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Let $\mathbb{C}\{z\}$ be the ring of convergent series in one variable over $\mathbb{C}$, $K$ the fraction field of $\mathbb{C}\{z\}$, $E$ a Galois extension of $K$ and $\mathcal{O}_{E}$ the integral closure of $\mathbb{C}\{z\}$ in $E$.

We have naturally the diagram induce by the inclusion maps:

$$\begin{array}{ccc} \mathbb{C}\{z\} & \rightarrow & \mathcal{O}_{E} \\ \downarrow & & \downarrow \\ K & \rightarrow & E. \end{array}$$

Now I'm requested to prove that the dual of the inclusion $\mathbb{C}\{z\} \hookrightarrow \mathcal{O}_{E}$, namely the map of topological spaces $Spec(\mathcal{O}_{E}) \rightarrow Spec(\mathbb{C}\{z\})$, is ramified only over $0$.

As a first step I calculated $Spec(\mathbb{C}\{z\})$, which should be the set $\{(0),(z)\}$, where $(z)$ is the maximal ideal. In specific this means that $\mathbb{C}\{z\}$ is a Discrete Valuation Ring and I have to compute the ramification explicitly only over two points. Moreover another consequence is that $\mathcal{O}_{E}$ is a Dedekind domain.

Unfortunately I have no idea on how to go further. I computed the valuation induced by $\mathbb{C}\{z\}$ on $K$, but it doesn't seem to bring anywhere.

I thank you previously for your help.

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I made your list of inclusions into a diagram. I hope that's what you intended, but you should be able to adjust it to your needs now. – t.b. Jun 8 '11 at 20:34
I'm not entirely sure I'm understanding the setup, but if so: you're looking at ramification in a Galois extension of DVRs with algebraically closed residue field. So the ramification index will just be the degree of the extension. Geometrically, this the ramification index over the closed point. At the generic point, you have a finite Galois, hence separable (anyway you are in characteristic zero) field extension, which is by definition unramified. – Pete L. Clark Jun 8 '11 at 20:36
@Theo, thak you very much for the correction, it is exactly what I was thinking about. @Pete, following your comment I have only a couple of questions. 1) Do you think it's correct that $\mathbb{C}\{z\}$ is a DVR? (I'm not completely sure of my conclusion). 2) Where can I look for some references to the results you exposed (and maybe also for similar results regarding generic fields instead of $\mathbb{C}$)? Thank you both for your time. – Giovanni De Gaetano Jun 8 '11 at 21:24
Student73: You're welcome. For the future: only one ping works per comment, see here for more details. @Pete: the OP has some follow-up questions directed at you. – t.b. Jun 8 '11 at 21:29
yes, $\mathbb{C} \{z\}$ is a DVR: the valuation of an element is $n$ if you can write it as $z^n g(z)$ with $g(0) \neq 0$. See also the wikipedia article on DVRs for confirmation. – Pete L. Clark Jun 9 '11 at 1:24

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