In modern valuation theory, one studies not just absolute values on a field, but also Krull valuations. The motivation is easy enough:
If $k$ is a field, a valuation ring of $k$ is a subring $R$ such that for every $x \in k^{\times}$, at least one of $x, x^{-1}$ is an element of $R$. (It follows of course that $k$ is the fraction field of $R$.) If $| \ |$ is a non-Archimedean norm on a field, then the set $\{x \in k \ | \ |x| \leq 1 \}$ is a valuation ring. However, the converse does not hold, since if $R$ is a valuation ring, then $k^{\times}/R^{\times}$ need not inject into $\mathbb{R}$: rather it is (under a straightforward extension of the divisibility relation on $R$) a totally ordered abelian group. Moreover, a certain formal power series construction shows that for any totally ordered abelian group $\Gamma$, there exists $k$ and $R$ with $k^{\times}/R^{\times} \cong \Gamma$.
My question is this: what are some instances where having the generality of Krull valuations is useful for solving some problem (which is not a priori concerned with valuation theory)? How do Krull valuations arise in algebraic geometry?
I can almost remember one example of this. I believe it is possible to give a quick proof of the Lang-Nishimura Theorem — that having a smooth $k$-rational point is a birational invariant among complete [hmm, valuative criterion!] $k$-varieties. I think I saw this in some of Bjorn Poonen's lecture notes, but I forget where. [Last year at this time, I would have emailed Bjorn. I am trying out this new approach on the theory that Bjorn can reply if he wishes, and if not someone else will surely be eager to tell me the answer.]
Are there other nice examples? Maybe something to do with resolution of singularities?
Best Answer
Since you asked for it, here is a little bit about the role of valuations in the Lang-Nishimura theorem, one version of which is as follows (my version implies yours):
Theorem (Lang-Nishimura): Let $X \to \to Y$ be a rational map between $k$-varieties, where $X$ is integral and $Y$ is proper. If $X$ has a smooth $k$-point $x$, then $Y$ has a $k$-point.
Sketch of proof: Let $K$ be the function field of $X$. The rational map gives a $K$-point on $Y$. If $\dim X=1$, then $\mathcal{O}_{X,x}$ is a valuation ring, so the valuative criterion for properness gives an $\mathcal{O}_{X,x}$-point of $Y$, which reduces to a $k$-point of $Y$. For $\dim X=n>1$, modify the argument by embedding $K$ into a valued field with value group $\mathbb{Z}^n$ and residue field $k$, namely the iterated Laurent series field $F:=k((t_1))((t_2))\cdots((t_n))$, where the $t_i$ are local parameters at $x$. $\square$
1) If one prefers, one can replace this use of the valuative criterion for a rank $n$ discrete valuation with $n$ uses of the valuative criterion for rank $1$ discrete valuations: prove the lemma that if a proper variety has an $L((t))$-point, then it has an $L$-point, and apply the lemma $n$ times. So for this particular application, you don't really need the fancy valuations.
2) Geometrically, the Lang-Nishimura theorem can be understood as follows: Find a smooth irreducible curve $C$ in $X$ through $x$ such that $C$ is not entirely contained in the locus of indeterminacy of $\phi \colon X \to\to Y$. Then $\phi|_C$ extends to a morphism, and the image of $x$ is a $k$-point of $Y$. (The existence of $C$ is not completely obvious, though, so the valuation-theoretic proof is cleaner.)