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I'm reading through Marcus's wonderful book Number Fields, and I had two questions on his proof of quadratic reciprocity in chapter $4$.

Marcus states first that if $p$ is an odd prime, and $q$ is any prime, that in the field $\mathbb{Q}[\omega]$ where $\omega = e^{2\pi i /p}$ that $q$ will split into $r$ primes, where $r = (p-1)/f$ and $f$ is the multiplicative order of $q \mod p$. He says this like it's obvious, so maybe I'm missing something simple, but I do not understand this line.

My second question is that Marcus also claims that the following are equivalent

$1)$ $q$ is a $d^{th}$ power $\mod p$

$2)$ $f|(p-1)/d$

$3)$ $d|r$

$4)$ $F_d \subset F_r$

Where $F_d$ is the unique subfield of $\mathbb{Q}[\omega]$ of degree $d$ over $\mathbb{Q}$ guaranteed by the fact that the Galois group of $\mathbb{Q}[\omega]/\mathbb{Q}$ is cyclic.

I think $1\implies 2\implies 3 \implies 4$ are pretty trivial, as well as $4\implies 3 \implies 2$, but I can't quite see the equivalence of $1$ with the rest of these statements.

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    $\begingroup$ Recall that if the ring of integers of $\mathbb{Q}(\alpha)$ is given by $\mathbb{Z}[\alpha]$, then you can see the splitting of the prime $(q)$ by taking the irreducible polynomial for $\alpha$ (which is monic having integer coefficients), reducing it modulo $q$, and factoring the resulting polynomial over $\mathbb{F}_q$. You can do that with $\omega$ and $q$ in the case under consideration. And this also relates to part (4), if I’m not mistaken... $\endgroup$
    – Arturo Magidin
    Commented Mar 1, 2019 at 4:48
  • $\begingroup$ I see. So this comes down to how the polynomial $x^{p-1}+x^{p-2}+\cdots + 1$ factors mod $q$. I know that it will have a factor of degree $f$ but it's still not immediately obvious to me why $x^{p-1}+x^{p-2}+\cdots + 1$ splits into $(p-1)/f$ factors mod $q$. $\endgroup$
    – user413766
    Commented Mar 1, 2019 at 5:03
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    $\begingroup$ Oh never mind, I see this now, it's because the extension is normal, so all the primes have to be to the same power, and hence the factor of degree $f$ is the same for all the factors. So we have that $f = (p-1)/r$. Thank you for the hint. $\endgroup$
    – user413766
    Commented Mar 1, 2019 at 5:14
  • $\begingroup$ Dear @Arturo Magidin , Why the multiplicative order of $q$ module $p$ is equal to inertia degree $f$? $\endgroup$
    – Davood
    Commented Apr 27, 2020 at 15:09
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    $\begingroup$ @Davood KHAJEHPOUR, I think you can find your answer here math.stackexchange.com/questions/981438/… $\endgroup$
    – user413766
    Commented Apr 28, 2020 at 23:53

1 Answer 1

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Your first question has its answer in the Corollary of Theorem 26. Observe that as $q$ does not divide $p$, so it should split into exactly $\frac{p-1}{f}$ factors, which he has denoted by $r$, and where $f$ is the multiplicative order of $q$ modulo $p$. So, $r = \frac{p-1}{f}$. In your second question, understanding the equivalence between two statements just as you have already mentioned will suffice to make sure that all are equivalent.

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  • $\begingroup$ Great answer!!! $\endgroup$
    – Sayan Dutta
    Commented May 20 at 8:36

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