Semigroup Completions of Lattices
D. H. Adams · Proceedings of the London Mathematical Society · 1970
1. The proof of Theorem 15 is incorrect. The fault, in the ‘if’ part of the proof, lies in the assertion that l(x)∩lr(x) = {0}. The theorem is true if the conditions are altered slightly; indeed it may be enlarged to become the following result. THEOREM 15′. Let S be a semigroup with 0 such that l(S) = r(S) = {0}. Then the following conditions are equivalent: (i) S has no non-zero nilpotent elements; (ii) L(S) is Boolean with complementation l(·); (iii) L(S) is orthocomplemented by l(·); (iv) L{S) is Boolean with complementation r(·); (v) L{S) is semi-completed by r(·). Proof. The proof that (i) ⇒ (ii) stands as in the original paper. It follows that (i) ⇒ (iv) by Lemma 14 in the original paper. The implications (ii) ⇒ (iii) and (iv) ⇒ (v) are trivial. To complete the proof, we need only show that (iii) ⇒ (i) and (v) ⇒ (i). (iii) ⇒ (i). (This part of the proof is due to M. F. Janowitz.) It is enough to show that x2 = 0 implies x = 0 for any x ɛ S. If x2 = 0, then x ɛ l(x) and ll(x) ⊆ l(x). Since ll(x)∩l(x) = {0}, it follows that ll(x) = {0} and so l(x) = S. Hence x ɛ rl(x) = r(S) = {0}. (v) ⇒ (i). If x2 = 0, then x ɛ l(x). As x ɛ rl(x) and as l(x)∩rl(x) = {0}, x ɛ l(x)∩rl(x) = {0}. 2. The conjecture in the conclusion that L(S) is the completion by cuts of L(S) whenever S is a Baer semigroup is correct. I am extremely grateful to Professor M. F. Janowitz for communicating this fact, for pointing out my mistake and for providing part of the proof of the above theorem.