# Simplices in the Endomorphism Semiring of a Finite Chain

Authors: Ivan Trendafilov
Journal: Algebra
 (2014)

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## Abstract

We establish new results concerning endomorphisms of a finite chain, if the cardinality of the image of such endomorphism is no more than some fixed number. The semiring of all such endomorphisms can be seen as a simplex whose vertices are the constant endomorphisms. We explore the properties of these simplices.


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## Body

## 1. Introduction and Preliminaries

It is well known that each simplicial complex has a geometric (continuous) interpretation as a convex set spanned byk geometrically independent points in some Euclidean space.Here, we present an algebraic (discrete) interpretation of simplicial complex as a subsemiring, containing (in some sense spanned by)k constant endomorphisms of the endomorphism semiring ℰ^𝒞n of a finite chain. The endomorphism semiring of a finite semilattice is well studied in [1–10].The paper is organized as follows. After the introduction and preliminaries, in Section2, we give basic definitions and obtain some elementary properties of simplices. Although we do not speak about any distance here, we define discrete neighborhoods with respect to any vertex of the simplex. In Section 3, we study discrete neighborhoods, left ideals, and right ideals of a simplex. The main results are Theorems 9 and 12, where we find two right ideals of simplex. In Section 4, Theorem 15 is the main result of the paper, where we show that important objects (idempotents, a-nilpotent elements, left ideals, and right ideals) of simplex (big semiring) can be constructed using similar objects of coordinate simplex (little semiring).Since the terminology for semirings is not completely standardized, we say what our conventions are. An algebraR=(R,+,·), with two binary operations + and · on R, is called asemiring if:(i)
(R,+) is a commutative semigroup;(ii)
(R,·) is a semigroup;(iii)
both distributive laws holdx·(y+z)=x·y+x·z and (x+y)·z=x·z+y·z, for any x,y,z∈R.LetR=(R,+,·) be a semiring. If a neutral element 0 of the semigroup (R,+) exists and 0x=0 or x0=0, it is called aleft or aright zero, respectively, for all x∈R. If 0·x=x·0=0, for all x∈R, then it is calledzero. An element e of a semigroup (R,·) is called aleft (right) identity provided that ex=x or xe=x, respectively, for all x∈R. If a neutral element 1 of the semigroup (R,·) exists, it is calledidentity.A nonempty subsetI of R is called anideal if I+I⊆I, RI⊆I, and IR⊆I.The facts concerning semirings can be found in [1].For a join-semilattice(ℳ,∨), set ℰℳ of the endomorphisms of ℳ to be a semiring with respect to the addition and multiplication defined as follows:(i)
h=f+g, when h(x)=f(x)∨g(x), for all x∈ℳ;(ii)
h=f·g, when h(x)=f(g(x)), for all x∈ℳ.This semiring is called the endomorphism semiring of ℳ.In this paper, all semilattices are finite chains. Following [2], we fix a finite chain 𝒞n=({0,1,…,n-1},∨) and denote the endomorphism semiring of this chain with ℰ^𝒞n. We do not assume that α(0)=0 for arbitrary α∈ℰ^𝒞n. So, there is not a zero in endomorphism semiring ℰ^𝒞n. Subsemirings ℰ𝒞n(a), where a∈𝒞n, of the semiring ℰ^𝒞n, consisting of all endomorphisms α with fixed point a, are considered in [3].Ifα∈ℰ^𝒞n such that f(k)=ik, for any k∈𝒞n, we denote α as an ordered n-tuple ≀i0,i1,i2,…,in-1≀. Note that the 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”. The identity i=≀0,1,…,n-1≀ and all constant endomorphisms k¯=≀k,…,k≀ are obviously (multiplicatively) idempotents.Leta∈𝒞n. For every endomorphism a¯=≀aa…a≀, the elements of
(1)𝒩n[a]={α∣α∈ℰ^𝒞n,𝒩n[a]=dαna=a¯forsomenaturalnumbernaℰ^𝒞n}
are called a-nilpotent endomorphisms. An important result for a-nilpotent endomorphisms is as follows.Theorem 1 (see [4, Theorem 3.3]).
For any naturaln, n≥2, and a∈𝒞n, the set of a-nilpotent endomorphisms 𝒩n[a] is a subsemiring of ℰ^𝒞n. The order of this semiring is |𝒩n[k]|=Ck·Cn-k-1, where Ck is the kth Catalan number.Another useful result is as follows.Theorem 2 (see [5, Theorem 9]).
The subset ofℰ^𝒞n, n≥3, of all idempotent endomorphisms with s fixed points k1,…,ks, 1≤s≤n-1 is a semiring of order ∏m=1s-1‍(km+1-km).For definitions and results concerning simplices, we refer the reader to [6, 7].


## 2. The Simplexσ(n){a0,…,ak-1}

Let us fix elementsa0,…,ak-1∈𝒞n, where k≤n, a0<⋯<ak-1, and let A={a0,…,ak-1}. We consider endomorphisms α∈ℰ^𝒞n such that Im⁡(α)⊆A. We denote this set by σ(n){a0,…,ak-1}.Let{b0,…,bℓ-1}⊆{a0,…,ak-1} and consider the set
(2)σ(n){b0,…,bℓ-1}={β∣β∈σ(n){a0,…,ak-1},Im⁡(β)={b0,…,bℓ-1}σ(n)}.Forβ1,β2∈σ(n){b0,…,bk-1}, let β1~β2, if and only if the sets Im⁡(β1) and Im⁡(β2) have a common least element. In this way, we define an equivalence relation. Any equivalence class can be identified with its least element which is the constant endomorphism bm¯=≀bm,…,bm≀, where m=0,…ℓ-1.Now take a simplicial complexΔ with vertex set V={a0¯,…,ak-1¯}. The subset {b0¯,…,bℓ-1¯} is a face of Δ. Hence, we can consider the set σ(n){b0,…,bℓ-1} as a face of Δ. In particular, when the simplicial complex Δ consists of all subsets of V, it is called a simplex (see [6]) and Δ=σ(n){a0,…,ak-1}.It is easy to see thatIm⁡(α)⊆A and Im⁡(β)⊆A imply Im⁡(α+β)⊆A and Im⁡(α·β)⊆A, and so we have proved the following.Proposition 3.
For any setA={a0,…,ak-1}⊆𝒞n, the simplex σ(n){a0,…,ak-1} is a subsemiring of ℰ^𝒞n.The numberk is called a dimension of simplex σ(n){a0,…,ak-1}. Any simplex σ(n){b0,b1,…,bℓ-1}, where b0,…,bℓ-1∈A, is aface of simplex σ(n){a0,…,ak-1}. If ℓ<k, face σ(n){b0,b1,…,bℓ-1} is called a proper face.The proper faces of simplexσ(n){a0,…,ak-1} are as follows:(i)
0-simplices are vertices a0¯,…,ak¯;(ii)
1-simplices are called strings; they are denoted by 𝒮𝒯ℛ(n){a,b}, where a,b∈A;(iii)
2-simplices are called triangles; they are denoted by Δ(n){a,b,c}, where a,b,c∈A;(iv)
3-simplices are called tetrahedra; they are denoted by 𝒯ℰ𝒯ℛ(n){a,b,c,d}, where a,b,c,d∈A;(v)
the last proper faces are simplicesσk-1(n){b0,…,bk-2}, where {b0,…,bk-2}⊂A.The boundary of the simplexσ(n){a0,…,ak-1} is a union of all its proper faces and is denoted by ∂(σ(n){a0,…,ak-1}). The set
(3)ℐ𝒩𝒯(σ(n){a0,…,ak-1})=σk(n)(A)∖∂(σ(n){a0,…,ak-1})
is called an interior of simplex σ(n){a0,…,ak-1}.It follows that the interior of simplexσ(n){a0,…,ak-1} consists of endomorphisms α, such that Im⁡(α)={a0,a1,…,ak-1}. So, we have the following.Proposition 4.
The interiorℐ𝒩𝒯(σ(n){a0,…,ak-1}) is an additive semigroup.Proposition 5.
Any face of simplexσ(n){a0,…,ak-1} is a left ideal.Proof.
Letσ(n){b0,…,bℓ-1} be a face of simplex σ(n){a0,…,ak-1}. Obviously the face is a subsemiring of σ(n){a0,…,ak-1}. Let α∈σ(n){b0,…,bℓ-1} and β∈σ(n){a0,…,ak-1}. Since for any i∈𝒞n we have φ(i)∈{a0,…,ak-1}, then (φ·α)(i)∈{b0,…,bℓ-1}. Thus φ·α∈σ(n){b0,…,bℓ-1}. Hence, σ(n){b0,…,bℓ-1} is a left ideal of simplex σ(n){a0,…,ak-1}.Note that any face of some simplex is not a right ideal of the simplex. For instance, take the vertexam¯∉σ(n){b0,…,bℓ-1}. Then, bs¯·am¯=am¯∉σ(n){b0,…,bℓ-1}, for all s=0,…,ℓ-1. From the last proposition consider the following.Corollary 6.
The boundary∂(σ(n){a0,…,ak-1}) is a multiplicative semigroup.The boundary and the interior of a simplex are, in general, not semirings.For any naturaln, endomorphism semiring ℰ^𝒞n is a simplex with vertices 0¯,…,n-1¯. The interior of this simplex consists of endomorphisms α, such that Im⁡(α)=𝒞n. Since the latter is valid only for identity i=≀0,1,…,n-1≀, it follows that ℐ𝒩𝒯(ℰ^𝒞n)=i.There is a following partial ordering of the faces of dimensionk-1 of simplex σ(n){a0,…,ak-1}: least face does not contain the vertex ak-1¯ and the biggest face does not contain the vertex a0¯.The biggest face of the simplexℰ^𝒞n is the simplex σ(n){1,…,n-1}. Now ℰ^𝒞n∖σ(n){1,…,n-1}=ℰ𝒞n(0) which is a subsemiring of ℰ^𝒞n. Similarly, the least face of ℰ^𝒞n is σ(n){0,…,n-2}. Then, ℰ^𝒞n∖σ(n){0,…,n-2}=ℰ𝒞n(n-1) which is also a subsemiring of ℰ^𝒞n. The other faces of ℰ^𝒞n, where n≥3, do not have this property. Indeed, one middle face is σ(n){0,…,k-1,k+1,…,n-1}. But set R=ℰ^𝒞n∖σ(n){0,…,k-1,k+1,…,n-1} is not a semiring because, for any n≥3 and any k∈{1,…,n-2}, if α=≀0,…,0,k≀∈R, then α2=0¯∉R.Let us fix vertexam¯, where m=0,…,k-1 of simplex σ(n){a0,…,ak-1}. The set of all endomorphisms α∈σ(n){a0,…,ak-1} such that α(i)=am just for s elements i∈𝒞n is called sth layer of the simplex with respect to am¯, where s=0,…,n-1. We denote the sth layer of the simplex σ(n){a0,…,ak-1} with respect to am¯ by ℒams(σ(n){a0,…,ak-1}). So, the 0-layer with respect to any vertex of the simplex σ(n){a0,…,ak-1} is a face of the simplex; hence, it is a semiring. In the general case, the sth layer ℒams(σ(n){a0,…,ak-1}), where s∈𝒞n, s=1,…,n-2, is not a subsemiring of simplex σ(n){a0,…,ak-1}.On the other hand, sinceℒams(σ(n){a0,…,ak-1}) consists of all endomorphisms α, such that α(i)=am just for s elements i∈𝒞n, it follows that this sth layer is closed under the addition. Hence, we have the following.Proposition 7.
Any layerℒams(σ(n){a0,…,ak-1}) of simplex σ(n){a0,…,ak-1} is an additive semigroup.Letam¯ be an arbitrary vertex of simplex σ(n){a0,…,ak-1}. From a topological point of view, the set 𝒟𝒩m1={am¯}∪ℒamn-1(σ(n){a0,a1,…,ak-1}) is a discrete neighborhood consisting of the “nearest points to point am¯.” Similarly, we define 𝒟𝒩m2=𝒟𝒩m1∪ℒamn-2(σ(n){a0,a1,…,ak-1}). More generally
(4)𝒟𝒩mt={am¯}∪⋃ℓ=n-tn-1‍ℒamℓ(σ(n){a0,a1,…,ak-1}),
where m=0,…,k-1 and t=1,…,n is calleddiscretet-neighborhood of the vertex am¯.


## 3. Subsemirings and Ideals of the Simplexσ(n){a0,…,ak-1}

Lemma 8.
Letam¯, where m=0,…,k-1, be a vertex of the simplex σ(n){a0,a1,…,ak-1} and ℒamn-1(σ(n){a0,a1,…,ak-1}) be the (n-1)th layer of the k-simplex with respect to am¯. Then, the set 𝒟𝒩m1={am¯}∪ℒamn-1(σ(n){a0,a1,…,ak-1}), where m=0,…,k-1, is a subsemiring of σ(n){a0,a1,…,ak-1}.Proof.
We consider three cases.
Case
1. Let m=0. Then, elements of 𝒟𝒩01 are endomorphisms:
(5)a0¯,(a0)n-1a1=≀a0,…,a0︸n-1,a1≀,…,(a0)n-1ak-1=≀a0,…,a0︸n-1,ak-1≀.
Since a0¯<(a0)n-1a1<⋯<(a0)n-1ak-1, it follows that set 𝒟𝒩01 is closed under the addition.
We find(a0)n-1ai·a0¯=a0¯·(a0)n-1ai=a0¯, for all i=1,…,k-1. Also we have (a0)n-1ai·(a0)n-1aj=(a0)n-1aj·(a0)n-1ai=a0¯, for all i,j∈{1,…,k-1}, with the only exception when ak-1=n-1. Now ((a0)n-1(n-1))2=(a0)n-1(n-1), (a0)n-1(n-1)·(a0)n-1ai=(a0)n-1ai, and (a0)n-1ai·(a0)n-1(n-1)=a0¯. Hence, 𝒟𝒩01 is a semiring.
Case
2. Let m=k-1. Then, elements of 𝒟𝒩k-11 are endomorphisms:
(6)a0(ak-1)n-1=≀a0,ak-1,…,ak-1︸n-1≀,…,ak-2(ak-1)n-1a0(ak-1)n-1=≀ak-2,ak-1,…,ak-1︸n-1≀,ak-1¯.
Since a0(ak-1)n-1<⋯<ak-2(ak-1)n-1<ak-1¯, it follows that the set 𝒟𝒩k-11 is closed under the addition.
We findai(ak-1)n-1·ak-1¯=ak-1¯·ai(ak-1)n-1=ak-1¯, for all i=1,…,k-1. Also we have ai(ak-1)n-1·aj(ak-1)n-1=aj(ak-1)n-1·ai(ak-1)n-1=ak-1¯, for all i,j∈{0,…,k-2}, with also the only exception when a0=0. We have (0(ak-1)n-1)2=0(ak-1)n-1, 0(ak-1)n-1·ai(ak-1)n-1=ai(ak-1)n-1, and ai(ak-1)n-1·0(ak-1)n-1=ak-1¯. Hence, 𝒟𝒩k-11 is a semiring.
Case
3. Let 0<m<k-1. Then elements of 𝒟𝒩m1 are endomorphisms:
(7)a0(am)n-1=≀a0,am,…,am︸n-1≀,…,am-1(am)n-1=≀am-1,am,…,am︸n-1≀,am¯,(am)n-1am+1=≀am,…,am︸n-1,am+1≀,…,(am)n-1ak-1=≀am,…,am︸n-1,ak-1≀.
Since a0(am)n-1<⋯<am-1(am)n-1<am¯<(am)n-1am+1<⋯<(am)n-1ak-1, it follows that set 𝒟𝒩m1 is closed under the addition.
Now there are four possibilities.
(
1
) Let 0<a0 and ak-1<n-1. Then
(8)ai(am)n-1·aj(am)n-1=aj(am)n-1·ai(am)n-1=am¯ai(am)n-1·aj(am)n-dddforanyi,j=0,…,m-1,(am)n-1ai·(am)n-1aj=(am)n-1aj·(am)n-1ai=am¯dddddddddddddddforanyi,j=m+1,…,k-1,ai(am)n-1·(am)n-1aj=(am)n-1aj·ai(am)n-1=am¯ssssforanyi=0,…,m-1,j=m+1,…,k-1.
Sinceai(am)n-1·am¯=am¯·ai(am)n-1=am¯, for i=1,…,m-1, and, in a similar way, (am)n-1aj·am¯=am¯·(am)n-1aj=am¯, for j=m+1,…,k-1, and also (am¯)2=am¯, it follows that 𝒟𝒩m1 is a commutative semiring.
(
2
) Let a0=0 and ak-1<n-1. Then, (0(am)n-1)2=0(am)n-1,
(9)0(am)n-1·ai(am)n-1=ai(am)n-1,ai(am)n-1·0(am)n-1=am¯ccccccccforanyi=1,…,m-1,0(am)n-1·(am)n-1aj=(am)n-1aj·0(am)n-1=am¯0(am)n-1·(am)n-1aforanyj=m+1,…,k-1.
We also observe thatam¯·0(am)n-1=0(am)n-1·am¯=am¯. All the other equalities between the products of the elements of 𝒟𝒩m1 are the same as in (1).
(
3
) Let a0>0 and ak-1=n-1. Then, ((am)n-1(n-1))2=(am)n-1(n-1),
(10)(am)n-1(n-1)·ai(am)n-1=ai(am)n-1·(am)n-1(n-1)=am¯aaaaaaaaaaajaaforanyi=1,…m-1,(am)n-1(n-1)·(am)n-1aj=(am)n-1aj,(am)n-1aj·(am)n-1(n-1)=am¯mmmmmforanyj=m+1,…,k-1.
We also observe thatam¯·(am)n-1(n-1)=(am)n-1(n-1)·am¯=am¯. All the other equalities between the products of the elements of 𝒟𝒩m1 are the same as in (1).
(
4
) Let a0=0 and ak-1=n-1. Now all equalities between the products of the elements of 𝒟𝒩m1 are the same as in (1), (2), and (3). So, 𝒟𝒩m1 is a semiring.Theorem 9.
The unionJ=⋃m=0k-1‍𝒟𝒩m1 of the discrete 1-neighborhoods with respect to all vertices of the simplex σ(n){a0,…,ak-1} is a right ideal of the simplex.Proof.
Letα∈𝒟𝒩m1. Then, α=am¯, α=ai(am)n-1, or α=(am)n-1aj, where i=0,…,m-1 and j=m+1,…,n-1. Let β∈𝒟𝒩s1. Then β=as¯, β=ap(as)n-1, or β=(as)n-1aq, where p=0,…,s-1 and q=s+1,…,n-1.
Suppose thats>m. Then, we find α+β=as¯ or α+β=(as)n-1at, where t=max⁡(j,q). So, in all cases α+β∈J, what means that J is closed under the addition.
Letα∈J and φ∈σ(n){a0,…,ak-1}. Then, α∈𝒟𝒩m1, for some m=0,…,k-1.
Ifα=am¯ and φ(am)=as, then it follows that α·φ=as¯∈J.
Ifα=ai(am)n-1, where i=0,…,m-1, φ(am)=as, and φ(ai)=ap, then α·φ=ap(as)n-1∈J.
Ifα=(am)n-1aj, where j=m+1,…,k-1, φ(am)=as, and φ(aj)=aq, it follows that α·φ=(as)n-1aq∈J.
Hence, in all casesα·φ∈J.Any simplexσ(n){b0,…,bℓ-1} which is a face of simplex σ(n){a0,…,ak-1} is called internal of the simplex σ(n){a0,…,ak-1}, if a0∉σ(n){b0,…,bℓ-1} and ak-1∉σ(n){b0,…,bℓ-1}.Similarly simplexσ(n){a0,…,ak-1}, which is a face of n-simplex ℰ^𝒞n, is called internal simplex, if 0∉σ(n){a0,…,ak-1} and n-1∉σ(n){a0,…,ak-1}.Immediately from the proof of Proposition4 consider the following.Corollary 10.
For any internal simplexσ(n){a0,a1,…,ak-1}, semirings 𝒟𝒩m1 are commutative and all their elements are am-nilpotent, where m=0,…,k-1.Lemma 11.
Letam¯, where m=0,…,k-1, be a vertex of internal simplex σ(n){a0,…,ak-1}. Then, the set 𝒟𝒩m2=𝒟𝒩m1∪ℒamn-2(σ(n){a0,…,ak-1}), where m=0,…,k-1, is a subsemiring of σ(n){a0,…,ak-1}.Proof.
Since𝒟𝒩m2=𝒟𝒩m1∪ℒamn-2(σ(n){a0,…,ak-1}), it follows that the elements of 𝒟𝒩m2 are endomorphisms:
(11)am¯ai(am)n-1,wherei=0,…,m-1,(am)n-1aj,wherej=m+1,…,k-1,apaq(am)n-2,wherep,q=0,…,m-1,p≤q,(am)n-2aras,wherer,s=m+1,…,k-1,r≤s,ap(am)n-2as,wherep=0,…,m-1,ap(am)n-2as,whs=m+1,…,k-1.
From Lemma8, we know that the discrete 1-neighborhood 𝒟𝒩m1 is closed under the addition. From Proposition 7, it follows that the layer ℒamn-2(σ(n){a0,…,ak-1}) also is closed under the addition. Hence, in order to prove that 𝒟𝒩m2 is closed under the addition, we calculate the following:
(12)ai(am)n-1+apaq(am)n-2={apaq(am)n-2,ifi≤p,aiaq(am)n-2,ifi>p,ai(am)n-1+(am)n-2aras=(am)n-2aras,(am)n-1aj+(am)n-2aras={(am)n-2aras,ifj≤s,(am)n-2araj,ifj>s,(am)n-1aj+apaq(am)n-2=(am)n-1aj,ai(am)n-1+ap(am)n-2as={ap(am)n-2as,ifi≤p,ai(am)n-2as,ifi>p,(am)n-1aj+ap(am)n-2as={ap(am)n-2as,ifj≤s,ap(am)n-2aj,ifj>s,am¯+apaq(am)n-2=am¯,am¯+(am)n-2aras=(am)n-2aras,am¯+ap(am)n-2as=(am)n-1as,
where i,p,q=0,1,…,m-1, p≤q, j,r,s=m+1,…,k-1, and r≤s. So, we prove that the discrete 2-neighborhood 𝒟𝒩m2 is closed under the addition.
Now we consider six cases, where, for the indices, the upper restrictions are fulfilled.
Case
1. Let am=1. We shall show that all endomorphisms of 𝒟𝒩12 are 1-nilpotent with the only exception when ak-1=n-2. When ak-1<n-2, since 1 is the least image of any endomorphism, there are only a few equalities: 1n-2aras·1n-2ar0as0=1¯,
(13)1n-1aj·1n-2aras=1n-2aras·1n-1aj=1¯,1¯·1n-2aras=1n-2aras·1¯=1¯.
Hence, it follows that 𝒟𝒩12 is a commutative semiring with trivial multiplication.
Ifak-1=n-2, it is easy to see that endomorphism 1n-2(n-2)2 is the unique idempotent of 𝒟𝒩12 (see [5]). Now, we find 1n-2(n-2)2·1n-2aras=1n-2(ar)2, 1n-1(n-2)·1n-2aras=1n-1ar, and 1n-2aras·1n-1(n-2)=1¯. Hence, 𝒟𝒩12 is a semiring.
Case
2. Let am=n-2. We shall show that all the endomorphisms of 𝒟𝒩n-22 are 1-nilpotent with the only exception when a0=1. When a0>1, we find
(14)apaq(n-2)n-2·ap0aq0(n-2)n-2=n-2¯,ai(n-2)n-1·apaq(n-2)n-2=apaq(n-2)n-2·ai(n-2)n-1=n-2¯,n-2¯·apaq(n-2)n-2=apaq(n-2)n-2·n-2¯=n-2¯.
Ifa0=1, the only idempotent is 12(n-2)n-2 and we find
(15)12(n-2)n-2·apaq(n-2)n-2=(aq)2(n-2)n-2,1(n-2)n-1·apaq(n-2)n-2=aq(n-2)n-1,apaq(n-2)n-2·1(n-2)n-1=n-2¯.
Hence, 𝒟𝒩n-22 is a semiring.
Case
3. Let 1<a0 and ak-1<n-2. We find the following trivial equalities, which are grouped by duality:
(16)apaq(am)n-2×ap0aq0(am)n-2=am¯,(am)n-2aras·(am)n-2ar0as0=am¯,apaq(am)n-2·ap0(am)n-2as0=ap0(am)n-2as0·apaq(am)n-2=am¯,(am)n-2aras·ap0(am)n-2as0=ap0(am)n-2as0·(am)n-2aras=am¯,apaq(am)n-2·(am)n-2aras=(am)n-2aras·apaq(am)n-2=am¯,ai(am)n-1·apaq(am)n-2=apaq(am)n-2·ai(am)n-1=am¯,(am)n-1aj·apaq(am)n-2=apaq(am)n-2·(am)n-1aj=am¯,ai(am)n-1·ap(am)n-2as=ap(am)n-2as·ai(am)n-1=am¯,(am)n-1aj·ap(am)n-2as=ap(am)n-2as·(am)n-1aj=am¯,ai(am)n-1·(am)n-2aras=(am)n-2aras·ai(am)n-1=am¯,(am)n-1aj·(am)n-2aras=a(am)n-2aras·(am)n-1aj=am¯,am¯·apaq(am)n-2=apaq(am)n-2·am¯=am¯,am¯·ap(am)n-2as=ap(am)n-2as·am¯=am¯,am¯·(am)n-2aras=(am)n-2aras·am¯=am¯.
Case
4. Let a0=1 and ak-1<n-2. Then, 12(am)n-2 is the only idempotent in 𝒟𝒩m2. Additionally, to the equalities of the previous case, we find
(17)12(am)n-2·apaq(n-2)n-2=(aq)2(am)n-2,1(am)n-1·apaq(am)n-2=aq(am)n-1.
Case
5. Let 1<a0 and ak-1=n-2. Now the only idempotent endomorphism in 𝒟𝒩m2 is (am)n-2(n-2)2. We additionally find the following equalities:
(18)(am)n-2(n-2)2·(am)n-2aras=(am)n-2(ar)2,(am)n-1(n-2)·(am)n-2aras=(am)n-1ar.
Case
6. Let a0=1 and ak-1=n-2. Now, in 𝒟𝒩m2, there are two idempotents: 12(am)n-2 and (am)n-2(n-2)2. Here, the equalities from Cases 4 and 5 are valid and also all the equalities from Case 3, under the respective restrictions for the indices, are fulfilled.
Hence,𝒟𝒩m2 is a semiring.From Lemma11, we find the following.Theorem 12.
The unionI=⋃m=0n-1‍𝒟𝒩m2 of the discrete 2-neighborhoods with respect to all vertices of internal simplex σ(n){a0,…,ak-1} is a right ideal of the simplex.Proof.
For the endomorphisms of𝒟𝒩m2 and 𝒟𝒩t2, where t>m, from (11), it follows as shown in Table 1
Letβi, i=2,3,4,5,6, be elements of 𝒟𝒩t2 (see Table 1). We denote by βi~ an endomorphism, which maps the same elements to at, but other images (one or two) are not in Im⁡(α). For example, β2~=aℓ0(at)n-1, where ℓ0=0,…,m-1 and ℓ0≠ℓ. Evidently, βi~∈𝒟𝒩t2, for i=2,3,4,5,6.
Now we calculate the following:(1)
α1+β1=β1, α1+β2=β2 or α1+β2=β2~, α1+β3=β3, and α1+β4=β4 or α1+β4=β4~, α1+β5=β5, and α1+β6=β6 or α1+β2=β6~;(2)
α2+β1=β1, α2+β2=β2 or α2+β2=β2~, α2+β3=β3, and α2+β4=β4 or α2+β4=β4~, α2+β5=β5, and α2+β6=β6 or α2+β2=β6~;(3)
α3+β1=β1 or α3+β1=β3~, α3+β2=β2 or α3+β2=β2~, α3+β3=β3 or α3+β3=β3~, α3+β4=β4 or α3+β4=β4~, α3+β5=β5 or α3+β5=β5~, and α3+β6=β6 or α3+β6=β6~;(4)
α4+β1=β1, α4+β2=β2 or α4+β2=β2~, α4+β3=β3, and α4+β4=β4 or α4+β4=β4~, α4+β5=β5 and α4+β6=β6 or α4+β6=β6~;(5)
α5+β1=β1 or α5+β1=β3~ or α5+β1=β5~, α5+β2=β2 or α5+β2=β2~ or α5+β2=β5~ or α5+β2=β6~, α5+β3=β3 or α5+β3=β3~ or α5+β3=β5~, α5+β4=β4 or α5+β4=β4~ or α5+β4=β5~ or α5+β4=β6~, α5+β5=β5 or α5+β5=β5~ and α5+β6=β6 or α5+β6=β5~ or α5+β6=β6~;(6)
α6+β1=β1 or α6+β1=β3~, α6+β2=β2 or α6+β2=β2~, α6+β3=β3 or α6+β3=β3~ or α6+β3=β5~, α6+β4=β4 or α6+β4=β4~ or α6+β4=β6~, α6+β5=β5 or α6+β5=β5~ and α6+β6=β6 or α6+β6=β5~ or α6+β6=β6~.
From these calculations and Lemma11 we conclude that I is closed under the addition.
Letα∈I and φ∈σ(n){a0,…,ak-1}. Then, α∈𝒟𝒩m2, for some m=0,…,k-1. Considering the cases where α=αi, i=1,2,3, we consider in the proof of Theorem 9. Now we consider three new cases.
Case
1. Let α=α4=apaq(am)n-2, where p,q=0,…,m-1. Let φ(am)=at, φ(ap)=ap0, and φ(aq)=aq0. Then, α·φ=ap0aq0(at)n-2∈I. The same is true when q0=t or p0=q0=t.
Case
2. Let α=α5=(am)n-2aras, where r,s=m+1,…,k-1. Let φ(am)=at, φ(ar)=ar0, and φ(as)=as0. Then α·φ=(at)n-2ar0as0∈I. The same is true when r0=t or s0=r0=t.
Case
3. Let α=α6=ap(am)n-2as, where p=0,…,m-1 and s=m+1,…,k-1. Let φ(am)=at, φ(ap)=ap0, and φ(as)=as0. Then, α·φ=ap0(at)n-2as0∈I. The same is true when p0=t, s0=t, or p0=s0=t.
Hence, in all cases,α·φ∈I.Table 1
𝒟
𝒩
m
2
𝒟
𝒩
t
2
α
1
=
a
m
¯
β
1
=
a
t
¯
α
2
=
a
i
(
a
m
)
n
-
1
,
i
=
0
,
…
,
m
-
1
β
2
=
a
ℓ
(
a
t
)
n
-
1
,
ℓ
=
0
,
…
,
m
-
1
α
3
=
(
a
m
)
n
-
1
a
j
,
j
=
m
+
1
,
…
,
k
-
1
β
3
=
(
a
t
)
n
-
1
a
u
,
u
=
m
+
1
,
…
,
k
-
1
α
4
=
a
p
a
q
(
a
m
)
n
-
2
,
p
,
q
=
0
,
…
,
m
-
1
β
4
=
a
h
a
g
(
a
t
)
n
-
2
,
h
,
g
=
0
,
…
,
m
-
1
α
5
=
(
a
m
)
n
-
2
a
r
a
s
,
r
,
s
=
m
+
1
,
…
,
k
-
1
β
5
=
(
a
t
)
n
-
2
a
u
a
v
,
u
,
v
=
m
+
1
,
…
,
k
-
1
α
6
=
a
p
(
a
m
)
n
-
2
a
s
,
p
=
0
,
…
,
m
-
1
,
s
=
m
+
1
,
…
,
k
-
1
β
6
=
a
ℓ
(
a
t
)
n
-
2
a
u
,
ℓ
=
0
,
…
,
m
-
1
,
u
=
m
+
1
,
…
,
k
-
1From Theorem9 and Theorem 12, we obtain the following.Corollary 13.
(a) Ifσ(n){a0,…,ak-1} is an internal simplex, then the union J=⋃m=0n-1‍𝒟𝒩m1 of the discrete 1-neighborhoods with respect to all vertices is an ideal of the simplex.
(b) Letσ(n){x0,…,xs-1} be an internal simplex and σ(n){a0,…,ak-1} is internal of the simplex σ(n){x0,…,xs-1}. Then, the union I=⋃m=0n-1‍𝒟𝒩m2 of the discrete 2-neighborhoods is an ideal of the simplex σ(n){a0,…,ak-1}.Proof.
(a) Ifσ(n){a0,…,ak-1} is an internal simplex and φ∈σ(n){a0,…,ak-1}, then φ(0)>0 and φ(n-1)<n-1. We calculate φ·am¯=φ·ai(am)n-1=φ·(am)n-1aj=am¯.
From Theorem9, it follows that J is an ideal of the simplex σ(n){a0,…,ak-1}.
(b) Ifσ(n){a0,…,ak-1} satisfies the condition of Theorem 12 and φ∈σ(n){a0,…,ak-1}, then φ(0)>1 and φ(n-1)<n-2. Thus (see the notations in the proof of Theorem 12) we obtain
(19)φ·α1=φ·α2=φ·α3=φ·α4=φ·α5=φ·α6=am¯.
From Theorem 12, it follows that I is an ideal of the simplex σ(n){a0,…,ak-1}.Proposition 14.
Letσk(n)(A)=σ(n){a0,a1,…,ak-1} be a simplex.(a)
For the least vertexa0¯, it follows that𝒟𝒩0n-a0-1=σk(n)(A)∩ℰ𝒞n(a0).(b)
For the biggest vertexak-1¯, it follows that𝒟𝒩k-1ak-1=σk(n)(A)∩ℰ𝒞n(ak-1).Proof.
(a) Sincea0¯ is the least vertex of the simplex, it follows that layer ℒa0a0+1(σ(n){a0,a1,…,ak-1}) consists of endomorphisms α=(a0)a0+1(a1)p1⋯(ak-1)pk-1, where a0+1+p1+⋯+pk-1=n; that is, α(0)=a0,…,α(a0)=a0. All the layers ℒa0ℓ(σ(n){a0,a1,…,ak-1}), where ℓ≥a0+1, consist of endomorphisms having a0 as a fixed point. So, 𝒟𝒩0n-a0-1⊆σk(n)(A)∩ℰ𝒞n(a0).
Conversely, letα∈σk(n)(A)∩ℰ𝒞n(a0). Then, α(a0)=a0. Since a0¯ is the least vertex of the simplex, we have α(0)=⋯=α(a0-1)=a0; that is, α∈ℒa0ℓ(σ(n){a0,a1,…,ak-1}), where ℓ≥a0+1. Hence, 𝒟𝒩0n-a0-1=σk(n)(A)∩ℰ𝒞n(a0).
(b) Sinceak-1¯ is the biggest vertex of the simplex, it follows that layer ℒak-1n-ak-1(σ(n){a0,a1,…,ak-1}) consists of endomorphisms α=(a0)p0…(ak-2)pk-2(ak-1)n-ak-1, where p0+⋯+pk-2+n-ak-1=n. So, p0+⋯+pk-2=ak-1 implies that the images of 0,…,ak-1-1 are not equal to ak-1, but α(ak-1)=ak-1. For all the endomorphisms of layers ℒak-1ℓ(σ(n){a0,a1,…,ak-1}), where ℓ≥n-ak-1, we have p0+⋯+pk-2=ak-1. Hence, the elements of these layers have ak-1 as a fixed point and 𝒟𝒩0ak-1⊆σk(n)(A)∩ℰ𝒞n(ak-1).
Conversely, letα∈σk(n)(A)∩ℰ𝒞n(ak-1). Then, α(ak-1)=ak-1. Since ak-1¯ is the biggest vertex of the simplex, we have α(ak-1+1)=⋯=α(n-1)=ak-1. Thus, it follows that α∈ℒak-1ℓ(σ(n){a0,a1,…,ak-1}), where ℓ≥n-ak-1.
Hence,𝒟𝒩k-1ak-1=σk(n)(A)∩ℰ𝒞n(ak-1).


## 4. A Partition of a Simplex

Letσ(n){a0,a1,…,ak-1} be a simplex. Then, α∈σ(n){a0,a1,…,ak-1} is called endomorphism of type 〈〈m0,…,mk-1〉〉, where mi∈{0,…,k-1}, mi≤mj for i<j, i,j=0,…,k-1, if α(ai)=ami.Obviously, the relationα~β, if and only if α and β are of the same type, is an equivalence relation. Any equivalence class is closed under the addition. But there are equivalence classes which are not closed under the multiplication. For example, consider α=(a0)a1+1(a1)a2-a1(a2)n-a2-1, where n>a3>a2+1. Since α(a0)=a0, α(a1)=a0, α(a2)=a1, and α(a3)=a2,…,α(ak-1)=a2, the type of α is 〈〈0,0,1,2,…,2〉〉. But α2=(a0)a2+1(a1)n-a2-1 is of type 〈〈0,0,0,1,…,1〉〉.Sometimes it is possible to describe the semiring structure of union of many equivalence classes, that is, blocks of our partition. For example, the union of endomorphisms from all the blocks of type〈〈0,*,…,*〉〉 is the set of α∈σ(n){a0,a1,…,ak-1} such that α(a0)=a0 and, from Proposition 14, it is the semiring 𝒟𝒩0n-a0-1.The type of any endomorphism is itself an endomorphism of a simplexℰ^𝒞k=σ(k){0,1,…,k-1}. The simplex ℰ^𝒞k is called a coordinate simplex of σ(n){a0,a1,…,ak-1}. From this point of view, the set of endomorphisms from all the blocks of type 〈〈0,*,…,*〉〉 really corresponds to the set of all endomorphisms ≀0,m1,…,mk-1≀∈ℰ^𝒞k, which is the semiring ℰ𝒞k(0). More generally, we can consider the semiring ℰ𝒞k(j), where j∈𝒞k, consisting of all φ∈ℰ^𝒞k such that φ(j)=j. Then, α(aj)=amj=aφ(j)=aj. So the union of endomorphisms from all the blocks of type 〈〈*,…,*,j,*,…,*〉〉 is semiring σ(n){a0,a1,…,ak-1}∩ℰ𝒞n(aj). Now, more generally again, we can consider the semiring ⋂s=1m‍ℰ𝒞k(js) consisting of all endomorphisms of ℰ^𝒞k having j1,…,jm as fixed points. Then, similarly, α(ajs)=ajs, for all s=1,…,m. So the union of endomorphisms from all the blocks of type 〈〈*,…,*,j1,*,…,*,jm,*,…,*〉〉 is semiring σ(n){a0,a1,…,ak-1}∩(⋂s=1m‍ℰ𝒞n(ajs)). The next results from this section are announced in [8].Theorem 15.
Letσ(n){a0,…,ak-1} be a simplex. Let R be a subsemiring of the coordinate simplex ℰ^𝒞k of σ(n){a0,…,ak-1}. The set R~ of endomorphisms of σ(n){a0,…,ak-1}, of type 〈〈m0,…,mk-1〉〉, where the endomorphism ≀m0,…,mk-1≀∈R, is a semiring. Moreover, when R is a (right, left) ideal of semiring ℰ^𝒞k, it follows that R~ is a (right, left) ideal of simplex σ(n){a0,…,ak-1}.Proof.
Letα,β∈σ(n){a0,…,ak-1}. Let α be of type 〈〈p0,…,pk-1〉〉, where the endomorphism φ=≀p0,…,pk-1≀∈R and similarly β are of type 〈〈q0,…,qk-1〉〉, where the endomorphism ψ=≀q0,…,qk-1≀∈R. Then, we find (α+β)(ai)=α(ai)+β(ai)=api+aqi=ami, where mi=max⁡{pi,qi}. But the endomorphism ≀m0,…,mk-1≀ is the sum φ+ψ. So we prove that endomorphism α+β is of type φ+ψ∈R; that is, α+β∈R~.
Now, let us assume thatφ and ψ are arbitrary endomorphisms of ℰ^𝒞k. Then, we find
(20)(α·β)(ai)=β(α(ai))=β(ami)=β(aφ(i))=aψ(φ(i))=a(φ·ψ)(i).
So, ifR is a right ideal of ℰ^𝒞k and φ∈R, R is a left ideal of ℰ^𝒞k and ψ∈R, or R is an ideal of ℰ^𝒞k and one of φ and ψ is from R, it follows that φ·ψ∈R. Hence, in each of the three cases α·β∈R~ and this completes the proof.Now, let endomorphismα be of type 〈〈ι0,…,ιk-1〉〉 and the endomorphism ι=≀ι0,…,ιk-1≀ from the coordinate simplex ℰ^𝒞k be an idempotent, different from constant endomorphisms j¯, where j=0,…,k-1, and the identity i. Then we say that α is of an idempotent type.Corollary 16.
The set of endomorphismsα∈σ(n){a0,…,ak-1} of a fixed idempotent type 〈〈ι0,…,ιk-1〉〉 is a semiring.Proof.
Obviously, since the setR={ι}, where ι is an idempotent, is a semiring, the semirings of an idempotent type are denoted by R~=ℐ(ι), where ι=≀ι0,…,ιk-1≀ is the corresponding idempotent.Letℓ∈𝒞k. For any ℓ¯=≀ℓ,…,ℓ≀∈ℰ^𝒞k, we consider (see the first section) the set
(21)𝒩k[ℓ]={φ∣φ∈ℰ^𝒞k,𝒩k[ℓ]=dφmℓ=ℓ¯forsomenaturalnumbermℓ}.From Theorem1, it follows that 𝒩k[ℓ], for k≥2 and ℓ∈𝒞k, is a subsemiring of ℰ^𝒞k.Now, let the endomorphismα∈σ(n){a0,…,ak-1} be of any type 〈〈m0,…,mk-1〉〉, where the endomorphism ≀m0,…,mk-1≀∈𝒩k[ℓ], for some fixed ℓ∈𝒞k. Then, we say that α is of an ℓ-nilpotent type.Corollary 17.
The set of endomorphismsα∈σ(n){a0,…,ak-1} of ℓ-nilpotent type, for some fixed ℓ∈𝒞k, is a semiring.Proof.
Immediately from the last theorem and Theorem1, the semirings of ℓ-nilpotent type are denoted by R~=𝒩[aℓ](σ(n){a0,…,ak-1}).Now, let the endomorphismα∈σ(n){a0,…,ak-1} be of any type 〈〈m0,…,mk-1〉〉, where the endomorphism φ=≀m0,…,mk-1≀ from the coordinate simplex ℰ^𝒞k is neither an idempotent nor an ℓ-nilpotent for some ℓ∈𝒞k. Then, according to [5], φ is a root of some idempotent ι=≀ι0,…,ιk-1≀. Since the roots of identity of semiring ℰ^𝒞k do not exist (see [2]), it follows that ι is an idempotent, different from ℓ¯, ℓ∈𝒞k, and identity. In this case, the endomorphism α is called a type related to idempotent type 〈〈ι0,…,ιk-1〉〉. To clarify the above definition, we give an example of endomorphisms of type related to some idempotent type.Example 18.
Let us consider the simplexσ(10){0,2,3,5,8}. The coordinate simplex consisting of all the types of endomorphisms of σ(10){0,2,3,5,8} is the simplex σ(5){0,1,2,3,4}. In this coordinate simplex, we chose the idempotent ι=≀0,0,0,0,4≀. From Theorem 19 of [5], it follows that the idempotent ι and its roots form a semiring of order C3=5. Let us note thatCatalan sequence is the sequence C0,C1,C2,C3,…, where Cn=1/(n+1)(2nn). So, the elements of the semiring, generated by ι contains 5 endomorphisms: φ1=≀0,0,0,1,4≀, φ2=≀0,0,0,2,4≀, φ3=≀0,0,1,1,4≀, φ4=≀0,0,1,2,4≀, and ι. Now, we choose endomorphisms from the simplex σ(10){0,2,3,5,8}: α=042284 and β=03223385. The endomorphism α is of type 〈〈0,0,0,1,4〉〉, related to idempotent type 〈〈0,0,0,0,4〉〉, and endomorphism β is of type 〈〈0,0,1,2,4〉〉, related to the same idempotent type. We compute α2=0684 which is an idempotent and also β2=052382 which is not an idempotent, but β3=0882 is an idempotent. Thus, we show that α and β are roots of different idempotents, but they are of types related to the same idempotent type. We can also show that α·β=0684 is an idempotent and β·α=0882 also is an idempotent.Corollary 19.
The set of endomorphismsα∈σ(n){a0,…,ak-1} of a type, related to some fixed idempotent type, is a semiring.Proof.
Immediately from the last theorem and Theorem 19 of [5], the semirings from the last corollary are called idempotent closures of type ι, where ι=≀ι0,…,ιk-1≀ is the corresponding idempotent and is denoted by R~=ℐ𝒞(ι).From the last theorem, we also find the following.Corollary 20.
The set of endomorphisms ofσ(n){a0,…,ak-1}, of type 〈〈m0,…,mk-1〉〉, where the endomorphism ≀m0,…,mk-1≀ belongs to some face of the coordinate simplex ℰ^𝒞k, is a left ideal.Now we describe the left ideals from the last corollary whenk=4.Example 21.
For a simplex𝒯ℰ𝒯ℛ(n){a0,a1,a2,a3}, the coordinate simplex is 𝒯ℰ𝒯ℛ(4){0,1,2,3} and its faces are as follows: triangles:I0=∆(4){1,2,3}, I1=∆(4){0,2,3}, I2=∆(4){0,1,3}, and I3=∆(4){0,1,2};strings:I0∩I1=𝒮𝒯ℛ(4){2,3}, I0∩I2=𝒮𝒯ℛ(4){1,3}, I0∩I3=𝒮𝒯ℛ(4){1,2}, I1∩I2=𝒮𝒯ℛ(4){0,3}, I1∩I3=𝒮𝒯ℛ(4){0,2}, and I2∩I3=𝒮𝒯ℛ(4){0,1};vertices:I0∩I1∩I2={3¯}, I0∩I1∩I3={2¯}, I0∩I2∩I3={1¯}, and I1∩I2∩I3={0¯}.
Then, the left idealI0~ of simplex 𝒯ℰ𝒯ℛ(n){a0,a1,a2,a3} consists of endomorphisms α, such that a0 is not a fixed point of α. Now, the triangle ∆(n){a1,a2,a3}, which is a left ideal of semiring 𝒯ℰ𝒯ℛ(n){a0,a1,a2,a3}, is contained in I0~. Moreover,
(22)I0~∖∆(n){a1,a2,a3}=⋃s=1a0‍ℒa0s(𝒯ℰ𝒯ℛ(n){a0,a1,a2,a3})={α=(a0)ℓ(a1)p1(a2)p2(a3)p3,wherewjwhere0<ℓ≤a0,ℓ+p1+p2+p3=nα=(a0)ℓ(a1)p1(a2)p2(a3)p3}.
Similarly, the left idealsI1~, I2~ И I3~ consist of endomorphisms α, such that a1, a2, and a3 are not fixed points of α, respectively.
The left idealsIp~∩Iq~, where p,q∈{0,1,2,3}, p≠q, consists of endomorphisms α, such that ap and aq are not fixed points of α. Let {ar,as}={a0,a1,a2,a3}∖{ap,aq} И  ar<as. Then, 𝒮𝒯ℛ(4){ar,as}⊂Ip~∩Iq~. Observe that in the interior of one of the triangles with vertices ap¯, ar¯, and as¯ and aq¯, ar¯, and as¯, respectively, there are endomorphisms α, such that ap and aq are not fixed points of α.
The left idealI~, where I=Ip∩Iq is actually the ideal Ip~∩Iq~.
The left idealI0~∩I1~∩I2~ consists of endomorphisms α such that a0, a1, and a2 are not fixed points of α. Hence, all elements of this left ideal have a3 as a fixed point. Similarly, we determine the left ideals I0~∩I1~∩I3~, I0~∩I2~∩I3~, and I1~∩I2~∩I3~.
From the last theorem, also we have two consequences.Corollary 22.
The set of endomorphisms of the simplexσ(n){a0,…,ak-1}, having a type 〈〈ai,…,ai〉〉, where i=0,…,k-1, is an ideal.Corollary 23.
The set of endomorphisms of the simplexσ(n){a0,…,ak-1}, having a type 〈〈m0,…,mk-1〉〉, where the endomorphism ≀m0,…,mk-1≀ belongs to the union J=⋃m=0k-1‍𝒟𝒩m1 of the discrete 1-neighborhoods of all vertices of coordinate simplex ℰ^𝒞k, is a right ideal.At last, we consider the endomorphismsα∈σ(n){a0,…,ak-1} of type 〈〈0,1,…,k-1〉〉. Now the corresponding endomorphism from the coordinate simplex is identity i. In order to find the set of endomorphisms R~ of this type, we need the following definition. Idempotent α∈σ(n){a0,…,ak-1} is called a boundary idempotent of the simplex, if α∈ℬ𝒟(σ(n){a0,…,ak-1}), similarly, an interior idempotent of the simplex, if α∈ℐ𝒩𝒯(σ(n){a0,…,ak-1}). Note that, in the coordinate simplex, the identity i is the unique interior idempotent.Theorem 24.
The set of endomorphisms ofσ(n){a0,…,ak-1}, which are right identities, is a semiring of order ∏i=0k-1‍(ai+1-ai).Proof.
Let us denote byℛℐ(σ(n){a0,…,ak-1}) semiring σ(n){a0,…,ak-1}∩(⋂i=0k-1‍ℰ𝒞n(ai)). Then, for any element x∈𝒞n and arbitrary endomorphism α∈ℛℐ(σ(n){a0,…,ak-1}), it follows that α(x)=ai, for some i=0,…,k-1. Then α2(x)=α(ai)=ai=α(x); that is, α is an idempotent. Since α=(a0)ℓ0…(ak-1)ℓk-1, where ℓi>0, for any i=0,…,k-1 (if we suppose ℓi=0, then α∉ℰ𝒞n(ai)), it follows that α is an interior idempotent.
Conversely, letα be an interior idempotent. Then, we can express α=(a0)ℓ0⋯(ak-1)ℓk-1, where ℓi>0, for any i=0,…,k-1 (if we suppose ℓi=0, then α∉ℐ𝒩𝒯(σ(n){a0,…,ak-1})). So, we prove that the semiring ℛℐ(σ(n){a0,…,ak-1}) consists of all interior idempotents of the simplex.
Letα∈ℛℐ(σ(n){a0,…,ak-1}) and β be an arbitrary element of the simplex. For x∈𝒞n, it follows that β(x)=ai, where i=0,…,k-1. Then (β·α)(x)=α(β(x))=α(ai)=ai=β(x). Hence, α is a right identity of the simplex.
Conversely, letα be a right identity of the simplex. Evidently, α is an idempotent. Assume that, for some i=0,…,k-1, we have ai∉Im⁡(α). Since for some β∈σ(n){a0,…,ak-1} and x∈𝒞n, it follows that β(x)=ai, where i=0,…,k-1, we find (β·α)(x)=α(β(x))=α(ai)≠ai=β(x); that is, β·α≠β, which contradicts that α is a right identity. So, Im⁡(α)={a0,…,ak-1}; that is, α is an interior idempotent or α∈ℛℐ(σ(n){a0,…,ak-1}).
Hence, we prove thatℛℐ(σ(n){a0,…,ak-1}) is a semiring of right identities of simplex σ(n){a0,…,ak-1}. Since elements of ℛℐ(σ(n){a0,a1,…,ak-1}) are all the idempotents with fixed points a0,…,ak-1, from Theorem 2 (Section 1), it follows that |ℛℐ(σ(n){a0,…,ak-1})|=∏i=0k-1‍(ai+1-ai).So, we can construct a partition of the simplexσ(n){a0,…,ak-1} such that all blocks of this partition are as follows:(1)
semiringsR~=𝒩[aℓ](σ(n){a0,…,ak-1}), where ℓ=0,…,k-1;(2)
semiringsR~=ℐ(ι), where ι=≀ι0,…,ιk-1≀ is an idempotent of ℰ^𝒞k;(3)
semiringsR~=ℐ𝒞(ι), where ι=≀ι0,…,ιk-1≀ is an idempotent of ℰ^𝒞k;(4)
semiringR~=ℛℐ(σ(n){a0,…,ak-1}).From the last theorem we also find the following.Corollary 25.
There is at least one right identity of the simplexσ(n){a0,…,ak-1}.If there are two or more right identities of simplexσ(n){a0,…,ak-1}, then there is not a left identity. Let us suppose that there is a left identity ω∈σ(n){a0,…,ak-1}. Let ɛ be a right identity of the simplex. Then, it follows that ɛ=ω·ɛ=ω. If there is a single right identity, from the last theorem, it follows that ai+1=ai+1, for any i=0,…,k-2. Then, this single right identity is ɛ=(a0)a0+1(a0+1)⋯(a0+k-2)(a0+k-1)n-a0-k+1. Let α=(a0)a0(a1)n-a0. We find ɛ·α=a1¯≠α, and so we have proved.Corollary 26.
There are not any left identities of the simplexσ(n){a0,…,ak-1}.



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