Alternative proof that $[\Bbb{Q}(\zeta_n) : \Bbb{Q} ]= \varphi(n)$ uses circular reasoning?

I am doing exercise 3.24 of Marcus which is the following.

Let $L,K$ be number fields with $L/K$ a finite extension (of degree $[L:K] = n$) with $R = \mathcal{O}_K$ and $S = \mathcal{O}_L$. A prime $P$ of $R$ is said to be totally ramified in $L$ iff $PS = Q^n$ for some prime $Q$ of $S$.

(a) Show that if $P$ is totally ramified in $L$ then it is totally ramified in $M$ for any intermediate field $M$, $K \subseteq M \subseteq L$.

(b) If $P$ is totally ramified in $L$ and unramified in another extension $L'$ of $K$ then $L \cap L'= K$.

(c) Give a new proof that $\Bbb{Q}(w)$ where $w = e^{2\pi i/m}$ has degree $\varphi(m)$ over $\Bbb{Q}$ (First do it for $m = p^r$ by using the fact that $(p) = (1- w)^{\varphi(m)}$ and then build up to any $m$ by using (b) above.

Now I have done (a) and (b) but when I come to (c) I fail to see how the problem is essentially circular. Even in the case $m = p^r$, I would have to use the fact that the ring of integers of $\Bbb{Q}(w)$ is $\Bbb{Z}[w]$. The proof of this fact already uses knowledge of $[\Bbb{Q}(w) : \Bbb{Q}]$ being equal to $\varphi(m)$. Am I missing something here?

Thanks.

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You can get the equality of ideals $(p)=(1-w)^{\varphi(p^r)}$ (the case $m=p^r$) in the ring of integers of $\mathbf{Q}(w)$ without knowing precisely what that ring is (just that $w$ is in it and that for $r,s$ relatively prime, $(1-w^r)/(1-w^s)$ is a unit in it). Since $w$ is a root of the cyclotomic polynomial, which has degree $\varphi(p^r)$, the degree $[\mathbf{Q}(w):\mathbf{Q}]$ is at most $\varphi(p^r)$. On the other hand, from the formula $[\mathbf{Q}(w):\mathbf{Q}]=efg$, where $e,f,g$ are the usual invariants attached to $p$ for the Galois extension $\mathbf{Q}(w)/\mathbf{Q}$, the number of prime ideal factors of $(p)$ in the ring of integers of $\mathbf{Q}(w)$ is at most $[\mathbf{Q}(w):\mathbf{Q}]$. From the equality above, the number of prime ideal factors is at least $\varphi(p^r)$. So $\varphi(p^r)\leq[\mathbf{Q}(w):\mathbf{Q}]$. So this gives you the prime-power case.

Note that this argument doesn't use (b), but you will use it when extending to the case of arbitrary composite $m$.

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Ah but even then in the non-cubic case wouldn't I need to know that $\Bbb{Q}(\zeta_m)\cap \Bbb{Q}(\zeta_n) = \Bbb{Q}(\zeta_{(m,n)})$? Also, since this was a homework problem perhaps you shouldn't have spelt it all out. Thanks anyway for your answer, I think I can go from here. –  user38268 Dec 23 '12 at 3:18
Dear @BenjaLim, Unless I'm mistaken, you should only need this in the case $(m,n)=1$, in which case you want to prove $\mathbf{Q}(\zeta_m)\cap\mathbf{Q}(\zeta_n)=\mathbf{Q}$. If you want to use (b) for this, I suppose you'll want to try and prove that for $\ell\neq p$, $\ell$ is unramified in $\mathbf{Q}(\zeta_{p^r})$. I'm not sure what you have at your disposal because I'm not familiar with the book, but you could do this by computing the discriminant of the basis $\{1,w,\ldots,w^{\varphi(p^r)-1}\}$, or just proving that it is a power of $p$. This can be done without proving the ring of –  Keenan Kidwell Dec 23 '12 at 3:39
integers is equal to $\mathbf{Z}[w]$...although it's almost enough to deduce that. –  Keenan Kidwell Dec 23 '12 at 3:40
I know that the discriminant of $\Bbb{Q}(\zeta_{p^r})$ divides $(p^r)^{\varphi(p^r)}$. –  user38268 Dec 23 '12 at 3:45
Thanks so much for all your help. I will post an answer soon ( made CW of course) to show if I have understood this problem or not. –  user38268 Dec 23 '12 at 4:08

So after the discussion with Keenan above here is how we settle the case when $m = p^r$. Now by definition the $m$ - th cyclotomic polynomial

$$\Phi_m(x) = \prod_{1\leq k \leq m-1 : (m,k) = 1}(x- \zeta_m^k)$$

and since $\zeta_m$ is a root of this polynomial this means that $[\Bbb{Q}(\zeta_m):\Bbb{Q}] \leq \varphi(m)$, the degree of the $m$- th cyclotomic polynomial. Now because $\Bbb{Q}(\zeta_m)$ is a Galois extension we have that the ramification indices of all primes that lie above $p$ in $\mathcal{O}_{\Bbb{Q}(\zeta_m)}$ are all equal, similarly for the inertia degrees. So we can say that

$$[\Bbb{Q}(\zeta_m):\Bbb{Q}] = efg$$

where $e$ is the ramification index of $p$, $f$ the inertia degree and $g$ the number of distinct prime ideals in the prime factorisation of $(p)$ in $\mathcal{O}_{\Bbb{Q}(\zeta_m)}$. Now it is clear that we have inclusions $$\Bbb{Z} \subseteq \Bbb{Z}[\zeta_m] \subseteq \mathcal{O}_{\Bbb{Q}(\zeta_m)}.$$

Since the ramification index is multiplicative in towers, we have

$$e = e(Q|(1-w))e((1-w)|p) = e(Q|(1-w))\varphi(m)$$

where $Q$ is any prime ideal of $\mathcal{O}_{\Bbb{Q}(\zeta_m)}$ lying over $(1-w)$ and hence over $p$. This means that the number $e$ and hence the number $[\Bbb{Q}(\zeta_m) : \Bbb{Q}] \leq \varphi(m)$ and so we conclude that

$$[\Bbb{Q}(\zeta_m) : \Bbb{Q}] = \varphi(m).$$

Now for the general case. By the counting formula that I invoked in one of my comments to Keenan's answer above, it is enough to prove that

1. $\Bbb{Q}(\zeta_{p^r}) \cap \Bbb{Q}(\zeta_{l^s}) = \Bbb{Q}$.
2. $\Bbb{Q}(\zeta_{p^r})\Bbb{Q}(\zeta_{l^s}) = \Bbb{Q}(\zeta_{p^rl^s}).$

I will only prove the first of these for the second is a standard fact concerning cyclotomic extensions. By (b) to prove the first of these we need to show that if $p$ is a prime not equal to $l$, then $(p)$ is unramified in $\Bbb{Q}(\zeta_{l^s})$ for any $s \geq 1$. To do this we recall the following fact from algebraic number theory:

Let $p$ be a prime in $\Bbb{Z}$ and suppose $p$ is ramified in a number ring $R$. Then $p |\textrm{disc}(R)$.

We already know that $\textrm{disc}(\zeta_{l^s})$ divides $l^{s\varphi(l^s)}$ and so $\textrm{disc}(\zeta_{l^s})$ is a power of $l$. Since $p$ does not divide any power of $l$ by the theorem above $p$ is unramified in $\Bbb{Q}(\zeta_{l^s})$.

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If you're talking about the theorem in the box, it doesn't use the computation of the ring of integers. You know the discriminant of the order $\mathbf{Z}[w]$ is a power of $p$, and the discriminant of the full ring of integers divides this (basic fact about discriminants), so the latter discriminant must also be a power of $p$. –  Keenan Kidwell Dec 23 '12 at 4:50
Also, the computation of the discriminant of the basis $\{1,w,\ldots,w^{\varphi(p^r)-1}\}$ does not use, and in fact is used in proving, that the ring of integers is $\mathbf{Z}[w]$. –  Keenan Kidwell Dec 23 '12 at 4:51
@KeenanKidwell Why must the discriminant of the full ring of integers divide the discriminant of $\Bbb{Z}[w]$? I have not seen it in Marcus' number fields, I will check Milne's notes. –  user38268 Dec 23 '12 at 5:05
Actually I think this will follow from the fact that the free abelian group of rank $\varphi(p^r)$ on $1,w,\ldots,w^{\varphi(p^r) -1}$ is contained in the full ring of integers, which is free abelian of the same rank as well. –  user38268 Dec 23 '12 at 5:08