Given a projective $R$-module $M$, how does one find another $R$-module $N$ such that $M \oplus N$ is a free $R$-module?
In my case, $R = {\mathbb{C}[x, y]}/{(y^2 - x^3 + x)}$, and $M$ is the ideal generated by $\{x, y\}$ over $R$. Since $M$ is projective I know such $N$ exists, but how do I find it?
Here is my initial attempt at solving this.
Since $M$ is generated by $\{x, y\}$, we have a function $F$ (the identity function) from $\{x, y\}$ to $M$, by universal property of free modules, we have a map $G$ from $R^2$ to $M$, and since $M$ is generated by $\{x, y\}$, $G$ is surjective. Further more, in a previous question we proved that $M \oplus \ker(G)$ is isomorphic to $R^2$, so I assumed one must find $\ker(G)$, but can't see how this could be done.
Second edit: Tried continuing in this direction, here's what I got so far. By definition, we must have $G((1,0))=x$ and $G((0,1))=y$ or vise versa ("G maps generators to generators"). If we defines $G(( f(x,y), g(x,y)))=x*f((x,y))+y*g((x,y))$ then we can show that G is indeed a surjective map, and by the uniqueness of G, we get that G must be as we defined it to be. Now all that is left is to find a description of KER($G$) in terms of $R^2$