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Given projective curves $C$ and $C'$ and a surjective morphism $\varphi\colon C\to C'$, such that $Q\in C'$ is a smooth point and its fibre $\varphi^{-1}(Q)$ consists of smooth points.

Then $\mathcal{O}_{C',Q}$ and $\mathcal{O}_{C,P}$, for $\varphi(P) = Q$, are discrete valuation rings.

My question is: Are all discrete valuation rings in $k(C)$ containing $\mathcal{O}_{C',Q}$ of the form $\mathcal{O}_{C,P}$, for some $P\in\varphi^{-1}(Q)$?

If this is not true under this conditions, can one make additional assumptions, such that this is true? And a counterexample, if my claim is wrong, would be nice.

I already asked this on math.stackexchange, but didn't get any answer.

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2 Answers 2

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There are DVRs between $k[x]_{(x)}$ and $k[[x]]$ that are transcendental over $k(x)$. For example, the $x$-adic valuation DVR of $k(x,y)$ with $y=\Sigma x^{n!}$ or $y=e^x$ (characteristic 0).

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  • $\begingroup$ First of all I want to thank you for the answer. I'm not sure if I understood your example but maybe I haven't made clear, that I want the DVR containing $\mathcal{O}_{C',Q}$ be itself contained in $k(C)$. Is this satisfied in your example? (I'm not sure but I think it is not) $\endgroup$
    – kesa
    Commented May 23, 2015 at 13:53
  • $\begingroup$ My examples are not in $k(C)$, $k(C)$ is finite algebraic over $k(C')$. Sorry, I missed that in your question. $\endgroup$
    – David Lampert
    Commented May 23, 2015 at 14:09
  • $\begingroup$ Well, I had forgotten to mention this explicitly. But still thanks for your answer. It is still interesting, that (at least) the assumptions I made are necessary. $\endgroup$
    – kesa
    Commented May 23, 2015 at 14:18
  • $\begingroup$ If deg $\phi$ isn't prime then there can be $C \rightarrow C'' \rightarrow C'$ with an intermediate DVR corresponding to the point on $C''$. Example: the $x$-adic DVRs of $k(x) \subset k(\sqrt{1+x}) \subset k(\sqrt[4]{1+x})$ (characteristic $\neq$ 2). $\endgroup$
    – David Lampert
    Commented May 24, 2015 at 1:40
  • $\begingroup$ I still don't think, that the resulting DVR is then in $k(C)$ or am I mistaken? Sorry but I'm not very familiar with x-adic DVRs so I need some time to understand your example anyway. $\endgroup$
    – kesa
    Commented May 24, 2015 at 10:46
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Answering my own question is weird to me, but still, I have come to this point (and still need some help):

Lemma 1 Let $F\subset F'$ be function fields (i.e. finite algebraic extensions of $k(X)$) over an algebraically closed field $k$, such that $F'/F$ is a (finite) algebraic extension. Let $(A,\mathfrak{m}_A)$ in $F$ and $(B,\mathfrak{m}_B)$ in $F'$ be discrete valuation rings. Then the following are equivalent:

  • $\mathfrak{m}_A\subset\mathfrak{m}_B$
  • $A\subset B$

Furthermore, we have $A = F\cap B$ and $\mathfrak{m}_A = F\cap \mathfrak{m}_B$.

A prove can be found in Stichtenoth's Algebraic function fields and codes

Lemma 2 If the local rings of a point on a variety contains the local ring of another point on the same variety, then the points coincide, i.e. $\mathcal{O}_P \subset \mathcal{O}_{P'} \Rightarrow P = P'$.

This is in Hartshorne's Algebraic Geometry I.6.4

Using these facts, we can conclude (using the notation from the question):

Let $R\subset k(C)$ a discrete valuation ring containing $\mathcal{O}_{C',Q}$. Then $R$ corresponds to a point $P$ in $C$. Therefore $\varphi$ maps $P$ to some $Q'$ in $C'$. But the local ring of this $Q'$ is contained in $k(C')\cap R = \mathcal{O}_{C',Q}$, hence $Q' = Q$, and thus $P\in\varphi^{-1}(Q)$.

The only thing, that is not very clear to me now is that I assumed, that this $R$ corresponds to a point $P$ on $C$. For smooth $C$, this would be ok! (Hartshorne I.6.?) But without this additional assumption, this is not yet clear to me.

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