# How does the curves-fields correspondence fail in higher dimensions?

It's embarassing how little I know about higher dimensional varieties.

Let $k$ be an algebraically closed field. Then, the category of smooth projective curves over $k$ and nonconstant morphisms is equivalent to the category of fields $K/k$ of transcendence degree one.

How does this fail in higher dimensions?

I'm looking for examples of smooth projective varieties $X,Y$ which have isomorphic function fields, but are not isomorphic.

Also, in higher dimensions, how can a rational map between smooth proj. varieties fail to be a morphism?

• Blowup a surface in one point. – MooS Mar 1 '17 at 8:44
• To add to MooS' pithy comment, the key word you are looking for is "birational". Good luck! – Nefertiti Mar 1 '17 at 9:13

Consider this rational map from a smooth curve to a projective space: $$f: \mathbb A^1 -\!\! \rightarrow\mathbb P^1, \ \ f(t) = [t:1/t^2]$$ At first glance, it looks like $f$ is not regular at $t = 0$. But that is an illusion. Clearing denominators, you see that $f$ can be rewritten as $$f(t) = [t^3:1]$$ which is manifestly regular.
So what is the algebraic property that makes the "clearing denominators" trick work? It's fact that the local ring on $\mathbb A^1$ at $t = 0$ is a discrete valuation ring. In this example, $t$ has valuation $1$ and $1/t^2$ has valuation $-2$ in the local ring, so the denominators are cleared by multiplying by $t^2$ which has valuation $+2$.
How does this generalise when you have a rational map $$f: X -\!\! \rightarrow \mathbb P^n$$ where $X$ is smooth, but of arbitrary dimension? Well, the local rings of $X$ in codimension 1 are discrete valuation rings. Using the same trick of "clearing denominators", you see that $f$ is regular everwhere except on a codimension 2 subspace of $X$. But as MooS points out, you can't do better than this - consider the blow up of a surface.