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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?

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2 Answers 2

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

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How can I extend this? –  max Jan 17 '13 at 18:27
Or alternatively: What is an argument, that i can really extend these valuations? –  max Jan 17 '13 at 18:42
Let $R$ be the integral closure of $\mathbb{C}[X]$ in $K$. Then by the Lying-over-Theorem for every $(X-a)$ there exists a prime $P$ of $R$ lying over $(X-a\mathbb{C}[X]$. Since $R$ is a Dedekind domain (Krull-Akizuki), the local ring $R_P$ is a discrete valuation ring. –  Hagen Jan 21 '13 at 7:06

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.

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