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

Let E An be the subset of ÊA nconsisting of all nonconstant endomorphisms andthe additively absorbing element ≀ m m . . . m ≀.Proposition 2.4 The set E An is a subsemiring of Ê An with element ∞ =≀ m m . . . m ≀ .Proof. Follows immediately from Proposition 2.1 - b and from equalities f ·∞ =∞ · f = ∞ for arbitrary f ∈ E An .∑n−1( ) 2 n − 1Proposition 2.5 The order of the semiring E An is |E An | =k!.kk=0Proof. Let f ∈ E An . Suppose that k elements a is , where s = 1, . . . , k, of theset ( {a 1 ,). . . , a n−1 } has images f (a is ) different( from m. ) We may choose them byn − 1n − 1ways. We put them on k places by k! ways. So the number ofkk( ) 2 n − 1endomorphisms with just k values different from m is equal to k!. Finallyk∑n−1( ) 2 n − 1the number of all endomorphisms of E An is |E An | =k!.kk=0Immediately followsCorollary 2.6 The order of the semiring ÊA nis∣∣ ∣∣∑n−1∣ÊA n =k=0( ) 2 n − 1k!+n−1.Consider for any i, j = 1, . . . , n − 1 the endomorphisms f ij ∈ E An defined by{aj , if i = kf ij (a k ) =m, if i ≠ k .These maps are called almost absorbing endomorphisms and the subsetof the semiring E An consisting of all almost absorbing endomorphisms and theabsorbing element ∞ = ≀ m, . . . , m ≀ is denoted by AA (E An ).Proposition 2.7 The set AA (E An ) is an ideal of the semiring E An .Proof. Let f ij , f kj , f il ∈ AA (E An ) where k ≠ i and l ≠ j. It is easy to see thatf ij + f ij = f ij , f ij + f kj = ∞, f ij + f il = ∞, f ij + ∞ = ∞.kFor arbitrary f = ≀ a k1 , . . . , a kn−1 , m ≀ ∈ E An , where some of elements a kibe equal to m, followsmay(f ij · f) (a i ) = f (f ij (a i )) = f(a j ) = a kj , and(f ij · f) (a k ) = f (f ij (a k )) = f(m) = m, where k ≠ i.So, either f ij · f = f ikj , when a kj ≠ m, or f ij · f = ∞, when a kj = m.23