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 ?

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.
– David SpeyerSep 25 '11 at 15:09

1

One can restrict to invertible ideals, and in that case, one still obtains I group. But that's a bit like cheating...
– felixSep 25 '11 at 15:22

1

That's not cheating, it's how you define ray class groups.
– Franz LemmermeyerSep 25 '11 at 19:25

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.
– KConradSep 26 '11 at 12:06

everyproper 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. – KConrad Sep 26 '11 at 12:06