Semigroup Completions of Lattices

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 x 2 = 0 implies x = 0 for any x ɛ S . If x 2 = 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 x 2 = 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.