1
$\begingroup$

Let $L/K$ be a finite separable field extension with $O_v$ valuation ring of $K$. Here I will not assume $O_v$ complete and similarly for $K$ as well.(i.e. There is a completion $\hat{O}_v$ of $O_v$ and $\hat{K}=Frac(\hat{O}_v)$ both of which are complete.)

Now consider integral closure of $O_v$ in $L$ which I will denote as ring $S$. Now since $O_v$ is valuation ring, it is clear that $S$ is Dedekind for sure.

$\textbf{Q:}$ Is $S$ DVR in general? Note that if $O_v$ is complete with $K$ complete, then it follows that $S$ is DVR from theorem of general local complete field extensions.

The reason I ask this question is the following. Consider a complex Riemann surface $X$ and $x\in X$. Denote $M_x$ the meromorphic germs at $x$ and set $z$ as the valuation parameter of $M_x$.(This is in particular DVR but not complete as not all cauchy against valuation converging. The completion is formal power series and formal laurent expansions respectively.) Let $L/M_x$ be any finite field extension. Since $char(M_x)=0$, it follows that the extension is separable. Now by Thm on Puisseux series(Forster Lectures on Riemann Surfaces Chpt 1), one deduces $L=M_x[T]/(T^d-z)$. Note this extension is always totally ramified and valuation ring as well.

$\textbf{Q':}$ It seems from Riemann surface persepctive, this is always DVR? This forces the extension always totally ramified.

$\textbf{Q'':}$ It seems that whether the valuation ring is complete is irrelevant here? I do recall that completeness is required to construct the non-trivial norm by inducting on the degree of extensions(Taylor, Frolich, Algebraic Number Theory Chpt 3) and there might be a way to bypass it. So for general valuation ring $O_v$ of field $K$ and $L/K$ finite separable extensions, I can still consider $L\otimes_O O_v$ which will split into product of "local fields"? The story goes along for complete DVR for sure but does it hold for non-complete DVR? What should I watch out for this procedure?

$\endgroup$
  • $\begingroup$ @reuns Maybe this is dumb. It seems that your assertion works in general setting without $L/K$ normal. Consider normal closure of $L$ over $K$ in algebraic closure. Then one apply your argument and restricts the valuation to $L$. Then your assertion is asserting $O_L$ is always DVR irregardless. Is this what you are implying here? If so, how do I see $O_L$ is DVR?(In other words, you are hinting the proof for complete local ring setting goes through without hitches.) $\endgroup$ – user45765 Mar 6 at 22:10
  • $\begingroup$ @reuns I guess, you mean to obtain all valuations over the prime of $O_K$ but there is no good reason to expect they are all the same if you do not have complete DVR for $O_K$. The uniqueness of extension I knew requires completeness of base field or base valuation ring. $\endgroup$ – user45765 Mar 6 at 22:13
  • 1
    $\begingroup$ If $O_K$ is a complete DVR and $L/K$ is a finite normal extension then $v$ extends uniquely to $L$ through $v(a) = v(b)$ whenever $a,b$ are $K$-conjugates so $O_L$ is a DVR and it is the integral closure of $O_K$. If $O_K$ is non-complete look at the completion to find the different ways to extend $v$. For a Riemann surface the holomorphic germs is (with a local chart sending $x$ to $0$) $O_{M_x}=\{\sum_{n \ge 0}c_nz^n\in\Bbb{C}[[z]],\exists r>0,|c_nr^n|\to0\}$ and $O_L=O_{M_x}[T]/(T^d-z)$ is the equivalent subring of $\Bbb{C}[[z^{1/d}]]$. The valuation extends uniquely. $\endgroup$ – reuns Mar 6 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.