# Splitting primes and quadratic reciprocity.

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.

• 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... Mar 1, 2019 at 4:48
• 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 $q$. Mar 1, 2019 at 5:03
• 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. Mar 1, 2019 at 5:14
• Dear @Arturo Magidin , Why the multiplicative order of $q$ module $p$ is equal to inertia degree $f$? Apr 27, 2020 at 15:09
• @Davood KHAJEHPOUR, I think you can find your answer here math.stackexchange.com/questions/981438/… Apr 28, 2020 at 23:53

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.