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(I) Suppose A is a finite commutative local algebra. Must every lattice of local subalgebras of A be a distributive lattice ?

By a subalgebra of A we mean an algebra contained in A that shares the same unity element. By a lattice of subalgebras of A we mean, as usual, a family of subalgebras of A that is partially ordered with respect to set inclusion, and each of whose non-empty finite subsets admits an infimum and supremum. As usual, by a distributive lattice of subalgebras we mean a lattice L of subalgebras whose meet distributes over its join; inf(A,sup(B,C))=sup(inf(A,B),inf(A,C)).

(II) It is very well known and easy to show that every finite commutative algebra can be uniquely decomposed into a direct sum of local algebras. However, who might I accredit this to ? Who first formulated this result ? Or, what early and general theory in commutative algebra is this a direct result of in the literature ?

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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... –  Goldstern Jan 29 '13 at 23:26
    
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. –  Goldstern Jan 29 '13 at 23:30
    
(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). –  Ralph Jan 29 '13 at 23:33
    
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. –  Oliver Kayende Jan 30 '13 at 13:40
    
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. –  Oliver Kayende Jan 30 '13 at 14:13
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1 Answer

(Building on Goldstern's comment:) If fields are ok, (and if you allow infinite algebras -- see comment by Mariano Suárez-Alvarez below) then the distributivity certainly does not hold.

Take e.g. a finite degree extension $F$ of $\mathbb{Q}$ with Galois group $G$. Subalgebras of $F$ are subfields, by undergraduate field theory, so the subalgebra lattice over $\mathbb{Q}$ is the field extension lattice of $F:\mathbb{Q}$. (I'm assuming that you're taking all your algebras over a fixed field, here $\mathbb{Q}$.) By the Galois correspondence, the field extension lattice is anti-isomorphic to the subgroup lattice of the Galois group.

And subgroup lattices certainly need not be distributive. Indeed, by a theorem of Ore, the subgroup lattice $L(G)$ of a finite group $G$ is distributive iff $G$ is cyclic.

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The first question seems to refer to finite local algebras (and for finite fields your argument does not work because the $G$s are cyclic!) –  Mariano Suárez-Alvarez Jan 30 '13 at 7:08
    
Doh! I'd missed the finite part! I'll edit to reflect. –  Russ Woodroofe Jan 30 '13 at 16:16
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