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 Answer 1

  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.


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