Timeline for If a polynomial f is irreducible then (f) is radical, without unique factorization?
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| Mar 9, 2011 at 11:49 | comment | added | Hailong Dao | @Qiaochu: by the way, your first example $k[x^2,xy,y^2]$ works, just take $(x^2)$. In fact, it is the same as Gerry's example. | |
| Mar 9, 2011 at 11:43 | comment | added | user9072 | Sorry to insist on this, but I am one of the 'some people' Pete L. Clark mentioned. Actually, one could even say that the notion 'irreducible' is only useful in non-UFDs, since in UFDs it coincides with the notion 'prime' and thus it is redundant. Teaching at least the differene between the two notions is I think already useful, and can be done in a short way in an elementary number theory context; for example, by considering all positive integers congruent $1$ modulo $4$ with multiplication (sometimes called Hilbert semigroup, as he is said to have used it as instructional example). | |
| Mar 9, 2011 at 6:34 | comment | added | Pete L. Clark | @Qiaochu: haven't you been told that a word should not appear in its own definition? Or are you rebelling against the Foundation Axiom? ( :) ) | |
| Mar 8, 2011 at 19:36 | comment | added | Qiaochu Yuan | @Pete: I should clarify what I mean by "useful." I mean "something we can easily prove useful theorems about in the context of an undergraduate course." | |
| Mar 8, 2011 at 17:51 | comment | added | Pete L. Clark | Irreducible elements in arbitrary domains are a huge topic of contemporary research among commutative algebraists. Because they behave in complicated ways, they are interesting to study (at least for some people). Just look up the notion of elasticity of a domain to get an idea of how much has been done here. | |
| Mar 8, 2011 at 12:52 | comment | added | Emil Jeřábek | Re irreducibility is not a useful idea in non-UFDs: UFD is a much stronger condition than necessary. The statement follows if every irreducible element is prime; this holds for example in every GCD domain, and more generally, in every pre-Schreier domain. | |
| Mar 8, 2011 at 11:53 | comment | added | Gerry Myerson | Is the goal an example with polynomials? In the ring $k[x,y,z]/(xy-z^2)$, $y$ is irreducible, $z^2$ is in $(y)$, but $z$ is not in $(y)$. | |
| Mar 8, 2011 at 11:20 | comment | added | Qiaochu Yuan | Okay, scratch that example then. | |
| Mar 8, 2011 at 11:19 | history | edited | Qiaochu Yuan | CC BY-SA 2.5 |
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| Mar 8, 2011 at 11:06 | comment | added | Gjergji Zaimi | But then why $x^2y^2\in (x^2y)$? | |
| Mar 8, 2011 at 11:01 | comment | added | Qiaochu Yuan | Ha, yes, let me modify the ring a little... | |
| Mar 8, 2011 at 11:01 | history | edited | Qiaochu Yuan | CC BY-SA 2.5 |
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| Mar 8, 2011 at 10:30 | comment | added | Harry Altman | But $x^2y$ doesn't lie in $k[x^2,xy,y^2]$... | |
| Mar 8, 2011 at 10:07 | history | edited | Qiaochu Yuan | CC BY-SA 2.5 |
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| Mar 8, 2011 at 9:51 | history | answered | Qiaochu Yuan | CC BY-SA 2.5 |