Timeline for Is the decomposition of an algebra into irreducible components essentially unique?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 17, 2010 at 10:31 | vote | accept | Andrej Bauer | ||
| Nov 17, 2010 at 9:21 | vote | accept | Andrej Bauer | ||
| Nov 17, 2010 at 10:31 | |||||
| Nov 17, 2010 at 1:56 | comment | added | Gerhard Paseman | If you can get it, Chapter 5 of "Algebras, Lattices, Varieties" has a lot to say about unique factorization, and contains theorems and examples similar to what Gerald Edgar posted. My advisor Ralph McKenzie was one of the authors. Lovasz's proof of the cancellation theorem is one of the most amazing results I have seen, and a version of it is in the chapter. Gerhard "Ask Me About System Design" Paseman, 2010.11.16 | |
| Nov 17, 2010 at 1:00 | answer | added | Gerald Edgar | timeline score: 3 | |
| Nov 17, 2010 at 0:06 | answer | added | Steve Lack | timeline score: 3 | |
| Nov 11, 2010 at 16:11 | comment | added | Todd Trimble | A side note is that the term "indecomposable" might be preferable to "irreducible". Usually "irreducible" means having no nontrivial quotients, e.g., for group theory, irreducibles are simple groups. The group $S_3$ is an indecomposable which is not irreducible. | |
| Nov 11, 2010 at 14:44 | comment | added | Andrej Bauer | It seems that the Krull-Schmidt theorem, see en.wikipedia.org/wiki/Krull%E2%80%93Schmidt_theorem, answers my question positively for groups. | |
| Nov 11, 2010 at 13:40 | comment | added | Andrej Bauer | In terms of groups the question is this: can a finite group be a direct product of irreducible groups in two essentially different ways? (A group is irreducible if it cannot be written as a non-trivial direct product.) I suspect Jordan–Hölder theorem is relevant here, but I am not an algebraist. Is a simple group the same thing as an irreducible group? | |
| Nov 11, 2010 at 9:39 | comment | added | Denis Serre | I'm not sure if it helps, but in finite group theory, the decomposition of a $\mathbb C$-representation into irreducible ones is not unique, unless all their multiplicities are $0$ or $1$. For instance, the decomposition of the regular representation of a finite group $G$ is not unique, unless $G$ is abelian. | |
| Nov 11, 2010 at 8:59 | history | asked | Andrej Bauer | CC BY-SA 2.5 |