All valuation rings below mean discrete valuation ring.

Let $L/K$ be finite separable extension.(I do not know whether separability here is necessary.) Let $B$ be an valuation ring where $B$ is integral closure of $A$ in $L$ with $Frac(A)=L$. Since $B$ is a valuation ring and $A\to B$ integral closure, it is clear that $A$ is valuation ring. Let $l,k$ be the corresponding residue fields of valuation rings $B,A$. Suppose $l/k$ is separable. Then $B=A[x]$ for some $x\in B$.

$\textbf{Q:}$ Any reason to expect $B=A[x]$ here? I feel this is a very strange statement. If $L/K$ is finite separable extension of local fields with total ramification, then it is clear that $B=A[x]$ by Eisenstein and residue field identification. The other possibility is that for the simple generator of $L=K(x)$, $L/K$ is non-ramified. I do not see any real good reason that $B=A[x]$ in general as $A[x]$ normally should coincide with dual module of $B$.(i.e. $B^D=\{x\in L\vert Tr(xb)\in B\forall b\in B\}$) Correct me if I am wrong here. The separability of local fields does play a role for picking out particular uniformizer.

Ref: Serre Local Fields. Chpt 3, Sec 6, Prop 12.

  • $\begingroup$ Do you mean $Frac(A) = K$? $\endgroup$ – user355183 Oct 12 '18 at 4:32
  • $\begingroup$ I think you need to assume $A$ henselian (e.g., complete). Otherwise the integral closure inside $L$ need not be local. $\endgroup$ – user355183 Oct 12 '18 at 4:46
  • $\begingroup$ @stupid_question_bot Serre's proof seems correct to me and there is no need for henselian condition. However, it might be requirement $B$ DVR and $l/k$ separable allowing this happening. I do not see why these 2 conditions is good enough to hope for simple generation. $l/k$ is always separable for number field case. Since there is 1-1 correspondence between localization and completion, I think the statement should also apply to localized number rings. $\endgroup$ – user45765 Oct 12 '18 at 12:29
  • $\begingroup$ Simple generation does not imply local. Take a map of curves over an algebtaically closed field, and take the integral closure of the local ring at an unramified point of the base curve. It’s integral closure in the larger function field is not local. $\endgroup$ – user355183 Oct 12 '18 at 16:06
  • $\begingroup$ @stupid_question_bot The question is not regarding simple generation implying local. Simple generation in general will not say anything about even completeness not even local fields. There would be too many counter examples. I am interested in why larger ring being DVR and residue field separability are sufficient to deduce simple generation by naively guessing or intuition of such phenomena. $\endgroup$ – user45765 Oct 12 '18 at 16:23

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