I know that $C$ is an irreducible curve and it is the projective closure of the twisted cubic.
The divisor can be calculated by $$\operatorname{div}(f) = \sum_{P \in C} \operatorname{ord}_P(f) P$$
And since rational function only has finitely many zeroes and poles, and points that are not zeros or poles have an order of 0.
We only need to consider the zeroes and poles.
But I wasn't able to find finite poles explicitly.
$$\frac{w}{z} = \frac{x^2}{zy}$$ $$\frac{w}{z} = \frac{xy}{z^2}$$ $$\frac{w}{z} = \frac{wx}{y^2}$$
To find the pole, we let the denominator of all the expressions to be 0. $$z = zy = z^2 = y^2 = 0$$ I get $$z = y = 0$$ However, this does not seem to give me finite points.
Is there anything that I did wrong? If you, could you point it out and give me a correct approach?
Thanks
Thanks to anyone who commented. I can't believe that I forgot to look for the poles inside $C$...
I feel embarrassed to even ask this question...
Anyway, I should post my attempt to pay for my sins. Below is my attempt to this question. There are certainly some parts where I am being lazy and didn't fully explain what I am doing...
I assume that the point is ordered as follows: $$[w: x: y: z]$$
From $C$, we have: $$\frac{w}{z} = \frac{x^2}{zy}$$ $$\frac{w}{z} = \frac{xy}{z^2}$$ $$\frac{w}{z} = \frac{wx}{y^2}$$
To calculate $\operatorname{div}(w / z)$, we need to find the zeros and poles.
We first find the poles:
$$z = zy = z^2 = y^2 = 0$$ We have, $z = 0, y = 0$.
Substitute $z = 0, y = 0$ back to the polynomial, we get: $$0 = x^2$$ $$x = 0$$
We get the pole is $[w: x: y: z] = [w: 0: 0: 0] = [1: 0: 0: 0]$.
To find the tangent, we dehomogenize the polynoimals.
Let $w = 1$, we effectively work with the affine piece $U_w$.
$$V(y - x^2, z - xy, y^2 - xz)$$
We calculate the tangent is at $(0, 0, 0)$:
$$J_V(x, y, z) = \left[ \begin{array}{ccc} -2 x & 1 & 0 \\ -y & -x & 1 \\ -z & 2 y & -x \\ \end{array} \right] $$ $$J_V(0, 0, 0) = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array} \right]$$ $$\ker J_V(0, 0, 0) = t[1, 0, 0]$$ We can choose $x$ to be a uniformizer.
From the first polynomial, we get: $$y = \frac{x^2}{w}$$ $$y^2 = \frac{x^4}{w^2}$$
Hence, \begin{align*} \frac{w}{z} & = \frac{wx}{y^2} \\ & = \frac{w^3x}{x^4} \\ & = \frac{w^3}{x^3} \\ & = x^{-3} w^3 \end{align*}
Hence, we can see that $\operatorname{ord}_{[1:0:0:0]}\left( \frac{w}{z} \right) = -3$.
To find the zeros, we write the following: $$w = x^2 = xy = wz = 0$$ $$w = x = 0$$
Substitute $x = w = 0$ back to the polynomial, we get: $$y = 0$$
Hence, the zero is give as: $$[0: 0: 0: 1]$$
We use a similar strategy to find the tangent space
We dehomogenize the polynomial.
Let $z = 1$. We effectively work with the affine piece $U_z$.
$$J_V(0, 0, 0) = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & -1 & 0 \\ \end{array} \right]$$
$$\ker J_V(0, 0, 0) = t[0, 0, 1]$$
We can choose $y$ to be a uniformizer.
Still, from the third polynomial: $$\frac{y^2}{z} = x$$
\begin{align*} \frac{w}{z} & = \frac{xy}{z^2} \\ & = \frac{y^3}{z^3} \\ & = y^3 \frac{1}{z^3} \end{align*}
Hence, we can conclude that $\operatorname{ord}_{[0: 0: 0: 1]}(\frac{w}{z}) = 3$.
Hence, we can write out $\operatorname{div}(w / z)$:
\begin{align*} \operatorname{div}(w / z) & = \operatorname{ord}_{[0: 0: 0: 1]}(w / z) [0: 0: 0: 1] + \operatorname{ord}_{[1: 0: 0: 0]}(w / z)[1: 0: 0: 0] \\ & = 3[0: 0: 0: 1] - 3[1: 0: 0: 0] \end{align*}