Non-trivial valuation of $\mathbb R$ 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.
 A: 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?
A: I think Theorem 3.2.11 of the book Valued Fields by Antonio J. Engler and Alexander Prestel, Springer, 2005, can be helpful.
