Volume 61 Issue 2 (2011) - Годишник на ТУ - София - Технически ...

Let us consider the endomorphisms f ∈ E An such that for any i = 1, . . . , n − 1either f(a i ) = a i , or f(a i ) = m. Then follows that either f 2 (a i ) = f(a i ) = a i , orf 2 (a i ) = f(m) = m, i.e. f 2 = f.Conversely, let f ∈ E An be an idempotent endomorphism. If suppose thatf(a i ) = a j , then f 2 (a i ) = f(a j ). But f 2 = f, hence f 2 (a i ) = f(a i ). Thus wehave f(a j ) = f 2 (a i ) = f(a i ) = a j . Since f is an injection in the set {a 1 , . . . , a n−1 }it means that a j = a i or a j = m.So, the set of endomorphisms f such that for any i = 1, . . . , n − 1 eitherf(a i ) = a i , or f(a i ) = m is the set of all idempotent endomorphisms in thesemiring E An . We denote this set by ID (E An ).For every element a of a finite (multiplicative) semigroup there is a positiveinteger n = n(a) such that a n is multiplicatively idempotent, see [8]. So, thefollowing is trueProposition 3.3 For every f ∈ E An there exists the natural number n suchthat for any i = 1, . . . , n − 1 either f n (a i ) = a i , or f n (a i ) = m.Theorem 3.4 For every n ≥ 2 the set ID (E An ) is a commutative subsemiringof the semiring E An . The order of this semiring is |ID (E An )| = 2 n−1 .Proof. Let f, g ∈ ID (E An ). Let for some i = 1, . . . , n − 1 follows f(a i ) = a iand g(a i ) = a i . Then (f + g)(a i ) = f(a i ) ∨ g(a i ) = a i ∨ a i = a i .Let for some i = 1, . . . , n − 1 we have f(a i ) = a i but g(a i ) = m. Then(f + g)(a i ) = f(a i ) ∨ g(a i ) = a i ∨ m = m.The last case is when f(a i ) = m and g(a i ) = m for some i = 1, . . . , n − 1. Now(f + g)(a i ) = f(a i ) ∨ g(a i ) = m ∨ m = m.Hence f + g ∈ ID (E An ). If f(a i ) = g(a i ) = a i for some i = 1, . . . , n − 1 then(f · g)(a i ) = g(f(a i )) = g(a i ) = a i .If f(a i ) = a i and g(a i ) = m for some i = 1, . . . , n − 1 it follows that(f · g)(a i ) = g(f(a i )) = g(a i ) = m.When f(a i ) = m and g(a i ) = m for some i = 1, . . . , n − 1 it follows that(f · g)(a i ) = g(f(a i )) = g(m) = m.Hence f · g = f + g = g + f = g · f ∈ ID (E An ).Arbitrary endomorphism f ∈ E An is ordered n – tuple ≀ a k1 , . . . , a kn−1 , m ≀. Theelement m is an image of all a i , i = 1, . . . , n − 1, by one way and so we constructthe endomorphism ∞ = ≀ m, . . . , m ≀. The element m is an image of k elements26