Timeline for Useless question on rank
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 13, 2010 at 16:42 | comment | added | ashpool | Abstract Algebra, Dummit & Foote, 2nd ed. 1. p.319, Example(3): "The module $R^{n}$ is called the free module of rank $n$ over $R$" 2. p.336, Ex.2: "Assume $R$ is commutative. Prove that $R^{n}\cong R^{m}$ if and only if $n=m$, i.e., two free $R$-modules of finite rank are isomorphic if and only if they have the same rank." As Tom Goodwillie mentioned, I think Dummit just defines a "free $R$-module of rank $n$" to be just $R^{n}$, and has nothing to do with linear independence. | |
| Jul 13, 2010 at 9:54 | history | edited | darij grinberg |
edited tags
|
|
| Jul 13, 2010 at 6:09 | comment | added | Robin Chapman | Do many careless authors really assert that for the zero ring, the zero module has rank $n$? Any citations? | |
| Jul 12, 2010 at 22:59 | comment | added | Tyler Lawson | One might just say that the zero ring doesn't have the invariant dimension property (it's the unique such commutative ring, but there are interesting noncommutative examples). Or, one might take the K-theoretic perspective that rank of a projective module (or even a nonprojective one) should take values in some group dependent on the ring, or that there are many possible ranks associated to the primes of A. Asking what rank really means and how to generalize it does lead to interesting mathematical questions. | |
| Jul 12, 2010 at 22:29 | vote | accept | ashpool | ||
| Jul 12, 2010 at 21:58 | comment | added | Jack Huizenga | Many of the "many authors" that you claim to be mistaken probably don't consider a ring with 0=1 to be a ring at all. If it is an elementary textbook they might have been careful enough to announce this somewhere, but even if they didn't it's really not a big deal. | |
| Jul 12, 2010 at 21:56 | answer | added | Tom Goodwillie | timeline score: 11 | |
| Jul 12, 2010 at 21:46 | comment | added | Yemon Choi | "Not a real question?" or am I being misguidedly harsh here? | |
| Jul 12, 2010 at 21:43 | comment | added | Pete L. Clark | Or better, avoid defining ranks of modules over the zero ring. There is clearly nothing to gain from assigning a numerical invariant to a class of objects that are all isomorphic. | |
| Jul 12, 2010 at 21:41 | comment | added | Mariano Suárez-Álvarez | The rank can be defined as $0$, $-\infty$, or $-7$... The only way to evaluate a definition is in the context of a specific body of results (how many exceptions will there be in the propositions proved about modules due to a particular choice? &c) | |
| Jul 12, 2010 at 21:36 | history | asked | ashpool | CC BY-SA 2.5 |