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Nov 29, 2013 at 0:22 comment added Gerry Myerson So, maybe you could edit "square-free" into the question? People shouldn't have to trawl through the comments to find it.
Nov 28, 2013 at 13:31 comment added Michael Barr Yes, I should have said that $n$ was square free.
Nov 28, 2013 at 4:17 comment added Qiaochu Yuan Yes, we need $n$ squarefree.
Nov 28, 2013 at 4:09 comment added Gerry Myerson If $n=-4$, your ring, ${\bf Z}[2i]$, is not integrally closed, is it?
Nov 28, 2013 at 3:32 review First posts
Nov 28, 2013 at 3:44
Nov 28, 2013 at 3:26 answer added Qiaochu Yuan timeline score: 7
Nov 28, 2013 at 3:21 comment added Qiaochu Yuan Your ring is a Dedekind domain with finitely many prime ideals and any such thing is a PID.
Nov 28, 2013 at 3:16 history asked Michael Barr CC BY-SA 3.0