Timeline for Irreducible Polynomials in UFD and corresponding Quotient Field
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 4, 2010 at 22:37 | comment | added | Charles Siegel | Yeah, once you actually know what the contrapositive is, this is entirely trivial. | |
| Feb 4, 2010 at 21:53 | comment | added | user3795 | Qiaochu, I meant the easier direction (the forward direction can be adapted from a textbook or the Wikpedia page mentioned below). Charles, thanks. I had been playing around with that kind of idea, but got turned around about what the contrapositive actually was in this case. Too many negatives with IRreducible. | |
| Feb 4, 2010 at 21:32 | answer | added | Victor Miller | timeline score: 1 | |
| Feb 4, 2010 at 21:26 | comment | added | Charles Siegel | Shanest, here's a quick argument for you, if you want irreducible in $F[x]$ implies irreducible in $D[x]$. Let $f\in D[x]$. Assume it factors as $f=gh$ in $D[x]$. Then it factors as $f=gh$ in $F[x]$. This is actually true regardless of primitivity, which you need for the other direction, as Qiaochu mentioned. The proof I'm giving you proves the contrapositive, instead of irred in $F[x]$ implies irred in $D[x]$ it argues that reducible in $D[x]$ implies reducible in $F[x]$ which is equivalent. | |
| Feb 4, 2010 at 20:49 | comment | added | Qiaochu Yuan | Being irreducible in the quotient field automatically implies being irreducible in the domain. Are you sure you don't mean the other direction? In either case, this is a standard exercise (first show that the product of two primitive polynomials is primitive) called Gauss's lemma. | |
| Feb 4, 2010 at 20:46 | history | asked | user3795 | CC BY-SA 2.5 |