Let $C$ be a smooth geometrically integral affine curve. This question concerns the smooth completion $C\_1$ of $C$, both defined over a number field $k$.


We know that given any smooth projective geometrically integral curve $X$, and for a finite number of points $\{p\_1,...,p\_r\} \subset X$, the curve $X\backslash \{p\_1,...,p\_r\}$ is affine. However, is $C\_1 \backslash C$ always a finite number of points?


Furthermore, we have the inclusion of adelic points $C(\mathbb{A}\_k) \subset C\_1(\mathbb{A}\_k)$. Is it possible that this inclusion is an equality? Or is it possible that this inclusion is in fact false to begin with? The latter question arised because I was worried about the archimedean places, even though I'm not certain if that will be a problem.



The smooth completion of a curve
Yes, in fact whenever $C$ is a non-empty Zariski open inside a smooth connected curve $C\_1$, then the complement $C\_1\setminus C$ is finite. This is easy to see in the case that $C$ is affine, since in that case $C\_1=D(f)$ is the locus of non-vanishing of a non-zero algebraic function $f$ on $C\_1$. So the complement $C\_1\setminus C$ is the (reduced) vanishing locus of $f$, which has dimension $0$ and hence is finite.


The general case is easy to deduce from this, either by covering $C\_1$ with open affines, or by removing a single closed point $c$ of $C\_1$ and using that $C\_1\setminus\{c\}$ is affine, as you remark.


Regarding your final question, $C(\mathbb A\_k)\to C\_1(\mathbb A\_k)$ is always injective, but will not be bijective in general. For injectivity, you have the general fact that if $C$ is a Zariski-open subscheme of a scheme $C\_1$, then a morphism of schemes $f\colon S\to C\_1$ factors through $C$ if and only if the image of $f$ is contained in the open subset $C\subseteq C\_1$. (I thought this fact was in Hartshorne, but couldn't find it on a quick search.) This says that the map $C(S)\rightarrow C\_1(S)$ is injective in general.


Now the map $C(\mathbb A\_k)\to C\_1(\mathbb A\_k)$ certainly can be bijective in degenerate cases. For example, if $C\_1(k\_v)=\emptyset$ for some place $v$ of $k$, then both sets are empty and so any map between them is a bijection.


But you shouldn't suppose that this map will be bijective in any generality. For example, if $C\_1\setminus C$ contains a $k$-rational point $c$, then the map $C(S)\hookrightarrow C\_1(S)$ is not bijective for *any* non-empty $k$-scheme $S$. Indeed, the morphism $f\colon S\to C\_1$ collapsing all of $S$ to the point $c$ is an element of $C\_1(S)$ that does not lie in $C(S)$, since the image of $f$ is not contained in $C$.


