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Let $A$ be a complete DVR with fraction field $K$ and residue field $k$, let $l$ be a finite separable extension of $k$, then $l=k(\bar\alpha)$ for some $\bar\alpha\in l$ whose minimal polynomial $\bar g\in k[x]$ is monic, irreducible and separable. Let $g\in A[x]$ be any lift of $\bar g$ that is monic, then $g$ is also irreducible and separable.

Define $L:= K[x]/(g(x))=K(\alpha)$, where $\alpha$ is the image of $x$ in $K[x]/(g(x))$, then $L$ has valuation ring $B$ which equals to $A[\alpha]$.

Why is $B$, the valuation ring of $L = K(\alpha)$, equal to $A[\alpha]$?

I know since $A$ is a complete DVR, $B$ is the integral closure of $A$ is $L$, so $A[\alpha]\subset B$, but I don't know the other inclusion.

p.s. It is found in one step in the proof in a course manual, that if $A$ is a complete DVR with fraction field $K$ and residue field $k$, then there's a correspondence between unramified extensions $L/K$ and separable extensions $l\,/\,k$.

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The reason is quite simple.

First we show $l$ is indeed the residual field $B/\mathfrak{P}$, temporarily let $l'$ denote the residual field. Then $[l':k]\leq [L:K]$. Also $$\alpha \in L\implies \bar{\alpha} \in l' \implies l=k(\bar{\alpha}) \subset l'$$ Hence $[l':l]\geq 1$. If $[l':l]>1$, then $[l:k] \leq [L:K]/[l':l] < [L:K]$, contradicting to $[L:K]=[l:k]$. Therefore $l=l'$.


Let $n=[L:K]=[l:k]$, $\mathfrak{P}$ be the prime ideal in $L$ and $v$ the valuation on $K$. Suppose $$k_0 + k_1 \alpha + \cdots + k_{n-1} \alpha^{n-1} \in B \qquad k_i\in K$$ If there is some $k_i$ such that $v(k_i)<0$, let this $i$ be such that $v(k_i)$ is minimal. Dividing both sides by $k_i$ gives $$a_0 + a_1 \alpha + \cdots + a_{n-1} \alpha^{n-1} \in \mathfrak{P} \qquad v(a_j)\geq 0,a_i=1$$ this says $\{1,\bar{\alpha},\cdots,\bar{\alpha}^{n-1}\}$ are linearly dependent over $k$, a contradiction.

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  • $\begingroup$ It needs $v(\alpha)=0$ right?($v$ extends from $K$ to $L$) Why is it the case? $\endgroup$
    – CYC
    Sep 5, 2018 at 8:21
  • $\begingroup$ @CYC If $v(\alpha) > 0$, then $\alpha \in \mathfrak{P} \implies \bar{\alpha} = 0$ in $l$. $\endgroup$
    – pisco
    Sep 5, 2018 at 8:52
  • $\begingroup$ But $l$ is a given field and $L$ is what we constructed from it, at given point we don't know $l$ isomorphic to $B/\mathfrak B$. $\endgroup$
    – CYC
    Sep 5, 2018 at 9:07
  • $\begingroup$ I suppose you meant $\mathfrak{P}$ rather than $\mathfrak{B}$. $l = B/\mathfrak{P}$ is true because $\mathfrak{P}$ is defined as the prime ideal of $B$. $\endgroup$
    – pisco
    Sep 5, 2018 at 9:16
  • $\begingroup$ I mean $L$ is newly constructed from a given $l$, then $B$ is defined from $L$, at bare construction we don't know yet $l$ is its residue field(though in fact they will be isomorphic.), and I think in fact we need $B=A[\alpha]$ first in order to prove $l$ is the residue field of $L$. $\endgroup$
    – CYC
    Sep 5, 2018 at 9:22
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Thanks for @pisco that done most of the work for me, though due to subtle detail of notations(like different definitions of $\bar\alpha$), I will accept his and write my commentary here.

Let $\hat\alpha$ be the image of $\alpha$ in $B/\mathfrak P$, then by construction of $g$, $\bar g(\hat\alpha) = 0$, since $\bar g$ is irreducible by assumption, $\hat\alpha\neq 0$ in $B/\mathfrak P$, so $\alpha\notin \mathfrak P$ hence $B=A[\alpha]$.

p.s. We have $n=[k(\hat\alpha):k] \leq [B/\mathfrak P : k]\leq [L:K] = n$, so $B/\mathfrak P =k(\hat\alpha) \approx k(\bar\alpha) = l.$

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    $\begingroup$ There are quite a few obscurities in your argument which I don't understand. First, why is $\alpha$ in $B$? It is so only if you take a lift $g$ of $\bar g$ which is monic. But this is a punctilious remark. My second objection is more serious. In my opinion, the proof in the first part of the answer of @pisco only shows that $A[\alpha]$ , as a $\mathbf Z$-module, has the same rank $n$ as $B$, i.e. the index $(B : A[\alpha])$ is finite, but not automatically 1. To show equality, it will be crucial to know that $L/K$ is unramified ... $\endgroup$ Sep 5, 2018 at 19:28
  • $\begingroup$ ... as shown in your p.s. (= the second part of pisco's answer). For what follows, the main reference is Serre’s « Local Fields », chap.3, §§5-6. Write $C= A[\alpha]$. On the one hand, in the general case, the different $\mathfrak D_{B/A}$ divides $(g’(\alpha))$, and these two ideals are equal iff $B=C$ (§5, coroll. 2). On the other hand, the calculation of the different shows that $L/K$ is unramified iff $B=C$ (§6, lemma 3). This solves your problem. $\endgroup$ Sep 5, 2018 at 19:30
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    $\begingroup$ @nguyenquangdo First, I missed a detail that $g$ is required to be monic, thanks for noting. Second, I don't use the conclusion of pisco's answer, only as a proof that $B$ is a generated by $\alpha$ as an A-module,(For any $b\in B$, since $b$ is in $L$, it is generated by $\alpha$ with coefficients in $K$, then since $\alpha$ has valuation $=0$, pico's work shows the result that each coefficient is in fact in $A$.), it follows $B=A[\alpha]$. $\endgroup$
    – CYC
    Sep 6, 2018 at 1:55
  • $\begingroup$ @nguyenquangdo $L/K$ is unramified follows from the assumption, and in my proof, I indeed used the extension is unramified. Also, I proved $B=A[\alpha]$, not merely the same rank. Of course, your proof using different also sounds. $\endgroup$
    – pisco
    Sep 6, 2018 at 4:09
  • $\begingroup$ Thanks for the clarifications. I missed the fact that $\alpha$ is invertible, which explains the second (ex- first) part of pisco's answer. Everything's OK now. $\endgroup$ Sep 6, 2018 at 6:33

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