Page 213 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press

VI: Skew Lattices in Rings

Theorem 6.1.3. Given the lattice T as above, for each e ∈ T set
Se = {eʹ ∈ Re⎮ ∀f ∈T, eʹf ∈ Ref }.

Then S = e∈T Se is the unique maximal right-handed skew lattice in R having section T.

(S is called the right extension of T and each Se is called the S-set of e relative to T.)
Proof. Given a right-handed skew lattice Sʹ in E(R) having T as a lattice section, clearly Sʹ ⊆ S.
We need only show that S is closed under both skew lattice operations and thus is a right-handed
skew lattice. To begin, given eʹ ∈ Se and fʹ ∈ Sf,

eʹfʹ = eʹffʹ ∈ Ref
since eʹf ∈ Ref by definition of Se and thus (eʹf)fʹ ∈ Ref also. Next, given g ∈ T, fʹg ∈ Rfg with
fʹg = fgfʹg. Thus

(eʹfʹ)g = eʹ(fʹg) = eʹ(fgfʹg) = (eʹfg)fʹg ∈ Refg
because eʹfg ∈Refg and Rfg ∪ Refg is a right primitive skew lattice. Hence eʹfʹ ∈ Sef and S is at
least a right regular band under multiplication. As such S generates a right-handed skew lattice
Sʹ. But since R is the D-congruence on Sʹ, T must be a lattice section on Sʹ forcing Sʹ ⊆ S, that is
Sʹ = S. £

Examples 6.1.1 from M4(F) for any field F. The lattice section in each case is the
lattice T of all diagonal matrices in the skew lattice, one from each D-class. Notice that any D-
class that is comparable to all other classes is a full R-set. This is true for all right extensions of
lattices in rings. Incomparable pairs of D-classes, however, in some way create “interference”
with each other. Both classes are properly less than full R-sets. That this is always the situation
for matrix rings over fields is justified later.

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