DERIVATIONS IN SOME FINITE ENDOMORPHISM SEMIRINGS

Discussiones Mathematicae General Algebra and Applications 32 (2012) 77 100 doi:10.7151/dmgaa.1191 DERIVATIONS IN SOME FINITE ENDOMORPHISM SEMIRINGS Ivan Dimitrov Trendafilov Technical University of Sofia Faculty of Applied Mathematics and Informatics Department of Algebra and Geometry e-mail: ivan d trendafilov@tu-sofia.bg Abstract The goal of this paper is to provide some basic structure information on derivations in finite semirings. Keywords: endomorphism semiring, derivations, differential algebra. 2010 Mathematics Subject Classification: 16Y60, 12H05, 13N15. 1. Introduction Over a period of sixteen years differential algebra went from being an approach that many people mistrusted or misunderstood to being a part of algebra that enjoys almost unquestioned acceptance. This algebra has been studied by many authors for the last 60 years and especially the relationships between derivations and the structure of rings. The notion of the ring with derivation is quite old and plays an important role in the integration of analysis, algebraic geometry and algebra. In the 1940 s it was discovered that the Galois theory of algebraic equations can be transferred to the theory of ordinary linear differential equations (the Picard-Vessiot theory). In the 1950 s the differential algebra was initiated by the works of J.F. Ritt and E.R. Kolchin. In 1950, Ritt [11] and in 1973, Kolchin [9] wrote the classical books on differential algebra. The theory of derivations plays a significant role not only in ring theory, but also in functional analysis and linear differential equations. For instance, the classical Noether-Skolem theorem yields the solution of the problem for finite dimensional central simple algebras (see the well-known Herstein s book [6]). One of the natural questions in algebra and analysis is whether a map can be defined

78 I. Trendafilov by its local properties. For example, the question whether a map which acts like a derivation on the Lie product of some important Lie subalgebra of prime rings is induced by an ordinary derivation was a well-known problem posed by Herstein [5]. The first result in this direction was obtained in an unpublished work by Kaplansky, who considered matrix algebras over a field. Herstein s problem was solved in full generality only after the powerful technique of functional identities was developed, see [1]. In 1950 s, Herstein ([4, 5, 7]) started the study of the relationship between the associative structure and the Jordan and Lie structures of associative rings. An additive mapping D from R to R, where R is an associative ring, is called a Jordan derivation if D(x 2 ) = D(x)x + xd(x) holds for all x R. Every derivation is obviously a Jordan derivation and the converse is in general not true. It is important to note that the definition of Jordan derivation presented in the work by Herstein is not the same as the one given above. In fact, Herstein constructed, starting from the ring R, a new ring, namely the Jordan ring R, defining the product in this a b = ab + ba for any a, b R. This new product is well-defined and it can be easily verified that (R, +, ) is a ring. So, an additive mapping D, from the Jordan ring into itself, is said by Herstein to be a Jordan derivation, if D(a b) = D(a) b+a D(b), for any a, b R. So, in 1957, Herstein proved a classical result: If R is a prime ring of a characteristic different from 2, then every Jordan derivation of R is a derivation. During the last few decades there has been a great deal of works concerning derivations D i in rings, in Lie rings, in skew polynomial rings and other structures, which commute, i.e., D i D j = D j D i. What do we know about derivations in semirings? Nothing, or almost nothing except the definition in Golan s book [3] and a few propositions. This paper is an attempt to start a study of derivations in finite semirings. Following Herstein s idea of multiplication in Jordan ring, we construct derivations in the endomorphism semiring of a finite chain. The paper is organized as follows. After the second section of preliminaries, in Section 3 we introduce a semiring consisting of endomorphisms having an image with two fixed elements called a string. In such a string we consider the arithmetic and some kinds of nilpotent elements and subsemirings. In Section 4 we construct a mapping D from the given string into itself and prove that D is a derivation in one subsemiring of the string. Then we show that the semiring is a maximal differential subsemiring of this string. Section 5 is devoted to the construction of maps δ α from given string into itself. They are Jordan multiplications, and we are studying their properties. The main results are that δ α are derivations which commute and that the set of all derivations is a multiplicative semilattice with an identity and an absorbing element. In Section 6 we generalize the notion

Derivations in some finite endomorphism semirings 79 of string and consider the arithmetic in such strings. In Section 7 we give some counterexamples and show that the maps δ α, where α are from the whole string, are derivations in an ideal of this string. Finally in this section we consider a class of maps which are the derivations in the whole string. 2. Preliminaries An algebra R = (R, +,.) with two binary operations + and on R, is called semiring if: (R, +) is a commutative semigroup, (R, ) is a semigroup, both distributive laws hold x (y +z) = x y +x z and (x+y) z = x z +y z for any x, y, z R. Let R = (R, +,.) be a semiring. If a neutral element 0 of semigroup (R, +) exists and satisfies 0 x = x 0 = 0 for all x R, then it is called zero. If a neutral element 1 of semigroup (R, ) exists, it is called one. An element a of a semiring R is called additively (multiplicatively) idempotent if a + a = a (a a = a). A semiring R is called additively idempotent if each of its elements is additively idempotent. An element a of a semiring R is called an additively (multiplicatively) absorbing element if and only if a + x = a (a x = x a = a) for any x R. The zero of R is the unique multiplicative absorbing element; of course it does not need to exist. Following [10], an element of a semiring R is called an infinity if it is both additively and multiplicatively absorbing. Facts concerning semirings, congruence relations in semirings and (right, left) ideals of semirings can be found in [3] and [10]. An algebram with binary operation such as a (b c) = (a b) c for any a, b, c M; a b = b a for any a, b M; a a = a for any a M. is called semilattice (join semilattice). Another term used for M is a commutative idempotent semigroup see [15]. For any a, b M we denote a b a b = b. In this notation, if there is a neutral element in M, it is the least element. For a semilattice M the set E M of the endomorphisms of M is a semiring with respect to the addition and multiplication defined by: h = f + g when h(x) = f(x) g(x) for all x M,

80 I. Trendafilov h = f g when h(x) = f(g(x)) for all x M. This semiring is called the endomorphism semiring of M. It is important to note that in this paper all semilattices are finite chains. Following [12] and [13] we fix a finite chain C n = (0, 1,..., n 1}, ) and denote the endomorphism semiring of this chain with ÊC n. We do not assume that α(0) = 0 for arbitrary α ÊC n. So, there is not a zero in endomorphism semiring ÊC n. Subsemiring E Cn = EC 0 n of ÊC n consisting of all maps α with property α(0) = 0 has zero and is considered in [12] and [15]. If α ÊC n such that f(k) = i k for any k C n we denote α as an ordered n tuple i 0, i 1, i 2,..., i n 1. Note that mappings will be composed accordingly, although we shall usually give preference to writing mappings on the right, so that α β means first α, then β. For other properties of the endomorphism semiring we refer to [8, 12, 13] and [15]. In the following sections we use some terms from book [2] having in mind that in [14] we show that some subsemigroups of the partial transformation semigroup are indeed endomorphism semirings. 3. Strings of type 2 By ST Ra, b} we denote subset of ÊC n consisting of endomorphisms with image a, b} (either a}, or b}) where a, b C n. This set is called a string of type 2. So, for fixed a, b C n and a < b, a string of type 2 is ST Ra, b} = a,..., a, a,..., a, b,..., a, b,..., b, b,..., b }. In the semiring ÊC n there is an order of the following way: For any two endomorphisms α = k 0, k 1,..., k n 1 and β = l 0, l 1,..., l n 1 the relation α β means that k i l i for all i = 0, 1,..., n 1, i.e., α + β = β. With regard to this order each of sets ST Ra, b} is an n + 1 element chain with the least element a,..., a and the biggest element b,..., b. Hence string ST Ra, b} is closed under the addition of semiring ÊC n. It is easy to see that the composition of two endomorphisms with images a, b} is an endomorphism of such type. So, we have Proposition 1. For any a, b C n string ST Ra, b} is a subsemiring of semiring Ê Cn. Note that string ST Ra, b} is not a subsemiring of semiring E (a) C n (see [13]), namely 2, 2, 3, 3 E (3) C 4, but 2, 2, 3, 3 / E (2) C 4. E (b) C n

Derivations in some finite endomorphism semirings 81 Let us denote the elements of semiring ST Ra, b} by α a,b k, where k = 0,..., n is the number of the elements of C n with an image equal to b, i.e., α a,b k = a,..., a, b,..., b. }} k When a and b are fixed, we replace α a,b k by α k. Proposition 2. Let a, b C n and ST Ra, b} = α 0,..., α n }. For every k = 0,..., n follows α k α s = α 0, where 0 s n b 1 α k α s = α k, where n b s n a 1 α k α s = α n, where n a s. Proof. Let us multiply α k α s, where 0 s n b 1. This means that at least b + 1 elements of C n have images under endomorphism α s equal to a and then b is not a fixed point of α s. Hence, for all i C n follows αs (a), if i n k 1 (α k α s )(i) = α s (α k (i)) = α s (b), if i n k = a which means that α k α s = α 0. Let us multiply α k α s, where n b s n a 1. Now a and b are both fixed points of the endomorphism α s. Hence, for all i C n follows αs (a), if i n k 1 (α k α s )(i) = α s (α k (i)) = α s (b), if i n k a, if i n k 1 = = α k (i). b, if i n k which means that α k α s = α k. Let us multiply α k α s, where n a s n. This means that at least n a elements of C n have images under endomorphism α s equal to b and then a is not a fixed point of α s. Hence, for all i C n follows αs (a), if i n k 1 (α k α s )(i) = α s (α k (i)) = α s (b), if i n k = b which means that α k α s = α n. The endomorphism α is called a nilpotent if for some natural k follows that α k = α 0 and respectively, b nilpotent if α k = α n. From the last proposition follows that in both cases k = 2 and also

82 I. Trendafilov Corollary 3. For any a, b C n the set: a. N a = α 0,..., α n b 1 } of all a-nilpotent elements is a subsemiring of semiring ST Ra, b}. b. N b = α n a,..., α n } of all b-nilpotent elements is a subsemiring of semiring ST Ra, b}. c. Id a,b = α n b,..., α n a 1 } of all idempotent elements is a subsemiring of semiring ST Ra, b}. Note that all idempotent elements in semiring ÊC n do not form a semiring. The constant endomorphisms a,..., a and b,..., b are called centers of semirings E (a) C n and E (b) C n, respectively. So, we may imagine that string ST Ra, b} connects the two centers a,..., a and b,..., b. Semiring N a is a subsemiring of E (a) C n, analogously N b is a subsemiring of E (b) C n and semiring Id a,b is a subsemiring of E (a) C n E (b) C n. Let us consider the subset S a,b = α 0,..., α n a 1 } of the ST Ra, b}. Obviously, the set S a,b is closed under the addition. So, immediately from Proposition 2 follows Corollary 4. For any a, b C n the set S a,b is a subsemiring of semiring ST Ra, b}. Note that semiring S a,b is a disjoint union of semirings N a and Id a,b. From dual point of view we consider the subset T a,b = α n b,..., α n } of string ST Ra, b}. This set is also closed under the addition and from Proposition 2 we have Corollary 5. For any a, b C n the set T a,b is a subsemiring of semiring ST Ra, b}. Note that semiring T a,b is a disjoint union of semirings Id a,b and N b. 4. The derivation D Now we consider set DS a,b = α 0,..., α n b }. Clearly this set is closed under the addition. Let α k, α l DS a,b and k, l n b. Then from Proposition 2 follows that α k α l = α 0. Also we have α n b α s = α 0, α k α n b = α k and α n b α n b = α n b. Thus we prove Proposition 6. For any a, b C n set DS a,b is a subsemiring of semiring S a,b. Note that in semiring DS a,b endomorphism α 0 is the zero element and endomorphism α n b is the unique right identity. Semiring DS a,b consist of all a nilpotent endomorphisms and the least idempotent endomorphism.

Derivations in some finite endomorphism semirings 83 Now we define a mapping D : ST Ra, b} ST Ra, b} by the rules D(α k ) = α k 1 for any k = 1,..., n, and D(α 0 ) = α 0. Let α k, α l ST Ra, b} and k > l. Then D(α k + α l ) = D(α k ) = α k 1 = α k 1 + α l 1 = D(α k ) + D(α l ), that means D is a linear mapping. Let α k, α l N a. Then from Proposition 3.2 follows that D(α k α l ) = D(α 0 ) = α 0 and also D(α k )α l = α k 1 α l = α 0, α k D(α l ) = α k α l 1 = α 0. So, we have D(α k α l ) = D(α k )α l + α k D(α l ). Since the same equalities are hold if we replace α k with α n b, then it follows D(α n b α l ) = D(α n b )α l + α n b D(α l ). For any α k N a we compute D(α k α n b ) = D(α k ) = α k 1, D(α k )α n b = α k 1 α n b = α k 1 and α k D(α n b ) = α k α n b 1 = α 0. So, we have D(α k α n b ) = D(α k )α n b + α k D(α n b ). Also we compute D(α 2 n b ) = D(α n b) = α n b 1, D(α n b )α n b = α n b 1 α n b = α n b 1 and α n b D(α n b ) = α n b α n b 1 = α 0. So, it follows Thus, we prove D(α 2 n b ) = D(α n b)α n b + α n b D(α n b ). Proposition 7. For any a, b C n mapping D is a derivation in semiring DS a,b. Now we shall prove that there are not differential semirings (under derivation D) containing DS a,b, which are subsemirings of ST Ra, b}. Let α k Id a,b, where k > n b. We compute D(αk 2) = D(α k) = α k 1 and from Proposition 3.2 follows D(α k )α k = α k 1 α k = α k 1 and α k D(α k ) = α k α k 1 = α k. This means that D(αk 2) D(α k)α k + α k D(α k ). Let α k N b. We compute D(αk 2) = D(α n) = α n 1, D(α k )α k = α k 1 α k = α n αn a, if k = n a and α k D(α k ) = α k α k 1 = α n, if k > n a. Hence D(α k)α k +α k D(α k ) = α n α n 1 = D(αk 2). Thus we prove that D is not a Jordan derivation, see [5], in any subset of ST Ra, b}, which contains DS a,b, that means D is not a derivation. So, follows Proposition 8. The semiring DS a,b is the maximal differential subsemiring (under the derivation D) of string ST Ra, b}.

84 I. Trendafilov Note that each of subsets I 0 = α 0 }, I 1 = α 0, α 1 },..., I n b 1 = N a is a subsemiring of N a with trivial multiplication α k α l = α 0 for any 0 k, l n b 1. Since these semirings I k, 0 k n b 1, are closed under derivation D, then I k are differential subsemirings of DS a,b. But from Proposition 2 follows that I k are ideals in semiring DS a,b. Hence Proposition 9. In semiring DS a,b there is a chain of differential ideals I 0 I k = α 0,..., α k } I n b 1 = N a. Let R be an arbitrary differential semiring with derivation d and I is a differential ideal of R. We consider I = x x R, n N 0}, d n (x) I}. R Let x, y R I where dm (x) I and d n (y) I. If m < n we have d n (x) = d n m (d m (x)) I. Then, using that d is a linear map, follows d n (x + y) = d n (x) + d n (y) I, that is x + y R I. On the other hand, d m+n (xy) = m+n ( m+n ) k=0 k d m+n k (x)d k (y) I, which means that xy R I. It is clear that x R I implies d(x) R I. Thus, we prove Proposition 10. Let R be a differential semiring with derivation d and I is a differential ideal of R. Then R I is a differential subsemiring of R. Using Propositions 9 and 10 we can describe the differential structure of semiring DS a,b. Corollary 11. For every differential ideal I k of semiring DS a,b, where k = 0,..., n b 1, follows DS a,b = DS a,b I k. 5. Derivations in string ST Ra, b} Now we use the well known Jordan multiplication in associative rings to define some new derivations in strings ST Ra, b}. Let α ST Ra, b}. We define a mapping δ α : ST Ra, b} ST Ra, b} by the rule δ α (α k ) = αα k + α k α for any k = 0, 1,..., n. The main result in this section is Theorem 12. For any a, b C n and arbitrary α ST Ra, b} mapping δ α is a derivation in string ST Ra, b}.

Derivations in some finite endomorphism semirings 85 Proof. From δ α (α k + α l ) = α(α k + α l ) + (α k + α l )α = αα k + α k α + αα l + α l α = δ α (α k ) + δ α (α l ) where k, l 0,..., n} follows that mapping δ α is a linear. Now we prove equality (1) δ α (α k α l ) = δ α (α k )α l + α k δ α (α l ). Case 1. Let α N a. If α k N a, then δ α (α k ) = αα k + α k α = α 0 + α 0 = α 0. If α k Id a,b, then δ α (α k ) = αα k + α k α = α + α 0 = α. If α k N b, then δ α (α k ) = αα k + α k α = α n + α 0 = α n. 1.1. Let α l N a. Then δ α (α k α l ) = δ α (α 0 ) = α 0, δ α (α k )α l = α 0 and α k δ α (α l ) = α k α 0 = α 0. So, (1) holds. 1.2. Let α l N b. Then δ α (α k α l ) = δ α (α n ) = α n, δ α (α k )α l = α n and α k δ α (α l ) = α k α n = α n. So, (1) holds. 1.3. Let α k N a, α l Id a,b. Then δ α (α k α l ) = δ α (α k ) = α 0, δ α (α k )α l = α 0 α l = α 0 and α k δ α (α l ) = α k α 0 = α 0. So, (1) holds. 1.4. Let α k, α l Id a,b. Then δ α (α k α l ) = δ α (α k ) = α, δ α (α k )α l = αα l = α and α k δ α (α l ) = α k α = α 0. So, (1) holds. 1.5. Let α k N b, α l Id a,b. Then δ α (α k α l ) = δ α (α k ) = α n, δ α (α k )α l = α n α l = α n and α k δ α (α l ) = α k α = α 0. So, (1) holds. Case 2. Let α Id a,b. If α k N a, then δ α (α k ) = αα k + α k α = α 0 + α k = α k. If α k Id a,b, then δ α (α k ) = αα k + α k α = α + α k. If α k N b, then δ α (α k ) = αα k + α k α = α n + α k = α n. 2.1. Let α l N a. Then (1) holds after the same equalities like in 1.1. 2.2. Let α l N b. Then (1) holds after the same equalities like in 1.2. 2.3. Let α k N a, α l Id a,b. Then δ α (α k α l ) = δ α (α k ) = α k, δ α (α k )α l = α k α l = α k and α k δ α (α l ) = α k (α + α l ) = α k α + α k α l = α k + α k = α k. So, (1) holds. 2.4. Let α k, α l Id a,b. Then δ α (α k α l ) = δ α (α k ) = α + α k, δ α (α k )α l = (α + α k )α l = α + α k and α k δ α (α l ) = α k (α + α l ) = α k. So, (1) holds. 2.5. Let α k N b, α l Id a,b. Then δ α (α k α l ) = δ α (α k ) = α n, δ α (α k )α l = α n α l = α n and α k δ α (α l ) = α k (α + α l ) = α k. So, (1) holds. Case 3. Let α N b. If α k N a, then δ α (α k ) = αα k + α k α = α 0 + α n = α n. If α k Id a,b, then δ α (α k ) = αα k + α k α = α + α n = α n. If α k N b, then δ α (α k ) = αα k + α k α = α n + α n = α n.

86 I. Trendafilov 3.1. Let α l N a. Then δ α (α k α l ) = δ α (α 0 ) = α n, δ α (α k )α l = α 0 and α k δ α (α l ) = α k α n = α n. So, (1) holds. 3.2. Let α l N b. Then (1) holds after the same equalities like in 1.2. 3.3. Let α k N a, α l Id a,b. Then δ α (α k α l ) = δ α (α k ) = α n, δ α (α k )α l = α n α l = α n and α k δ α (α l ) = α k α n = α n. So, (1) holds. 3.4. Let α k, α l Id a,b. Then δ α (α k α l ) = δ α (α k ) = α n, δ α (α k )α l = α n α l = α n and α k δ α (α l ) = α k α n = α n. So, (1) holds. 3.5. Let α k N b, α l Id a,b. Then δ α (α k α l ) = δ α (α k ) = α n, δ α (α k )α l = α n α l = α n and α k δ α (α l ) = α k α n = α n. So, (1) holds and this completes the proof. Proposition 13. All the derivations δ α, where α ST Ra, b} commutes. Proof. Let α, β N a. For arbitrary α k N a follows δ α (α k ) = α 0 and δ β (α k ) = α 0 what implies δ α δ β = δ β δ α. For arbitrary α k Id a,b we have δ α (α k ) = α and δ β (α k ) = β. Then δ β (δ α (α k )) = δ β (α) = α 0 and δ α (δ β (α k )) = δ α (β) = α 0 what implies δ α δ β = δ β δ α. Let α N a, β Id a,b. For arbitrary α k N a we have δ α (α k ) = α 0 and δ β (α k ) = α k. So, δ α (δ β (α k )) = α 0 = δ β (δ α (α k )), i.e., δ α δ β = δ β δ α. For arbitrary α k Id a,b we have δ α (α k ) = α k and δ β (α k ) = β + α k. Then δ β (δ α (α k )) = β + α k and δ α (δ β (α k )) = δ α (β + α k ) = β + α k, that is δ α δ β = δ β δ α. Let α, β Id a,b. For arbitrary α k N a follows δ α (α k ) = α k and δ β (α k ) = α k what implies δ α δ β = δ β δ α. For arbitrary α k Id a,b we have δ α (α k ) = α+α k and δ β (α k ) = β + α k. Then δ β (δ α (α k )) = δ β (α + α k ) = β + α + α k and δ α (δ β (α k )) = δ α (β + α k ) = α + β + α k, that is δ α δ β = δ β δ α. Let β N b. For any α k follows δ β (α k ) = α n. Then for any α ST Ra, b} we have δ α (δ β (α k )) = δ α (α n ) = α n and δ β (δ α (α k ) = α n, i.e., δ α δ β = δ β δ α. When α k N b, then δ α (α k ) = α n and δ β (α k ) = α n for any α, β ST Ra, b} what implies δ α δ β = δ β δ α. Now we ask what structure has the set of all derivations δα, where α ST Ra, b}. First, we consider two examples. Example 14. Let us fix a = 0 and examine string ST R0, b}. Consider the set of derivations = δ α α Id 0,b }. Here only α n is an element of semiring N b. For any b, 0 < b n 1, we compute δ αn b (α k ) = α k for all α k N a. It is clear that δ αn b (α n b ) = α n b and δ αn b (α k ) = α n b α k + α k α n b = α n b + α k = α k for all α k Id 0,b. Moreover, δ αn b (α n ) = α n. Hence, δ αn b is an identity map. Let l, m n b,..., n 1}, that is α l, α m Id 0,b, and l m. Then follows δ αl (α k ) = δ αm (α k ) = α k for all α k N a and also δ αl (α n ) = δ αm (α n ) = α n. For arbitrary α k Id 0,b we have δ αl (α k ) = α l, where α k α l and δ αl (α k ) = α k for any k l + 1,..., n 1}. Hence, δ αm (δ αl (α k )) = δ αm (α k ) for every α k Id 0,b. Thus, δ αm δ αl = δ αm when l m.

Derivations in some finite endomorphism semirings 87 So, we conclude that the set of derivations = δ αn b,..., δ αn 1 } is a commutative idempotent semigroup, i.e. semilattice with identity δ αn b. Example 15. Let us fix a = n 2 and b = n 1. Now we compute the values of derivations: δ α0 (α 0 ) = α 0, δ α0 (α 1 ) = α 0, δ α0 (α 2 ) = α n,..., δ α0 (α n ) = α n, δ α1 (α 0 ) = α 0, δ α1 (α 1 ) = α 1, δ α1 (α 2 ) = α n,..., δ α1 (α n ) = α n, δ α2 (α 0 ) = α n, δ α2 (α 1 ) = α n, δ α2 (α 2 ) = α n,..., δ α2 (α n ) = α n,............................................................................... δ αn (α 0 ) = α n, δ αn (α 1 ) = α n, δ αn (α 2 ) = α n,..., δ αn (α n ) = α n. Hence, δ α2 = = δ αn. It is easy to verify that δα 2 0 = δ α0, δ α0 δ α1 = δ α0, δ α0 δ α2 = δ α2, δα 2 1 = δ α1, δ α1 δ α2 = δ α2 and δα 2 2 = δ α2. So, we conclude that the set of all derivations in string ST Rn 2, n 1} is the commutative idempotent semigroup = δ α0, δ α1, δ α2 } with multiplication table δ α0 δ α1 δ α2 δ α0 δ α0 δ α0 δ α2 δ α1 δ α0 δ α1 δ α2 δ α2 δ α2 δ α2 δ α2. Theorem 16. For any a, b C n the set of derivations = δ α0,..., δ αn a } in string ST Ra, b} is a semilattice with an identity δ n b and an absorbing element δ n a. Proof. Using Proposition 2 and reasonings similar to those in proof of Theorem 12, we consider three cases. Case 1. Let α l N a. It follows α l α k = α k α l = α 0, where α k N a and then δ αl (α k ) = α 0 ; α l α k = α l and α k α l = α 0, where α k Id a,b and then δ αl (α k ) = α l ; α l α k = α n and α k α l = α 0, where α k N b and then δ αl (α k ) = α n. Case 2. Let α l Id a,b. It follows α l α k = α 0 and α k α l = α k, where α k N a and then δ αl (α k ) = α k ; α l α k = α l and α k α l = α k, where α k Id a,b and then δ αl (α k ) = α l + α k ; α l α k = α n and α k α l = α k, where α k N b and then δ αl (α k ) = α n.

88 I. Trendafilov Case 3. Let α l N b. It follows α l α k = α 0 and α k α l = α n, where α k N a and then δ αl (α k ) = α n ; α l α k = α l and α k α l = α n, where α k Id a,b and then δ αl (α k ) = α n ; α l α k = α k α l = α n, where α k N b and then δ αl (α k ) = α n. From the last case we can conclude that δ n a = = δ n. Using the equalities in the cases above follows δ α0 (α 0 ) = α 0,..., δ α0 (α n b 1 ) = α 0, δ α0 (α n b ) = α 0,..., δ α0 (α n a 1 ) = α 0, δ α0 (α n a ) = α n,..., δ α0 (α n ) = α n, δ α1 (α 0 ) = α 0,..., δ α1 (α n b 1 ) = α 0, δ α1 (α n b ) = α 1,..., δ α1 (α n a 1 ) = α 1, δ α1 (α n a ) = α n,..., δ α1 (α n ) = α n,............................................................................... δ αn b 1 (α 0 ) = α 0,..., δ αn b 1 (α n b 1 ) = α 0, δ αn b 1 (α n b ) = α n b 1,..., δ αn b 1 (α n a 1 ) = α n b 1, δ αn b 1 (α n a ) = α n,..., δ αn b 1 (α n ) = α n, δ αn b (α 0 ) = α 0, δ αn b (α 1 ) = α 1,..., δ αn b (α n b 1 ) = α n b 1, δ αn b (α n b ) = = α n b,..., δ αn b (α n a 1 ) = α n a 1, δ αn b (α n a ) = α n,..., δ αn b (α n ) = α n,............................................................................... δ αn a 1 (α 0 ) = α 0, δ αn a 1 (α 1 ) = α 1,..., δ αn a 1 (α n b 1 ) = α n b 1, = δ αn a 1 (α n b ) = α n a 1,..., δ αn a 1 (α n a 1 ) = α n a 1, δ αn a 1 (α n a ) = α n,..., δ αn a 1 (α n ) = α n, δ αn a (α 0 ) = α n,..., δ αn a (α n b 1 ) = α n, δ αn a (α n b ) = α n,..., δ αn a (α n a 1 ) = α n, δ αn a (α n a ) = α n,..., δ αn a (α n ) = α n. Let us consider δ αl, where l n a 1. Then for k n a 1 we compute δ αl (δ α0 (α k )) = δ αl (α 0 ) = α 0 and for k n a follows δ αl (δ 0 (α k )) = δ αl (α n ) = α n. Also we compute δ αn 2 (δ α0 (α k )) = α n for arbitrary α k. Thus we prove that δ α0 δ αl = δ α0 and δ α0 δ αn a = δ αn a. We find δ α1 (δ α1 (α k )) = δ α1 (α 1 ) = α 0 where k n a 1 and δ α1 (δ α1 (α k )) = δ α1 (α n ) = α n for n a k n. So, we prove δα 2 1 = δ α0. Now we compute for 2 l n b 1 elements δ αl (δ α1 (α 0 )) = δ αl (α 0 ) = α 0 and δ αl (δ α1 (α 1 )) = δ αl (α 0 ) = α 0, and also for 2 k n a 1 elements δ αl (δ α1 (α k )) = δ αl (α 1 ) = α 0. We have δ αl (δ α1 (α k ) = δ αl (α n ) = α n for all k n a. So, we prove δ α1 δ αl = δ α0 for all 2 l n b 1.

Derivations in some finite endomorphism semirings 89 Now we compute for n b l n a 1 elements δ αl (δ α1 (α 0 )) = δ αl (α 0 ) = α 0 and δ αl (δ α1 (α 1 )) = δ αl (α 0 ) = α 0. Also for 2 k n a 1 elements δ αl (δ α1 (α k )) = δ αl (α 1 ) = α 1. Similarly, we have δ αl (δ α1 (α k ) = δ αl (α n ) = α n for all k n a. So, we prove δ α1 δ αl = δ α1 for all n b l n a 1. For arbitrary α k we compute δ αn a (δ α1 (α k )) = α n, so, we prove δ α1 δ αn 2 = δ αn 2. Thus, using the similar and clear reasonings, and Proposition 13, we can construct the following table δ α0 δ α1 δ αn b 1 δ αn b δ αn b+1 δ αn a 1 δ αn a δ α0 δ α0 δ α0 δ α0 δ α0 δ α0 δ α0 δ αn a δ α1 δ α0 δ α0 δ α0 δ α1 δ α1 δ α1 δ αn a........ δ αn b 1 δ α0 δ α0 δ α0 δ αn b 1 δ αn b 1 δ αn b 1 δ αn a δ αn b δ α0 δ α1 δ αn b 1 δ αn b δ αn b+1 δ αn a 1 δ αn a δ αn b+1 δ α0 δ α1 δ αn b 1 δ αn b+1 δ αn b+1 δ αn a 1 δ αn a........ δ αn b 1 δ α0 δ α1 δ αn b 1 δ αn a 1 δ αn a 1 δ αn a 1 δ αn a δ αn a δ αn a δ αn a δ αn a δ αn a δ αn a δ αn a δ αn a This completes the proof that set = δ α0,..., δ αn a } is a semilattice with identity δ n b absorbing element δ n a. Let R be a differential semiring with set of derivations = δ 1,..., δ m } and I be a differential ideal of R that is closed under each derivation δ i. For any i = 1,..., m we denote δi R I = x x R, n N 0}, δ n i (x) I}. From Proposition 2 follows that the set I = α 0, α n } is an ideal in the string ST Ra, b}. From the proof of Theorem 12 we conclude that I is closed under each derivation δ αi where α i ST Ra, b}. An easy consequence of Proposition 10 and the proof of Theorem 16 is the following Corollary 17. For any a, b C n subsemirings of the string R = ST Ra, b} are δα0 R I = = δαn b 1 R I = δαn a R I = R, δαn b R I = δαn a 1 R I = I, where δ αi = δ α0,..., δ αn a } and I = α 0, α n }.

90 I. Trendafilov 6. Strings of arbitrary type For any a 1,..., a m C n, where a 1 < a 2 <... < a m, m = 2,..., n set ST Ra 1,..., a m } = m 1 i=1 ST Ra i, a i+1 } is called a string of type m. Let α ST Ra i, a i+1 } and β ST Ra j, a j+1 }. If i = j, then either α β, or β α. If i < j, then α a i+1,..., a i+1 a j,..., a j β. Hence, string ST Ra 1,..., a m } is a (m 1)n + 1 element chain with the least element a 1,..., a 1 and the biggest element a m,..., a m, so this string is closed under the addition of semiring ÊC n. On the other hand, for α ST Ra i, a i+1 } and β ST Ra j, a j+1 }, where i < j, it is easy to show that α β ST Ra j, a j+1 }. Thus we prove Proposition 18. For any a 1,..., a m C n string ST Ra 1,..., a m } is a subsemiring of semiring ÊC n. Immediately follows Corollary 19. For arbitrary subset b 1,..., b l } a 1,..., a m } string ST Rb 1,..., b l } is a subsemiring of string ST Ra 1,..., a m }. The elements of semiring ST Ra 1,..., a m }, using the notations from Section 3, are α a l,a l+1 k, where l = 1,..., m 1 and k = 0,..., n is the number of elements of C n with image equal to a l+1. We can simplify this notations if we replace α a l,a l+1 k with α k,l, where l = 1,..., m 1. This means that α k,l is the element of ST Ra l, a l+1 } defined in the same way as in Section 3. But using these notations we must remember that α 0,2 = α n,1,..., α 0,l+1 = α n,l,..., α o,m = α n,m 1. The next proposition is a generalization of Proposition 2. Proposition 20. Let a 1,..., a m C n and ST Ra 1,..., a m } = α 0,1,..., α n,1, α 1,2,..., α n,2,..., α n,m 1, α 1,m,..., α n,m }. For k = 0,..., n for the endomorphisms α k,l ST Ra l, a l+1 } and α s,r ST Ra r, a r+1 } follows

Derivations in some finite endomorphism semirings 91 α k,l α s,r = α 0,r, where 0 s n a l+1 1 α k,l α s,r = α k,r, where n a l+1 s n a l 1 α k,l α s,r = α n,r, where n a l s n. Proof. Let 0 s n a l+1 1. For arbitrary i C n follows αs,r (a l ), if i n k 1 (α k,l, α s,r )(i) = α s,r (α k,l (i)) = α s,r (a l+1 ), if i n k = ar, if i n k 1 a r, if i n k],. = α 0,r (i) So, α k,l α s,r = α 0,r. Let n a l+1 s n a l 1. For all i C n follows (α k,l, α s,r )(i) = α s,r (α k,l (i)) = αs,r (a l ), if i n k 1 α s,r (a l+1 ), if i n k = ar, if i n k 1 a r+1, if i n k. = α k,r (i) So, α k,l α s,r = α k,r. Let n a l s n. For all i C n follows αs,r (a (α kl, α s,r )(i) = α s,r (α k,l (i)) = l ), if i n k 1 α s,r (a l+1 ), if i n k = ar+1, if i n k 1 a r+1, if i n k. = α n,r (i) So, α k,l α s,r = α n,r and this completes the proof. Endomorphisms α 0,l where l = 1,..., m and α n,m are called constant endomorphisms in [8]. According to [13], we denote the set of all constant endomorphisms of ST Ra 1,..., a m } by CO(ST Ra 1,..., a m }). From Proposition 3.1 of [13] follows Corollary 21. Set CO(ST Ra 1,..., a m }) is an ideal of semiring ST Ra 1,..., a m }.

92 I. Trendafilov 7. Derivations in strings of arbitrary type We now proceed with the construction of derivations in strings ST Ra 1,..., a m }. Suppose that there was l = 1,..., m 1 such as a l+1 a l 2. For this l the set of endomorphisms α s,l, where n a l+1 s n a l 1 has 2 or more elements. So, if we put p = n a l+1, then α p and α p+1 are from the semiring of the idempotent endomorphism in string ST Ra l, a l+1 }. Now, using Proposition 20 we compute α p+1,l α p,l = α p+1,l, α p,l α p+1,l = α p,l. For r l + 1 we compute α p,l α p,r = α p,r, α p,l α p+1,r = α p,r, α p+1,l α p,r = α p+1,r. To find composition α p,r α p,l we consider two possibilities: 1. If n a r+1 p n a r 1, then follows α p,r α p,l = α p,l. 2. If n a r p n, then α p,r α p,l = α n,l. For composition α p+1,r α p,l there are two similar possibilities, so we have either α p+1,r α p,l = α p+1,l, or α p+1,r α p,l = α n,l. Consider a mapping δ α : ST Ra 1,..., a m } ST Ra 1,..., a m } defined by the rule (2) δ α (α k,l ) = αα k,l + α k,l α, where α and α k,l are arbitrary elements of the string ST Ra 1,..., a m }. Then we find δ αp,l (α p+1,l ) = α p+1,l. Now we compute δ αp,l (α p,r ) = α p,l α p,r + α p,r α p,l = α p,r + α p,r α p,l. But α p,r > α n,l α p,l, hence δ αp,l (α p,r ) = α p,r. In such a way δ αp,l (α p+1,r ) = α p,l α p+1,r + α p+1,r α p,l = α p,r + α p+1,r α p,l. Using the inequalities α p,r > α n,l α p+1,l follows δ αp,l (α p+1,r ) = α p,r. So, we have δ αp,l (α p+1,l α p,r ) = δ αp,l (α p+1,r ) = α p,r. On the other hand, follows δ αp,l (α p+1,l ) α p,r = α p+1,l α p,r = α p+1,r and also α p+1,l δ αp,l (α p,r ) = α p+1,l α p,r = α p+1,r. Hence δ αp,l (α p+1,l ) α p,r + α p+1,l δ αp,l (α p,r ) = α p+1,r α p,r = δ αp,l (α p+1,l α p,r ). Thus, we show that mapping δ α defined by (2), where α ST Ra 1,..., a m }, in the general case, is not a derivation. Note that there is not a counterexample when r = l because δ αp,l (α p+1,l α p,l ) = δ αp,l (α p+1,l ) = α p+1,l = δ αp,l (α p+1,l ) α p,l +α p+1,l δ αp,l (α p,l ).

Derivations in some finite endomorphism semirings 93 So, the last arguments do not contradict the proof of Theorem 12. To avoid possibility a l+1 a l 2 we fix m = n and a 1 = 0, a 2 = 1,..., a n = n 1. Now we denote ST R0, 1,..., n 1} = ST R(ÊC n ). Example 22. In string ST R(ÊC 4 ) we consider endomorphisms α 0,2 = 2, 2, 2, 2, α 2,1 = 0, 0, 1, 1 and α 2,3 = 2, 2, 3, 3. Since α 2,1 α 2,3 = α 0,2 it follows that δ α0,2 (α 2,1 α 2,3 ) = δ α0,2 (α 0,2 ) = α 0,2. Now we compute δ α0,2 (α 2,1 ) = 2, 2, 2, 2 0, 0, 1, 1 + 0, 0, 1, 1 2, 2, 2, 2 = α 0,2. Then δ α0,2 (α 2,1 ) α 2,3 = α 0,2 α 2,3 = α 0,3. Hence δ α0,2 (α 2,1 α 2,3 ) δ α0,2 (α 2,1 ) α 2,3 + α 2,1 δ α0,2 (α 2,3 ). Thus, we showed that, even in the simplest case, in the string of type m, where m > 2, the mappings defined by (2) in general are not derivations. Now we consider the ideal of constant endomorphisms. It is clear that CO(ST R(ÊC n )) = CO(ÊC n ). So, CO(ÊC n ) = κ 0,..., κ n 1 } where κ i = i,..., i for i = 0,..., n 1. κi, if i j Since κ i κ j = κ j it follows that δ κi (κ j ) = κ j, if i < j. More generally, for arbitrary α s,r ST Rr 1, r}, where r = 1,..., n 1 follows that (3) κ i α s,r = κ r, κr 1, if i n s 1 if i n s. Obviously α s,r κ i = κ i. Hence, there are three cases: 1. If i r, then δ αs,r (κ i ) = κ i. 2. If i r 1 and i n s 1, then δ αs,r (κ i ) = κ r 1. 3. If i r 1 and i n s, then δ αs,r (κ i ) = κ r. We can now establish a result concerning the semiring CO(ÊC n ) using Corollary 21. Proposition 23. Let δ αs,r : CO(ÊC n ) CO(ÊC n ), where α s,r ST R(ÊC n ), be a mapping defined by δ αs,r (κ i ) = α s,r κ i + κ i α s,r for arbitrary κ i CO(ÊC n ). Then for any κ i, κ j CO(ÊC n ) follows (4) δ αs,r (κ i κ j ) = δ αs,r (κ i ) κ j + κ i δ αs,r (κ j ).

94 I. Trendafilov Proof. From equality κ i κ j = κ j follows that δ αs,r (κ i κ j ) = δ αs,r (κ j ), κ i δ αs,r (κ j ) = δ αs,r (κ j ) and δ αs,r (κ i ) κ j = κ j. So (4) is equivalent to equality δ αs,r (κ j ) = δ αs,r (κ j ) + κ j. If j r from 1 follows δ αs,r (κ j ) = κ j, so, (4) holds. If j r 1 and j n s 1, then from 2 follows δ αs,r (κ j ) = κ r 1 and from κ r 1 = κ r 1 + κ j equality (4) holds. When j r 1 and j n s, from 3 follows δ αs,r (κ i ) = κ r which implies κ r = κ r + κ j, so, (4) holds. Is there a semiring which contain semiring CO(ÊC n ) and is invariant under all mappings δ αs,r, where α s,r ST R(ÊC n ), so that equality (3) holds for all elements of this semiring? Studying this question, we consider the set S = CO(ÊC n ) ST Rn 2, n 1}. Proposition 24. Set S is a subsemiring of semiring ST R(ÊC n ) and is invariant under all mappings δ αk,l, where α k,l ST R(ÊC n ). Proof. Let x, y S. If either x, y CO(ÊC n ), or x, y ST Rn 2, n 1}, then from Corollary 21 and Proposition 1 follows that x + y and x y are from the same semiring. Let x = α s,n 1 ST Rn 2, n 1} and y = κ i CO(ÊC n ), where s, i = 0,..., n 1. Then α s,n 1 + κ i = α s,n 1 if i n 2 and α s,n 1 + κ i = κ i if κn 2, if i n s 1 i = n 1. Since α s,n 1 κ i = κ i and κ i α s,n 1 = κ n 1, if i n s follows that x y, y x S. So, we prove that S is a subsemiring of ST R(ÊC n ). Let α k,l ST R(ÊC n ). For any κ i CO(ÊC n ) from 1, 2 and 3, just before Proposition 23, follows that δ αk,l (κ i ) CO(ÊC n ). Let α s,n 1 ST Rn 2, n 1}. From Proposition 6.3 follows α k,l α s,n 1 = κ n 2 if 0 s n l 2, α k,l α s,n 1 = α k,n 1 if s = n l 1 and α k,l α s,n 1 = κ n 1 if n l s n 1. Since α s,n 1 α k,l is equal to κ l 1, α s,l and κ l in similar cases we can conclude that when l n 2 follows κ n 2, if 0 s n l 2 (5) δ αk,l (α s,n 1 ) = α k,n 1, if s = n l 1 κ n 1, if n l s n 1. If l = n 1 obviously δ αk,n 1 (α s,n 1 ) S. So, S is invariant under arbitrary δ αk,l and this completes the proof. is not a deriva- The following counterexample shows that, in general, map δ αk,l tion. Example 25. In string ST R(ÊC 4 ) we consider endomorphisms α 3,2 = 1, 2, 2, 2, κ 1 = 1, 1, 1, 1 CO(ÊC 4 ) and α 2,3 = 2, 2, 3, 3 ST R2, 3}.

Derivations in some finite endomorphism semirings 95 Since κ 1 α 2,3 = κ 2 follows that δ α3,2 (κ 1 α 2,3 ) = δ α3,2 (κ 2 ) = = 1, 2, 2, 2 2, 2, 2, 2 + 2, 2, 2, 2 1, 2, 2, 2 = 2, 2, 2, 2 = κ 2. Now we compute δ α3,2 (κ 1 ) = 1, 2, 2, 2 1, 1, 1, 1 + 1, 1, 1, 1 1, 2, 2, 2 = κ 2. Then δ α3,2 (κ 1 ) α 2,3 = 2, 2, 2, 2 2, 2, 3, 3 = 3, 3, 3, 3 = κ 3. Hence δ α3,2 (κ 1 α 2,3 ) δ α3,2 (κ 1 ) α 2,3 + κ 1 δ α3,2 (α 2,3 ). Now we present a class of maps of type δ α which are derivations in the whole semiring ST R(ÊC n ). We need some preliminary lemmas. Using that δ α (β) = δ β (α) and formulas (5) we obtain Lemma 26. For any l = 1,..., n 2 and k, s = 0,..., n follows κ n 2, if 0 s n l 1 δ αs,n 1 (α k,l ) = α k,n 1, if s = n l κ n 1, if n l + 1 s n. Let us make one necessary observation. Lemma 27. For any q = 1,..., n 2, k, p, s = 0,..., n and arbitrary l follows δ αs,n 1 (α k,l ) α p,q + α k,l δ αs,n 1 (α p,q ) = α k,l δ αs,n 1 (α p,q ). Proof. Let l n 2. Using Lemma 26 follows δ αs,n 1 (α k,l ) κ n 1. Then δ αs,n 1 (α k,l ) α p,q κ n 1 α p,q = κ q. The last equality follows from (3). Let l = n 1. Then (6) κ n 2, if k = s = 0, k = 0 and s = 1, k = 1 and s = 0 δ αs,n 1 (α k,n 1 ) = α 1,n 1, if k = s = 1 κ n 1, if k 2 or s 2. Analogously, we have δ αs,n 1 (α k,n 1 ) α p,q κ n 1 α p,q = κ q. On the other hand, from Lemma 26 we have δ αs,n 1 (α p,q ) κ n 2. Then for arbitrary l follows that α k,l δ αs,n 1 (α p,q ) α k,l κ n 2 = κ n 2. So, we have δ αs,n 1 (α k,l ) α p,q κ q κ n 2 α k,l δ αs,n 1 (α p,q ). Now we shall prove the main result of this section.

96 I. Trendafilov Theorem 28. For any s 2 the map δ αs,n 1 : ST R(ÊC n ) ST R(ÊC n ) defined by equality δ αs,n 1 (α) = α s,n 1 α + α α s,n 1, where α, α s,n 1 ST R(ÊC n ) is a derivation of ST R(ÊC n ). Proof. A. First, we shall consider endomorphisms α k,l and α p,q so that l, q = 1,..., n 2. For k, p = 0,..., n we verify that (7) δ αs,n 1 (α k,l α p,q ) = δ αs,n 1 (α k,l ) α p,q + α k,l δ αs,n 1 (α p,q ). From Proposition 20 follows (8) α k,l α p,q = κ q 1, if 0 p n l 1 α k,l α p,q = α k,q, if p = n l α k,l α p,q = κ q, if n l + 1 p n. Case 1. Let 0 s n q 1. Then, using Lemma 26, for arbitrary p follows δ αs,n 1 (α k,l α p,q ) = κ n 2. Now from Lemma 26 and Lemma 27 we find δ αs,n 1 (α k,l ) α p,q + α k,l δ αs,n 1 (α p,q ) = α k,l κ n 2 = κ n 2. So, equality (7) holds. Case 2. Let s = n q. 2.1. If 0 p n l 1, then from (8) follows α k,l α p,q = κ q 1. Hence δ αs,n 1 (α k,l α p,q ) = δ αs,n 1 (κ q 1 ) = κ n 2. Using Lemma 27 first and then Lemma 26 we have δ αs,n 1 (α k,l ) α p,q + α k,l δ αs,n 1 (α p,q ) = α k,l α p,n 1 = κ n 2. So, again, equality (7) holds. 2.2. If p = n l, then from (6) follows α k,l α p,q = α k,q. Thus follows δ αs,n 1 (α k,l α p,q ) = δ αs,n 1 (α k,q ) = α k,n 1.

Derivations in some finite endomorphism semirings 97 Now Lemma 26 and Lemma 27 yields δ αs,n 1 (α k,l ) α p,q + α k,l δ αs,n 1 (α p,q ) = α k,l α p,n 1 = α k,n 1. So, equality (7) holds. 2.3. If n l + 1 p n, then from (8) follows α k,l α p,q = κ q. So, we have δ αs,n 1 (α k,l α p,q ) = δ αs,n 1 (κ q ) = κ n 1. Now Lemma 26 and Lemma 27 imply δ αs,n 1 (α k,l ) α p,q + α k,l δ αs,n 1 (α p,q ) = α k,l κ n 1 = κ n 1. So, equality (7) holds. Case 3. Let n q + 1 s n. Then, from Lemma 26, for arbitrary p follows δ αs,n 1 (α k,l α p,q ) = κ n 1. From Lemma 26 and Lemma 27 we obtain δ αs,n 1 (α k,l ) α p,q + α k,l δ αs,n 1 (α p,q ) = α k,l κ n 1 = κ n 1. So, finally equality (7) holds. B. The second possibility is when α k,n 1, α p,n 1 ST Rn 2, n 1}. Now from Theorem 12 follows that δ αs,n 1 satisfies the Leibnitz s rule. C. Another possibility is to calculate δ αs,n 1 (α k,l α p,n 1 ), where l n 2. Now equalities (8) for q = n 1 are (9) α k,l α p,n 1 = κ n 2, if 0 p n l 1 α k,l α p,n 1 = α k,n 1, if p = n l α k,l α p,n 1 = κ n 1, if n l + 1 p n. Hence, using that s 2, for arbitrary p we obtain δ αs,n 1 (α k,l α p,n 1 ) δ αs,n 1 (κ n 2 ) = κ n 1. The last equality follows from (6). From (6) and (9) we have α k,l δ αs,n 1 (α p,n 1 ) = α k,l κ n 1 = κ n 1. Thus we prove that δ αs,n 1 (α k,l α p,n 1 ) = δ αs,n 1 (α k,l ) α p,n 1 + α k,l δ αs,n 1 (α p,n 1 ). D. The last possibility is the same as in C., but we shall prove equality (10) δ αs,n 1 (α p,n 1 α k,l ) = δ αs,n 1 (α p,n 1 ) α k,l + α p,n 1 δ αs,n 1 (α k,l ), where l n 2.

98 I. Trendafilov Now from equalities (8) we obtain (11) α p,n 1 α k,l = κ l 1, if k = 0 α p,n 1 α k,l = α p,l, if k = 1 α p,n 1 α k,l = κ l, if 2 k n. Case 1. Let 0 s n l 1. Then, using Lemma 7.5, for arbitrary k follows δ αs,n 1 (α p,n 1 α k,l ) = κ n 2. Now Lemma 26 and Lemma 27 imply δ αs,n 1 (α p,n 1 ) α k,l + α p,n 1 δ αs,n 1 (α k,l ) = α p,n 1 κ n 2 = κ n 2. So, equality (10) holds. Case 2. Let s = n l. 2.1. Let k = 0. Now from (11) follows α p,n 1 α k,l = κ l 1. Therefore, from Lemma 7.5, follows δ αs,n 1 (α p,n 1 α k,l ) = δ αs,n 1 (κ l 1 ) = κ n 2. Using Lemma 27 first and then Lemma 26 we have δ αs,n 1 (α p,n 1 ) α k,l + α p,n 1 δ αs,n 1 (α k,l ) = α p,n 1 κ n 2 = κ n 2. So, equality (10) holds. 2.2. Let k = 1. Now from (11) follows α p,n 1 α k,l = α p,l. Thus δ αs,n 1 (α p,n 1 α k,l ) = δ αs,n 1 (α p,l ) = α p,n 1. Now from Lemma 26 and 27 7.6 we obtain δ αs,n 1 (α p,n 1 ) α k,l + α p,n 1 δ αs,n 1 (α k,l ) = α p,n 1 α k,n 1 = α p,n 1. So, equality (10) holds. 2.3. Let 2 k n. From (11) follows α p,n 1 α k,l = κ l. Hence δ αs,n 1 (α p,n 1 α k,l ) = δ αs,n 1 (κ l ) = κ n 1.

Derivations in some finite endomorphism semirings 99 Now Lemma 26 and Lemma 27 imply δ αs,n 1 (α p,n 1 ) α k,l + α p,n 1 δ αs,n 1 (α k,l ) = α p,n 1 α k,n 1 = κ n 1. The last equality follows from the condition k 2. So, equality (10) holds. Case 3. Let n l + 1 s n. Then, using Lemma 7.5, for arbitrary k follows δ αs,n 1 (α p,n 1 α k,l ) = κ n 1. From Lemma 26 and Lemma 27 we find δ αs,n 1 (α p,n 1 ) α k,l + α p,n 1 δ αs,n 1 (α k,l ) = α p,n 1 κ n 1 = κ n 1. So, equality (10) holds and this completes the proof. After Theorem 16 we consider set δ i R I, where R is a differential semiring with a set of derivations = δ 1,..., δ m }, δ i, and I is a differential ideal of R, closed under each derivation δ i. From Proposition 10 follows that δ i R I is a subsemiring of R. In Theorem 28 we prove that string R = ST R(ÊC n ) is a differential semiring with set of derivations = δ αs,n 1 α s,n 1 ST Rn 2, n 1}, 2 s n}. From Proposition 23 we know that I = CO(ÊC n ) is a differential ideal of R, closed under all derivations of. Now from Lemma 26 and equalities (6) easily follows Proposition 29. If δ αs,n 1 then δ αs,n 1 R I = R. References [1] K.I. Beidar, W.S. Martindale III and A.V. Mikhalev, Rings with generalized identities (Marcel Dekker, 1996). [2] O. Ganyushkin and V. Mazorchuk, Classical Finite Transformation Semigroups: An Introduction (Springer-Verlag London Limited, 2009). doi:10.1007/978-1-84800-281-4 [3] J. Golan, Semirings and Their Applications (Kluwer, Dordrecht, 1999). [4] I.N. Herstein, Jordan derivations of prime rings, Proc. Amer. Math. Soc. 8 (1957) 1104 1110. doi:10.1090/s0002-9939-1957-0095864-2

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