Timeline for Ring of Integers as subring with most irreducibles
Current License: CC BY-SA 3.0
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| Feb 14, 2012 at 22:03 | comment | added | Pace Nielsen | You could make "most" less fuzzy by asking: Does the ring of integers satisfy the property that if there is a subring $S$ of $L$ in which $x\in S$ is irreducible, then $x\in R$ and is irreducible or a product of irreducibles. One problem is that, unless the class number is 1, the ring of integers is not a UFD. So irreducibles behave badly. This is why, historically, one uses prime ideals rather than irreducibles. We have unique factorization of prime ideals. | |
| Feb 14, 2012 at 13:27 | comment | added | François Brunault | It seems you're looking for the universal property of normalization, which is the geometric version of "taking the integral closure". See mathoverflow.net/questions/46/… You might also want to replace "irreducibles" by "prime ideals" in your question. | |
| Feb 14, 2012 at 12:09 | history | asked | user11863 | CC BY-SA 3.0 |