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 $k$?


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 $k$.
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