# Integral closure of complete DVR in an algebraic extension of its fraction field

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, $$\operatorname{tr}_{L/K}(x) \in R$$.

• 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 – user741314 Jan 13 at 0:07
• 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 $a_i \in L$? – user741314 Jan 13 at 0:07
• @MortyPB I expanded my answer. – Lukas Kofler Jan 13 at 0:49
• 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 $Q$? – user741314 Jan 13 at 2:57
• dpmms.cam.ac.uk/~jat58/antm2019/ANT_2019.pdf corollary 3.14 – Lukas Kofler Jan 13 at 8:05