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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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    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.2011-06-09

0 Answers 0