I've been reading the Andreas Gathmann's notes https://agag-gathmann.math.rptu.de/class/curves-2018/curves-2018.pdf about plane algebraic curves, and I'm trying to solve this exercise:
Exercise 2.17. (a) For the curves $F=y-x^3$ and $G=y^3-x^4$, find a polynomial representative of $\frac{1}{x+1}$ in $\mathscr{O}_0 /\langle F, G\rangle$, i. e. compute a polynomial $f \in K[x, y]$ whose class equals that of $\frac{1}{x+1}$ in $\mathscr{O}_0 /\langle F, G\rangle$ (b) Prove for arbitrary coprime $F$ and $G$ and any $P \in \mathbb{A}^2$ that every element of $\mathscr{O}_0 /\langle F, G\rangle$ has a polynomial representative. Is this statement still true if $F$ and $G$ are not coprime?
For the part (a), I understand that the class of $\frac{1}{x+1}$ in the quotient $\mathscr{O}_0 /\langle F, G\rangle$ is $\frac{1}{x+1}+(y-x^3)f+(y^3-x^4)g$ where $f,g\in\mathscr{O}_0$. I've seen some answers using localization of rings, but I want a more elementary answer. I want to find $f,g$ explictly such that the previous combination is a polynomial.
To prove this in general (part (b)) my idea is the following: If $F,G$ are coprimes we know that $\dim\mathscr{O}_0 /\langle F, G\rangle$ as $\mathbb{C}-$vector space is finite, and we have a finite basis. Moreover, we can assume that the elements of this base have the same denominator, because denominators are units in $\mathscr{O}_0$ so multiplying basis by units is still a basis. Then, we can have a basis with only polynomials and therefore all elements have a polynomial representation.
this is basically @KReiser nice answer, just lengthy: I don't think you can find easily an explicit representation without using localisation. Namely, the ring $\mathcal{O}_0/(F,G)$ is just $A=k[x,y]/(y-x^3,y^3-x^4)$ localised at $(x,y)$. As quotients commutes with localisation, compute first $A$: this is simply $A=k[x,x^3]/(x^9-x^4)=k[x]/(x^4(x^5-1))$. Now $A_0$ is this ring localized at $(x=0)$, so $x^5-1$ is a unit, i.e. $A_0=k[x]/(x^4)$. then use the above mentioned computation of that gives you $1/(x+1)=1-x+x^2-x^3$.