Divisor of a rational section in Ravi Vakil's notes

Question

The following is from Exercise 14.2.A of Ravi Vakil's algebraic geometry notes (page 401 here). The exercise asks us to consider the rational section $\frac{x^2}{x+y}$ of the sheaf $\mathcal{O}(1)$ on $\mathbb{P}_{k}^1$ and to compute the corresponding Weil divisor of poles and zeroes. I have a solution but it seems to go against what I think should be intuitively true so I was hoping someone could check it for me.

First of all, it seems to me that intuitively the resulting Weil divisor should be $2[(x)] - [(x+y)]$. However, let me explain my reasoning that led to a different result.

To compute the divisor of a rational section of an invertible sheaf we first have to choose a trivialisation. The obvious choice here is to trivialise on the open subset $D_+(x+y)$ of $\mathbb{P}^{1}$. This gives a trivialisation,

$$ \Psi: \mathcal{O}(1)|_{D_{+}(x+y)} \stackrel{\times \frac{1}{x+y}}{\longrightarrow} \mathcal{O}|_{D_{+}(x+y)}. $$ On sections, we obtain a map on the ring of homogenous polynomials, $$ \Psi_{D_{+}(x+y)}: k \Big[ \frac{x}{x+y}, \frac{y}{x+y} \Big] \cdot (x+y) \longrightarrow k \Big[\frac{x}{x+y} , \frac{y}{x+y}\Big]. $$ Then under this trivialisation, we obtain a section of the field of rational functions on $\mathbb{P}^{1}$. Namely we obtain the quotient of homogenous polynomials of the same degree, $$ \frac{x^2}{(x+y)^2}. $$ This then of course gives a Weil divisor of poles and zeroes of $2[(x)] - 2[(x+y)]$.

My confusion is that Ravi seems to say that you need to choose a trivialisation to compute this Weil divisor. But then of course a result of this is that ever single rational section of a line bundle on $\mathbb{P}^{1}$ gives a Weil divisor of degree $0$, since after choosing a trivialisation we will always have a quotient of homogenous polynomials of the same degree. I thought that was only true for principle Weil divisors, not locally principal.

So which answer is correct, my initial intuition, or the answer I obtained by doing the calculation through a trivialisation?

Answer 1

Your initial intuition is right. Your calculation has issues.

First, you need to find an open cover of $\Bbb P^1$ so that $\mathcal{O}(1)$ is trivialized on each piece. You've only selected one affine open, $D_+(x+y)$, which doesn't cover $\Bbb P^1$. You'd want to add in any affine open containing the point you've left out: $[1:-1]$.

Next, your calculation in the affine patch you have picked contains some incorrect arguments. Let's recap your work: you've identified that the coordinate algebra of $D_+(x+y)$ is $k[\frac{x}{x+y},\frac{y}{x+y}]$, that the trivialization map from sections of $\mathcal{O}(1)$ to elements of $k[\frac{x}{x+y},\frac{y}{x+y}]$ is division by $(x+y)$, the image of your section $\frac{x^2}{x+y}$ is $\frac{x^2}{(x+y)^2}$, and that this has a divisor of zeroes and poles $2[(x)]-2[(x+y)]$ inside this patch.

The first step could be slightly clearer: there is a relation between $\frac{x}{x+y}$ and $\frac{y}{x+y}$: they add to one. Instead, I'll write down the coordinate algebra of this patch as $k[\frac{x}{x+y}]$. This means that $\frac{x^2}{(x+y)^2}$, which you've correctly identified as the image of your rational section under trivialization, vanishes to order $2$ at the point $(\frac{x}{x+y})$ and has no other zeroes or poles in this open set, contrary to your claim (the key thing here is that $(x+y)$ isn't inside your trivializing open set, so you can't say anything about what happens here). One way to help resolve this confusion may be renaming the variables: writing $u$ for $\frac{x}{x+y}$, you're dealing with the divisor of $u^2$ on $\operatorname{Spec} k[u]$, which I hope you believe has a zero of order 2 at $(u)$ and no other zeroes/poles inside $\operatorname{Spec} k[u]$. So the divisor of zeroes and poles in this open subset is given by $2[(\frac{x}{x+y})]$, using your notation.

From here, if you added in another trivializing open set containing $[1:-1]$, you'd find that the divisor of zeroes and poles of $\frac{x^2}{x+y}$ under the trivialization on this open set would have a pole at $[1:-1]$, which would complete your computation. For instance, if we took $D_+(x)$ as our trivializing open set, then your section has image $\frac{x}{x+y} = (\frac{y}{x}+1)^{-1}$ in the coordinate algebra $k[\frac{y}{x}]$ under the trivialization map, so it has a pole at $\frac{y}{x}=-1$ of order one and no other zeroes/poles inside $D_+(x)$. This combines with the previous work to verify that $\operatorname{Div}(\frac{x^2}{x+y})=2[(x)]-[(x+y)]$.