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In a valued fields book on page $82$ there is a question: "show that every non-trivial valuation of $\mathbb R$ has divisible value group and algebraically closed residue class field." How do I approach this problem? Thanks.

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In general if a field $K$ is algebraically closed then the corresponding value group is divisible. Here, we have $K = \mathbb R$. This is making hard time for me. –  Sharma Sharma Jan 16 '13 at 3:17
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Just observe that $x$ and $-x$ have the same value, and one of the two will have an $n^{\text{th}}$ root. –  Brett Frankel Jan 16 '13 at 14:24
    
Thanks Brett this is perfect. –  Sharma Sharma Jan 16 '13 at 17:18
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It's not very useful to give a page number without saying which book it refers to... –  Hans Lundmark Jan 16 '13 at 18:21
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2 Answers

Let $v$ be a valuation on $\mathbb{R}$ and assume that its residue field $k$ is not algebraically closed. Then there exists a monic irreducible polynomial $f\in k[X]$ of degree $>1$. Its lift to $O[X]$, where $O$ is the valuation ring of $v$, is irreducible over $O$ and hence over $\mathbb{R}$ -- this follows from the so-called Gauss-Lemma. Consequently the degree of $f$ is $2$. By Artin's theorem it follows that $k$ is real closed. It therefore carries a unique ordering; the positive elements being precisely the sums of non-zero squares of $k$. Note that the same holds for $\mathbb{R}$. Consequently the residue map $O\rightarrow k$ preserves the orderings of $\mathbb{R}$ and $k$. This property directly implies

$ 0\leq x < y \Rightarrow v(x)\geq v(y) (*) $

Since $k$ has characteristic $0$ one knows that $v$ is trivial on $\mathbb{Q}$. The equation (*) therefore implies that for every positive $x\in M$, $M$ the maximal ideal of $O$, one has $x<q$ for every positive rational number $q$, contradicting the density of the rationals in the reals.

Is there a simpler proof?

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I think Theorem 3.2.11 of the book Valued Fields by Antonio J. Engler and Alexander Prestel, Springer, 2005, can be helpful.

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@ Azadeh Thanks. –  Sharma Sharma Jan 20 '13 at 16:28
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