I found interesting problem set http://www.math.lsa.umich.edu/~kesmith/593hmwk2-2014.pdf and I noted Problem 3-3.
And I found another version:
Let $\mathcal{U}$ be the subset of all open sets of $\mathbb{C}$ containing $0$ and define a partial ordering on $\mathcal{U}$ by $U \leq V$ iff $V \subseteq U$. For $U \in \mathcal{U}$ let $\mathcal{O}_U$ be the set of all analytic functions on $U$. Prove that the ring $$\mathcal{O}= \lim_{\rightarrow}\mathcal{O}_U$$ is a discrete valuation ring.
As written on problem set, I studied Dummit & Foote Excercise 7.8 and 7.9 (concept of germs), but I have no idea why $\mathcal{O}$ is a discrete valuation ring.
As definition of discrete valuation ring, First we should find a Field $K$, s.t. $R\backslash \{0\} \subseteq K^{\times}$ (What's the field contatins germs of analytic functions?)
And I have no idea to find a discrete valution $\nu:K^\times \to \mathbb{Z}$ satisfying three axioms.
What is the explicit discrete valution of field containg of those germs in to integer? I have completely no idea. I have no instinct to find this map on concept of germs.
Please give me hint to how to construct this map.
Every element of a ${\cal O}_U$ can be identified with a power series centered at $0$ with positive radius of convergence, and vice-versa. So you should think of $\varinjlim{\cal O}_U$ as the ring of power series with positive radius of convergence. This is reminiscent of the formal power series ring $\Bbb C[[z]]$.
You should already know how $\Bbb C[[z]]$ is a DVR - let this inspire you.