I've been stuck on Exercise II.1.21(e) from Hartshorne's book for quite a while. It concerns the projective line $\mathbb{P}^1$ over an algebraically closed field $k$: write $\mathscr{H}$ for the constant sheaf with values in the function field $K$ and $\mathcal{O}$ for the structure sheaf. The content of the exercise is that map on global sections $\Gamma(X,\mathscr{H}) \to \Gamma(X,\mathscr{H}/\mathcal{O})$ is surjective. In the previous subexercise one shows that $\mathscr{H}/\mathcal{O} \cong \bigoplus_{p \in \mathbb{P}^1} i_p(K/\mathcal{O}_p)$ where $i_p$ is the skyscraper sheaf construction at $p$.

So the following will suffice: given $f \in K$ and $p \in \mathbb{P}^1$, we need to produce $g \in K$ such that $f - g \in \mathcal{O}_p$ and $g \in \mathcal{O}_q$ for all $q \neq p$. But I'm at a loss as to how to do such a thing. I assume the first step is to remove some point besides $p$, so $\mathbb{A}^1 \subset \mathbb{P}^1$ is what's left over, and write $f$ explicitly in terms of the coordinate on $\mathbb{A}^1$. Could someone point me in the right direction? I would especially grateful for a more conceptual/less ad hoc hint or explanation.


Fix $f\in K$ and $p\in\mathbf P^1$. You are looking for a $g\in K$ such that

  • $g\in\mathcal O_q$ for all $q\neq p$, and

  • $f-g\in\mathcal O_p$.

The first condition means that $g$ has poles only possibly at $p$. The second, that the difference $f-g$ does not have poles at $p$. In other words, $g$ is the singular part of $f$ at $p$. So to construct $f$ we may write the partial fraction decomposition of~$f$ and drop all terms with a pole at a point different from $p$.

  • $\begingroup$ I haven't decoded Arturo's answer, but he is probably doing the same thing in more elaborate language :) $\endgroup$ – Mariano Suárez-Álvarez Apr 27 '11 at 20:09
  • $\begingroup$ Excellent! This I agree with. I've managed to convince myself that Arturo's answer is incorrect. Maybe a good exercise at this point would be to cook up a counterexample... $\endgroup$ – Justin Campbell Apr 27 '11 at 20:14
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    $\begingroup$ @Justin: notice that the result you want to prove is in fact essentially that partial fraction decompositions exist. Of course, when you say it like this it loses much of its appeal :) $\endgroup$ – Mariano Suárez-Álvarez Apr 27 '11 at 20:18
  • $\begingroup$ @Justin: It's possible there was a mistake there, though the grader for that course was pretty good at spotting such. I'll try to reconstruct my thinking later if I have the time and figure it out, but it's been well over 15 years since I last thought about it. (-: $\endgroup$ – Arturo Magidin Apr 27 '11 at 21:30

Here's what my solution was when I took the course:

We know that $\Gamma(X,\mathcal{O})=\mathcal{O}(X) = k$ by Theorem I.3.4; we also have $\Gamma(X,K) = K(X) = k(x)$, also by Theorem I.3.4. Finally, again by Theorem I.3.4, with $k(x)/k[x^{-1}]_{(x^{-1})}$ corresponding to the point at infinity, we have $$\Gamma(X,K/\mathcal{O}) = k(x)/k[x^{-1}]_{(x^{-1})} \oplus \bigoplus_{\alpha\in k}k(x)/k[x]_{(x-\alpha)}.$$ The claim is that $$0\to k\to k(x) \to k(x)/k[x^{-1}]_{(x^{-1})}\oplus\left(\bigoplus_{\alpha\in k} k(x)/k[x]_{(x-\alpha)}\right)\to 0$$ is exact; and this just means showing the last map is surjective.

Note: The basic idea was okay, but it was poorly phrased here. Part of it seems to have been some previous exercises in that homework. I'm rephrasing.

Fix $P$ and let $f+\mathcal{O}_{P}\in K/\mathcal{O}_{P}$ be a nonzero entry. Using a partial fraction decomposition we can write $$f + \mathcal{O}_P = \sum_{i=1}^n f_i + \mathcal{O}_P$$ where each $f_i$ is such that $$\frac{1}{f_i} = \lambda_i\pi^{n_i}$$ with $\lambda_i\in K$, $\pi$ a uniformizer of $\mathcal{O}_P$, and $n_i\gt 0$. E.g., for $\mathcal{O}_0$, you can find scalars $a_0,\ldots,a_k$ such that $f+\mathcal{O}_p = \frac{a_0}{x} + \cdots + \mathcal{a_k}{n^k} + \mathcal{O}_p$. Now let $$ g = \sum_{i=1}^n \frac{1}{\lambda_i}\pi_i^{-n_i}\in K(X).$$ Then $g$ maps to $0$ in $K/\mathcal{O}_Q$ for any $Q\neq P$, and since $$\lambda_i\pi_i^{-n_i} = \frac{1}{f_i}$$ then $$g\equiv \sum_{i=1}^n f_i \equiv f\pmod{\mathcal{O}_p}$$ so $g\in K(x)$ maps to the element that has $f$ in the $P$-component, and zeroes elsewhere. Since these elements generate $$k(x)/k[x^{-1}]_{(x^{-1})}\oplus\left(\bigoplus_{\alpha\in k} k(x)/k[x]_{(x-\alpha)}\right)$$ this shows the last map is surjective.

  • $\begingroup$ I agree that $f_i$ can be written as a unit of $\mathcal{O}_{P_i}$ times a power of a uniformizer, but most units in $\mathcal{O}_{P_i}$ are not scalars. So why $\lambda_i \in k$? $\endgroup$ – Justin Campbell Apr 27 '11 at 19:31
  • $\begingroup$ @Justin: I confess that I just copied what my solution was at the time (over 15 years ago); I'd need to think it through more carefully than I have the time for right now... $\endgroup$ – Arturo Magidin Apr 27 '11 at 20:27
  • $\begingroup$ Consider the element of $\bigoplus_{p \in \mathbb{P}^1} i_p(K/\mathcal{O}_p)$ whose only nonzero entry is $\frac{1}{x(x-1)} + \mathcal{O}_0$. If I understand your solution, you are claiming that $\frac{1}{x(x-1)} \in K$ maps to this element, which is false. For instance, $\frac{1}{x(x-1)}$ is nonzero in $K/\mathcal{O}_1$. But $-\frac{1}{x}$ works, since $\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1}{x-1} \in \mathcal{O}_0$. $\endgroup$ – Justin Campbell Apr 27 '11 at 20:28
  • $\begingroup$ @Justin: Like I said, I'm not on top of this material anymore. I'll probably erase this answer shortly, but somehow I got full credit on this question at the time (the course was being taught, though not graded by, Hartshorne). $\endgroup$ – Arturo Magidin Apr 27 '11 at 20:31
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    $\begingroup$ @Justin: And you might understand why the grader used to call me 'Arturo "Treekiller" Magidin' in that course... $\endgroup$ – Arturo Magidin Apr 28 '11 at 2:11

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