0
$\begingroup$

Let $X$ be a curve, therefore proper , one-dimensional $k$-scheme. Two questions:

  1. How exactly is $K(X)$ defined. I heard it's the fraction field of the stalk at generic point of $X$, but why is this generic point unique? Or is $K(X)$ independent of the choice of the generic point?

  2. Why $K(X)$ has as field extension exactly transcendence degree $1$ over $k$?

$\endgroup$
3
$\begingroup$

Firstly, for $K(X)$ to be reasonably defined, $X$ needs to be integral, i.e. reduced and irreducible.

  1. An irreducible variety has a unique generic point $\eta\in X$ whose closure is $X$. However, an easier and equivalent definition for $K(X)$ is to take the fraction field of any ring $A$ where $\operatorname{Spec}(A)$ is any open, affine subscheme of $X$. Indeed, you can define $\eta$ to be the unique point which is contained in all open affine subschemes of $X$, and in each of these affine subschemes it corresponds to the prime ideal $(0)\subseteq A$.
  2. This is sometimes used as the definition of the dimension of $X$. You are most likely using the definition that the only irreducible subvarieties of $X$ are points, which is paraphrasing the fact that "the maximal length of any chain of irreducible subvarieties of $X$ is one". This is equivalent to the following: For any open subscheme $\operatorname{Spec}(A)$ of $X$, the ring $A$ has Krull Dimension one. By this question, this means that $K(X)=\operatorname{Frac}(A)$ has transcendence degree one over $k$.
| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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