Timeline for Hermite normal form in families
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 19, 2018 at 15:00 | comment | added | Joshua Grochow | You don't need a Euclidean ring (nor Noetherian!) to get Hermite normal form, just a Hermite ring (Kaplansky, "Elementary Divisors and Modules", Trans. AMS, 1949). I believe it is almost trivial to show that your ring $B$ is Hermite: you just need that for any $(a,b)$, there is an invertible 2x2 matrix $Q$ over $B$ such that $(a,b)Q = (d,0)$ for some $d$. The key thing to show is that this is true for quasipolynomials; I think the trichotomousness (trichotomosity?) almost follows for free if $a,b$ are. | |
| Dec 1, 2009 at 16:31 | comment | added | Danny Calegari | OK, I will take away your "tick" and then put it back again. Thanks. | |
| Nov 29, 2009 at 7:10 | history | edited | Greg Kuperberg | CC BY-SA 2.5 |
Repaired argument
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| Nov 27, 2009 at 17:53 | vote | accept | Danny Calegari | ||
| Nov 27, 2009 at 6:44 | history | answered | Greg Kuperberg | CC BY-SA 2.5 |