For every function field $L$ there is a smooth projective curve $C$ with $L\simeq k(C)$

Question

Let $k$ be an algebraically closed field. It is well-known that:

The category of smooth (i.e., non-singular) projective curves with dominant morphisms is equivalent to the category of functions fields over $k$ in one variable with morphisms of $k$-algebras.

I'm trying to formalize the bijective correspondence between smooth projective curves and fields in one variable.

The map $C\mapsto k(C)$ which sends a smooth projective curve $C$ to its function field is the natural way to do it. It is injective (up to isomorphism), because $k(C)\simeq k(C')\Rightarrow C\simeq^{\text{birr}} C'\Rightarrow C\simeq^{\text{isom}} C'$.

But surjectivity is tricky. Let $L$ be a function field over $k$ in one variable. Then $\exists\, x\in L$ such that $L\mid k(x)$ is a finite extension. Letting $A$ be the integral closure of $k[x]$ in $L$, then $A$ is a finitely generated $k$-algebra and a domain, so $A\simeq k[x_1,...,x_n]/I(X)$ for some affine variety $X\subset \mathbb{A}^n$.

Now since $k(X)\simeq\text{Frac}(A)=L$, which has transcendence degree $1$, we have $\dim X=1$, so $X$ is a curve. Besides, since $A$ in integrally closed by construction, $X$ must be normal. But $\dim X=1$, so $X$ must be non-singular.

Since the projective closure $\overline{X}$ is birrationally equivalent to $X$, it seems like I'm almost there. The problem is that $\overline{X}$ may be singular at some point.

How do I treat this generally?

Answer 1

You're almost there. For each curve $X$ there exists a proper birational morphism $Y \to X$ where $Y$ is smooth, hence this finishes the proof. Such a map is called a resolution of singularities. In general existence of such maps is still open (but true in characteristic zero).

Answer 2

For simplicity say $k = k^{alg}$.

Let $L$ be a finite extension of $k(t)$ and $R$ the integral closure of $k[t]$ in $L$. Then $R = k[f_1,\ldots,f_n]= k[F_1,\ldots,F_n]/I$ where $V(I)$ is a smooth affine curve.

Let $P_1,\ldots,P_i$ be the discrete valuations on $L$ extending the one of $k[t^{-1}]$. It is very possible that the projective closure of $V(I)$ is singular at the $P_j$ (try with $R = k[u^3,u]$ then $[u^3:u:1] = [1:(u^{-1})^2:(u^{-1})^3]$ is singular at $u^{-1}=0$)

To remedy that, for $j = 1\ldots i$, let $M_p =\sup_{m \le n+j-1} v_P(f_m^{-1})$ the order of the largest pole at $p$, add iteratively to the $f_1,\ldots,f_{n+j-1}$ a function $f_{n+j}\in L$ with a pole at $P_j$ of order $v_{P_j}(f_{n+j}^{-1})=1+M_{P_j}$ and with poles of order at most $M_p+1$ everywhere else.

Then $k[f_1,\ldots,f_{n+i}] = k[F_1,\ldots,F_{n+i}]/V(J)$ where $V(J)$ is a smooth affine curve $\subset \Bbb{A}_k^{n+i}$ and its projective closure $\{ [1:f_1(p):\ldots:f_{n+j}(p)]\}\subset \Bbb{P}_k^{n+i}$ is a smooth projective curve.

(locally at $p$ with $f_l$ the one with largest pole the affine chart for the curve is $(\frac{1}{f_l},\frac{f_1}{f_l},\ldots,\frac{f_{n+i}}{f_l})$ and one of the $\frac{f_m}{f_l} -a$ has a simple zero at $p$)