Timeline for lattice of subalgebras of a finite commutative algebra
Current License: CC BY-SA 3.0
10 events
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| Jan 30, 2013 at 16:24 | comment | added | Russ Woodroofe | Ok, so my prior attempt missed a condition, but the answer still seems to me like it should be "no". Won't the group algebra of (Z_p)^2 over the field Z_p do as a counterexample? According to Wikipedia this is a local ring. It has at least 3 subalgebras of dimension 2 (corresponding to subgroups of order p), whose meet is Z_p, and whose join is the whole algebra. (And this contradicts distributivity.) | |
| Jan 30, 2013 at 14:15 | comment | added | Oliver Kayende | In response to Woodroofe, thank you but the question is about finite local algebras. Given a unital commutative ring R, by an R-algebra it is meant, as usual, a ring A with unity for which there exists a ring homomorphism f:R->Z(R) that maps the unity of R to the unity of A, where Z(R) is the center of A. So, R need not be a field. Moreover, by a subalgebra of A it is meant a unital subring of A whose unity is the same as the unity of A. Finally, as I believe Suarez-Alvarez was pointing out, the subfield lattice of a finite field is indeed distributive. | |
| Jan 30, 2013 at 14:13 | comment | added | Oliver Kayende | Perhaps, to apply said theorem of Ore, it would help to ask whether every such lattice is isomorphic to the subgroup lattice of a finite group. | |
| Jan 30, 2013 at 13:40 | comment | added | Oliver Kayende | Thank you, Ralph, for the reference. In response to Goldstern, yes, thank you, (a) and (b) is precisely what I meant if in (b) by "the family" it is meant the lattice L (a priori) and not necessarily the entire family of all subalgebras as in (a). In any case, these two axioms are basic parts of the definition of a lattice and I hope I have stated the problem clearly enough. | |
| Jan 30, 2013 at 3:16 | answer | added | Russ Woodroofe | timeline score: 1 | |
| Jan 29, 2013 at 23:33 | comment | added | Ralph | (II) I don't know about the history, but it immediately follows from the structure theorem of Artinian rings: Any Artinian ring is a finite direct product of local Artinian rings (Eisenbud, Cor. 2.16). | |
| Jan 29, 2013 at 23:30 | comment | added | Goldstern | I see, by "admits a supremum" you probably mean that (a) a supremum exists in the family of all subalgebras, and (b) this sup is a member of the family. | |
| Jan 29, 2013 at 23:26 | comment | added | Goldstern | Aren't fields local algebras? By your definition, $GF(2)$, $GF(2^2)$, $GF(2^3)$, $GF(2^5)$, $GF(2^{30})$ form a sublattice of $GF(2^{30})$. I must have misunderstood something... | |
| Jan 29, 2013 at 23:18 | history | edited | Goldstern | CC BY-SA 3.0 |
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| Jan 29, 2013 at 23:00 | history | asked | Oliver Kayende | CC BY-SA 3.0 |