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Given $R$ dedekind domain, it is clear that every ideal $I\subset R$ has $I^{-1}$. Furthermore, any principal ideal is clearly invertible. In $R$, all ideals forms an abelian group G. And principal ideals form a subgroup $H$ of this abelian group G. One define the class group by $G/H$. What is motivation for modding out this $H$? It looks like I am ignoring the blob generated by $(0)$ ideal. I found a pure number theoretical answer ( Motivation behind the definition of ideal class group). Can someone kindly provide a geometrical interpretation?

user45765
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  • I expect this t have been asked on this site a few times already, and answered. Have you looked? – Mariano Suárez-Álvarez Jun 08 '17 at 00:16
  • @MarianoSuárez-Álvarez I checked other answers. I only found number theory answers which I do not have a lot intuition on number theory. Was there a good reference one can use as self study to work out problems on computation of class groups? – user45765 Jun 08 '17 at 00:21
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    I think you have several imprecise statements, which may (or may not) be contributing to your confusion. 1) The zero ideal $(0)$ has no inverse. 2) The set of ideals of $R$ does not form a group, only a monoid. Actually not even that, because we again need to exclude the zero ideal so that $R$ is the multiplicative identity. Anyway, the set of nonzero fractional ideals is a group under multiplication. – Viktor Vaughn Jun 08 '17 at 00:30
  • For a more geometric approach, you might start by reading about divisors and the Picard group. – Viktor Vaughn Jun 08 '17 at 00:36
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    @Quasicoherent Yep. Thanks for the references. – user45765 Jun 08 '17 at 00:41

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Geometrically, a fractional ideal of a domain $R$ is a line bundle $L$ on $X=\operatorname{Spec}X$ together with a fixed embedding $L\to\mathcal K_X$ into the sheaf of rational functions on $X$. Modding out the principal ideals is the same, under this correspondence, to forgetting the embedding into $\mathcal K_X$

Mariano Suárez-Álvarez
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Since a Dedekind domain is a UFD iff it is a PID, the class group measures how much unique factorization fails in $R$ by measuring how far the ideals are from being principal.

lhf
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