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.

Advertisements
  1. Paul Erdős
    6 May 2011 at 12:03 pm

    You forgot to carry the three in the denominator…

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: