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

  • 17
    $\begingroup$ 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? $\endgroup$
    – joriki
    Commented Jul 25, 2012 at 11:48
  • 8
    $\begingroup$ I was hoping to suggest, not to discuss. Does your answer imply that you currently don't intend to follow my suggestion? $\endgroup$
    – joriki
    Commented Jul 25, 2012 at 11:59
  • 13
    $\begingroup$ "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. $\endgroup$ Commented Jul 25, 2012 at 12:12
  • 12
    $\begingroup$ @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. $\endgroup$
    – DonAntonio
    Commented Jul 25, 2012 at 12:25
  • 8
    $\begingroup$ "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." $\endgroup$ Commented Jul 26, 2012 at 0:25

1 Answer 1


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).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .