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

Let $$k$$ be an algebraically closed field. It is a well-known result 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 $$k(C)\simeq k(C')\Rightarrow C\simeq^{\text{birr}} C'\Rightarrow C\simeq^{\text{isom}} C'$$.

Surjectivity is trickier. 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?

• how do you concluded that $C\simeq^{\text{birr}} C'\Rightarrow C\simeq^{\text{isom}} C'$? is there some extension theorem in the game? intuitively it looks like a generalization of "cleaning denominators" but I'm not sure if this extending $C \vert _U \to C'$ to $C \to C'$ is always allowed and how far this concept can be extended to non affine schemes – Tim Grosskreutz Jul 19 at 17:20

## 2 Answers

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).

• So my line of argument doesn't work for positive characteristic? I didn't see that coming – rmdmc89 Jul 19 at 21:06
• @rmdmc89 : curves can be resolved over any field I believe, but in higher dimension that's an open question. I think it's also true for surfaces. – Nicolas Hemelsoet Jul 19 at 21:34
• Normal curves are smooth, and normalisation works in any characteristic. Similarly surfaces have minimal resolutions of singulariteis over any field. – user45878 Jul 19 at 22:30

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 $$p$$)