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I'm reading Gathmann's notes on algebraic curves. Here is exercise $2$.$17$(b):

If $F,G\in k[x,y]$ are coprime and $P\in\Bbb{A}^2$, then every element in $\mathscr{O}_P/\langle F,G\rangle$ has a polynomial representative.

In item (a), he asks us to find a polynomial representative in a concrete case, namely find a representative for $\frac{1}{x+1}$ when $F=y-x^3$, $G=y^3-x^4$ and $P=(0,0)$.

Here I've managed to show that $\mathscr{O}_P/\langle y-x^3,y^3-x^4\rangle=\mathscr{O}_P/\langle y,x^4\rangle=\text{span}_k(1,x,x^2,x^3)$. Findind the right coefficients, we get that:

$$\frac{1}{x+1}=1-x+x^2-x^3\,\,\text{ in }\mathscr{O}_P/\langle F,G\rangle.$$

For the general case, I imagine it's possible to write $\mathscr{O}_P/\langle F,G\rangle$ as the $k$-vector space generated by finitely many monomials, but I can't see how to show that.

Any suggestions?

rmdmc89
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1 Answers1

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If $F,G$ are coprime, then $V(F,G)$ is zero-dimensional and thus a finite collection of points with the discrete topology. So the global sections of its structure sheaf is just the product of the stalks, and thus the composite $k[x,y]\twoheadrightarrow k[x,y]/(F,G)\twoheadrightarrow \mathcal{O}_p/(F,G)$ where the first map is the natural surjection and the second map is the projection on to a factor of the product is surjective.

You can also use the fact that such a ring is Artinian to make hands-on constructions of inverses. As a zero-dimensional local ring like we have here is Artinian, the maximal ideal is nilpotent and therefore for any element $1-a$ where $a$ is in the maximal ideal, we can write $\frac{1}{1-a}=1+a+a^2+\cdots$ which makes sense because eventually $a^n=0$.

KReiser
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  • Can you please explain, why second map is surjection (in the case when sheaf is just product of the stalks)? – Donich Jul 01 '22 at 14:04
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    @DanilSkuridin Suppose $\mathcal{F}$ is a sheaf on a discrete space $X$. Then ${{x}}{x\in X}$ is an open covering of $X$, and by the equalizer exact sequence defining the sheaf condition, $\mathcal{F}(X)=\prod{x\in X} \mathcal{F}({x})$. But $\mathcal{F}({x})$ is also the stalk at $x$, since it's the smallest open neighborhood of $x$. So the projection from the product is a surjection from $\mathcal{F}(X)$ to $\mathcal{F}_x$. – KReiser Jul 01 '22 at 14:11