Bourbaki

Exercise 11, § 2

Algebra, I, 127

Bourbaki marks this one as harder than the rest.

E is called a left semi-group if every element xx of E is left cancellable.

(a) If uu is an idempotent (§1§ 1, no. 4), then ux=xux = x for all x∈Ex \in E (use Exercise 9(a)); uu is identity element for the law induced on EuEu.

(b) If uu and vv are two distinct idempotents of E, then Eu∩Ev=∅Eu \cap Ev = \varnothing and the sets EuEu and EvEv (with the laws induced by that on E) are isomorphic.

(c) Let R be the complement of the union of the sets EuEu, where uu runs through the set of idempotents of E. Show that ER⊂RER \subset R; it follows that R is a stable subset of E. If R is non-empty, then aR≠RaR \neq R for all a∈Ra \in R (use Exercise 9(a) to prove that, in the contrary case, R would contain an idempotent); in particular, R is then infinite. R is called the residue of the left semi-group E.

(d) If R is non-empty and there exists at least one idempotent uu in E (that is E≠RE \neq R), for all x∈REu,xEu≠Eux \in REu, xEu \neq Eu; in particular, no element of REuREu is invertible in EuEu and REuREu is an infinite set (use Exercise 8).

(e) If E has an identity element ee, ee is the only idempotent in E and R is empty (note that E=EeE = Ee).

(f) If there exists a right cancellable element aa in E (in particular if the given law on E is commutative), either E has an identity element or E=RE = R (note that if there exists an idempotent uu, then xa=xuaxa = xua for all x∈Ex \in E). For example, if E is the set of integers ≥1\geq 1, with addition, E is a commutative semi-group such that E=RE = R.

(g) If there exists a∈Ea \in E such that δa\delta_a is surjective, either E has an identity element or E=RE = R (examine separately the case where a∈Ra \in R and the case where a∈Eua \in Eu for an idempotent uu).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.