Infinite algebraic extensions of $\mathbb{Q}$

Question

I need help to solve the following exercise:

If $K$ is an algebraic extension of $\mathbb{Q}$ (finite or infinite), then we let $O_{K}$ denote the ring of algebraic integers in $K.$ If $\frak{P}$ is a prime ideal of $O_{K}$ then $\frak{P}∩\mathbb{Z}$ is a prime ideal of $\mathbb{Z}.$ Moreover, if $\frak{P}$ is non-zero, then so is $\frak{P}∩\mathbb{Z}$ . Thus $\frak{P}∩\mathbb{Z}=$$ p\mathbb{Z}$ for some prime $p.$ Show that $O_{K}/\frak{P}$ is a field, and is an algebraic extension of $\mathbb{F}_{p}$; then show that it is Galois over $\mathbb{F}_{p}$ with abelian Galois group.

Now suppose that $K/F$ is a (finite or infinite) Galois extension. Let $\frak{P},$ $O_{K}$ be as before and put $\frak{p}=\frak{P} \cap $$O_{F}$, a prime ideal of $O_{F}.$ Show that $\mathrm{Gal}(K/F)$ acts transitively on the set of prime ideals of $O_{K}$ above $\frak{p}.$

Thanks in advance

Answer 1

In the infinite case, any nonzero element $a$ of $O_K/P$ is algebraic over $Z/pZ$ (lift $a$ to $O_K$, find its minimal polynomial, and reduce it mod $p$.) Since $O_K/P$ is an integral domain, the finite subalgebra $Z/pZ[a]$ is a finite integral domain, hence a field. So $a$ has an inverse. This shows that $O_K/P$ is a field even in the infinite case.

For the second part, mixedmath's argument shows that the primes are permuted transitively at any finite level. Given $P$ and $P'$, choose a tower of finite subfields $K_n$ whose union is $K$, and let $P_n$, $P'_n$, and $O_n$ be the intersections of the ideals and $O_K$ with these subfields. Using the finite result you can construct a system of elements $\sigma_n$ in the Galois group of $K/Q$ that satisfies $\sigma_n(P_n)=P'_n$.

This sequence of elements $\sigma_n$ converges to an element $\sigma$ of the Galois group of the infinite extension which restricts to $\sigma_n$ on each $K_n$, and this $\sigma$ does what you want.

Answer 2

First, we show the following: $O_K / \mathfrak{p}$ is a field.

Recall that the norm $N$ of an the ideal $\mathfrak{p}$ is the same as the index of $\mathfrak{p}$ in $O_K$. So we know that the index is finite. So $O_K/\mathfrak{p}$ is a finite integral domain, and is therefore a field. $\diamondsuit$

We know that $O_K/\mathfrak{p}$ has characteristic $p$ by what you've stated, and it is a finite field. The finite fields of characteristic $p$ are all given by extensions of $\mathbb{F}_p$, and so is Galois. Further, the Galois group is cyclic, generated by the Froebenius automorphism $x \mapsto x^p$.

To see that the Galois group acts transitively, you should first show that if $\sigma \in G$ the Galois group, then $\sigma$ takes prime ideals to prime ideals.

Once we have that, we proceed by contradiction:

Suppose $\mathfrak{p}$ and $\mathfrak{p}'$ are two ideals above $(p)$, but $\mathfrak{p'} \neq \sigma (\mathfrak{p})$ for all $\sigma \in G$. By the Chinese Remainder Theorem, there is an element $x$ that is in $\mathfrak{p}$ but which is $\equiv 1 \mod \sigma (\mathfrak{p}')$ for all $\sigma$. Since $x$ is in $\mathfrak{p}$, the norm $N$ of $x$ is in $(p)$. On the other hand, $x \not \in \sigma (\mathfrak{p}')$ for all $\sigma$, or rather $\sigma^{-1} x \not \in \mathfrak{p}'$ for all $\sigma$. The set $\{\sigma^{-1} (x)\}$ is the same as the set $\{ \sigma(x)\}$, since each element of the Galois group will appear once in each. So we know that $\sigma (x) \not \in \mathfrak{p}'$ for all $\sigma$. But then the norm $N(x)$ cannot be in $(p) \subset \mathfrak{p}'$, since $\mathfrak{p}'$ is a prime ideal (a product of things not in a prime ideal will not be in the prime ideal).

$N(x)$ can't both be in and not in $(p)$. So we have a contradiction, and this concludes the proof. $\diamondsuit$