Timeline for Why do I get a morphism $f_P: Spec \mathcal{O}_K \to \mathcal{X}$ for every point $P\in X(K)$? How does this morphism look like?

6 events

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Dec 19, 2012 at 11:27	comment	added	ACL		As Martin said, this requires $\mathcal X$ to be projective. Anywayt, the adjective projective is probably implicit in the use of the word model, and hopefully included in its definition.
Dec 19, 2012 at 9:38	comment	added	Linda		I deleted the tag "arakelov-theory".
Dec 19, 2012 at 9:37	history	edited	Linda		edited tags
Dec 19, 2012 at 9:37	comment	added	Linda		Wow, thanks. This is probably everything I needed. Though I can't find where in the lecture notes that I am reading it is required that $\mathcal{X}$ should be proper over $\mathcal{O}_K$, but maybe I just missed that.
Dec 19, 2012 at 8:51	comment	added	Martin Bright		This is nothing specifically to do with Arakelov theory, and you want $\mathcal{X}$ to be proper over $\mathcal{O}_K$. That said, which part do you not understand? Are you happy that a $K$-point of $X$ is the same as a morphism $\textbf{Spec }K \to X$, and an $\mathcal{O}_K$-point of $\mathcal{X}$ is the same as a morphism $\textbf{Spec }\mathcal{O}_K \to \mathcal{X}$? If so, you just need to think about why, on a projective variety, having a $K$-point is the same as having an $\mathcal{O}_K$ point. Then look up the valuative criterion of properness.
Dec 19, 2012 at 8:24	history	asked	Linda	CC BY-SA 3.0