I'm looking for a solution to Hartshorne Exercise 1.1.1(c), that uses no, or as little as possible, coordinates transformations, and using only material from section 1 of Hartshorne.
The question is to show that if $f\in\mathbb{C}[x,y]$ is irreducible quadratic, then $\mathbb{C}[x,y]/(f)$ is isomorphic to $\mathbb{C}[x]$ or $\mathbb{C}[x,x^{-1}]$.
Here is the outline I have in mind:
If $\mathbb{C}\subsetneq A(W)^\times$, then let $u\in A(W)^\times\setminus \mathbb{C}$ be such that it has no root in $A(W)$ (i.e. there is no element $v$ s.t. $u=v^k$ for some $k>1$). Take the map $\mathbb{C}[x,x^{-1}]\to A(W),x\mapsto u$. This map is an isomorphism, the problem is how to prove it.
It is injective since the kernel is prime, and thus if nonzero corresponds to a maximal ideal in $\mathbb{C}[x,x^{-1}]$, contradicting the assumption that $u\not\in\mathbb{C}$.
I'm having trouble showing that this map must be surjective without some annoying explicit computations. I think we could start by noting that $\mathrm{Frac}(A(W))$ is an algebraic extension of $\mathrm{Frac}(\mathbb{C}[x,x^{-1}])$. Then we assume that the degree of the extension is greater then $1$. Then this means that $u$ (any representative in $\mathbb{C}[x,y]$) is at least quadratic. But $V(u)\cap W=\emptyset$, since $u$ is a unit in $A(W)$. Now I do not really how to proceed without resorting to doing some explicit computation anyway. So does someone know how to finish this in an elegant way?
In case $\mathbb{C}= A(W)^\times$ we would want to take an irreducible nonunit $u$, and construct the map $\mathbb{C}[x]\to u$. Again it is easy to see that this map is injective. In this case the map is surjective, since it having degree higher than $1$ contradicts the irreducibility of $u$.
