Exercise 5, § 2

Let $A$ be a lattice ordered ring. A convex ideal (VI, p. 30, Ex. 4) of $A$ is called irreducible if it is not the intersection of two convex ideals distinct from itself. a) Show that the intersection of the irreducible convex ideals of $A$ is 0 (if $a \in A$ is nonzero, consider a convex ideal of $A$ maximal among those which do not contain $a$). Deduce that $A$ is isomorphic to a subring $A'$ of a product $\prod A_i$, such that each $A_i$ is lattice ordered, $\operatorname{pr}_i(A') = A_i$, for all $i$ and the ideal $\{0\}$ is irreducible in $A$, (as a convex ideal) for all $i$. b) Show that the following conditions are equivalent in $A$: α) $|xy| = |x| \cdot |y|$ for all $x, y$; β) $x \cdot \sup(y, z) = \sup(xy, xz)$ for all $x \geqslant 0, y, z$; γ) $x \cdot \inf(y, z) = \inf(xy, xz)$ for all $x \geqslant 0, y, z$; δ) $\inf(y, z) = 0$ implies $\inf(xy, z) = 0$ for all $x \geqslant 0$. (To see that γ) implies δ), observe that $xy \leqslant y \cdot \sup(x, 1)$). When these conditions are satisfied, then $A$ is said to be strongly lattice ordered. c) Show that a ring $A$ is strongly lattice ordered if and only if it is isomorphic to a subring of a product $\prod A_i$, of totally ordered rings. (Using a), reduce to showing that if $\{0\}$ is irreducible in a strongly lattice ordered ring $A$, then $A$ is totally ordered; for this, note that if $B$ is a strongly lattice ordered ring then for all $b \in B$ the set $m$ of $x \in B$ such that $\inf(|x|, |b|) = 0$ and the set $n$ of $y \in B$ such that $|y| \leqslant |zb|$ for some $z \in B$ are two convex ideals of $B$ such that $m \cap n = \{0\}$. d) Show that in a strongly lattice ordered ring $A$ the relation $\inf(x, y) = 0$ implies $xy = 0$; for all $z \in A$ we have $z^2 \geqslant 0$; for all $x, y \in A$ we have $xy \leqslant \sup(x', y')$. e) Let $A$ be a lattice ordered ring with no nilpotent element $> 0$; show that if $\inf(x, y) = 0$ implies $xy = 0$ then $A$ is strongly lattice ordered (check condition δ) of b)).