Characterizations of finite dimensional Archimedean vector lattices

In this paper, we give some necessary and sufficient conditions for an Archimedean vector lattice $A$ to be of finite dimension. In this context, we give three characterizations. The first one contains the relation between the vector lattice $A$ to be of finite dimension and its universal completion $A^u$. The second one shows that the vector lattice $A$ is of finite dimension if and only if one of the following two equivalent conditions holds : (a) every maximal modular algebra ideal in $A^u$ is relatively uniformly complete or (b) $\mathrm{Orth}\,(A,A^u)=Z(A,A^u)$ where $\mathrm{Orth}\,(A,A^u)$ and $Z(A,A^u)$ denote the vector lattice of all orthomorphisms from $A$ to $A^u$ and the sublattice consisting of orthomorphisms $\pi$ with $|\pi(x)|\leq\lambda|x|$ $(x\in A)$ for some $0\leq\lambda\in\mathbb{R}$, respectively. It is well-known that any universally complete vector lattice $A$ is of the form $C^\infty (X)$ for some Hausdorff extremally disconnected compact topological space $X$. The point $x\in X$ is called $\sigma$-isolated if the intersection of every sequence of neighborhoods of $x$ is a neighborhood of $x$. The last characterization of finite dimensional Archimedean vector lattices is the following. Let $A$ be a vector lattice and let $A^{u}(=C^{\infty}\left(X\right))$ be its universal completion. Then $A$ is of finite dimension if and only if each element of $X$ is $\sigma$-isolated. Bresar in [3] raised a question to find new examples of zero product determined algebras. Finally, as an application, we give a positive answer to this question.