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There are 2 ways to approach function fields: the algebraic approach, i.e. looking at finite extensions of $K(s)$, where $s$ is transcendental. The other is geometric, i.e. considering functions over a curve: $K[s,t]/(q)$ where $q \in K[s,t]$ represents a plane curve.

My question is - are these 2 approaches basically the same? Specifically:

  1. Is $K(s)[t]/(q)$ naturally isomorphic to $K[s,t]/(q)$?
  2. Can every finite extension of $K(s)$ be represented as $K[s,t]/(q)$ for some $q$?
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up vote 2 down vote accepted

The two approaches are basically the same, but you're missing some details.

I believe the relevant theorem is:

There is a one-to-one correspondence between isomorphism classes
of finite extensions of K(s) and isomorphism classes of complete,
non-singular curves over K

I couldn't find a reference, so I may have some small technical detail wrong on what sort of isomorphism classes; e.g. maybe the correspondence involves isomorphism classes of extensions $E/K$. Or maybe it's between isomorphism classes of extensions $E/K(s)$ with isomorphism classes of bundles $X \to \mathbb{P}^1_K$?

The correspondence sends a curve $X$ not to its coordinate ring $\mathcal{O}(X)$, but to its function field $K(X)$, which is the fraction field of $\mathcal{O}(X)$. The reverse correspondence, IIRC, can be constructed by considering the discrete valuations of the field.

In the case of an affine plane curve $X$ defined by $f(s,t) = 0$, where $f$ is irreducible and not contained in $K(s)$, we have:

  • The ring of coordinate functions is $\mathcal{O}(X) \cong K[s,t] / f(s,t)$
  • The field of rational functions is $K(X) \cong K(s)[t] / f(s,t)$

One can say similar things with curves in higher dimensions; e.g. the curve in three dimensions defined by a pair of functions of three variables. If $X$ is non-singular, then $\mathcal{O}(X)$ is a Dedekind domain. The theory of curves, in fact, is extremely closely related to the theory of algebraic number fields -- localizations of the ring of integers of an algebraic number field are also Dedekind domains.

Note that the aforementioned correspondence tells us there's a unique way to modify our affine curve $X$, extending it to a complete curve and desingularizing it.

Every finite algebraic extension of a field can be obtained by iterating the construction $$ K \mapsto K[t] / f(t) $$ whether or not you can do it in just one step is the subject of the primitive element theorem. Doing it in $n$ steps merely corresponds to a curve in $n+1$ dimensions.

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Thanks, it made things a bit clearer in my head. – Ofir Mar 24 '12 at 12:02

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