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Let $C$ be a smooth projective geometrically integral curve over a number field $k$, and let $X = C \backslash \{p_1,...,p_n\}$ be the affine curve obtained from removing $n$ closed points of $C$. For each $i$, we let $Z_i$ denote the closed subscheme of $C$ corresponding to the point $p_i$, i.e., $Z_i = \{p_i\} \hookrightarrow C$.

  1. Can we then simply write $C = X \sqcup \bigsqcup _i Z_i$, a disjoint union of subschemes? Or are there things that are missing or unclear?

Suppose now that $p_i$ is a $k$-rational point for some $i$, then we have $Z_i = \mathrm{Spec}\,k$, so $Z(k)$ contains only a point, the isomorphism $\mathrm{Spec}\,k \rightarrow \mathrm{Spec}\,k$.

  1. However, if $p_i$ is not a $k$-rational point, but a degree $m$ closed point for some $m > 1$, i.e., a $K$-rational point where $K/k$ is of degree $m$ for some $K$, is it right to say that $Z_i = \mathrm{Spec}\,K$ and so $Z_i(k) = \emptyset$?

EDIT. Apparently the subschemes $Z_i$ should be written as $\mathrm{Spec}\,(k(p_i))$, where $k(p_i)$ is the finite field extension of $k$ containing $p_i$ as a rational point. And from this question I realised that a disjoint union of affine schemes is always affine. How do I then recover the whole projective curve $C$ by writing it as a disjoint union of subschemes?

oleout
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    In $X \coprod (\coprod_i Z_i)$, the closure of the generic point of $X$ would miss those $Z_i$. – David Lui Dec 26 '22 at 02:46
  • @DavidLui I don't get it. Wouldn't the closure of the generic point of $X$ be the whole of $X$? So by taking disjoint union with $Z_i$, we are adding back the closed points $p_i$ to give us the whole projective curve $C$, no? – oleout Dec 26 '22 at 02:52
  • @oleout yes, but it wouldn't be $C$ since the $Z_i$ are not there. – David Lui Dec 26 '22 at 04:23

2 Answers2

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How do I then recover the whole projective curve C by writing it as a disjoint union of subschemes?

You can't. An irreducible scheme cannot be written as a non-trivial (neither piece is empty) disjoint union. If it's $X \coprod Y$, then both $X$ and $Y$ are closed, and $X \cup Y$ is the entire space, so $X \coprod Y$ is not irreducible.

David Lui
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No, this is not true, even if you forget the scheme structure and remember only the topology. In that disjoint union, the special points are open. In the projective curve, they are not.

hunter
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