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?
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 $a^n=0$.