1. The product of two von Neumann n -frames, its characteristic, and modular fractal lattices.
- Author
-
Gábor Czédli
- Subjects
SET theory ,MATHEMATICS ,GEOMETRY ,LATTICE theory - Abstract
Abstract. Let L be a bounded lattice. If for each a1 b1 ∈ L and a2 b2 ∈ L there is a lattice embedding ψ: [a1, b1] → [a2, b2] with ψ(a1) = a2 and ψ(b1) = b2, then we say that L is a quasifractal. If ψ can always be chosen to be an isomorphism or, equivalently, if L is isomorphic to each of its nontrivial intervals, then L will be called a fractal lattice. For a ring R with 1 let $${\mathcal{V}}(R)$$ denote the lattice variety generated by the submodule lattices of R-modules. Varieties of this kind are completely described in [16]. The prime field of characteristic p will be denoted by Fp. Let $$\mathcal{U}$$ be a lattice variety generated by a nondistributive modular quasifractal. The main theorem says that $$\mathcal{U}$$ is neither too small nor too large in the following sense: there is a unique $$p = p(\mathcal{U})$$, a prime number or zero, such that $${\mathcal{V}}(F_p) \subseteq \mathcal{U}$$ and for any n ≥ 3 and any nontrivial (normalized von Neumann) n-frame $$(\vec{a},\vec{c}) = (a_1, . . . , a_n, c_{12}, . . . , c_{1n})$$ of any lattice in $$\mathcal{U}$$, $$(\vec{a},\vec{c})$$ is of characteristic p. We do not know if $$\mathcal{U} = \mathcal V(F_p)$$ in general; however we point out that, for any ring R with 1, $$\mathcal V(R) \subseteq \mathcal{U}$$ implies $${\mathcal{V}}(R) = {\mathcal{V}}(F_p)$$. It will not be hard to show that $$\mathcal{U}$$ is Arguesian. The main theorem does have a content, for it has been shown in [2] that each of the $${\mathcal{V}}(F_p)$$ is generated by a single fractal lattice Lp; moreover we can stipulate either that Lp is a continuous geometry or that Lp is countable. The proof of the main theorem is based on the following result of the present paper: if $$(\vec{a},\vec{c})$$ is a nontrivial m-frame and $$(\vec{u},\vec{v})$$ is an n-frame of a modular lattice L with m, n ≥ 3 such that $$u_1 \vee \cdot \cdot \cdot \vee u_n = a_1$$ and $$u_1 \wedge u_2 = a_1 \wedge a_2$$, then these two frames have the same characteristic and, in addition, they determine a nontrivial mn-frame $$(\vec{b},\vec{d})$$ of the same characteristic in a canonical way, which we call the product frame. [ABSTRACT FROM AUTHOR]
- Published
- 2009