I am trying to do the following exercise:
Consider the algebraic curve $C=\left\{((x: y),(s: t)) \mid x^2 s=y^2 t\right\} \subseteq \mathbb{P}^1 \times \mathbb{P}^1$ and consider the rational function $f=\frac{x s}{y t} \in k(C)$.
- Prove that $C$ is smooth and irreducible.
- Compute $\operatorname{div}(f)$. Hint: use the four standard charts.
- Compute $\operatorname{div}(d f)$ and determine the genus of $C$.
I have succesfully done the first part, have doubts about the second one and little to no clue on the third one. Here's my reasoning for the last two parts.
We know that $\operatorname{div}(f)=\sum_{P\in C}v_P(f)P$. We also know that in a smooth curve:
- $v_P(f)<0 \Longrightarrow f$ will have a pole in $P$ and $v_P(f)$ is the order of that pole.
- $v_P(f)>0 \Longrightarrow f$ is defined at $P$ and has a $0$ at $P$ of order $v_P(f)$.
- $v_P(f)=0 \Longrightarrow f$ is defined at $P$ and not $0$ at $P$.
I believe that with this information I could compute the different values of $v_P(f)$ and eventually the value of $\operatorname{div}(f)$. I would very much appreciate if someone could tell me if this is a correct approach, and in case it is not, tell me how to proceed. Also, I am not using the hint in the problem, which makes me think that my approach is probably incorrect (hints don't lie).
For the third part, I started by computing $df$. I get to the following expression: $$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial s}ds+\frac{\partial f}{\partial t}dt=\frac{s}{yt}dx-\frac{xs}{y^2t}dy+\frac{x}{yt}ds-\frac{xs}{yt^2}dt$$ However, I don't know what else to do. Regarding your answer, please beware that I am not familiar with schemes.
Computing valuations of $f$ is what we need to do but it's only straightforward in the one-dimensional case. Strictly speaking it doesn't make much sense to apply the same techniques to $f$, since writing $f$ in that form already involves some level of abuse of notation.
So, as the hint indicates, we can do the computations locally in appropriate affine subsets of $C$ (after all, the valuation is a local computation). We have the standard open affine cover $D(x)$ and $D(y)$ in $\mathbb{P}^1$, so in total we have to consider four opens: $D(x) \times D(s)$, $D(x) \times D(t)$, $D(y) \times D(s)$, and $D(y) \times D(t)$. To compute the valuation at a point, we first check which of these opens it falls in, then find a uniformizer in that local setting, and check $f$ with that uniformizer.
For example, here's how we can compute the valuation of $f$ at points in $D(x) \times D(s)$. These points will be of the form $P = ((1 : a), (1 : b))$. Let $g$ denote the polynomial that defines $C$. Note that $g$ in this local setting can be identified with $y^2t - 1$. To compute the uniformizer, we see that: $$\frac{\partial}{\partial t} g = y^2 $$ So as long as $a$ is nonzero, a uniformizer at $P$ is $(y - a)$. But if $a$ were zero, $P$ wouldn't satisfy the relation imposed by $g$, so we don't need to consider this case anyway. Now, in this local setting, $f$ can be written as $\frac{1}{yt}$. Therefore $v_P(f) = 0$.
Now you can do the same for the other three cases.
To compute $\mathrm{div}(\mathrm{d}f)$ is more of the same, but now you need to also be able to get a relation between $\mathrm{d}y$ and $\mathrm{d}s$ or whatever the variables are.
For instance, in the same setting as above, we have: $$0 = dg = d(y^2t - 1) = td(y^2) + y^2dt = 2ytdy + y^2dt$$ So: $$df = d\frac{1}{yt} = \frac{1}{t}d\frac{1}{y} + \frac{1}{y}d\frac{1}{t} = \frac{1}{t}\frac{-1}{y^2}dy + \frac{1}{y}\frac{-1}{t^2}dt = \frac{-1}{y^2t}dy + \frac{-1}{yt^2}\frac{-2yt}{y^2}dy = \frac{1}{yt^2}dy$$ And since $dy = d(y - a)$ we have that the valuation is zero everywhere in $D(x) \times D(s)$ again.