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Let $k$ be a subfield of ${\mathbb C}$. Let $f$ be a continuous ring homomorphism $k \to {\mathbb C}$ which sends 1 to 1. Is $f$ the identity map on $k$ or its complex conjugate?

The motivation is as follows. Let $f, g:K \rightarrow \mathbb{C}$ be embeddings of a field $K$. $f$ defines a valuation $v$ on $K$ by defini9ng $v(x) = |f(x)|$. Similarly $g$ defines a valuation $w$. Suppose $v$ and $w$ define the same topology on $K$. I wonder if $f = g$ or $f = \bar g$.

Makoto Kato
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    @WilliamCurtis The "missing context or other details" off-topic notice means "show what you've tried or where you're coming from with this" and weeds out e.g. questions which are simply cut-and-paste homework problems. Unfortunately, tidbits of obvious merit like your question are often flagged with this. Just update it with some of your thoughts and it should be reopened. – aes Feb 28 '15 at 01:53
  • @aes Just curious. What's wrong with homework? This is not homework though. – Makoto Kato Mar 02 '15 at 01:54
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    @WilliamCurtis See here. The policy here is that homework questions are fine, but bare questions with no context given or work shown (commonly homework problems which are simply cut and paste into the question box) are not. There's an obvious stigma to doing someone's homework for them, especially if their input consists only of stating the problem. Take this up further on meta if you're interested. – aes Mar 02 '15 at 04:12
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    @WilliamCurtis I'm merely reporting on the practicalities of this site. I support your question being reopened and was providing advice on how to get it reopened (edit your post with a bit of your thoughts). I'm not a moderator or site representative, merely someone who saw your question and thought it worthwhile. If you're interested in discussing the workings of this site, please take them up on meta (click help in the upper right, then meta). – aes Mar 04 '15 at 01:17
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    How about these steps: $k$ is dense in $\mathbb C$ or in $\mathbb R$. A continuous ring homomorphism $f$ is uniformly continuous. Therefore $f$ extends continuously to the closure, and is still a homomorphism. – GEdgar Mar 15 '15 at 00:10

1 Answers1

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Hints:

For an "abstract" solution.

  1. A uniformly continuous map $f$ from a metric space $X$ into a complete metric space $Y$ extends uniquely to a continuous map $\bar f$ from the completion $\bar X$ of $X$ into $Y$
  2. If $f$ is a homomorphism (of an algebraic structure on $X$ compatible with its topology), then so is $\bar f$.
  3. A continuous homomorphism between topological groups is automatically uniformly continuous (in fact, it only needs to be continuous at a single point).
  4. What is the completion of $k$?

I belive you don't need $X,Y$ to be metric spaces, they might as well be uniform spaces (and then the completeness and completion are in the uniform sense.) Of course, the spaces you consider here are metric.

For a more field-theoretic one:

  1. What can you say about the restriction of $f$ to ${\bf Q}$? What about the restriction to its closure $\bar {\bf Q}$ (in $k$)?
  2. What are the possible degrees of $k$ over $\bar{\bf Q}$? (Consider a nonreal element of $k$, if it exists, and its minimal polynomial over ${\bf R}$.)
tomasz
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