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**Example text**

Iii) The above map ' is the universal homomorphism to an algebraic group. On Algebraic Semigroups and Monoids 41 Proof. (i) By Proposition 3 (iii), it suffices to show that e is the unique idempotent of eSe. eSe/, then xy D xf y for all x; y 2 S , by Lemma 2. Taking x D y D e yields e D ef e D f . eye/ for all x; y 2 S . (iii) Let G be an algebraic group and let W S ! G be a homomorphism of algebraic semigroups. exe/ for all x 2 S . Thus, factors uniquely as the homomorphism ' followed by some homomorphism of algebraic groups eSe !

Then X has a structure of algebraic monoid with unit group G if and only if the Albanese morphism ˛ W X ! X / is affine. Proof. In view of Proposition 16, it suffices to show that X is an algebraic monoid if ˛ is affine. X /. X / Š G=K equivariantly for the left (or right) action of G. G/ Y ! G/; for the left (or right) G-action, this yields the fiber bundle G K Y ! G=K. Since ˛ is affine, so is Y . G/ by conjugation. Thus, the group scheme K is quasi-affine, and hence is affine. We now show that the group law K W K K !

On Algebraic Semigroups and Monoids 37 Step 1: we show that every idempotent of S is either a neutral or a zero element. S /. Since Se is a closed irreducible subvariety of S , it is either the whole S or a single point; in the latter case, Se D feg. Thus, one of the following cases occurs: (i) Se D eS D S . , e is the neutral element. (ii) Se D feg and eS D S . Then for any x; y 2 S , we have xe D e and ey D y. Thus, xy D xey D ey D y. So D r in the notation of Example 1 (i), a contradiction since is assumed to be nontrivial.