3
$\begingroup$

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?).

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ 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. $\endgroup$
    – Fan Zheng
    Commented Sep 28, 2015 at 0:51
  • $\begingroup$ Why did you say that for nonarchimedean valuations over number fields this is trivial? $\endgroup$
    – user82531
    Commented Nov 8, 2015 at 16:31
  • 2
    $\begingroup$ About your comment at the end, I do not see a pun there. $\endgroup$
    – KConrad
    Commented Nov 8, 2015 at 19:00
  • $\begingroup$ Doesn't this just follow from the fact that $L_w = L\otimes_K K_w$? $\endgroup$
    – Will Chen
    Commented Nov 8, 2015 at 19:30

1 Answer 1

Reset to default
7
$\begingroup$

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.

You can find a proof of the unique extension of absolute values from a complete field to a finite extension in Cassels's book Local Fields. It is also in Gouvea's book p-Adic Numbers: An Introduction. Koblitz's book on p-adic numbers and zeta-functions has a proof that assumes the base field is locally compact, but the other proofs avoid such a hypothesis.

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

Improve this answer
$\endgroup$
2
  • $\begingroup$ Just curious, can we have a finer control on $[L_w:K_v]$? For example, can we expect that $[L_w:K_v]$ divides $[L:K]$ or something like that? $\endgroup$
    – Yijun Yuan
    Commented Oct 4, 2023 at 2:21
  • 1
    $\begingroup$ Divisibility is not going to happen in general. When $K = \mathbf Q$ and $v = p$ for a prime $p$, with $w = \mathfrak p$ being a prime ideal lying over $p$ in $\mathcal O_L$, $[L_w:\mathbf Q_p] = e(\mathfrak p|p)f(\mathfrak p|p)$. Try $L = \mathbf Q(\sqrt[3]{2})$, so $[L:\mathbf Q] = 3$. Take $p = 5$. In $\mathcal O_L$, which is $\mathbf Z[\sqrt[3]{2}]$, we have $(p) = \mathfrak p\mathfrak q$ where $f(\mathfrak p|5) = 2$ and $f(\mathfrak q|p) = 1$. Thus $[L_\mathfrak p:\mathbf Q_5] = 2$ $\endgroup$
    – KConrad
    Commented Oct 4, 2023 at 7:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .