# Basis of primitive nth Roots in a Cyclotomic Extension?

While reading one of Keith Conrad's great blurbs, Linear Independence of Characters, there is a footnote at the bottom of page 2 saying

In general, the primitive $$n$$th roots of unity in the $$n$$th cyclotomic field form a normal basis over $$\mathbf{Q}$$ if and only if $$n$$ is squarefree.

A little bit of research didn't turn up any results, except apparently the result is in the paper K. Johnsen, Lineare Abhängigkeiten von Einheitswurzeln. Elem. Math. 40 (1985), 57–59 which I can't find anywhere. (J.M. found it here.)

Could someone be kind enough to give a proof of why this statement is true?

I made this attempt at the if direction. I try induction on $$n$$. If $$n$$ is prime, then the result follows from field theory, since the powers of $$\alpha$$ form a $$F$$-basis of $$F(\alpha)$$ over $$F$$. So let $$n=mp$$ where $$m$$ is coprime to $$p$$. If $$n$$ is squarefree, so is $$m$$, so by induction, the $$m$$th primitive roots of units form a basis for $$\mathbb{Q}(\zeta_m)$$ over $$\mathbb{Q}$$.

Then I think the primitive $$p$$th roots of unity form a basis of $$\mathbb{Q}(\zeta_m,\zeta_p)$$ over $$\mathbb{Q}(\zeta_m)$$, so taking pairwise products of the two bases gives a basis of primitive $$n$$th roots of $$\mathbb{Q}(\zeta_m,\zeta_p)$$ over $$\mathbb{Q}$$? Does $$\mathbb{Q}(\zeta_m,\zeta_p)=\mathbb{Q}(\zeta_n)$$? I know $$\mathbb{Q}(\zeta_m,\zeta_p)\subseteq\mathbb{Q}(\zeta_n)$$, but am not sure of the other containment, or if this argument is oversimplifying things.

• It's what I figured! Nonetheless, I like to read a precise question. :) Commented Dec 1, 2011 at 6:18
• The only if part seems to be easy. If $p^2\mid n$, and $\zeta$ is a primitive root of unity of order $n$, then $\xi=\zeta^{n/p}$ is a primitive root of unity of order $p$, and the sum of the roots $\zeta\xi^j, 0\le j<p,$ equals $\zeta\phi_p(\xi)=0$. But $\zeta\xi^j=\zeta^{1+(n/p)j}$ is a primitive root of unity of order $n$, because all the prime factors of $n$ are also factors of $n/p$. Thus there are linear dependences over the rationals among the primitive roots. I don't know about the other direction. Commented Dec 1, 2011 at 6:39
• I looked up the link given by @J.M. The other direction (if) seems to follow from the tensor product decomposition based on the fact that any cyclotomic field is a tensor product (over the rationals) of cyclotomic fields of a prime power conductor. If you can read German at all, please look at the article, and post a summary as an answer. If you don't know any German, I can try and post an outline later. Alas, RL duties won't allow me to put enough time towards doing that until later. Commented Dec 2, 2011 at 10:41
• @KCd: the moderators can remove bounties (I think). But is there a reason why you would want the bounty removed, as opposed to awarded to you? (See discussion: meta.math.stackexchange.com/q/3319/1543 If you don't feel like commenting publicly, you can reach me by e-mail.) Commented Dec 9, 2011 at 10:06
• Another method to show the "only if " direction is to use the fact that the trace of $\zeta_n$ is equal to zero if n is not square free, while by definition, the trace of $\zeta_n$ in this case is exactly the same as the sum of all the primitive n-th roots of unity, so we have a linearly dependent relation over $\mathbb{Q}$ for all the primitive n-th roots, so they could not form a basis, see the details in Dummit and Foote's "Abstract Algebra, 3rd Edition" page 603, exc. 6& 11. Commented Dec 17, 2012 at 13:49

In the comments to the question, Jyrki showed that if $$n$$ has a square factor bigger than $$1$$ then there are nontrivial $${\mathbf Q}$$-linear relations among $$\{\zeta_n^a : (a,n) = 1\}$$, so this set can only be a $${\mathbf Q}$$-basis of $${\mathbf Q}(\zeta_n)$$ when $$n$$ is squarefree. Another way to see this is that for all $$n \geq 1$$, the sum of the primitive $$n$$-th roots of unity is the Moebius value $$\mu(n)$$, and this is $$0$$ if $$n$$ has a square factor bigger than $$1$$.

To show that when $$n$$ is squarefree the set $$\{\zeta_n^a : (a,n)=1\}$$ is a $${\mathbf Q}$$-basis of $${\mathbf Q}(\zeta_n)$$ we will use induction on the number of prime factors of $$n$$.

Suppose first that $$n = p$$ is a prime. The usual $${\mathbf Q}$$-basis of $${\mathbf Q}(\zeta_p)$$ is $$\{1,\zeta_p,\zeta_p^2,\cdots,\zeta_{p}^{p-2}\}$$. Since $$\zeta_p^{p-1} = -1-\zeta_p - \cdots - \zeta_p^{p-2}$$, if we replace $$1$$ with $$\zeta_p^{p-1}$$ we still have a $${\mathbf Q}$$-basis, so $$\{\zeta_p,\zeta_p^2,\cdots,\zeta_p^{p-1}\}$$ is a basis of $${\mathbf Q}(\zeta_p)/{\mathbf Q}$$.

Now suppose we've proved the result when $$n$$ is any product of $$r$$ primes and consider a squarefree positive integer $$n$$ that is a product of $$r+1$$ primes. Write $$n = mp$$ where $$p$$ is one of the prime factors of $$n$$, so we know $$\{\zeta_m^i : 1 \leq i \leq m, (i,m) = 1\}$$ is a $${\mathbf Q}$$-basis of $${\mathbf Q}(\zeta_m)$$ and $$\{\zeta_p^{j} : 1 \leq j \leq p, (j,p) = 1\}$$ is a $${\mathbf Q}$$-basis of $${\mathbf Q}(\zeta_p)$$. From Galois theory, if $$E$$ and $$F$$ are (finite) Galois extensions of a common field $$L$$ and $$E \cap F = L$$ then as a basis of $$EF$$ over $$L$$ one can use the set of products $$\{e_if_j\}$$ where $$\{e_i\}$$ is any $$L$$-basis of $$E$$ and $$\{f_j\}$$ is any $$L$$-basis of $$F$$. We can apply this to the fields $$E = {\mathbf Q}(\zeta_m)$$, $$F = {\mathbf Q}(\zeta_p)$$, and $$L = {\mathbf Q}$$. (That the intersection of those two cyclotomic fields is $${\mathbf Q}$$ is a special case of a general formula $${\mathbf Q}(\zeta_r) \cap {\mathbf Q}(\zeta_s) = {\mathbf Q}(\zeta_{(r,s)})$$, which becomes $${\mathbf Q}$$ when $$r$$ and $$s$$ are relatively prime.) Therefore a $${\mathbf Q}$$-basis of $$EF = {\mathbf Q}(\zeta_m,\zeta_p) = {\mathbf Q}(\zeta_n)$$ is the set of products $$\{\zeta_m^i\zeta_p^j\}$$ with $$i$$ and $$j$$ running over the integers listed earlier. Inside $${\mathbf Q}(\zeta_n)$$ we can use $$\zeta_m := \zeta_n^{n/m}$$ and $$\zeta_p := \zeta_n^{n/p}$$, so a basis is $$\{\zeta_n^{(n/m)i + (n/p)j}\} = \{\zeta_n^{pi + mj}\}$$ where $$i$$ runs over integers from 1 to $$m$$ which are relatively prime to $$m$$ and $$j$$ runs over integers from 1 to $$p$$ which are relatively prime to $$p$$ (that is, $$1 \leq j \leq p-1$$).

Which $$n$$th roots of unity are in the set $$\{\zeta_n^{pi + mj}\}$$ and how many are there? The integers $$pi+mj$$ are all relatively prime to $$n = mp$$ (just reduce them mod $$m$$ and mod $$p$$ to check they are relatively prime to $$m$$ and $$p$$ separately), so the set consists of primitive $$n$$th roots of unity. If $$pi + mj \equiv pi' + mj' \bmod n$$ (where $$i'$$ and $$j'$$ are just second choices of parameters) then by reducing mod $$m$$ and mod $$p$$ we get $$i \equiv i' \bmod m$$ and $$j \equiv j' \bmod p$$, so $$i = i'$$ and $$j = j'$$ on account of the ranges of these parameters. Therefore the number of roots of unity in this set is $$\varphi(m)\varphi(p) = \varphi(n)$$, so our set is exactly the set of all primitive $$n$$th roots of unity. This completes the proof that the primitive $$n$$th roots of unity are a $${\mathbf Q}$$-basis of $${\mathbf Q}(\zeta_n)$$ when $$n$$ is squarefree.

Although the question has now been answered, let me indicate a place where it naturally fits into a broader picture within algebraic number theory. The result is related to a theorem of Emmy Noether on normal integral bases.

For any (finite) Galois extension $$K/{\mathbf Q}$$, a normal integral basis is a normal basis for the field extension which consists of a $${\mathbf Z}$$-basis of $${\mathcal O}_K$$. For example, $${\mathbf Q}(\sqrt{5})/{\mathbf Q}$$ has normal integral basis $$\{(1+\sqrt{5})/2,(1-\sqrt{5})/2\}$$. Another example is $${\mathbf Q}(\zeta_p)$$ for any odd prime $$p$$: the ring of integers is $${\mathbf Z}[\zeta_p]$$ and the usual $${\mathbf Z}$$-basis you may want to use is $$\{1,\zeta_p,\cdots,\zeta_p^{p-2}\}$$, but that's not a normal basis because it has 1 in it. Instead you can use $$\{\zeta_p,\zeta_p^2,\cdots,\zeta_p^{p-1}\}$$; that is a normal basis and it's also a $${\mathbf Z}$$-basis of $${\mathbf Z}[\zeta_p]$$, so $${\mathbf Q}(\zeta_p)/{\mathbf Q}$$ has a normal integral basis.

Here's an example without a normal integral basis. If $$d$$ is squarefree and not $$1 \bmod 4$$, the ring of integers of $${\mathbf Q}(\sqrt{d})$$ is $${\mathbf Z}[\sqrt{d}]$$. A normal basis over $${\mathbf Q}$$ has the form $$\{a+b\sqrt{d},a-b\sqrt{d}\}$$ for rational $$a$$ and $$b$$ with $$b \not= 0$$. If this basis is in $${\mathbf Z}[\sqrt{d}]$$ then it can't be a $${\mathbf Z}$$-basis of $${\mathbf Z}[\sqrt{d}]$$ because $${\mathbf Z}(a+b\sqrt{d}) + {\mathbf Z}(a-b\sqrt{d})$$ has index $$2|ab| \geq 2$$ inside $${\mathbf Z}[\sqrt{d}]$$.

The ring of integers of $${\mathbf Q}(\zeta_n)$$ is $${\mathbf Z}[\zeta_n]$$. Check that if $$\{\zeta_n^a : (a,n) = 1\}$$ were a basis of $${\mathbf Q}(\zeta_n)/{\mathbf Q}$$ then it would be a $${\mathbf Z}$$-basis of $${\mathbf Z}[\zeta_n]$$, hence the $$n$$-th cyclotomic field would have a normal integral basis.

Emmy Noether proved that if a Galois extension $$K/{\mathbf Q}$$ has a normal integral basis then $$K$$ is tamely ramified over $${\mathbf Q}$$. So being tamely ramified is a necessary (although generally not sufficient) condition for a Galois extension of $${\mathbf Q}$$ to have a normal integral basis. For example, if $$d$$ is squarefree and $$d \not\equiv 1 \bmod 4$$ then 2 is not tamely ramified in $${\mathbf Q}(\sqrt{d})$$, so Noether's theorem implies that this quadratic field has no normal integral basis over $${\mathbf Q}$$, which we already proved by a direct computation above. An example more relevant for us here is $${\mathbf Q}(\zeta_{p^2})$$. The prime $$p$$ is not tamely ramified in this field, so any cyclotomic field $${\mathbf Q}(\zeta_n)$$ with $$n$$ divisible by the square of a prime is not tamely ramified at that prime, hence it doesn't have a normal integral basis. Therefore a necessary condition for $$\{\zeta_n^a : (a,n) = 1\}$$ to be a normal basis of the $$n$$-th cyclotomic field is that $$n$$ is squarefree. Noether's theorem has provided a conceptual explanation for why $$n$$ must be squarefree.

I suggest looking at Robert Long's book "Algebraic Number Theory" for more information on tamely ramified extensions of $${\mathbf Q}$$ and the relation to normal integral bases. I don't have the book in front of me right now and Google Books is not giving good views of it, but I'm pretty sure he has a chapter on this topic in it since Galois module structure was one of his areas of interest.

But the solution would be trying on induction, I am not sure but I looked for some help and found this :

Solution : Consider $n$ to be a prime number, and we know that powers of the adjoined element in a simple extension form a basis. But when $n=kp$, $k$ is squarefree, so by the induction, the primitive $k$-th roots of unity form a basis for $\mathbb{Q}(\zeta_{k})/\mathbb{Q}$ . By the result of simple extensions, the primitive $p$-th roots of unity form a $\mathbb{Q}(\zeta_{k})$ basis of $\mathbb{Q}(\zeta_{k},\zeta_{p})$ over $\mathbb{Q}(\zeta_{k})$ , so letting the basis run over products of primitive roots of $\zeta_{p}$ and $\zeta_{k}$ gives a basis of primitive $n$-th roots of $\mathbb{Q}(\zeta_{k},\zeta_{p})$ over $\mathbb{Q}$.

I never want a bounty even though if this answer may be right,

Thank you .

• I think you have the right idea, but I'd like to see more details. Is it useful that $\phi(kp)=\phi(k)\phi(p)$ and that $\mathbb{Q}(\zeta_p,\zeta_k)=\mathbb{Q}(\zeta_{pk})$? Commented Dec 5, 2011 at 7:26