Timeline for Freeness of a Z[x]-module
Current License: CC BY-SA 3.0
13 events
when toggle format | what | by | license | comment | |
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Jul 13, 2014 at 17:04 | answer | added | Thomas | timeline score: 3 | |
Jul 13, 2014 at 15:26 | comment | added | Thomas | I've published results about this and functions similar on other domain on my website at thomasoandrews.com/math/congruent | |
Jul 13, 2014 at 14:35 | comment | added | Bill Dubuque | Similar question (and answer) on MSE: math.stackexchange.com/questions/33521/… | |
Mar 12, 2014 at 6:12 | answer | added | joaopa | timeline score: 7 | |
Jun 13, 2013 at 13:54 | vote | accept | Stefan Kohl♦ | ||
Jun 11, 2013 at 7:46 | answer | added | François Brunault | timeline score: 15 | |
Jun 10, 2013 at 20:33 | vote | accept | Stefan Kohl♦ | ||
Jun 13, 2013 at 13:54 | |||||
Jun 10, 2013 at 19:49 | answer | added | David E Speyer | timeline score: 18 | |
Jun 10, 2013 at 19:47 | comment | added | Stefan Kohl♦ | Let me add a remark: the values of a generalized polynomial $f$ on an interval $\{a,a+1, \dots, b\}$ determine $f(b+1)$ modulo ${\rm lcm}(1, \dots, b-a+1)$ -- apart from this, the value can be anything. | |
Jun 10, 2013 at 19:42 | comment | added | Stefan Kohl♦ | @Steven: No. -- However I guess a spanning set would (roughly) have to contain a set of representatives of "growth classes", i.e. equivalence classes consisting of generalized polynomials which have, up to multiplication by polynomials, the same rate of growth. | |
Jun 10, 2013 at 18:23 | comment | added | Todd Trimble | Not a precise comment, but my gut says it's not free, and not even free as a $\mathbb{Z}$-module. The same gut suggests trying to find a $\mathbb{Z}$-submodule which is obviously not free, such as a countable product of copies of $\mathbb{Z}$ or something along similar lines. | |
Jun 10, 2013 at 17:33 | comment | added | Steven Landsburg | Do you know a spanning set? | |
Jun 10, 2013 at 17:00 | history | asked | Stefan Kohl♦ | CC BY-SA 3.0 |