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This question comes from exercise I-10 of The Geometry of Schemes by Joe Harris (although this question is not about schemes). It is translated to less abstract language below:

Consider the Riemann sphere $\hat {\mathbb C} = \mathbb C \cup \{\infty\}.$ Find an example of an open set $U\subseteq \hat {\mathbb C}$ and an analytic function $f$ on $U$ such that $f$ cannot be written as a sum of two analytic functions $f_0 + f_{\infty},$ where $f_0$ vanishes at $0$ and $f_\infty$ vanishes at $\infty.$ (If $U$ does not include $0,$ then the "vanish at $0$" condition is vacuously true and $f_0$ can just be any analytic function on $U$. Similar for $\infty.$)

(The sheaf-theoretic version is this: Let $\mathcal F_0,\mathcal F_\infty$ be the sheaf of ring of analytic functions vanishing at $0, \infty.$ Let $\mathcal F = \mathcal F_0 \oplus \mathcal F_\infty,$ and define $\varphi: \mathcal F \to \mathcal A : (f_0, f_\infty )\mapsto f_0+ f_\infty,$ where $\mathcal A$ is the sheaf of analytic functions. Then although the morphism $\varphi$ is surjective at every stalk of $\mathcal F,$ it is not surjective for every open set $U,$ and the ring $\mathcal F(U).$ )

I am thinking about how to construct such an example. Let $U$ be an open set that contains both $0$ and $\infty,$ and let $m$ be a $C_c^\infty$ function supported in a compact neighborhood of $0.$ Then for any smooth $u$ vanishing at both $0$ and $\infty,$ we have $$ f= [mf+u] +[(1-m)f-u] $$ If we want both $[]$ to be analytic, then we need to satisfy the Cauchy-Riemann equation $$ \frac{\partial}{\partial \bar z} (mf+u) =0, \frac{\partial}{\partial \bar z} (-mf-u)=0. $$ For what I remember, the CR equation is quite nice (as square root of Laplacian), and a smooth solution for $u$ should exist with the boundary conditions. So such a decomposition $f_0 + f_{\infty}$ exists. For $U$ that does not contain both $0$ and $\infty,$ such a decomposition is is trivial.

So maybe the example I am looking for in the exercise does not exist? Is there a flaw in my argument?

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  • $\begingroup$ Try taking $U$ to be the whole of $\hat{\mathbb C}$. $\endgroup$
    – André Henriques
    Commented May 25, 2021 at 13:18
  • $\begingroup$ @AndréHenriques That's what I am thinking. So the CR PDE actually behaves differently on the entire Riemann sphere than in bounded domains? $\endgroup$
    – Ma Joad
    Commented May 25, 2021 at 13:19
  • $\begingroup$ I think that the Weierstras-Malgrange Preparation theorem provides an obstruction $\endgroup$
    – Nagaraj Iyengar
    Commented May 25, 2021 at 15:31

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If $z_0\not\in U$, then the function $\frac{1}{z-z_0}$ takes distinct values at $0$ and $\infty$.

You can then write $f=f_0+f_\infty$ for $f_0=f+\frac{a}{z-z_0}+b$ and $f_0=-\frac{a}{z-z_0}-b$ for suitable constants $a$ and $b$.

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  • $\begingroup$ I am looking for an example where $f$ cannot be written in this form! $\endgroup$
    – Ma Joad
    Commented May 25, 2021 at 13:33
  • $\begingroup$ This answer implies that such an example can't exist if $U$ is not all of $\hat\mathbb C$. $\endgroup$
    – Wojowu
    Commented May 25, 2021 at 13:54
  • $\begingroup$ But the focus is - why it exists if $U=\hat {\mathbb C}?$ (If I have explained clearly enough.) $\endgroup$
    – Ma Joad
    Commented May 25, 2021 at 14:04
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    $\begingroup$ The only functions analytic on the entire Riemann sphere are constant. $\endgroup$
    – Wojowu
    Commented May 25, 2021 at 14:04
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    $\begingroup$ I don't know what you mean with it, but usually it means "analytic on $\mathbb C$ and analytic at $\infty$". If you allow meromorphic functions, then by the same argument as in the answer there will be no examples on all of $\hat{\mathbb C}$ either. $\endgroup$
    – Wojowu
    Commented May 25, 2021 at 14:16

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