Finding a Uniformizer of a Discrete Valuation Ring Suppose I have a discrete valuation ring. Then what are some techniques for explicitly finding a uniformizer? I'm especially interested in situations where the ring is given similarly to the following example.
Suppose we take the projective variety defined by the equation $C: X^2+Y^2-Z^2$, Then if $P \in C$ is a point on this curve, then $\mathbb{C}[C]_P$ is a discrete valuation ring. Thus, we should be able to exhibit a uniformizer. 
If we scale $Z$ to $1$ for the moment, we can think of this as a circle of radius $1$. A line through the point $(0, \frac{1}{2})$ will intersect the circle at exactly one other point (except when it is horizontal and intersects $(0, \frac{1}{2})$ with multiplicity $2$). Thus, if we let $t$ be the slope of such a line, we can parametrize any point on the curve as $(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2}, 1)$ and scaling back, $(2t, 1-t^2, 1+t^2)$. We see that this satisfies $C$ for all complex $t$. Now, I would like to say that for a given point, something like $(1+t^2)X-(2t)Z$ should be a uniformizer but I don't know of a good way to tell if this is true or how to prove it. 

As the  local ring at a point $P$ is defined as $\mathcal{O}_{C,P} = \varinjlim_{P \in U} \mathcal{O}_C(U)$, given any point $P \in C$ we can first pass to whichever affine open that contains $P$, and then calculate the direct limit over all opens in that affine $U$. As passing to an affine open corresponds to dehomogenizing, this is a very powerful idea which I will now illustrate.
Say without loss of generality that $P = [a_0 : b_0: 1] \in U_2$, the affine open set where $z\neq 0$. Then
$$\mathcal{O}_{C,P} = \left(\frac{k[X,Y]}{X^2 + Y^2 - 1}\right)_{(X-a_0, Y-b_0)}$$
with $a_0,b_0$ satisfying $a_0^2 + b_0^2 - 1 = 0$.
The ultimate point 
If $b_0 \neq 0$, then  a uniformizing parameter for $\mathcal{O}_{C,P}$ is given by the line $X-a_0$. If $b_0 = 0$, then $a_0 \neq 0$ for $[0:0:1] \notin C$ and so a uniformizing parameter would then be given by $Y- a_0$.
Proof 
Let us assume we are in the case $b_0 \neq 0$. The case $b_0 = 0,a_0 \neq 0$ follows along similar lines. We will now show $(X-a_0,Y -b_0) = (X-a_0)$. Note the inclusion $\supseteq $ is clear. For the other inclusion, as $X^2 + Y^2 + 1 = 0$ we have 
$$\begin{eqnarray*} X^2 + Y^2 - 1 &=& X^2 + (Y+b_0)(Y-b_0) +b_0^2 - 1 \\
&=& X^2 + (Y+b_0)(Y-b_0) - a_0^2 \end{eqnarray*}$$
which implies that $Y-b_0 = (X^2 - a_0^2)/(Y +b_0)$. Note we can invert $Y+b_0$ using the hypothesis $b_0 \neq 0$ as $Y+b_0 \notin (X-a_0,Y-b_0)$. We have found a generator for the local ring at $P$ which then must be the uniformizing parameter for the DVR. 
