Assume we have a free ring $S$ of rank 2 and $K= S \otimes Q$. Now if we define an equivalent relation similar to the case $S=O_{K}$ then is it true that the $S$-ideal classes form a group ?
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3$\begingroup$ If I understand you correctly, then no. Take $S = \mathbb{Z}[\sqrt{8}]$. Let $I = \langle 2 \rangle$ and let $J = \langle 2, \sqrt{8} \rangle$. Then $I \neq J$, but $IJ = J^2$. The integral closure condition shows up in number theory precisely to avoid cases like this. $\endgroup$– David E SpeyerCommented Sep 25, 2011 at 15:09
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1$\begingroup$ One can restrict to invertible ideals, and in that case, one still obtains I group. But that's a bit like cheating... $\endgroup$– felixCommented Sep 25, 2011 at 15:22
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1$\begingroup$ That's not cheating, it's how you define ray class groups. $\endgroup$– Franz LemmermeyerCommented Sep 25, 2011 at 19:25
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4$\begingroup$ Worst title ever! $\endgroup$– Kevin O'BryantCommented Sep 26, 2011 at 1:09
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1$\begingroup$ To emphasize the point raised by David, every proper subring $S$ of the integers of a quadratic field, besides the subring ${\mathbf Z}$, is free of rank 2 as a ${\mathbf Z}$-module and contains a non-invertible ideal, so the $S$-ideal classes are not a group under multiplication. $\endgroup$– KConradCommented Sep 26, 2011 at 12:06
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