Suppose that $X$ is a projective variety, and that $v$ is a discete valuation on $K(X)$
(trivial on $k$) whose corresponding valuation ring we will denote by $R$. The valuative criterion shows that
the map Spec $K(X) \rightarrow X$ extends to a map Spec $R \rightarrow X$. If I have the terminology correct, the image of the closed point of Spec $R$ is called the centre of
the valuation $v$ on $X$. It has codimension anywhere between $1$ and dim $X$.
Note that if $x \in X$ is the centre, then $R$ dominates $\mathcal O_x$ in $K(X)$
(i.e. we have a local inclusion of local rings $\mathcal O_x \subset R$).
Let's suppose for a moment that $X$ is a smooth surface. If the centre $x$ is codimension 1,
then both $\mathcal O_x$ and $R$ are (discrete) valuation rings. Since valuation rings
are (characterized by being) maximal for the partial order of dominance, $R$ and $\mathcal O_x$ coincide, and so the discrete valuation $v$ is just that given by the divisor of which
$x$ is the generic point.
Suppose instead that $x$ is a closed point.
Now we can blow up $x$ in $X$, to get a projective variety $X_1$, and the centre of $v$ in $X_1$ will now be contained in the exceptional divisor of $X_1$ (i.e. the preimage of $x$). If it coincides with the exceptional divisor,
then we have found a curve on $X_1$ giving rise to $v$; otherwise it is a point
$x_1$, which we can blow up again.
Either we eventually obtain a divisor on some iterated blow-up of $X$, or we
obtain a sequence of points $x \in X, x_1 \in X_1, \ldots,$ with each $X_n$ a blow-up
of the previous. In this case one sees that $R = \bigcup \mathcal O_{x_n}.$
There are a couple of exercises related to this issue in Hartshorne, namely II.4.5, II.4.12, and V.5.6. If I understand them correctly, any such sequence of $x\_n$ gives
a valuation ring $R$ in this way, and $R$ is a discrete valuation ring unless one constructs the sequene $x_n$ in the following manner: choose an irreducible curve $C$ in $X$ and define $x_n$ to
be the intersection of the proper transform of $C$ in $X_n$ with the exceptional
divisor. For a sequence $x_n$ constructed in this latter manner, one obtains not
a discrete valuation ring, but rather a rank 2 valuation ring: the valuation is determined
by first taking the valuation at the generic point of $C$, and then (for those functions
which are defined and non-zero at this generic point) restricting to $C$ and computing
the order of zero or pole at $x$.
What is the geometric intuition for the discrete valuation rings that correspond
to an infinite sequence $x_n$ rather than to some curve on $X$? One can think of
them as a transcendental curve on $X$, passing through $x$.
Indeed, imagine you had such a curve.
Then you could restrict a rational function to it; since it is transcendental,
a non-zero rational function would not have a zero or pole along this curve, and so
would restrict to give a non-zero meromorphic function on the curve. We could then
compute the order of the zero or pole of this meromorphic function at $x$. In other
words, because the curve is transcendental, we get a rank one valuation, in contrast to the rank two valuations that arise when we apply this process with an algebraic curve
$C$ passing through $x$.
I'm not sure about the details of the higher dimensional case. (Among other things, I am worried about the possibility of the center being codim > 1, but singular, which seems like it could complicate the analysis.) Does anyone here know how it goes?
