31
$\begingroup$

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.

$\endgroup$
11
  • 5
    $\begingroup$ It's what I figured! Nonetheless, I like to read a precise question. :) $\endgroup$ Commented Dec 1, 2011 at 6:18
  • 7
    $\begingroup$ 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. $\endgroup$ Commented Dec 1, 2011 at 6:39
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Dec 2, 2011 at 10:41
  • 2
    $\begingroup$ @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.) $\endgroup$ Commented Dec 9, 2011 at 10:06
  • 4
    $\begingroup$ 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. $\endgroup$
    – ougao
    Commented Dec 17, 2012 at 13:49

2 Answers 2

39
+50
$\begingroup$

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.

$\endgroup$
0
5
$\begingroup$

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 .

$\endgroup$
1
  • 1
    $\begingroup$ 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 $\mathbb{Q}(\zeta_p,\zeta_k)=\mathbb{Q}(\zeta_{pk})$? $\endgroup$ Commented Dec 5, 2011 at 7:26

You must log in to answer this question.

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