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Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$. Let $K = \mathbb{Q}(\zeta)$. Is it true that any root of unity in $K$ is of the form $\pm\zeta^k$ where $k$ is an integer?

Motivation: A root of unity in $K$ is an invertible element of the ring of algebraic integers in $K$. The determination of the group of invertible elements of this ring is important for several reasons. For example, it is used in the computation of the class number of $K$.

I came up with two different ideas each of which might solve this problem.

(1) Use the fact that the group of roots unity in $K$ is finite. Hence this group is cyclic. Let $\omega$ be its generator. Compare [$\mathbb{Q}[\omega] : \mathbb{Q}$] with $l - 1$ = [$K : \mathbb{Q}$].

(2) Use the fact that the only prime number which ramifies in a cyclotomic number field of prime power order $p^n$ is $p$, except $p = 2$ and $n = 1$.

Related question: The group of roots of unity in an algebraic number field

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Makoto Kato
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  • Yes, of course. What have you tried so far? – Pete L. Clark Jul 25 '12 at 11:46
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    Looking at the main page these last few days, I've seen quite a large number of questions from you on cyclotomic number fields, integers and polynomials; most of the time one of them was quite far up on the page. I know you're well aware of the issue of economic use of space on the main page due to the discussions about your self-answered questions on commutative algebra without choice. I'm wondering whether it might not be possible to aggregate all these questions somewhat, thus not only economizing on a scarce public good but also perhaps exhibiting the connections between them to others? – joriki Jul 25 '12 at 11:48
  • @Makoto: Okay, that's a good start. Where does that get you? – Pete L. Clark Jul 25 '12 at 11:57
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    @joriki Please open a meta thread to discuss non-mathematical issue. – Makoto Kato Jul 25 '12 at 11:57
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    I was hoping to suggest, not to discuss. Does your answer imply that you currently don't intend to follow my suggestion? – joriki Jul 25 '12 at 11:59
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    "I think I solved it, but I prefer not to answer my question for some reason." I'm sorry, that's not a response which motivates me to continue. – Pete L. Clark Jul 25 '12 at 12:12
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    @MakotoKato , I think you're exaggerating, at least this time. You could, say, write down your solution to your question and ask for some possible simplification of this or that step. Stating that you've the solution but you prefer not to show it makes the whole issue JamesBondian but not mathematically interesting and, I'd dare to say, even a little annoying, like you're daring us or teasing us. – DonAntonio Jul 25 '12 at 12:25
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    @DonAntonio I prefer not to answer my question just because some people don't like it. This is no secret. I didn't make it clear because I didn't want to start the argument again! – Makoto Kato Jul 25 '12 at 13:30
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    @joriki Some of my questions are related, but they are different questions. I think merging them into one thread is confusing. For example, the title would be misleading. – Makoto Kato Jul 25 '12 at 15:08
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    It's not only about merging, @MakotoKato, but also about not giving background, explanation, showing no effort in your own question, in particular after you stated you've already solved it yet you won't publich it, and I still cannot understand why. I don't think that's the way to get some appreciation from the forum, in particular after you've asked many closely related questions. – DonAntonio Jul 25 '12 at 15:39
  • @DonAntonio I don't think this question needs background and explanation. It's obvious for people who have basic knowledge of algebraic number theory. For people who don't, this question is not for them. As for my not answering my question, it's only because some people don't like or hate it. We discussed about it a lot in meta. – Makoto Kato Jul 25 '12 at 16:01
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    "@MakotoKato, as you wish. BTW, if in your question you meant "Any $,\mathcal l-,$th root of unity..." , then it has a completely trivial answer not only for anyone who has a basic knowledge of alg. number theory but for anyone with a basic knowledge of group theory. – DonAntonio Jul 25 '12 at 16:08
  • @DonAntonio I said any root of unity, but I didn't say any $l$-th root of unity. Those questions are totally different. I meant the former. – Makoto Kato Jul 25 '12 at 16:37
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    "I don't think this question needs background and explanation..." Translation: "I can't be bothered; if you get it, you're clearly smart; if you don't get it, you are ignorant and I don't want to hear from you anyway." – Arturo Magidin Jul 26 '12 at 00:25
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    @ArturoMagidin I added motivation. – Makoto Kato Jul 26 '12 at 00:34
  • Please let me know the reasons for the downvotes.Unless you make them clear, it is hard to improve my question. – Makoto Kato Jul 28 '12 at 03:14
  • @joriki and others, Please reset your downvotes, since the problem of frequently posting questions was solved(the limit was introduced). I'm not asking this for myself, but for the readers who might think this question is useless because of the downvotes. – Makoto Kato Sep 19 '12 at 00:49
  • I don't know how you inferred that I downvoted; I didn't. Also, I get the impression you still haven't understood what the problem was with your self-answered questions. It was made quite clear in the discussions about them that answering your own questions is not in and of itself a problem. – joriki Sep 19 '12 at 07:01
  • @joriki Please tell me what the problem was with my questions. – Makoto Kato Sep 19 '12 at 11:21
  • I have nothing to add to this answer, which received the most upvotes in the thread discussing your questions at the time. – joriki Sep 19 '12 at 11:29
  • @joriki "However, from these three statements it doesn't follow that posting partial answers to one's own questions and then editing them a lot is welcome." This problem was solved 2 months ago when the sandbox thread in meta was made. Please notice that this question was asked after that. – Makoto Kato Sep 19 '12 at 20:44
  • @Makoto: Perhaps that was a misunderstanding. By "your self-answered questions" I was referring to the questions you asked before the sandbox was created. I had assumed that you had incorrectly inferred from the reaction to those questions that "some people don't like or hate" answering your own question. If that's not how you inferred this, I'd be interested to know how you did infer it. – joriki Sep 19 '12 at 20:49
  • @joriki The problems of editing and asking a lot of question were solved. They no longer exist. So if the reasons for the downvotes were related to these issues, I ask the members who donvoted to reset. – Makoto Kato Sep 19 '12 at 21:10
  • @joriki I apologize for inferring that you downvoted. If you want me to delete the comment, I will do so. – Makoto Kato Sep 19 '12 at 21:45

1 Answers1

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The OP has already suggested two ways to solve this:

(a) Let $\zeta_n$ generated the group of roots of unity in $\mathbb Q(\zeta_l)$. Then $2l$ divides $n$, and also $\mathbb Q(\zeta_l) = \mathbb Q(\zeta_n)$. A consideration of degrees shows that $\varphi(n) = \varphi(l)$, and combining this with the fact that $2l$ divides $n$, elementary number theory implies that in fact $n = 2l$.

(b) Ramification theory rules out the possibility of $\zeta_n$ lying in $\mathbb Q(\zeta_l)$ if $n$ is divisible by an odd prime $p \neq l$ or by a power of $2$ greater than the first.

Here are some other arguments (I continue to let $\zeta_n$ be the generator of the roots of unity in $\mathbb Q(\zeta_l)$):

(c) Galois theoretic: since $\mathbb Q(\zeta_l) = \mathbb Q(\zeta_n)$, passing to Galois groups over $\mathbb Q$, we find that the units in $\mathbb Z/n$ project isomorphically onto the units in $\mathbb Z/l$. Given that $2l | n$, we deduce from the Chinese remainder theorem that $n = 2l$.

(d) Discriminants: Since $\mathbb Q(\zeta_l) = \mathbb Q(\zeta_n)$, a consideration of the standard discriminant formulas shows that $n = 2l$.

(e) Looking at the reduction modulo split primes: Choose $p$ prime to $n$ and congruent to $1$ mod $l$. Then the group of $n$th roots of unity injects into the residue field of any prime lying over $p$. Since $p \equiv 1 \bmod l$, this residue field is just $\mathbb F_p$, and so we find that $n | p-1$ if $p > n$ (say) and $p \equiv 1 \bmod l$. Dirichlet's theorem then gives that the units in $\mathbb Z/n$ project isomorphically onto the units in $\mathbb Z/l$, from which we deduce that $n = 2l$.

(f) Working locally at l: it is not hard to check that the roots of unity in $\mathbb Q_l(\zeta_l)$ are precisely $\mu_{l(l-1)}$. So we have to show that the only $(l-1)$st roots of $1$ in $\mathbb Q(\zeta_l)$ are $\pm 1$. Actually I don't see how to do this right now without reverting to one of the other arguments, but there's probably a pithy way.

Note that (c) is just a fancy version of (a), while (d) is a more concrete form of (b) (which uses less theory). It may seem that (e) is overkill, and it certainly is for this question, but the method can be useful, and it has an obvious connection to (c) via reciprocity laws. Method (f) (unfortunately incomplete) is related to (b).

Matt E
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