Home > algebraic geometry > Sheaves on the projective line and partial fractions

## Sheaves on the projective line and partial fractions

This post constitutes a solution to Exercise II.1.21(e) from Hartshorne’s book with a bit of extra discussion. Section II.1 is very important technically (and not just to algebraic geometers) but I thought this was the only genuinely interesting exercise. Thanks to Mariano Suárez-Alvarez over at the Mathematics Stack Exchange for a helpful hint.

Here’s the setup: write $\mathbb{P}^1$ for the projective line over an algebraically closed field $k$ and $K$ for the function field. We denote by $\mathscr{H}$ the constant sheaf on $\mathbb{P}^1$ with values in $K$, which is the same as the constant presheaf because $\mathbb{P}^1$ is irreducible.  Then the structure sheaf $\mathcal{O}$ is naturally a subsheaf of $\mathscr{H}$, and it is not hard to see that the quotient $\mathscr{H}/\mathcal{O}$  is identified with $\bigoplus_{p \in \mathbb{P}^1} i_p(K/\mathcal{O}_p)$, where $i_p(K/\mathcal{O}_p)$ denotes the skyscraper sheaf at $p$ with stalk $K/\mathcal{O}_p$.

Now for the interesting part. I claim that the map on global sections $K \to \bigoplus_{p \in \mathbb{P}^1} K/\mathcal{O}_p$ induced by the projection $\mathscr{H} \to \mathscr{H}/\mathcal{O}$ is actually surjective.

It suffices to show that this map surjects onto each summand, so given $f \in K$ and $p \in \mathbb{P}^1$ we need to find $g \in K$ such that $f-g \in \mathcal{O}_p$ and $g \in \mathcal{O}_q$ for all $q \neq p$. Intuitively, we need $g$ to have at most a pole at $p$ (but nowhere else), and we require that $f-g$ has no pole at $p$. So $g$ is the “singular part of $f$ at $p$.”

For this, remove some point $\infty \in \mathbb{P}^1$ besides $p$ where $f$ does not have a pole, and without loss of generality suppose $f(\infty) = 0$. So we are left with $\mathbb{A}^1 \subset \mathbb{P}^1$ and we can write

$f(x) = c\frac{(x-a_1)\cdots(x-a_m)}{(x-b_1)\cdots(x-b_n)}$

where $x$ is the coordinate on $\mathbb{A}^1$. Note that $m < n$ since $f$ vanishes at infinity. If necessary, reorder the $b_j$‘s so that $b_1 = \cdots = b_r = x(p)$ but $b_j \neq x(p)$ for $j > r$, then write out the partial fraction decomposition

$f(x) = \frac{c_1}{x-b_1} + \cdots + \frac{c_n}{x-b_n}.$

I claim that

$g(x) = \frac{c_1}{x-b_1} + \cdots + \frac{c_r}{x-b_r}$

is the desired element of $K$.  Indeed, since $b_1 = \cdots = b_r = x(p)$ we have $g \in \mathcal{O}_q$ for all $q \neq p, \infty$, and clearly $g$ vanishes at $\infty$ so $g \in \mathcal{O}_{\infty}$ also. Since $b_j \neq x(p)$ for any $j > r$, we have $f - g \in \mathcal{O}_p$ as desired.

One interesting consequence is that that $H^1(\mathbb{P}^1,\mathcal{O}) = 0$. To prove this, consider the short exact sequence

$0 \to \mathcal{O} \to \mathscr{H} \to \mathscr{H}/\mathcal{O} \to 0$

and pass to the long exact sequence in sheaf cohomology

$\cdots \to H^0(\mathscr{H}) \to H^0(\mathscr{H}/\mathcal{O}) \to H^1(\mathcal{O}) \to H^1(\mathscr{H}) \to \cdots$.

Since the first map is surjective, the second is zero, whence the third map is injective. But $\mathscr{H}$ is flasque, so $H^1(\mathscr{H}) = 0$, which implies $H^1(\mathcal{O}) = 0$.