# Splitting of primes and other properties of $\mathbb{Q}[\omega]$ for $\omega=e^{2\pi i/m}$

Reading through Marcus I came to this exercise part of which already have answers in this same site (Splitting of primes in real cyclotomic field ) but no complete answer can be found and I'm having some troubles based on my little knowledge about Galois theory.

The text is the following

1. Let K be a subfield of $$\mathbb{Q}[\omega]$$ for $$\omega=e^{\frac{2\pi i}{m}}$$. Indentify $$\mathbb{Z}*_m$$ with the Galois group of $$\mathbb{Q}[\omega]$$ over $$\mathbb{Q}$$ in the usual way (which is?), and let H be the subgroup of $$\mathbb{Z}*_m$$fixing K pointwise. For a prime $$p\in \mathbb{Z}$$ not dividing m, let f deonte the least positive integer such that $$\overline{p^f}\in H$$, where the bar denotes the congruence class module m.

Show that f is the inertial degree $$f(P|p)$$ for any prime P of K lying over p. (Hint: f(P|p) is the order of the Frobenius automprhism \phi(P|p). Use the fact that if $$M\supset L$$ and they are both normal over K then \phi(Q|P) is the restriction of \phi(U|P) to L for Q, U primes lying over P)

2. Let p be a prime not dividing m, determin how it splits in $$\mathbb{Q}[\omega+\omega^{-1}]$$ (Hint: wht is H?)
3. Let p be a prime not dividing m, and let K be any quadratic subfield $$\mathbb{Q}[\sqrt{d}]\subset \mathbb{Q}[\omega]$$. With the same notation as in the first point show that if p is odd then $$\overline{p}\in H$$ $$\iff$$ d is a square module p, and if $$p=2$$ then $$\overline{p}\in H$$ $$\iff$$ $$d\equiv 1\; (mod\; 8)$$ (Hint: use theorem 25, note that $$p\not| m$$ implies that p is unramified in $$\mathbb{Q}[\omega]$$ hence also in \$\mathbb{Q}[\sqrt{d}]. )

Theorem 25 is the following

We now consider in detail the way in which primes p $$\in \mathbb{Z}$$ split in quadratic fields.

Let $$R=A \cap \mathbb{Q}[\sqrt{m}]$$, m squarefree.

Recall that R has integral basis $$\{1, \sqrt{m}\}$$ and discriminant 4m when $$m\equiv 2\; or\; 3\; (mod\; 4)$$, and integral basis $$\{1,\frac{1+\sqrt{m}}{2}\}$$ and discriminant m when $$m\equiv 1\; (mod\; 4)$$.

Let p be a prime in $$\mathbb{Z}$$. Theorem 21 shows that there are just three possibilities: $$pR=\begin{cases} P^2&\Leftarrow f(P|p)=1\\ P&\Leftarrow f(P|p)=2\\ P_1P_2 &\Leftarrow f(P_1|p)=f(P_2|p)=1. \end{cases}$$

Theorem 25 With notation as above, we have:

If p | m, then $$pR=(p,\sqrt {m})^2.$$

If m is odd, then $$2R= \begin{cases} (2,1+\sqrt {m})^2&\text{if m\equiv 3\pmod4}\\ \left(2,\frac{1+\sqrt{m}}{2}\right)\left(2,\frac{1-\sqrt{m}}{2}\right) & \text{if m\equiv 1\pmod8}\\ \text{prime if m\equiv 5\pmod8.} \end{cases}$$

If p is odd, $$p\not| m$$ then $$pR=\begin{cases} (p,n+\sqrt{m})(p,n-\sqrt{m})\; \text{if m\equiv n^2 \pmod p}\\ \text{prime if m is not a square mod p} \end{cases}$$ where in all relevant cases the factors are distinct.

1. The "usual way" refers to the isomorphism $$\varphi\colon (\mathbb Z/m\mathbb Z)^*\to Gal(\mathbb Q(\omega)/\mathbb Q)$$ that sends $$a$$ to the unique automorphism of $$\mathbb Q(\omega)$$ such that $$\omega\mapsto \omega^a$$. Now you see immediately that if $$p\nmid m$$ is a rational prime and $$\mathfrak p$$ is a prime of $$K$$ lying above it, the Frobenius for $$\mathfrak p/p$$ is simply $$\varphi(p)$$, because the map that sends $$\omega\mapsto \omega^p$$ becomes the map $$x\mapsto x^p$$ in the quotient ring $$\mathcal O_K/\mathfrak p$$. Notice how the Frobenius does not depend on $$\mathfrak p$$, because the extension is abelian. The hint tells you that the Frobenius at $$p$$ in $$K$$ is simply the restriction of the Frobenius at $$p$$ in $$\mathbb Q(\omega)$$. Hence, the Frobenius at $$p$$ in $$K$$ is just the restriction of $$\varphi(p)$$ to $$K$$. But then its order in $$Gal(K/\mathbb Q)$$ is just the order of $$p$$ in $$(\mathbb Z/m\mathbb Z)^*/H$$. On the other hand, the order of the Frobenius at $$p$$ in $$K$$ is precisely the inertia degree of $$p$$ in $$K$$, simply by definition.
2. Here you have to notice that $$K=\mathbb Q(\omega+\omega^{-1})=\mathbb Q(\omega)^H$$ where $$H=\{\pm1\}$$ (because if $$\varphi(a)$$ fixes $$\omega+\omega^{-1}$$, then $$a=\pm1$$). Now part 1. tells you the inertia degree $$f(p)$$ of $$p$$ in $$K$$: if the order $$o(p)$$ of $$p$$ modulo $$m$$ is odd, then $$f(p)=o(p)$$, otherwise $$f(p)=o(p)/2$$. Now just use the fact that in a Galois extension $$K/\mathbb Q$$ every unramified prime $$p$$ decomposes as $$\mathfrak p_1\ldots\mathfrak p_r$$, where the $$\mathfrak p_i$$'s all have the same inertia degree (and hence $$r=[K:\mathbb Q]/f(p)$$).
3. By point 1., $$p\in H$$ if and only if the inertia degree of $$p$$ in $$K$$ is 1, which is equivalent to say that $$p$$ splits in $$K$$. Now just apply the theorem.