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Timeline for Freeness of a Z[x]-module

Current License: CC BY-SA 3.0

13 events
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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