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The only examples I know of valuations of rank $n>1$ are of the following type:

Let $K$ be a field and consider the field of rational functions on $n$ variables $K(x_1,\dots,x_n)$ for some $n\in\mathbb{N}$. Define $|\cdot|:K(x_1,\dots,x_n)\to\mathbb{Z}^n$ by $|a|=(0,0,\dots,0)$ for all $a\in\mathbb{K}$, $a\neq0$ and $|x_k^m|=(r_1,\dots,r_n)$ where $r_k=m$ and $r_i=0$ for $i\neq k$.

Is it possible to define a valuation of rank $n>1$ in a smaller field like $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{Q_p}$?

Chilote
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    IIRC, assuming the axiom of choice, there are isomorphisms from $\mathbb{C}$ to the algebraic closure of $\mathbb{C}(x_1, ..., x_n)$. So whatever valuation one has on the latter can be transported to the former and its subfields. (With the axiom of choice, one has to be careful with the notion of "smaller field".) – Torsten Schoeneberg Dec 20 '17 at 01:19
  • In you example, with $K=\mathbb{C}(x_1,\dots,x_n)$, why such isomorphism exist? Can we apply the same reasoning to any field $K$ that contains $\mathbb{C}$ as a subfield? – Chilote Dec 20 '17 at 02:14
  • Assuming choice, every field of characteristic 0 and with cardinality less or equal to that of $\mathbb{C}$ can be embedded into $\mathbb{C}$. Compare https://mathoverflow.net/questions/23264/ and https://math.stackexchange.com/questions/338148/ (especially Mercio's answer). (Note that this is also how one proves the isomorphism of $\mathbb{C}_p$ and $\mathbb{C}$ we talked about yesterday, which is just as non-explicit and counterintuitive as the existence of embeddings $\mathbb{C}(x) \hookrightarrow \mathbb{C}$.) – Torsten Schoeneberg Dec 20 '17 at 05:19
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    Note, however, that function fields $K(x_1, ..., x_n)$ for $K = \mathbb{C}, \mathbb{R}, \mathbb{Q}_p, \mathbb{Q}, ...$ just embed into $\mathbb{C}$, of course non-surjectively; only their algebraic closures are really isomorphic to $\mathbb{C}$. So in order to answer your question via this route, one would have to know whether there are higher rank valuations on the algebraic closures of such $K(x_1, ..., x_n)$. About this, you probably know more than I do. – Torsten Schoeneberg Dec 20 '17 at 05:26
  • Actually, I don't know how exactly the elements of the algebraic closure of $K(x_1,\dots,x_n)$ look like. Also, I don't know if there is an extension of the valuation of a field to its algebraic closure in case rank >1. – Chilote Dec 20 '17 at 18:07

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