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Let us work over a fixed algebraically closed field $k$ and consider a non-singular projective curve $X$ and $\varphi : X \to \Bbb{P}^1$ a non-constant morphism.

My question is: For $P \in X$, do we have an isomorphism $$\mathcal{O}_{P,X} \cong \mathcal{O}_{\varphi(P),\Bbb{P}^1}?$$

The reason I ask this question is because I want to prove that $\varphi$ is surjective. I believe I have almost done this, and this is the last part in the proof that I basically need. Now I have determined that $\varphi$ is actually a dominant morphism (by topological considerations and using that the cardinality of $X$ is necessarily infinite). So actually I already know that

$$\varphi_P^\ast : \mathcal{O}_{\varphi(P),\Bbb{P}^1} \to \mathcal{O}_{P,X}$$

is injective. How can I prove that it has to be surjective? Do I know that $\mathcal{O}_{P,X}$ is finitely generated (as a module) over the image of $\mathcal{O}_{\varphi(P),\Bbb{P}^1}$?

  • $\varphi_p^*$ is not surjective at any branch point of $\varphi$. – Yuchen Liu Jun 16 '13 at 14:08
  • For your last sentence, the finiteness of $\varphi$, see this question: http://math.stackexchange.com/questions/138922/why-morphism-between-curves-is-finite – Yuchen Liu Jun 16 '13 at 14:16
  • @YuchenLiu: Dear Yuchen, $\varphi_P^*$ will not be surjective at any point unless $\varphi$ is an isomorphism. (It is injective, as the OP notes, and if it is also surjective, it is an isomorphism. But then, as noted in my answer, it would induce an isomorphism on function fields and hence an isomorphism of the underlying smooth projective curves.) Regards, – Matt E Jun 16 '13 at 14:31
  • @MattE: Thanks for pointing out my mistake:) I was thinking about the analytic local ring... Big difference here~ – Yuchen Liu Jun 16 '13 at 14:35

2 Answers2

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These local rings are not isomorphic, unless $\varphi$ itself is an isomorphism. The situation, from an algebraic perspective, is similar to the inclusion of $\mathbb Z$ into $\mathbb Z[i]$. This is not an isomorphism, and does not become one if you localize at $2$ and at the prime above $2$.

One way to see it is that if this were an isomorphism, it would induce an isomorphism on fraction fields, i.e. an isomorphism $K(\mathbb P^1) \cong K(X)$, but such an isomorphism implies that $X$ itself is isomorphic to $\mathbb P^1$ (since a smooth projective curve is determined by its function field).

Matt E
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To complement Matt E's nice answer (+1), let me just observe that you know $\phi$ is surjective because its domain is proper, and hence its image is closed---you know it is dense already.

Here, in more detail, is the proof that a map whose domain is a projective variety is closed (this is the consequence of properness that you need): after unwinding the definitions, Thm 5.7A says precisely that the projection $$\mathbb{P}^n \times \mathbb{A}^m \rightarrow \mathbb{A}^m$$ is a closed map. It follows that the same is true for the projection $$\mathbb{P}^n \times X \rightarrow X$$ onto any affine variety $X$. Since an arbitrary variety has a covering by open affines, it follows also for an arbitrary variety $X$. Furthermore, we may replace $\mathbb{P}^n$ by a closed subvariety $V$. Now given a map $\phi:V \rightarrow X$ from a projective variety $V$ to a variety $X$, we may factor it as $$V \rightarrow V \times X \rightarrow X$$ where the first map is the graph $v \mapsto (v,\phi(v))$ of $\phi$ (a closed embedding, and hence closed) and the second is the projection (a closed map by the above). It follows that $\phi$ is closed. The same argument works for projective schemes, given the appropriate statement of the elimination theorem.

Stephen
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  • Dear Steve, I am only in chapter 1 of Hartshorne; I don't know what properness means yet :D –  Jun 16 '13 at 14:14
  • Ah, sorry! OK, doesn't he prove (or maybe just state? I don't remember) that the projection $\mathbb{P}^n \times X \rightarrow X$ is a closed map for any variety $X$? This is all you need. – Stephen Jun 16 '13 at 14:16
  • Dear Steve, I looked through Hartshorne chapter 1 sections 1-6 and can't find anything like that. Regards, –  Jun 16 '13 at 14:19
  • It should follow from Elimination Theory, Theorem 5.7A. – Stephen Jun 16 '13 at 14:23
  • @BenjaLim: Dear BenjaLim, From the point of view of Hartshorne I.6, the idea will be as follows: you have an inclusion of function fields $K(\mathbb P^1) \subset K(X)$ (this is what $\varphi$ gives). A point on $\mathbb P^1$ corresponds to a discrete valuation on $K(\mathbb P^1)$ that is trivial on $k$, and now you have to extend it to a discrete valuation on $K(X)$ (which will then correspond to a point on $X$ mapping to the given point on $\mathbb P^1$). The existence of such an extension is part of the general theory of discrete valuations, and (although I don't remember it as well ...) – Matt E Jun 16 '13 at 14:27
  • ... as I used too) it is the kind of thing that should be discussed in Hartshorne I.6. You can also look in the first chapter or two of Silverman's book on elliptic curves, where he also develops this "discrete valuation" approach to points on curves and maps between curves. Cheers, – Matt E Jun 16 '13 at 14:28
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    @BenjaLim P.S. For later reference, the way to reconcile this discrete valuation approach to the more geometric point of view of properness that Steve mentions is via the valuative criterion for properness. – Matt E Jun 16 '13 at 14:29
  • @MattE, Good point! I agree. On the other hand, I think that elimination is more useful in general than valuation, and it's not very hard. I suppose a proof of Thm 5.7A appears in Eisenbud. – Stephen Jun 16 '13 at 14:30
  • @MattE Dear Matt, that is precisely my idea. You see what I had in my mind was this. Choose a point $Q \in \Bbb{P}^1$. Then I can consider $R$, the integral closure of $\mathcal{O}{Q,X}$ inside of $k(X)$ via my injective map $\varphi^\ast$. Now I know that there is at least one DVR that dominates $\mathcal{O}{Q,X}$ (for example by localizing $R$ at a maximal ideal lying over the maximal ideal of $\mathcal{O}_{Q,X}$ - such a maximal ideal exists by finiteness of the integral closure. –  Jun 16 '13 at 14:30
  • @BenjaLim: Dear Benjamin, I think you mean choose a point $Q \in \mathbb P^1$. Then your argument looks good; the DVR dominating $\mathcal O_{Q,\mathbb P^1}$ will be a point $P$ of $X mapping to the point $Q$ of $\mathbb P^1$. You will be done, and you don't need the (false) surjectivity that you are asking about. Cheers, – Matt E Jun 16 '13 at 14:33
  • Dear Steve, Sorry to clutter up your answer with all these comments! The only reason I took the valuation point of view is that I think this is how Hartshorne presents things in I.6, and so is probably how the OP is thinking about the problem. I agree that elimination theory is more useful and general (and that your properness argument is the natural one). Cheers, – Matt E Jun 16 '13 at 14:35
  • @MattE Thanks for your last comment, sorry indeed I meant a point $Q \in \Bbb{P}^1$. The reason why I asked this question is because I don't know why the DVR dominating $\mathcal{O}_{Q,\Bbb{P}^1}$ should be a point of $X$ mapping to $P$. That is a point of $X$ is clear to me from the identification of $X$ with an abstract non-singular projective curve. But why should $Q$ map to $P$? That's why I asked about the surjectivity of $\varphi$ because my idea is to use the following result: –  Jun 16 '13 at 14:36
  • Let $Y$ be a quasi projective variety. Let $P,Q \in Y$ and suppose that $\mathcal{O}_P \subseteq \mathcal{O}_Q$ as subrings of $K(Y)$. Then $P = q$. –  Jun 16 '13 at 14:38
  • @BenjaLim: Dear Benjamin, Think about what dominating means: it says that any rational function on $\mathbb P^1$ defined and vanishing at $Q$ pulls back to a rational function on $X$ defined at and vanishing at $P$. This wouldn't be possible unless $P$ maps to $Q$ (since otherwise you could just choose a rational function defined at both $Q$ and the image of $P$, but vanishing at $Q$ but not at the image of $P$). Cheers, – Matt E Jun 16 '13 at 14:38
  • @BenjaLim: Dear Benjamin, You are thinking along the right lines, but your precise idea doesn't work. Think more about my previous comment, and combine it with what you were already doing, and you should get there. Cheers, – Matt E Jun 16 '13 at 14:40
  • @MattE, no worries! Again, I agree that Hartshorne most likely intended this to be done with valuations. Anyway, I'll edit my answer to use Thm 5.7 A. – Stephen Jun 16 '13 at 14:40
  • @Steve I accepted Matt E's answer, because I felt it answered directly the question I asked. Though, when I read about valuative criteria I will certainly look at your answer again! Regards, –  Jun 16 '13 at 16:44
  • @BenjaLim, No problem! Take care – Stephen Jun 16 '13 at 19:23
  • @BenjaLim, I should have added---the method I suggest for solving this problem is somehow orthogonal to the valuation theory, and in no way requires the schemes machinery or the valuative criterion for properness. Indeed, I strongly suggest you learn elimination theory (e.g., from Eisenbud's commutate algebra book) before learning about properness in general (and anyway, elimination theory is IMO the best way to prove that projective space is proper). – Stephen Jun 17 '13 at 00:49