Function field has infinitely many valuations.

Question

Suppose we have a field $K$ which is a finite algebraic extension of the field $\mathbb{C}(X)$. Can you give me an argument that $K$ admits infinitly many discrete valuations?

Answer 1

Let $a$ be a complex number. $\mathbb{C}(X)$ has a discrete valuation at $X-a$. $K$ has a valuation extending it.

Answer 2

I think you can use one idea that for a given field $k$, $k[x]_{(x - a)}$ is a DVR for every $a\in k$ and choose $k = \mathbb C$. Now, you can extend from $\mathbb C(x)$ to any algebraic extension, which is again DVR.