Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $q$?
share|improve this question

1 Answer 1

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.