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. 
 A: 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. 
A: Lets look at the part $U$ of the curve given by $Z\neq 0$, so that we can use the affine coordinates $x=X/Z$ and $y=Y/Z$. Then $\mathbb{C}[U]=\mathbb{C}[x,y]$. For every point $P=(a,b)\in U$ the ideal $m$ generated by $x-a$ and $y-b$ is a maximal ideal of $\mathbb{C}[x,y]$ and $\mathbb{C}[C]_P=\mathbb{C}[x,y]_m$. Now $m\mathbb{C}[x,y]_m$ is generated by $x-a$ and $y-b$ and we have: $1=(x-a+a)^2+y^2=(x-a)^2+2a(x-a)+a^2+y^2$ leading to $b^2-y^2=(x-a)(x+a)$ and thus
$y-b=\frac{(x-a)(x+a)}{y+b}$
where $y+b\not\in m$, provided that $b\neq 0$. Consequently $x-a$ is a uniformizer for the point $P=(a,b)$, $b\neq 0$.
In the case $b=0$ we can write
$(1+x)(1-x)=y^2$
with the consequence that $y$ is a uniformizer for the points $(1,0)$ and $(-1,0)$.
It remains to treat the points $P\in C\setminus U$. In fact there are only two such points $Q_1$, $Q_2$, namely $Q_1=[i:1:0]$ and $Q_2=[-i:1:0]$ in projective coordinates. We can use the affine coordinates $x=X/Y$ and $z=Z/Y$. Then $\mathbb{C}[C]_{Q_i}=\mathbb{C}[x,z]_{m_i}$, where $m_i$ is generated respectively by $x-i,z$ and $x-i,z$. The relation $x^2+1-z^2=0$ yields
$(x-i)(x+i)=z^2$
which shows that $z$ is a uniformizer for $Q_1$ and $Q_2$.
In principal one can do such computation whenever the curve $C$ considered is regular and plane (= given by one equation). Of course in higher degrees things can become tedious ...
