I just worked through a proof in Daniel Marcus' book Number Fields that if $p\nmid n$, the inertial degree of any prime ideal of $\mathbb{Q}(\zeta_n)$ lying over $p$ is equal to the order of $p$ in $(\mathbb{Z}/n\mathbb{Z})^\times$ (p.77). The argument rests on the equation $(1-\zeta_n)(1-\zeta_n^2)\dots(1-\zeta_n^{n-1})=n$. I tried to reorganize the proof and was surprised to end up with what seems to me to be a complete proof that doesn't make explicit use of this fact. My question is, where did I hide it? (Or, is my proof missing something? Or, is this fact actually extrinsic to the result?)

The setup: We have a prime $p$ not dividing $n$, and the discriminant of $\mathcal{O}_{\mathbb{Q}(\zeta_n)}=\mathbb{Z}[\zeta_n]$ divides a power of $n$ so is prime to $p$, so $(p)$ splits into distinct prime ideals in $\mathbb{Z}[\zeta_n]$:

$$ p\mathbb{Z}[\zeta_n] = P_1\dots P_r $$

The field extension is normal and degree $\varphi(n)$, so we have $fr=\varphi(n)$ where $f$ is the inertial degree of the $P_i$. By definition, $f$ is the degree of the field extension $[\mathbb{Z}[\zeta_n]/P_i:\mathbb{Z}/(p)]$. This field extension is cyclic, with Galois group generated by the Frobenius automorphism $\sigma$, which therefore has order $f$. Meanwhile, $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong (\mathbb{Z}/n\mathbb{Z})^\times$ has an automorphism $\tau$ mapping $\zeta_n\mapsto \zeta_n^p$, since $p$ is prime to $n$. The order of $\tau$ is the order of $p$ in $(\mathbb{Z}/n\mathbb{Z})^\times$. The problem is to show that $\tau$ and $\sigma$ have the same order.

Marcus' argument: Consider the action of $\tau$ on the field generator $\zeta_n$ and of $\sigma$ on its image in $\mathbb{Z}[\zeta_n]/P_i$. $\tau^m=\mathrm{id.}$ if and only if $\tau^m(\zeta_n)=\zeta_n$ because it is a generator. Likewise, $\sigma^m=\mathrm{id.}$ if and only if $\sigma^m(\bar\zeta_n)=\bar\zeta_n$, with calculation taking place in $\mathbb{Z}[\zeta_n]/P_i$, i.e. mod $P_i$. The problem is thus to show that $\zeta_n^{p^m}=\zeta_n$ if and only if $\zeta_n^{p^m}=\zeta_n\mod P_i$. The "only if" is trivial. If $\zeta_n^{p^m}=\zeta_n\mod P_i$ then $\zeta_n^{p^m-1}=1\mod P_i$ because $\zeta_n$ is a unit. So $1-\zeta_n^{p^m-1}\in P_i$. However,

$$(1-\zeta_n)(1-\zeta_n^2)\dots (1-\zeta_n^{n-1}) = n\notin (p)\subset P_i$$

Therefore, $p^m-1$ can't be equal to $1,2,\dots,n-1$ mod $n$. The only possibility is $p^m-1=0\mod n$, and then $\zeta_n^{p^m-1}=1$, so we have our "if" statement.

My argument: $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ acts transitively on the $P_i$, of which there are $r$, so by the orbit-stabilizer theorem, $f=|\operatorname{Stab}P_i|$. Let $\pi:\mathbb{Z}[\zeta_n]\rightarrow\mathbb{Z}[\zeta_n]/P_i$ be the canonical homomorphism. $\tau$ acts trivially on $\mathbb{Z}$ and sends $\zeta_n$ to its $p$th power; $\sigma$ sends the image of $\zeta_n$ to its $p$th power (since it is the $p$th power map on everything) and acts trivially on the image of $\mathbb{Z}$. By consideration of their action on $\mathbb{Z},\zeta_n$ which generate the ring $\mathbb{Z}[\zeta_n]$, we conclude that $\pi\circ\tau$ and $\sigma\circ\pi$ are equal. It follows that $\tau\in\operatorname{Stab}P_i$ and that $\tau\mapsto \sigma$ under the homomorphism $\operatorname{Stab}P_i\rightarrow Gal((\mathbb{Z}[\zeta_n]/P_i)/(\mathbb{Z}/(p)))$ induced by $\pi$. It is immediate that this map is surjective because $\sigma$ generates the image. Since domain and image both have order $f$, this means it is an isomorphism, and it follows that $\sigma,\tau$ have the same order.

To reiterate my question: where did I hide Marcus' calculation $(1-\zeta_n)\dots(1-\zeta_n^{n-1})=n?$ Or, is my proof wrong? Or, is the calculation actually extrinsic to the conclusion?

  • 1
    $\begingroup$ seems right to me. $\endgroup$ – user29743 Oct 11 '12 at 1:58
  • $\begingroup$ Actually I prefer your proof ! $\endgroup$ – user18119 Apr 15 '13 at 22:18

Your proof is correct. The crucial part in both proofs consist in proving the map $$\mathrm{Stab}(P_i)\to \mathrm{Gal}((\mathbb Z[\zeta_n]/P_i)/\mathbb F_p)$$ is injective. This map is known to be always surjective. The difference between the two proofs:

  • You use implicitely the fact that the inertia group at $P_i$ is trivial (when you say the stablizer has order $f$).

  • Marcus is proving directly the injectivity, using the identity on the product of the $(1-\zeta_n^k)$'s. The same identity can be used to show the extension is unramified above $p$ (hence the inertia at $P_i$ trivial).

  • $\begingroup$ Thanks! What's the idea for using the identity to show that $p$ is unramified? $\endgroup$ – Ben Blum-Smith Apr 16 '13 at 22:32
  • $\begingroup$ The identity implies that mod $p$, $\Phi_n(X)$ is separable because the difference of any two distinct roots $\zeta_n^i-\zeta_n^j=\zeta_n^i(1-\zeta_n^{i-j})$ is non-zero mod any prime ideal lying over $p$. Hence the cyclotomic extension is unramified over $p$. But I notice that in Marcus the non-ramification is proved earlier and differently. $\endgroup$ – user18119 Apr 17 '13 at 20:17
  • $\begingroup$ Yes, he had already proved non-ramification a different way based on the discriminant, which is why I was comfortable depending on that for the argument. By the way, your answer here is just the kind of thing I was looking for! Thanks. (Accepted and +1) $\endgroup$ – Ben Blum-Smith Apr 18 '13 at 16:39

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