# Completions of number fields

I would like to prove a statement about completions of number fields, but I'm running into a problem. The statement I want to prove is

Let $L/K$ be a Galois extension of number fields, $p$ a prime of $K$. Suppose $P_1$, $P_2$ are primes of $L$ lying over $p$. Then $L_{P_1}$ and $L_{P_2}$ are isomorphic as extensions of $K_p$ (the subscript denotes completion at a prime).

I am trying to construct an isomorphism $L_{P_1}\to L_{P_2}$ preserving $K_p$. Because $L$ is dense in $L_{P_1}$ and every $K_p$-linear map is continuous, a map of $K_p$ extensions from $L_{P_1}$ to $L_{P_2}$ is determined by its restriction to $L$, and this restriction must take $P_1$ to $P_2$. The Galois group of $L/K$ acts transitively on the primes over $p$, so we can find such a $\sigma\in\mathrm{Gal}(L/K)$ taking $P_1$ to $P_2$.

Problem: some elements of $L$ are already in $K_p$ (by which I mean there are elements of $K_p\backslash K$ whose image in $L_{P_1}$ is actually in $L$), and these elements are not necessarily fixed by $\sigma$. So it seems that $\sigma$ need not induce an isomorphism of $K_p$-extensions.

Where did I go wrong? Is the statement even true?

Let’s look at an example. There’s nothing peculiar to the finite-$p$ situation here, the same thing happens in the real-complex situation. I think that we all have a better feel for that.
Our prime now will be “$\infty$”, and $\Bbb Q_\infty$ is $\Bbb R$, canonically. We consider the $\Bbb Q$-irreducible polynomial $F=X^4+1$, that’s the $8$-cyclotomic polynomial, and our extension field $L$ is $\Bbb Q(\zeta_8)$. Now $F$ factors over the reals as $(X^2+\alpha X+1)(X^2-\alpha X+1)$, where $\alpha^2=2$. Thus there are two primes of $L$ above infinity, namely the one where $\alpha$ is positive and the one where it’s negative. The corresponding completions, call them $L_p$ and $L_n$, are both isomorphic to $\Bbb C$.
Now back to your question: “some elements of $L$ are already in $\Bbb Q_\infty$”. True, for instance $\alpha\in\Bbb R$. But in two possible ways! This is just what happens when the decomposition group acts. Can you take it from there?
• I see. A map $L_{P_1}\to L_{P_2}$ is a map of $K_p$ extensions if it commutes with the inclusions of $K_p$, and since these inclusions can take a given element of $K_p$ to different elements of $L$, this is not the same as fixing $K_p$ pointwise. – User Mar 26 '16 at 16:29
• Just so. Maybe it’s helpful to think of the field(s) $L\cap K_p$. Even to give such a thing meaning, you need a map of $L$ into some suitably large extension $\Omega$ of $K_p$. And the fact that there are several extensions of $p$ to $L$ means that there are several maps of $L$ into $\Omega$. In our example, this extension of $\Bbb Q$ is $L\cap\Bbb R=\Bbb Q(\sqrt2\,)$, and there are two embeddings of this field into $\Bbb R$, each of which extends to embeddings of $L$ into $\Bbb C$. – Lubin Mar 27 '16 at 0:19