# Completion of a finite field extension is also finite?

Let $(L,w)/(K,v)$ be a finite extension of valuation fields, and let $L_w$, $K_v$ be the respective completions of $(L,w)$, $(K,v)$. Is the field extension $L_w/K_v$ finite?

For nonarchimedean valuations on number fields it is trivial, as everything is explicit in terms of prime ideals. For archimedean valuations this is also trivial if we know what they are, but I'm thinking about turning this around and deriving the archimedean valuations on number fields from this (seemingly easy) lemma.

Also, I'm wondering how far can we push it towards nondiscrete, non-Noetherian domains (pun?).

• This result is implicit in the proof of Theorem 8.1 in Chapter 2 of Neukirch, Algebraic Number Theory, but I'm not sure if it's really trivial or not. Sep 28 '15 at 0:51
• Why did you say that for nonarchimedean valuations over number fields this is trivial?
– user82531
Nov 8 '15 at 16:31
• About your comment at the end, I do not see a pun there. Nov 8 '15 at 19:00
• Doesn't this just follow from the fact that $L_w = L\otimes_K K_w$? Nov 8 '15 at 19:30

Since $L/K$ is finite, we can write $L = K(a_1,\ldots,a_r)$ for some $a_i$. These same $a_i$ will generate $L_w$ over $K_v$. You can't hope for anything simpler than that! Well, you could hope that a $K$-basis of $L$ is a $K_v$-basis of $L_w$, but that is WRONG in general. For instance, $K_v$ can sometimes equal $L_w$ even if $[L:K] > 1$, e.g., the $(1+2i)$-adic completion of $\mathbf Q(i)$ is the $5$-adic completion of $\mathbf Q$.
To prove $L_w = K_v(a_1,\ldots,a_r)$, one containment is easy: since $K_v$ and each $a_i$ is in $L_w$, we get $K_v(a_1,\ldots, a_r) \subset L_w$.
Since $K$ is in $K_v$ we get $L = K(a_1,\ldots,a_r) \subset K_v(a_1,\ldots,a_r)$. Thus $$L \subset K_v(a_1,\ldots,a_r) \subset L_w.$$ The $w$-completion of $L$ is $L_w$, so $$(K_v(a_1,\ldots,a_r))_w = L_w,$$ where the field on the left is the $w$-completion (or closure) of $K_v(a_1,\ldots,a_r)$ inside $L_w$. We are NOT done because we need to explain why the $w$-completion of $K_v(a_1,\ldots,a_r)$ is itself, in other words why this field is complete with respect to the absolute value $w$ on $L_w$. We will use a theorem about extending absolute values to finite extensions of a complete field.
Theorem. If $F$ is a complete valued field then any finite extension of $F$ admits a unique extension of the absolute value on $F$ and it is complete with respect to that extension.
We can apply this theorem because $K_v(a_1,\ldots,a_r)$ is a finite extension of $K_v$: a nonzero polynomial in $K[x]$ with $a_i$ as a root also has coefficients in $K_v$, so each $a_i$ is algebraic over $K_v$. Thus $K_v(a_1,\ldots,a_r)$ is a finite extension of $K_v$ and the absolute value $w$ on it from $L_w$ extends the absolute value $v$ on $K_v$, so the theorem implies that $K_v(a_1,\ldots,a_r)$ is complete with respect to $w$. QED