the cyclotomic polynomials are solvable, and the first paragraph of my answer addresses that question and only that question. Have you read my comment on Paul Garrett's answer? – 2011-12-18
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0@Qiaochu, of course I am reading carefully everything you tell me, I'm just not as fast as you. I know about the causus irreducibilis and I don't mind those expressions at all: I think they are much more in the spirit of "algebraic solutions" than the nth roots of unity. I have to grant that you may have anwered my question in your first paragraph except for a small point that wasn't explicitly clear to me, and that's the point that Paul cleared up. The main thing is you surely ought to realize that I'm not smart enough for you to be wasting your time arguing this point any further with me. – 2011-12-18
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0@Marty In Qiaochu's first sentence whenever a polynomial is said to be "solvable" it means "solvable in radicals" rather than referring to the existence of roots in the complex numbers. The existence of roots is guaranteed for any polynomial by the fundamental theorem of algebra. – 2011-12-19
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0I am quite sure that I am correct in identifying an ambiguity in Qiaochu's first sentence. I can technically solve the 5th order cyclotomic polynomial by writing x=(1)^(1/5). Is this or is this not a solution in radicals? I have great respect for Qiaochu but he simply fails to address this question; perhaps because the answer is too obvious for him. Yet that was, is, and remains my question. – 2011-12-20
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0More to the point, I suspect that the proof of solvability by radicals referred to by Qiaochu, namely the solvability of all abelian groups, specifically admits solutions of the type I am trying to avoid...the trivial nth roots of unity. I suspect that further proof is required to show that non-trivial solutions are available. Paul Garrett shows this by Langrangian resolvents, but it is not at all clear to me that the success of this method is already guaranteed by the general solvability of abelian groups. – 2011-12-20
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0@Marty: yes, that is a solution in radicals. It is not a very _interesting_ solution in radicals. That is not the question you asked in the OP. You did ask it in the third comment to this answer, and _I answered it_ in the seventh comment. The fact that casus irreducibilis shows up when describing $\zeta_7 + \zeta_7^{-1}$ shows that you cannot avoid "the trivial nth roots of unity" in general (you need $\zeta_3$ at least in this case, or $\zeta_4$). – 2011-12-20
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0I wonder how that is not the question I originally asked? What do you think I meant when I said "obviously we can write (them)...as the nth root of unity...but aren't there usually "better" solutions"...(rather than)...just saying the nth root of unity is a solution? I understood from your very first comment that you understood that I was looking for "interesting" solutions. – 2011-12-20
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0@Marty: the question you originally asked had no confusion about the definition of "in terms of radicals." It asked for more interesting expressions in terms of radicals. The question you are currently asking is whether $\sqrt[n]{1}$ counts as a solution in terms of radicals _at all_, which is a different question. Again, you asked this in the third comment to this answer, _not in the original post_, and again, I answered in the seventh comment. I don't see the point of arguing this any further. – 2011-12-20
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0Qiaochu, I posted a follow-up question at http://math.stackexchange.com/questions/92870/non-trivial-solutions-for-cyclotomic-polynomials which I hope you will want to look at. – 2011-12-20