4 ms·
Ok, but rings for which every nonzero element is a power of a fixed nonzero nonunit element up to associates are exactly discrete valuation rings. So your state
by tprice7 12y ago
Ok, but rings for which every nonzero element is a power of a fixed nonzero nonunit element up to associates are exactly discrete valuation rings. So your statement boils down to "The only UFDs that are discrete valuation rings are (rank 1) discrete valuation rings". And I'm not sure what you mean by rank 1 here, but I can deduce that all discrete valuation rings are rank 1 from what you've said, since all discrete valuation rings are UFDs.
So what you've said boils down to:
"The only UFDs that are discrete valuation rings are discrete valuation rings", which again doesn't make UFDs more relevant than any other condition you might add.
- tprice7 12y agoCorrection: the _integral domains_ for which every nonzero element is a power of a fixed nonzero nonunit element up to associates are exactly discrete valuation rings. So my final statement should read: "...which again doesn't make UFDs more relevant than any other condition you might add _that is stronger than the much more general condition of being an integral domain_"
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