Remark 15.111.6. Let $A$ be a discrete valuation ring with fraction field $K$. Let $L/K$ be a finite separable field extension. Let $B \subset L$ be the integral closure of $A$ in $L$. Picture:

$\xymatrix{ B \ar[r] & L \\ A \ar[u] \ar[r] & K \ar[u] }$

By Algebra, Lemma 10.161.8 the ring extension $A \subset B$ is finite, hence $B$ is Noetherian. By Algebra, Lemma 10.112.4 the dimension of $B$ is $1$, hence $B$ is a Dedekind domain, see Algebra, Lemma 10.120.17. Let $\mathfrak m_1, \ldots , \mathfrak m_ n$ be the maximal ideals of $B$ (i.e., the primes lying over $\mathfrak m_ A$). We obtain extensions of discrete valuation rings

$A \subset B_{\mathfrak m_ i}$

and hence ramification indices $e_ i$ and residue degrees $f_ i$. We have

$[L : K] = \sum \nolimits _{i = 1, \ldots , n} e_ i f_ i$

by Algebra, Lemma 10.121.8 applied to a uniformizer in $A$. We observe that $n = 1$ if $A$ is henselian (by Algebra, Lemma 10.153.4), e.g. if $A$ is complete.

Comment #5071 by comment_bot on

It would be nice to say that $B_{\mathfrak{m}_i}$ give all the possible extensions of the valuation of $K$ ot a valuation of $L$. Is this mentioned somewhere else? I couldn't find it.

Comment #5072 by on

A valuation of $L$ which extends the valuation on $K$ given by $A$ would have center on $B$ (as $B$ is integral over $A$) and hence have to be one of the valuations given by the $B_{\mathfrak m_i}$. The reason this isn't stated is that we have almost no material on valuations so everything is done directly using valuation rings. For example, your statement would translate to: every valuation ring $\mathcal{O} \subset L$ with fraction field $L$ dominating $A$ is equal to $B_{\mathfrak m_i}$ for some $i$. The proof is the same. Yes, it might be a good idea to add this to the discussion.

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