Hartshorne II Prop 6.8

Question

My weaknesses with commutative algebra are really slowing down my progress through Hartshorne. I hope someone can help me understand some statements in the proof of the proposition below.

Prop 6.8: Let $X$ be a complete nonsingular curve over $k$, let $Y$ be any curve over $k$, and let $f:X\to Y$ be a morphism. Then either (1) $f(X)=$ a point, or (2) $f(X)=Y$. In case (2), $K(X)$ is a finite extension field of $K(Y)$, $f$ is a finite morphism, and $Y$ is also complete.

I am confused about two statements in the proof, when we have case (2): First, how do we know that $K(X)$ and $K(Y)$ finitely generated extension fields of $k$? Second, why is it clear that $U=f^{-1}V$?

Also, Hartshorne then goes on to define the degree of any finite morphism $f:X\to Y$ of curves as the degree of the field extension $[K(X):K(Y)]$. But how do we know that $K(X)$ contains $K(Y)$?

Answer 1

It is written (pretty clearly, to my surprise) in the book: the given morphism is dominant and thus $\;K(Y)\subset K(X)\;$ , and since both fields are simple, finite transcendental extensions of $\;k\;$ of trans. degree one, $\;K(X)/K(Y)\;$is algebraic.

This is usually studied in a course in algebraic structures: fields extensions, Galois theory and etc.

Answer 2

1) The fact that they are finitely generated extensions over $k$ is proved in Chapter I, Theorem 4.4. The point is that the function field of a variety $X$ is always determined by a dense open subset, so you can reduce to the affine case where it is pretty easy to see $K(X)$ is a finitely generated extension of $k$.

2) If $f:X\to Y$ is dominant, it induces an injection $K(Y)\subset K(X)$, and thus $K(X)$ is a finite extension of $K(Y)$.

3) To see that $U=f^{-1}(V)$, define $U$ as $f^{-1}(V)$ and note that it is integrally closed (by smoothness assumption).