Extension of Completions of Number Fields

Question

On p. 116 of Milne's notes on Algebraic Number Theory, he gives the following construction.

Let $K$ be a field with a valuation $|\cdot|$ (archimedean or discrete nonarchimedean), and let $L$ be a finite separable extension of $K.$

Write $L = K[\alpha]$, and let $f(X)$ be the minimum polynomial of $\alpha$ over $K.$ Let $|\cdot|'$ be an extension of $|\cdot|$ to L. Then we can form the completion $\hat{L}$ of $L$ with respect to $|\cdot|'$. Then $\hat{L} = \hat{K}[\alpha].$ In particular, $\hat{L}$ does not depend seem to depend on which extension $|\cdot|'$ of $|\cdot|$ we choose.

For example, if $L/K$ is a separable extension of number fields, $\mathfrak{p}$ is a prime ideal of $K$ and $\mathcal{P}_1,\dotsc,\mathcal{P}_n$ are primes of $L$ dividing $\mathfrak{p}$ then this seems to show that all completions $L_{\mathcal{P}_i}$ of $L$ with respect to $|\cdot|_{\mathcal{P}_i}$ are equal.

My question is: is this reasoning correct? Are all such completions indeed equal? This seems strange to me, but possibly I just don't have the correct intuition here.

Answer 1

After an enlightening talk with my friend Ben today, I think I can answer my own question now. Thanks to Ben for pointing out this solution!

As was pointed out by Lubin in his answer to this question, the completions may not be equal or even isomorphic. So the question is now: where is the flaw in my reasoning? Still $\hat{K}[\alpha]$ does not seem to depend on which completion $\hat{L}$ of $L$ we took?

This is where I am wrong. At the start we have $\alpha\in L$. However, when we consider $\hat{K}[\alpha]$, we are taking $\alpha\in \hat{L}$. Hence, quietly, we are not really considering $\alpha$ but rather the image of $\alpha$ under the embedding $L\hookrightarrow\hat{L}$. Let's call the distinct completions $L_i$ for a moment and let's refer to the image of $\alpha$ in $L_i$ as $\alpha_i$ to clearly emphasise its dependence on $L_i$. It is not so hard to see that the $\alpha_i$ can now be very distinct elements of the algebraic closure of $\hat{K}$ and hence the $\hat{K}[\alpha_i]$ - this is what we called $\hat{K}[\alpha]$ before - can be different fields depending on the completion $L_i$.

Answer 2

Since Milne is reliable, I wonder whether you’ve quoted him correctly. Consider a nonnormal extension like $\Bbb Q(\sqrt[4]p)$ over the rationals for any prime $p$ and the Archimedean place on $\Bbb Q$. There are two real places (the root is positive or negative), inequivalent, but the completions are both $\Bbb R$, while there is one complex place, and the completion is $\Bbb C$. With a suitable choice of a different prime number $q$, you can get a $q$-adic example for the same extension.