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I am reading Proposition 10.3 of Neukirch which I have appended below:

Proposition 10.3 (Neukirch): Let $n = \prod_{p} p^{\nu_p}$ be the factorisation of the positive integer $n$ into prime numbers, and for every prime number $p$ denote by $f_p$ the multiplicative order of $p$ mod $n/p^{\nu_p}$. Then one has in $\Bbb{Q}(\zeta_n)$ the factorisation $$p = (\mathfrak{p}_1\ldots \mathfrak{p}_r) ^{\varphi(p^{\nu_p})} \pmod{p}$$ where the $\mathfrak{p}_i$ are distinct prime ideals of $\mathcal{o} = \Bbb{Z}[\zeta_n]$ lying over $(p)$, all of ramification index $f_p$.

The idea of the proof is that it will suffice to show that $\phi_n(X)$, the minimal polynomial of $\zeta_n$ factors as

$$\phi_n(X) = (p_1(X)\ldots p_r(X))^{\varphi(p^{\nu_p})} \pmod{p}.$$

Now let us write $n = p^{\nu_p}m$ with $(m,p) = 1$. I get how he deduces the congruence

$$\phi_n(X) = \phi_m(X)^{\varphi(p^{\nu_p})} \pmod{p}.$$

The part where I don't understand next is where he says:

Observing that $f_p$ is the smallest positive integer such that $p^{f_p} \equiv 1 \pmod{m}$, it is obvious that this congruence reduces us to the case where $p$ does not divide $n$, and hence $\varphi(p^{\nu_p}) = \varphi(1) = 1.$

Why is it that in the case $p|n$ then the result follows? It seems to me he is claiming that when $p|n$ we have the polynomial $x^m-1$ having no multiples roots mod $p$. I don't understand the link between $p$ multiplicative order $f_p$ mod $m$ and the separability of $x^m - 1$ in the case $p|n$

Thanks.

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2 Answers 2

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The factorisation $p = (\mathfrak p_0 \ldots \mathfrak p_{r_n})^{\phi(p^{\nu_p})} \pmod p$ is equivalent to the factorisation $\phi_n(X) = (p_1(X)\ldots p_{r_n}(X))^{\phi(p^{\nu_p})} \pmod p$ where the $p_i$ are pairwise distinct, irreducibles, have degree $f_p$, and where $r_n f_p \phi(p^{\nu_p}) = \deg \phi_n = \phi(n)$.

Writing $n=p^{\nu_p}m, \phi(n) = \phi(p^{\nu_p})\phi(m)$ and noticing that the the multiplicative order of $p$ mod $m$ is again $f_p$, we want to deduce the wanted factorisation from the factorisation for $m$, which is $\phi_m(X) = (p_1(X)\ldots p_{r_m}(X))^1 \pmod p$, where the $p_i$ are pairwise distinct, irreducibles, have degree $f_p$, and where $r_m f_p = \phi(m)$.

Using $\phi_n(X) = \phi_m(X)^{\phi(p^{\nu_p})} \pmod p$, we get $\phi_n(X) = (p_1(X)\ldots p_{r_m}(X))^{\phi(p^{\nu_p})} \pmod p$. And since the $p_i$ are pairwise distinct, irreducibles, and have degree $f_p$, they are the $p_i$ we were originally looking for, and there is the right number of them (as $r_m = \phi(m)/f_p = \phi(n)/(f_p \phi(p^{\nu_p})) = r_n$). So this is the factorisation we needed.

Thus the theorem for general $n$ is an easy consequence of the case where $n$ and $p$ are coprime.

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I don't know how Neurkirch reduced the proof to $p$ prime to $n$. But if $p\mid n$, then it is known that $$\phi_m(X^{p^{v-1}})\phi_n(X)=\phi_m(X^{p^v}), \quad v=\nu_p.$$ Now we have $F(X^{p})\equiv F(X)^{p} \mod p$ for all polynomials with coefficients in $\mathbb Z$. So $$ \phi_m(X)^{p^{v-1}}\phi_n(X) \equiv \phi_m(X)^{p^v} \mod p$$ and $\phi_n(X)\equiv \phi_m(X)^{p^v-p^v}=\phi_m(X)^{\varphi(p^v)} \mod p$.

It is true that $x^m-1$ has no multiple roots mod $p$ because its derivative $mx^{m-1}$ is prime to $x^m-1$ mod $p$.

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Thanks for your answer. I have some questions though, how did you get the first identity above? Also, in your question above it seems you have proven $\phi_n(X) = \phi_m(X)^{\varphi(p^v)} \pmod{p}$ which I already know. –  user38268 Dec 17 '12 at 12:54
    
@BenjaLim: you can find the first identity for instance in yimin-ge.com/doc/cyclotomic_polynomials.pdf Cor. 4. I didn't understand you knew the identity mod p. Then to conclude, it is enough to see that $x^m-1$ is separable mod $p$. –  user18119 Dec 17 '12 at 13:11
    
Thanks for your link. Could you say a little more on the separability of $x^m - 1$ in the case that $p |n$? I don't understand where the bit on $p$ having order $f_p$ mod $m$ comes in. –  user38268 Dec 18 '12 at 9:38
1  
This is a typo in Neukirch, $f_p$ is not the ramification index but the degree of the residue extension $k(\mathfrak p_i)/\mathbb F_p$, see the answer of mercio. –  user18119 Dec 18 '12 at 13:44

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