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Q1) Suppose I have a smooth curve $C \subset \mathbb{A}^2$. Does there always exist another curve $C' \subset \mathbb{A}^2$ such that $C \cap C' = \{ p \}$ is a single point (topologically, scheme theoretically it may be nonreduced). Can I require that $C'$ be smooth as well?

In this case, we have $C = \mathrm{Spec}(k[x,y]/(f))$ and $A = k[x,y]/(f)$ is Dedekind so the question becomes does there exist $f' \in A$ such that $\sqrt{(f')} = \mathfrak{m}$ for some maximal idea. This means that $\mathfrak{m}$ is torsion in the class group (so the points which can occur will be limited).

By a result of Claborn, every abelian group can appear as the class group of some Dedekind domain. This leads to me to ask:

Q2) Is there a classification of groups which appear as the class group of $A = k[x,y]/(f)$? Surely these must have finite $n$-torsion?

However, haveing no torsion in the class group does not immeidately produce a counter example since we must also ensure that no point $\mathfrak{m} \subset A$ is trivial in the class group. Does there exist such an example?

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    $\begingroup$ Relevant: mathoverflow.net/questions/47783/… $\endgroup$
    – pinaki
    Oct 2, 2020 at 22:17
  • $\begingroup$ If $C=\bar{C}\smallsetminus\{p_1,\ldots ,p_s\} $, where $\bar{C}$ is smooth projective, the Picard group of $C$ (= class group of $A$) is the quotient of $\operatorname{Pic}(\bar{C}) $ by the subgroup generated by the classes of the $p_i$. If you are over $\mathbb{C}$, these groups are quotient of a complex torus by a finitely generated subgroup — "classifying" them does not make much sense. $\endgroup$
    – abx
    Oct 3, 2020 at 4:44

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