Timeline for How big can a commutative subalgebra of Weyl algebra be?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 26, 2013 at 18:10 | vote | accept | John | ||
| Aug 26, 2013 at 18:10 | vote | accept | John | ||
| Aug 26, 2013 at 18:10 | |||||
| Aug 22, 2013 at 19:19 | vote | accept | John | ||
| Aug 26, 2013 at 18:10 | |||||
| Aug 20, 2013 at 23:59 | answer | added | David E Speyer | timeline score: 6 | |
| Aug 20, 2013 at 16:21 | answer | added | David E Speyer | timeline score: 7 | |
| Aug 20, 2013 at 15:51 | comment | added | John | Totally agree that the notion of being big was not made precise enough. Actually, I still have no idea what is the best definition of the size. I mentioned pairs of commuting operators that obey elliptic curve type relations to emphasize that the subalgebra should be more rich/nontrivial than $p^n, p^{n+1},...$ in the example above. @David, thank you for a nice reference! | |
| Aug 20, 2013 at 15:33 | comment | added | David E Speyer | @QiaochuYuan The Krull dimension is $\leq 1$. Let $f$ be any nonscalar element of $A_1$ and let $C(f)$ be the set of $g \in A_1$ that commute $f$. Then $C(f)$ is a finitely generated $k[f]$ module. Since a nontrivial commutative subalgebra must be contained in some $C(f)$, this shows that commutative subalgebras are finite over one dimensional polynomial rings, and hence are Krull dimension one and finitely generated. In fact, more is true: $C(f)$ is, itself, a commutative subalgebra of $A_1$. See Amitsur ams.org/mathscinet-getitem?mr=95305 . | |
| Aug 20, 2013 at 5:00 | comment | added | Qiaochu Yuan | This seems like a bad notion of bigness. I think you want something like Krull dimension instead. | |
| Aug 20, 2013 at 0:06 | history | edited | user5810 | CC BY-SA 3.0 |
fixed title
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| Aug 19, 2013 at 22:44 | comment | added | David E Speyer | Dumb comment: You'll want to restrict yourself to maximal subalgebras in order to say anything. Otherwise, for any positive integer $n$, $k[p^{n+1}, p^{n+2}, \cdots, p^{2n}]$ is a commutative subalgebra which can't be generated by fewer than $n$ elements. | |
| Aug 19, 2013 at 22:16 | history | asked | John | CC BY-SA 3.0 |