Given some irreducible polynomial $p(x,y)$ in $\mathbb{C}^2$ (in this case, quadratic in both $x,y$), from what I understand there are two ways to more or less naturally construct a surface out of it:
- one can consider the zero locus of $p$ (after homogenization) and obtains an algebraic (projective) curve $\mathcal{C}$,
- one can consider a local solution of the form $(x,y(x))$, analytically continue this function element and thus get a Riemann surface $\mathcal{S}$.
If $p$ is smooth (in the sense that $\mathcal{C}$ is smooth, ie $p(x,y)=\frac{\partial p}{\partial x}(x,y)=\frac{\partial p}{\partial y}(x,y)=0$ has no solutions), as far as I understand, we must have $\mathcal{C}=\mathcal{S}$: by the smoothness of $p$, we can equip $\mathcal{C}$ with a complex structure, more or less immediately making it a Riemann surface. As each function element $(z,f(z))$ in $\mathcal{S}$ must satisfy $p(z,f(z))=0$, we have $\mathcal{S}\subseteq \mathcal{C}$, and as $\mathcal{C}$ is path-connected, by definition we can reach every point of $\mathcal{C}$ by analytic continuation, hence $\mathcal{C}\subseteq\mathcal{S}$.
However, I am not sure what happens if $\mathcal{C}$ has singular points. Clearly, I can find a local solution $y(x)$ as before, using the quadratic formula, of the form $y=f(x)\cdot (g(x)\pm\sqrt{\triangle(x)})$, where $f(x)$, $g(x)$ are holomorphic on $\overline{\mathbb{C}}$ and $\triangle(x)$ is a polynomial of degree at most $4$. Now, the genus of $\mathcal{S}$ should depend only on the zeros of $\triangle(x)$. If I knew $\triangle(x)$ to have $4$ distinct zeros, for instance, then I would know the genus of $\mathcal{S}$ to be $1$.
If I again consider the (now not necessarily smooth) curve $\mathcal{C}$, then as far as I know it is birationally equivalent to precisely one smooth algebraic curve. It would be very nice if this curve would just so happen to be the same as $\mathcal{S}$ again, for instance, that if $\triangle(x)$ has $4$ distinct zeros, that then $\mathcal{C}$ would be birationally equivalent to an elliptic curve.
This leads to my actual question:
Given a Riemann surface $\mathcal{S}$ obtained by analytic continuation of a local solution of the equation $p(x,y)=0$, $p$ a polynomial, and the smooth projective curve $\overline{\mathcal{C}}$ birationally equivalent to the (possibly singular) projective curve $\mathcal{C}$ induced by the zero locus of $p$, how are they related? Intuitively, it seems they should be fairly close to each other if not the same, but I don't really trust my intuition here.
Please do note that the above has popped up rather randomly out of a completely different context, and I am not very familiar with either Complex Analysis or Algebraic Geometry. So it is entirely possible that there is some gross error in understanding on my part, in which case (obviously) feel very free to point this out to me.
Your singular (affine) curve $C$ is birational to a smooth curve $S$ which is (locally up to isomorphism) unique. At singular points you may have weird stuffs such as $y^2 =x^2- x^3$ which is birational to the smooth curve $Y^2 = 1-X\cong \Bbb{A}^1$. The problem is that the points $Y=1$ and $Y=-1$ correspond to the same point $y=0,x=0$ of the singular curve, so it is not immediate how to get a chart for $S$ from those of $C$.
To solve this you can consider $k(C)$ the field of rational functions on your singular curve $C$, the smooth curve $S$ is found from the set of its discrete valuation rings (containing $k$ and whose fraction field is $k(C)$), equivalently you can take the integral closure of $k[x,y]/(y^2-x^2+x^3)$, the smooth curve is its points.
Your projective singular curve is covered by finitely many affine curves so it suffices to repeat this process with each of those affine curves.