The product of two von Neumann n-frames, its characteristic, and modular fractal lattices

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 F _p .

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 L _p ; moreover we can stipulate either that L _p is a continuous geometry or that L _p 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.