Let $R$ be a complete discrete valuation ring of characteristic zero with residue field $k$ of positive characteristic $p$. Let $K=\mathrm{Frac}(R)$ and $L$ be an finite algebraic extension of $K$. Let $A$ be the integral closure of $R$ in $L$.

Questions. Why $A$ is also a discrete valuation ring and why completeness of $R$ imply that $A$ is finite over $R$ (i.e., "finite" as $R$-module)?


Let $L = K(\alpha)$ and let $f(x)$ be the minimal polynomial of $\alpha$.

Fact: there is a bijection between non-zero prime ideals of $A$ and factors of $f(x)$ in $K_P[x]$, where $P$ is the unique prime ideal of $K$.

Note that $K=K_P$ since $K$ is a complete discrete valuation field.

Now since we can assume that $f(x)$ is irreducible, there exists exactly one prime ideal $Q$ in $A$. Hence $A$ is a DVR.

Lastly, $L_Q = K_P \cdot L = K \cdot L = L$, as required.

As to your second question: $B$ being a finitely generated $A$-module does not require completeness. It is a standard result proved using the trace function.

Edit: $P = Q \cap R$. Showing that there is such a $Q$ can be done by appealing to the Lying Over Theorem. The fact that $A$ is a DVR, i.e. that there is only one prime ideal in $A$, is harder to prove: the only way I know of involves what I quoted as a "fact".

About $a_i$ lying in $A$:

We prove that for $x \in A$, $\operatorname{tr}_{L/K}(x) \in R$. Let $M$ be the Galois closure of $L/K$. Since $x$ is integral over $R$, $\sigma_i(x)$ is integral too for all $\sigma_i \in \operatorname{Gal}(M/K)$. Hence $\sum \sigma_i(x) = \operatorname{tr}_{L/K}(x)$ is integral over $R$. As a trace (fixed by the Galois action), this must lie in $K$. Therefore $\operatorname{tr}_{L/K}(x)$ is an element of $K$ integral over $R$, and as $R$ is integrally closed, $\operatorname{tr}_{L/K}(x) \in R$.

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  • $\begingroup$ Hi, thank you for your answer. two questions: I not understand how you relate $P$ and $Q$. The $P$ is exactly the unique nonzero prime of $R$, the uniformizer ideal, right? Thus $R_P=R$ and $K_P=K$ by completeness assumption, right? What is $Q$? Naively I would say that it's the image of $P$ under map $R \to A$, but then in general $Q$ would be not a prime. At least the image of $P$ in $A$ decomposes in a unique product of primes of $A$. Or did I misunderstand your $P$ and $Q$? Secondly: The trace-argument, that proves the finiteness of $A$: essentially you take the $\endgroup$ – user741314 Jan 13 at 0:07
  • $\begingroup$ vector basis ${a_i}$ of $L$ over $K$ and show that if $a \in A$ and $a = \sum \lambda_i a_i$, then using trace (it is not trivial since char zero and thus $L/K$ separable) that the $\lambda_i \in R$, right? What I not understand is, why you can assume that $a_i \in A$? Firstly, we only know $a_i \in L$? $\endgroup$ – user741314 Jan 13 at 0:07
  • $\begingroup$ @MortyPB I expanded my answer. $\endgroup$ – Lukas Kofler Jan 13 at 0:49
  • $\begingroup$ Great, thank you again. One remark: do you maybe know source where I can look the Fact that you used to establish the existence of $Q$? $\endgroup$ – user741314 Jan 13 at 2:57
  • $\begingroup$ dpmms.cam.ac.uk/~jat58/antm2019/ANT_2019.pdf corollary 3.14 $\endgroup$ – Lukas Kofler Jan 13 at 8:05

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