aa r X i v : . [ m a t h . R A ] D ec UNITS IN NEAR-RINGS
TIM BOYKETT AND GERHARD WENDTA
BSTRACT . We investigate near-ring properties that generalize nearfield prop-erties about units. We study zero symmetric near-rings N with identity withtwo interrelated properties: the units with zero form an additive subgroup of ( N , +) ; the units act without fixedpoints on ( N , +) . There are many similari-ties between these cases, but also many differences. Rings with these propertiesare fields, near-rings allow more possibilities, which are investigated. Descrip-tions of constructions are obtained and used to create examples showing the twoproperties are independent but related. Properties of the additive group as a p -group are determined and it is shown that proper examples are neither simple nor J -semisimple.
1. I
NTRODUCTION
Within nearring theory, the subjects of planar nearrings and nearfields form astrong core of well-understood structure theory. Two important tools are used,namely (right) translation maps and the properties of units.In this paper we investigate generalisations of nearfields and planar nearringsin order to examine the structural implications. The first property relates to theadditive closure of the units in a nearring, the second property relates to the natureof the right translation maps as fixed point free. We see similar results proven usingsimilar techniques in each case.We commence with an introduction to the relevant aspects of nearring theory,namely nearfields, planar nearrings and fixed point free transformations. Then weconsider the case of rings and show that our properties lead to rings with eitherproperty being fields. We follow this up with a general construction for nearringsand show a construction technique developed from near-vector-space considera-tions for examples having both properties, plus two specific examples of near-ringshaving our two properties independently.The next section investigates the structure of the additive group for each of theseproperties. We then investigate several special classes of near-rings having theproperties, before concluding with some open problems that suggest themselvesfor future work. 2. B
ACKGROUND A near-ring ( N , + , ∗ ) is a group ( N , +) with identity 0 and a semigroup ( N , ∗ ) with the right distributive law ∀ a , b , c ∈ N , ( a + b ) ∗ c = a ∗ c + b ∗ c . We use 0-symmetric nearrings such that 0 ∗ n = ∗ n =
0. If there is no risk of confusion
Mathematics Subject Classification.
Key words and phrases. fixedpointfree automorphisms, nearfields, group of units, simple near-rings, endomorphism semigroups.This work has been supported by grant P23689-N18 of the Austrian National Science Fund(FWF). we will often omit the symbol ∗ . We will be mostly concerned with near-ringssatisfying the DCCN, the descending chain condition for N -subgroups of N . Ifa near-ring N is a ring, then the DCCN is the same as the DCCL for rings, thedescending chain condition on left ideals. However, some results only apply in thefinite case.Let U be the set of all units, i.e. invertible elements w.r.t multiplication ∗ . Then ( U , ∗ ) is a group, the group of units of the near-ring. A near-field is a near-ring inwhich the non-zero elements are all units, forming a group under multiplication.An automorphism f of a group ( N , +) is said to have a fixed point n ∈ N if f ( n ) = n . If 0 is the only fixed point of f , then f is called fixed point free (abbreviated byfpf in the following) on N . A group of automorphisms F acting on the group ( N , +) is called a fixedpointfree automorphism group if every non-identity automorphismin F is fpf.For u ∈ N let y u : N −→ N , n n ∗ u be the right translation map induced by u . Since the right distributive law holds, y u is an endomorphism of ( N , +) . If u is a unit, then y u is an automorphism. The converse also holds. Let y r be anautomorphism. So, there is an element r such that y r ( r ) = r ∗ r =
1. Moreover, r = ∗ r = r ∗ = r ∗ ( r ∗ r ) = ( r ∗ r ) ∗ r . Since y r is injective, 1 ∗ r = ( r ∗ r ) ∗ r implies 1 = r ∗ r and so r is a unit.For a set S ⊆ N we define Y S : = { y s | s ∈ S } . Therefore, ( Y U , ◦ ) is a group ofautomorphisms of ( N , +) , under function composition. Lemma 2.1.
Let N be a zero symmetric near-ring with identity and DCCN. Thenn ∈ N has a multiplicative inverse iff y n is injective.Proof. Suppose that n is such that y n is injective. By the DCCN there exists anatural number k such that Nn k = Nn k + . Thus, for all m ∈ N there exists a j ∈ N such that mn k = jn k + . Consequently, ( m − jn ) n k =
0. Since y n is injective, also y n k is injective which results in m = jn . Thus, there is an element l ∈ N such that1 = ln . Thus y l is injective and in the same way we again see that there is anelement h ∈ N such that 1 = hl . So we have h = hln = n = n and we see that l isthe inverse of n . The rest is clear. (cid:3) Let ( N , + , ∗ ) be a nearring. ( M , +) ≤ ( N , +) is an N -group if ∀ n ∈ N , m ∈ M , nm ∈ M . Then Aut N ( M ) = { f ∈ Aut ( M ) |∀ n ∈ N , f ( nm ) = n f ( m ) } .A planar nearring is a nearring ( N , + , ∗ ) such that every equation of the form xa = xb + c with y a = y b has a unique solution x and there are at least three distinctright translation maps. The right translation maps for planar nearrings are all fpfautomorphisms of the group ( N , +) .Similarly right translations y n for nonzero elements n of a nearfield are fpf au-tomorphisms. Note that a planar nearring with identity is a nearfield.With these notes we have set the ground for our investigations. In this paper westudy near-rings which have special properties concerning their units, which weintroduce as follows.In a near-field, the sum of two units is a unit or zero, while in general, the unitsof a near-ring are not closed w.r.t. addition. On one hand, we are interested innear-rings where ( U ∪ { } , +) is a subgroup of ( N , +) . Since ( U , ∗ ) is a group, thisis equivalent to saying that ( U ∪ { } , + , ∗ ) is a subnear-field of ( N , + , ∗ ) . Trivialexamples for such type of near-rings are near-fields but we will see that these arenot the only examples. NITS IN NEAR-RINGS 3
Definition 2.2. An f-near-ring N is a zero symmetric near-ring with identity andset of units U where ( Y U , ◦ ) acts as a fixedpointfree automorphism group on theadditive group of the near-ring. An a-near-ring N is a zero symmetric near-ringwith identity and set of units U such that U , the set of units of the near-ring ad-joined with the zero of the near-ring, forms a subnear-field of the near-ring w.r.t. thenear-ring operations. Near-rings with both properties will be called af-near-rings .Note that if the group of units is trivial, i.e. consists only of the identity, then theconcept of f-near-ring is vacuous. In the next section we show that an a-ring withDCCL and with only one unit is Z k , k a natural number. a-nearrings with only oneunit have elementary abelian 2-groups additively, but otherwise remain open.In the next section we look at a- and f-near-rings which are rings and showthat they are all fields. Following that, we look at the general structure of a- andf-near-rings and show that many examples exist.3. T HE RING CASE
We first study rings which are a-near-rings, f-near-rings respectively. In case werestrict to rings with DCCL we will see that such rings are fields. The authors arenot aware of a ring theoretic result which has a similar content as the followingTheorem. Therefore we also give a full proof.
Theorem 3.1.
Let ( R , + , ∗ ) be a ring with identity and group of units U such that | U | ≥ . Then the following are equivalent: a. R is a field, b. R is an a-near-ring with DCCL, c. R is an f-near-ring with DCCL.Proof. a ⇒ b : This is clear from the definition of an a-near-ring and the fact thatany field satisfies the DCCL. b ⇒ c : Let u be a non-identity unit in R . Suppose that ru = r for some r ∈ R .Thus, r ( − u ) =
0. By assumption, 1 − u ∈ U and consequently, r =
0. This showsthat R is an f-near-ring. c ⇒ a : Let a , b ∈ U and suppose a − b U . Since R has DCCL and a − b is nota unit, y a − b cannot be an injective map by Lemma 2.1. Thus, there is an element z ∈ R \ { } such that 0 = z ( a − b ) = za − zb . Consequently, za = zb and therefore, z = zba − . So, z is a fixed point. Since ba − ∈ U , by assumption, ba − = a = b . Therefore, given a , b ∈ U such that a = b , a − b ∈ U .Suppose there is a nilpotent element a ∈ R , so a n = n . Then, ( − a )( + a + . . . + a n − ) =
1, so 1 − a is a unit, u say. Consequently,1 − u = a U . Since 1 = u implies 1 − u ∈ U , as just shown, 1 = u and a = R does not contain non-zero nilpotent elements, so it is a reducedring.Thus, by [6, Theorem 9.41] R is a finite direct product of k fields, say. Supposethat k ≥
2. Then there exist orthogonal idempotents being the identities of thefields in the direct product such that 1 = e + . . . + e k . By assumption, there existsa non-identity unit u = f + . . . + f k ∈ R . W.l.o.g. we assume that f = e . Then w = f + . . . + e k is also a unit which does not equal the identity. Then, e w = e ( f + . . . + e k ) = e and therefore, e is a fixed point of y w which contradicts theassumption. Thus, k = R is a field. (cid:3) TIM BOYKETT AND GERHARD WENDT
What happens in rings with identity and with DCCL which have the identityelement as their single unit element answers the next proposition.
Proposition 3.2.
Let R be a ring with identity, which is the only unit element in R,R satisfying the DCCL. Then, R is isomorphic to a direct product of copies of thefield Z .Proof. Suppose R contains a nilpotent element a , say. Then, ( − a )( + a + . . . + a n − ) = n >
1. Hence, 1 − a is a unit and by assumption,1 − a =
1. Thus, a = R is a reduced ring. By [6, Theorem 9.41]a reduced ring with DCCL is isomorphic to a direct product of fields. Since wehave only 1 as unit element in R , any field in the direct composition has only oneunit element which shows that R is isomorphic to a direct product of copies of Z . (cid:3) The situation in near-rings with identity having precisely the identity as unitelement is more complex and a complete solution of how these type of near-ringslook like is not known to the authors. We will look at some concrete examplesusing Theorem 4.5.Theorem 3.1 tells us that within the class of rings with DCCL there do not exista-rings or f-rings other than fields. We have to look at proper near-rings in order toget nontrivial examples, which we do in the following section.4. C
ONSTRUCTIONS USING SEMIGROUPS OF GROUP ENDOMORPHISMS
In this section we present a method of constructing all a- and f-near-rings.
Theorem 4.1.
The following are equivalent: (1) ( N , + , · ) is an f-near-ring. (2) There exists (a) a group ( M , +) . (b) a semigroup S = G ∪ E of group endomorphisms of ( M , +) where Gis a group of fpf automorphisms of ( M , +) and E is a semigroup ofnon-bijective endomorphisms containing the zero map . (c) an element m ∈ M such that S ( m ) = M and for all n ∈ M there is aunique s n ∈ S with n = s n ( m ) .such that ( N , + , · ) ∼ = ( M , + , ∗ ) , where a ∗ b = s b ( a ) (a , b ∈ M).Proof. ( ) ⇒ ( ) : Let ( M , +) = ( N , +) . For all r ∈ N , y r : N −→ N , n nr isan endomorphism of ( N , +) . Let S : = { y r | r ∈ N } , then S is a semigroup ofgroup endomorphisms. G : = Y U = { y u | u ∈ U } is a subgroup of S acting withoutfixedpoints on ( N , +) . Let E : = { y z | z ∈ N \ U } . Then S = G ∪ E and no elementin E is a bijection (certainly, 0 ∈ E ). Let m : =
1. For each n ∈ N we have n = ∗ n = y n ( ) with y n ∈ S . Certainly, y n is the unique element s in S such that s ( ) = n . S ( ) = N and for each n ∈ N there is a unique s n ∈ S such that n = s n ( ) ,namely s n = y n . Note that for a , b ∈ N we have a · b = y b ( a ) = s b ( a ) = a ∗ b . ( ) ⇒ ( ) : We first show that ∗ is indeed a near-ring multiplication. By assump-tion, we know that for a given element b ∈ M there is a unique element s b ∈ S suchthat b = s b ( m ) , so multiplication is well defined. We first prove associativity of ∗ :On one hand, ( a ∗ b ) ∗ c = s b ( a ) ∗ c = s c ( s b ( a )) . Since b = s b ( m ) and c = s c ( m ) wehave a ∗ ( b ∗ c ) = a ∗ s c ( b ) = a ∗ s c ( s b ( m )) = s c ( s b ( a )) . NITS IN NEAR-RINGS 5
Using that s c is a group endomorphism for given c ∈ M we get ( a + b ) ∗ c = s c ( a + b ) = s c ( a ) + s c ( b ) = a ∗ c + b ∗ c . This shows that ∗ is a right distributivemultiplication and ( M , + , ∗ ) is a near-ring.As a group G contains the identity function id , we have m = id ( m ) and so wehave a ∗ m = a . On the other hand, m ∗ a = s a ( m ) = a which proves that m is theidentity element w.r.t ∗ . Since 0 ∈ E (0 being the zero map) we have 0 = ( m ) and consequently, n ∗ = ( n ) = n for all n ∈ M and hence, M is a zero symmetricnear-ring.Let U be the group of units of M . We now prove U = G ( m ) : G ( m ) ⊆ U , because for all elements g ( m ) ∈ G ( m ) we have that g ( m ) ∗ g − ( m ) = g − g ( m ) = m , m being the identity element, so g − ( m ) is the inverse w.r.t ∗ .Let u ∈ U , so there is an element u − ∈ M such that u − ∗ u = m . We knowthat there is a unique element s u ∈ S such that s u ( m ) = u and also s u − ( m ) = u − .Therefore, u − ∗ u = s u ( s u − ( m )) = m . Since S is a semigroup, we have s u ◦ s u − ∈ S .So, we have m = id ( m ) = s u s u − ( m ) and this implies s u ◦ s u − = id because thereis only one endomorphism in S mapping m to m , by assumption. Suppose s u ∈ E .Since id ∈ G , s u − E and consequently, s u − ∈ G . But then s u = s − u − ∈ G , acontradiction. This shows that s u ∈ G . This finally gives us U ⊆ G ( m ) .To finish the proof we have to show that Y U acts without fixedpoints on ( M , +) .Let m = b = s b ( m ) where s b ∈ G \ { id } . Suppose there is an element a ∈ M suchthat a ∗ b = a . Then we have that s b ( a ) = a . Since each non-identity automorphismin G acts without fixedpoints on M , we must conclude a = (cid:3) Theorem 4.1 will now be used to give an example of an f-near-ring which is notan a-near-ring. More examples for constructing f-near-rings will follow.
Example 4.2.
Let ( M , +) : = ( Z , +) be the -dimensional direct sum of the group ( Z , +) written as column vectors. In the following, the matrices and vectors willbe over the field Z . Let m : = ( , , , ) t .Let G = { ( , , , ) t , ( , , , ) t , ( , , , ) t , ( , , , ) t } ⊆ M, then defineG : = * − −
10 0 1 0 + ≤ GL ( , ) E : = { n n − n n n n − n n | ( n , n , n , n ) t ∈ M \ G } One checks that G acts without fixed points on M and is a subgroup of order of the general linear group GL ( , ) . By straightforward calculation we see that Eis a matrix semigroup. Also, one checks that for given matrix A ∈ G and B ∈ E, theproduct AB and BA of the matrices is contained in E. Let S : = G ∪ E. It followsthat S is a matrix semigroup. A matrix s ∈ S induces an endomorphism on ( M , +) simply by matrix multiplication. Thus (without distinguishing between the matrixand the induced endomorphism) S is a semigroup of group endomorphisms of M.Also, we see that G · m = G and E · m = M \ G . Thus, for each element n ∈ M, thereis a unique matrix s ∈ S such that n = s · m.We now define the multiplication ∗ on M as a ∗ b = a ∗ ( s · m ) : = s · a, ( a , b ∈ M ) . TIM BOYKETT AND GERHARD WENDT
By Theorem 4.1, ( M , + , ∗ ) is an f-near-ring. The units U of this near-ring areU : = G . We clearly have that U is not additively closed. A similar structure exists for a-near-rings.
Theorem 4.3.
The following are equivalent: (1) ( N , + , · ) is an a-near-ring. (2) There exists (a) a group ( M , +) with a subgroup H ⊆ M. (b) a semigroup S = G ∪ E of group endomorphisms of ( M , +) where Gis a group of automorphisms and E is a semigroup of non-bijectiveendomorphisms containing the zero map . (c) an element m ∈ M such that S ( m ) = M, G ( m ) = H \ { } and for alln ∈ M there is a unique s n ∈ S with n = s n ( m ) .such that ( N , + , · ) ∼ = ( M , + , ∗ ) , where a ∗ b = s b ( a ) (a , b ∈ M).Proof. ( ) ⇒ ( ) : As in the proof of Theorem 4.1 let ( M , +) = ( N , +) and S : = { y r | r ∈ N } . G : = Y U = { y u | u ∈ U } is a subgroup of S acting on ( N , +) . Let E : = { y z | z ∈ N \ U } . Then S = G ∪ E is a semigroup and no element in E is abijection (certainly, 0 ∈ E ). We now let H = U , so G ( ) = U . As in the proof of ( ) ⇒ ( ) of Theorem 4.1 we let m = ( ) ⇒ ( ) : To prove that ( M , + , ∗ ) is a near-ring runs as in the proof of Theorem4.1. Also as in the proof of Theorem 4.1 we see that G ( m ) = H \ { } is the set ofunits of the near-ring.It only remains to show that the units are additively closed. Let a = g ( m ) and b = g ( m ) be two units, a = b . Then we know that a − b = g ( m ) − g ( m ) ∈ G ( m ) since H is a subgroup of M . Thus, a − b is a unit. (cid:3) We construct an example of a proper a-near-ring.
Example 4.4.
We let M be the group which is given by the following multiplicativepresentation (the finite Heisenberg group on p = ).M : = (cid:10) x , y , z : x = y = z = , yz = zyx , xy = yx , xz = zx (cid:11) M is a non-abelian group of order . Let S : = { s a | a ∈ M } where s a : M → Mis a group endomorphism defined on the generators x , y , z of the group as follows.For a ∈ M \ { z , z } we define s a ( x ) : = , s a ( y ) : = , s a ( z ) : = a, otherwise s z ( x ) : = x , s z ( y ) : = y , s z ( z ) : = z and s z ( x ) : = x , s z ( y ) : = y , s z ( z ) : = z . The endomorphismss z and s z are automorphisms of the group, s z = id the identity. We now let G = { s z , s z } and E = { s a | a ∈ M \ { z , z }} . Hence, G ( z ) = { z , z } and consequently,G ( z ) ∪ { } is a subgroup H of M. By taking m = z we can now apply Theorem 4.3and see that ( M , + , ∗ ) is an a-near-ring. Example 4.4 is
LibraryNearRingWithOne ( GT W − , ) in the GAP packageSONATA [1].We now use Theorems 4.1 and 4.3 to construct af-near-rings which are not near-fields such that the units are isomorphic to a given near-field. Theorem 4.5.
Let ( F , + , ∗ ) be a near-field and ( V , +) = ( F k , +) the k dimensionaldirect sum of ( F , +) . For i ∈ { , . . . , k } let a i : F → F be zero preserving mapswhich are multiplicative automorphisms of the group ( F \ { } , ∗ ) where a k is the NITS IN NEAR-RINGS 7 identity map. For a , x ∈ V define the operation ∗ in V in the following way:x ∗ a : = ( x ∗ a ( a k ) , . . . , x k − ∗ a k − ( a k ) , x k ∗ a k ) if a = a = . . . = a k − = x ∗ a : = ( x k ∗ a , . . . , x k ∗ a k ) otherwiseThen ( V , + , ∗ ) is an af-near-ring with units U = { ( , . . . , , a ) | a ∈ F , a = } ,identity ( , . . . , , ) , and ( U ∪ { } , + , ∗ ) isomorphic to ( F , + , ∗ ) .Proof. Let s : V → End ( V ) be defined by s ( a )( x ) : = x ∗ a . By right distributivityof the near-field multiplication, s ( a ) is a homomorphism.We first show that S = { s ( v ) | v ∈ V } is closed under composition of functionsand is a semigroup. For the rest of this proof, we write a ∗ b as ab to aid readability.We now show S is closed with 4 cases depending upon the structure of a and b . Let Z = { a ∈ V | a = . . . = a k − = } .Case 1: Let a , b ∈ V \ Z . Then ( s ( a ) ◦ s ( b ))( x ) = s ( a )( s ( b ( x ))= s ( a )( x k b , . . . , x k b k )= ( x k b k a , . . . , x k b k a k ) If b k = s (( , . . . , )) , otherwise there is some nonzero b k a , . . . , b k a k − .Thus, ( x k b k a , . . . , x k b k a k ) = s (( b k a , . . . , b k a k ))( x , . . . , x k ) , so S is closed.Case 2: Now consider a , b ∈ Z . We get ( s ( a ) ◦ s ( b ))( x ) = s ( a )( x a ( b k ) , . . . , x k b k ) = ( x a ( b k ) a ( a k ) , . . . , x k b k a k ) From the fact that a i , i ∈ { , . . . , k } are multiplicative homomorphisms we have that ( x a ( b k ) a ( a k ) , . . . , x k b k a k ) = ( x a ( b k a k ) , . . . , ( x k − a k − ( b k a k ) , x k b k a k )= s (( , . . . , , b k a k ))( x ) so S is closed.Case 3: Let a ∈ V \ Z , b ∈ Z . We calculate: ( s ( a ) ◦ s ( b ))( x ) = s ( a )( x a ( b k ) , . . . , x k b k )= ( x k b k a , . . . , x k b k a k ) If b k =
0, this is the zero map s (( , . . . , )) . In case b k = b k a , . . . , b k a k − , so we get ( x k b k a , . . . , x k b k a k ) = s (( b k a , . . . , b k a k ))( x ) , so S isclosed.Case 4: Let a ∈ Z , b ∈ V \ Z . This gives: ( s ( , . . . , , a k ) ◦ s ( b ))( x ) = s ( , . . . , , a k )( x k b , . . . , x k b k )= ( x k b a ( a k ) , . . . , x k b k a k ) If a k = a i , i ∈ { , . . . , k } are zero preserving. If a k = a i , i ∈ { , . . . , k } are automorphisms to see that there issome nonzero b a ( a k ) , . . . , b k − a k − ( a k ) . So, we get ( x k b a ( a k ) , . . . , x k b k a k ) = s (( b a ( a k ) , . . . , b k a k ))( x ) , so S is also closed.Thus we see that ( S , ◦ ) is a semigroup.For all a ∈ F \ { } , s (( , . . . , , a )) is a bijection, thus an automorphism. Thisgives us our group G : = { s (( , . . . , , a )) | a ∈ F ∗ } . Since F is a near-field, s (( , . . . , , a )) is fpf or a =
1. For a ∈ V \ Z , s ( a ) is not surjective, thus a proper endomorphism.Also s (( , . . . , ))( x ) = ( , . . . , ) is the zero map. TIM BOYKETT AND GERHARD WENDT
For all x ∈ V , s ( x )( , . . . , , ) = x . Thus S = { s ( v ) | v ∈ V } satisfies the require-ments of Theorem 4.1, so we have an f-near-ring. Noting that G ( m ) = × . . . × × F ≤ V we have the requirements for Theorem 4.3 so we have an a-near-ring andthus an af-near-ring. (cid:3) The multiplication ∗ of Theorem 4.5 is similar to the action of near-vectorspaces, see [4] for details. We do not follow this line of discussion and possiblelinks to the near-vector space construction here.Theorem 4.5 can be efficiently used to construct examples of af-near-rings. Takea field F , k = a = id . Then ( V , + , ∗ ) as in Theorem 4.5 is an af-near-ring which is not a near-field and also not a ring. We might take a to be theFrobenius automorphism of the field F and we get another example of af-near-rings not being near-fields. We might also take k = a = id and a to be theFrobenius automorphism to get examples of af-near-rings of higher order. Also a = a can be taken. In this manner we see that we can construct af-near-rings ofall possible (finite) orders (that a finite f-near-ring must have an order of a powerof a prime follows from the next section).We can similarly take F to be a proper nearfield and use near-field automor-phisms to obtain yet more examples.To characterize when two af-near-rings are isomorphic is an open question.More examples using Theorem 4.1 and 4.3 will show up in the next section.Theorem 4.5 also allows us to explicitly construct near-rings with identity whichhave only the identity as single unit element. Take k a natural number and F = Z and a i = id for all i ∈ { , . . . , k − } . Then ( V , + , ∗ ) is a near-ring with one singleunit and it is not a ring.If each zero symmetric near-ring with identity which has only one unit is of theform as constructed by Theorem 4.5 is not known to the authors. At least we cansay that their additive groups are elementary abelian 2-groups and we can say moreabout their structure when they are 2-semisimple in Proposition 6.4. Proposition 4.6.
Let N be a zero symmetric near-ring with identity, which is theonly unit element in N. Then ( N , +) is an elementary abelian 2-group.Proof. Since ( − )( − ) = − − =
1. Thus, for any n ∈ N , n + n = ( + ) n = ( N , +) is an abelian group of exponent2, thus an elementary abelian 2-group. (cid:3) In the next section we look further at the additive group of a- and f-near-rings.5. T
HE ADDITIVE GROUP
Let ( N , +) be a group. As usual, the exponent of the group is the least commonmultiple of all the orders of the group elements. For k ∈ N , let ord ( k ) be the orderof k w.r.t. the group operation + . Then we have the following. Proposition 5.1. [6, Proposition 9.111]
Let N be a finite near-ring with identity .Let n be the exponent of ( N , +) . Then ord ( ) = n. Theorem 5.2.
Let ( N , + , ∗ ) be a finite a-near-ring. Then, ( N , +) is a group ofexponent p, ord ( ) = p, p a prime number. NITS IN NEAR-RINGS 9
Proof.
Let n be the exponent of ( N , +) . We have to show that n = p , p a primenumber. Consider the cyclic group ( h i , +) (additively) generated by 1. By as-sumption, ( h i ∪ { } ) ⊆ U ∪ { } . Let · be the usual product in the set of inte-gers and m , k be two positive integers. Let m · = + . . . + m times 1) and k · = + . . . + k times 1) be two elements of h i . By the right distributive lawin N and by using the fact that 1 is the identity we get ( m · ) ∗ ( k · ) = ( m · k ) · h i . Consequently, S : = ( h i , + , ∗ ) is a subnear-ringof N containing the identity (see also [5]), even more, it is a subnear-ring of thesubnear-field ( U ∪ { } , + , ∗ ) of N . Actually, S is a ring. It is easy to see that S isa ring isomorphic to Z n since the function y : S −→ Z n , m · [ m ] n is a near-ringisomorphism ( [ m ] n is the usual congruence modulo n ). For completeness we givea proof. y is well defined: Let m · = k · k > m . Then ( k − m ) · =
0. Since n is the additive order of 1 by Proposition 5.1, n | ( k − m ) and therefore [ k − m ] n = [ k ] n = [ m ] n and y is well defined. Furthermore, y ( m · + k · ) = y (( m + k ) · ) =[ m + k ] n = [ m ] n + [ k ] n = y ( m · )+ y ( k · ) . Also, y (( m · ) ∗ ( k · )) = y (( m · k ) · ) =[ m · k ] n = [ m ] n · [ k ] n . Consequently, y is a near-ring homomorphism. Let m · y . Then, y ( m · ) = [ m ] n = [ ] n and consequently n | m and we have m · = n is the exponent of ( N , +) . y certainly is surjective, so we see that S is isomorphic to Z n .Since S is contained in a finite near-field, every non-zero element s ∈ S has afinite multiplicative order m . But then s m − s = s m − ∈ S , so each elementin S is invertible w.r.t. multiplication. So, S is a field. Therefore, n = p for someprime p . (cid:3) The situation is even clearer for f-near-rings.
Theorem 5.3.
Let ( N , + , ∗ ) be a finite f-near-ring. Then, ( N , +) is an elementaryabelian group.Proof. Let n be the exponent of ( N , +) . If n =
2, then for all m ∈ N we have0 = ( + ) m = m + m , so ( N , +) is elementary abelian.Now we assume that n =
2. As in the proof of Theorem 5.2 we consider thesubring S , generated by the identity element 1. Let U S be the group of units of S . Since n = S is isomorphic to the ring Z n we have | U s |≥
2. Since S and N have the same identity, an element s being invertible in S is invertible in N .Therefore U S ⊆ U . By assumption, Y U does not have non-zero fixed points in theset N and consequently, Y U s does not have non-zero fixed points in S . Therefore, ( Y U s , ◦ ) is a group of fpf automorphisms of ( S , +) . Consequently, S fulfilles theassumptions of Theorem 3.1 and it follows that S is a field. Therefore, n = p forsome prime p .Since − ∈ S and 1 = −
1, we have that − ∈ U S ⊆ U . Therefore, y ( − ) ∈ Y U and by assumption y ( − ) is a fpf automorphism and the order of y ( − ) is two.Groups admitting fpf automorphisms of order 2 are known to be abelian (see forexample [3, Chapter 10, Theorem 1.4]). The proof is finished. (cid:3) From this result we know that Example 4.4 is a proper a-near-ring, because theadditive group is not abelian.While not every f-near-ring is additively closed, there is an additive closureproperty.
Corollary 5.4.
The set of units of a finite f-near-ring is a union of cyclic subgroups(all of the same prime order p) of the additive group.Proof.
Let N be a f-near-ring, u ∈ N be a unit. Note that by the proof of Theorem5.3, the subring S generated by 1 ∈ N is a prime field and thus all nonzero elementsin S are units. Then u + ... + u = ( + ... + ) u so u + ... + u is a unit, so U is closedunder the taking of cyclic subgroups. (cid:3)
6. S
PECIAL TYPES OF A - AND F - NEAR - RINGS
In this section we investigate several special classes of a- and f-near-rings.6.1.
Simple a-near-rings and f-near-rings.
We study simple and semisimple a-and f-near-rings with DCCN and see that apart from one exceptional case in theclass of a-near-rings, these are near-fields.
Theorem 6.1.
Let ( N , + , ∗ ) be a simple a-near-ring with DCCN and U its set ofunits. If N is a ring with | U | = , then N is isomorphic to a direct product of thefields Z . If N is a ring with | U | ≥ , then N is a field. If N is not a ring, then N isa near-field or N ∼ = M ( Z ) .Proof. From Theorem 3.1 and Proposition 3.2 the results concerning the ring casefollow. So we suppose that N is not a ring in the following.Due to the DCCN there exists a minimal N -subgroup M of N . Since N is as-sumed to be simple, N acts faithfully on M , so ( M ) : = { n ∈ N | ∀ m ∈ M : nm = } = { } . Due to the fact that N has an identity element, we have that Nm = { } for each non-zero element m ∈ M and so, Nm = M . Consequently, N acts 2-primitively on M (see [6, Corollary 4.47]). For m , m ∈ M we define m ∼ m iff ( m ) = ( m ) . ∼ is an equivalence relation in M (see [6, Remark4.20]). Since N has DCCN there is only a finite number of represantatives of ∼ by [6, Theorem 4.46]. Let m , m , . . . , m n ( n a natural number) be a complete setof representatives of the non-zero equivalence classes of ∼ in M . Since N has anidentity element only the zero 0 is equivalent to 0.We now assume that n >
2, so there exist two different non-zero elements m , m in M such that m m . By [6, Theorem 4.30] there is an element k ∈ N such that km = m , km = m and km i = m i for i ≥ k = m = m .Let a ∈ ( k ) : = { n ∈ N | nk = } . Thus, for each i ∈ { , . . . n } , am i = am = akm = m ∈ M \ { } . Therefore, there is an element j ∈ { , . . . , n } such that m ∼ m j . Since a ∈ ( m j ) , we now also have a ∈ ( m ) and so we must have am = m ∈ M . Since M is faithful, we have a = ( k ) = { } and consequently, the map y k is injective. Lemma 2.1 showsthat k is a unit.From km i = m i for i ≥ ( k − ) m i =
0. By assumption k − k − m i =
0. Thiscontradicts the fact that m , m , . . . , m n , n >
2, is a complete set of representativesof the non-zero equivalence classes of ∼ in M . Hence, n ≤ n =
1, so there is only one non-zero equivalence class w.r.t ∼ in M .Let m ∈ M \ { } and let a ∈ ( m ) . Then, for any other non-zero element n ∈ M we must have a ∈ ( n ) = ( m ) . Thus, aM = { } and by faithfulness of M weget that a =
0. Thus, y m is injective for any non-zero element m ∈ M . So, any m ∈ M \ { } is a unit in N by Lemma 2.1. Since NM ⊆ M we must have 1 ∈ M andtherefore N = M . Thus, N is a near-field. NITS IN NEAR-RINGS 11
Finally, suppose n = N is a 2-primitive near-ring with identity acting 2-primitively on M . Hence, by [6, Theorem 4.52] N ∼ = M G ( M ) where G : = Aut N ( M ) ∪{ } , 0 being the zero function on M and G : = Aut N ( M ) ( G may be just the set con-taining the identity map). G , the set of N -automorphisms, acts without fixed pointson M . Since ∀ m ∈ M \ { } : Nm = M , by [6, Proposition 4.21] we have that forall a , b ∈ M \ { } : a ∼ b ⇔ G ( a ) = G ( b ) . Thus, G has two non-zero orbits on M with orbit representatives e , e , say. Let f : { e , e } −→ M be a function. G acts without fixedpoints on M so by [6, Theorem 4.28, Proposition 7.8] this func-tion can be uniquely extended to a function f ∈ M G ( M ) by defining f ( ) : = f ( m ) : = g ( f ( e )) in case m = g ( e ) ∈ G ( e ) and in case m = g ( e ) ∈ G ( e ) , f ( m ) : = g ( f ( e )) .We now assume that | G | ≥ id = g ∈ G . Let h : { e , e } −→ M bethe function h ( e ) = e and h ( e ) = g ( e ) and h be its extension in M G ( M ) . Let i : { e , e } −→ M be the function i ( e ) = e and i ( e ) = g − ( e ) and i its extensionin M G ( M ) . Then, i is the inverse to f w.r.t. the near-ring multiplication (which isfunction composition) in M G ( M ) . On the other hand, ( id − h ) ∈ M G ( M ) and ( id − h )( e ) =
0. Thus, the non-zero function ( id − h ) is not injective and consequentlynot a unit in M G ( M ) . But this contradicts the assumption that N is an a-near-ring.From this contradiction it follows that | G | =
1, so G only contains the identitymap. But we have that G has two non-zero orbits in M and this implies that M is agroup with 3 elements. Hence, N ∼ = M ( Z ) . N has 9 = elements and two units,namely the identity function and the function switching 1 and 2. It is easy to seethat in this case the units with zero form a subfield of N . (cid:3) For simple f-near-rings we get similar results with similar methods.
Theorem 6.2.
Let ( N , + , ∗ ) be a simple f-near-ring with DCCN and U its set ofunits. If N is a ring with | U | = , then N is isomorphic to a direct product of thefields Z . If N is a ring with | U | ≥ , then N is a field. In case N is not a ring N isa near-field.Proof. From Theorem 3.1 and Proposition 3.2 the results concerning the ring casefollow. So, we suppose that N is not a ring in the following.As in the proof of Theorem 6.1 (by [6, Theorem 4.30]) there is an element k ∈ N such that km = m , km = m and km i = m i for i ≥
3, in case such an i ≥ m i , i ∈ { , . . . , l } ( l a natural number), are a set ofrepresentatives of the non-zero equivalence classes of ∼ in M . As in the proof ofTheorem 6.1 one gets that k is a unit in N . Also, by [6, Theorem 4.30] there is anon-zero element n ∈ N such that nm = m , nm = m , nm i = m i for i ≥
3, in casesuch an i ≥ l ≥ km = m , km = m and, if l ≥ km i = m i for i ≥ n we get the following:(1) nkm = nm = m = nm and so, nk − n ∈ ( m ) .(2) nkm = nm = m = nm and so, nk − n ∈ ( m ) .(3) for i ≥ nkm i = nm i and so, nk − n ∈ ( m i ) .Thus, for each i ∈ { , . . . , l } we have nk − n ∈ ( m i ) . By the fact that the elements m i are a full set of representatives of the non-zero equivalence classes w.r.t. ∼ weget nk − n ∈ ( M ) = { } . Thus, nk = n and since n = k is a unit we must have k =
1. Thus, m = m and this contradicts the fact that m = m . Hence we must have l =
1, so there is only one non-zero equivalence class w.r.t. ∼ in M . As done in the proof of Theorem 6.1 for the case of oneequivalence class one shows that N is a near-field. (cid:3) The results of the Theorems 6.1 and 6.2 can be extended to near-rings which are J -semisimple. For a near-ring N , J ( N ) is the intersection of all the annihilators ofthe N -groups of type 2 (see [6, Definition 5.1]) and is an ideal of the near-ring suchthat the factor near-ring N / J ( N ) becomes a subdirect product of 2-primitive near-rings. When N has DCCN this subdirect product will be a direct product of simplenear-rings by [6, Theorem 5.31]. Note that in case of a zero symmetric near-ring N with identity and DCCN we always have N = J ( N ) (see [6, Proposition 5.43]).We now study f- and a-near-rings N with DCCN and J ( N ) = { } and exclude thering case, for this situation is completely clear. Theorem 6.3.
Suppose that N is an f-near-ring or a-near-ring with DCCN whichis not a ring. If J ( N ) = { } then N is a near-field or N ∼ = M ( Z ) .Proof. Let J ( N ) = { } . By [6, Theorem 5.31] N = N / J ( N ) is a direct productof 2-primitive near-rings N i , i ∈ I , I a suitable index set and | I | = k , k a positiveinteger. By [6, Theorem 3.43], there are orthogonal idempotents e , . . . , e k suchthat e + . . . + e k =
1, each e i ∈ N i being the identity element in N i .First assume that the identity is the only unit in N . Each of the N i , i ∈ I , isa 2-primitive near-ring with DCCN and an identity element e i . Thus, for i ∈ I , N i ∼ = M G ( M ) where G : = Aut N ( M ) ∪ { } , 0 being the zero function on M and G : = Aut N ( M ) , by [6, Theorem 4.52]. Since we assume that N has a single unit, itfollows that N i has only the unit e i . The DCCN implies that G has finitely manyorbits on M , by [6, Corollary 4.59]. Let j , j , . . . , j n ( n a natural number) be acomplete set of orbit representatives of the action of G on M . Suppose that n ≥ f : { j , . . . , j n } −→ M be a function such that f ( j ) = j , f ( j ) = j and f ( j i ) = j i , else. Then, the extension f in M G ( M ) (for the construction of this extension seethe proof of Theorem 6.1 or [6, Theorem 4.28]) is again a unit in N i which is notthe identity. Consequently, we must have that G has a single orbit in M . Supposenow that | G | ≥ id = g ∈ G . Let h ( j ) = g ( j ) . Then the extension of h in M G ( M ) is a non-identity unit. This contradicts the fact that N i only has theidentity as single unit. From this we see that G = { id } and M G ( M ) = Z . Thus, N is a ring.So, we have that | U | ≥
2. Hence, there is a non-identity unit in N and so, there isa near-ring N j , j ∈ I with another unit element u j = e j . W.l.o.g. j =
1. Since N isthe direct product of the near-rings N i , i ∈ I , it is easily seen that u + e + . . . + e k is another unit in N .Now assume that N is an f-near-ring. Since N i N j = { } for i = j , for each j ≥
2, we use left distributivity over direct sums [6, Theorem 2.29] to see that e j ( u + e + . . . + e k ) = e j , so e j is a fixed point. By assumption, we now musthave e j = j ≥
2. Consequently, N = N is a 2-primitive near-ring. By [6,Corollary 4.47], N is simple. The result now follows from Theorem 6.2.Now assume that N is an a-near-ring. Let u + e + . . . + e k be a non-identityunit in N . Then ( u + e + . . . + e k ) − ( e + e + . . . + e k ) = u − e ∈ N is anotherunit in N by assumption. Since N is an ideal in N , containing the unit u − e itfollows that N = N . So, N is a 2-primitive and consequently a simple near-ring,so the result follows from Theorem 6.1. (cid:3) NITS IN NEAR-RINGS 13
The proof of Theorem 6.3 contains a result concerning near-rings with identitywith DCCN which have the identity element as their single unit.
Proposition 6.4.
Let N be a zero symmetric near-ring with identity and DCCNwhich contains only the identity element as its single unit. If J ( N ) = { } , then Nis a ring isomorphic to a direct product of the fields Z .Proof. This result follows from the proof of Theorem 6.3 where we considered thecase that a semisimple near-ring N with identity has only a single unit, where thefurther assumption that N is an a- or f-near-ring is not needed. (cid:3) The order of the group of units.
In this section we consider a situationwhere f-near-rings are a-near-rings.Let N be a finite f-near-ring and let 2 ≤ k : = | U | be the size of the group ofunits. By Theorem 5.3 we know that | N | = p n for some prime number p andnatural number n . Since U acts as a group of fpf automorphisms we must have that | U | divides | N | −
1, so k | p n − N is not a near-field. So, there must exist an ideal I in N by Theorem 6.2. Let m : = | I | . Since ( I , +) is a subgroup of ( N , +) we musthave m = p l for some natural number l < n . On the other hand, I ∗ U ⊆ I since I is an ideal of the near-ring N . This implies that U also acts as a group of fpfautomorphisms on ( I , +) . Consequently, we have that k | p l − k | gcd ( p n − , p l − ) = p gcd ( n , l ) − n is a prime we nowhave that k | p −
1. On the other hand, U contains at least p − | U | = p − Theorem 6.5.
Let N be a finite f-near-ring. We assume that N is not a near-fieldand | N | = p q , p , q being prime numbers. Then, | U | = p − and ( U ∪ { } , + , ∗ ) isa field isomorphic to Z p . So, N is an a-near-ring. Thus we can see that the near-ring constructed in Example 4.2 is a minimalexample of an f-near-ring that is not an a-near-ring. The lowest integers p q with q not prime are 2 =
16, 2 =
64 and 3 =
81. A search in SONATA showed thatno f-near-ring that is not an a-near-ring of order 2 exists. So Example 4.2 is thesmallest proper f-near-ring by “dimension” of the additive group. A proper f-near-ring of order 2 must have the group of units of order 3 or 7, as the units must alsoact fpf on a niontrivial ideal. However cyclic groups of units of these orders donot occur without being additively closed. Thus Example 4.2 is also the smallestproper f-near-ring by size.In general, the structure of U is not like in the theorem above. This followseasily from Theorem 4.5 which shows that we can construct f-near-rings of size | F k | , F a given near-field and k a natural number where the units have size | F \{ }| .In the next subsection we will construct all f-near-rings of size p which are notnear-fields, p a prime number, where clearly we have | U | = p −
1. We close thissubsection with a class of examples of af-near-rings of size p k , p a prime numberand k a natural number where we have | U | = p −
1. The examples are obtained bythe construction method of Theorems 4.1 and 4.3.
Example 6.6.
Let ( M , +) : = ( Z kp , +) be the k-dimensional direct sum of the group ( Z p , +) written as column vectors and m : = ( , . . . , , ) t . LetS : = { · a · · · ·· · · · · a k ∈ Z k × kp | ∃ i ∈ { , . . . k − } : a i = }∪ { diag ( l , . . . , l ) ∈ Z k × kp | l ∈ Z p } S is a semigroup of group endomorphisms of ( M , +) via matrix multiplication · using the field multiplication in Z p . For each element ( m , . . . , m k ) t ∈ M, there is aunique matrix s ∈ S such that ( m , . . . , m k ) t = s · m t .For elements ( m , . . . , m k ) t ∈ M and ( n , . . . , n k ) t ∈ M we define the multiplica-tion ∗ as ( n , . . . , n k ) t ∗ ( m , . . . , m k ) t = ( n , . . . , n k ) t ∗ ( s · m ) : = s · ( n , . . . , n k ) t By Theorem 4.1 ( M , + , ∗ ) is an f-near-ring, ( M , + , ∗ ) is not a near-field and theunits are the elements of the form ( , . . . , , a k ) t with non-zero a k , so there are p − many units in the near-ring. Also Gm = h m i ≤ ( M , +) so we have an af-near-ring. The example given in Example 6.6 can also be derived from Theorem 4.5. Wewill point this out, using the notation of Theorem 4.5. Take the field ( Z p , + , ∗ ) ,form the k -dimensional sum of ( Z p , +) to get the group ( V , +) . For i ∈ { , . . . , k } let a i be the identity map. Then, ( V , + , ∗ ) as constructed in Theorem 4.5 gives thesame near-ring as constructed in Example 6.6.6.3. Near-rings of order p . We study f-near-rings N of size p that are notnearfields. From Theorem 6.5 we know that this implies they are a-near-rings.We will see that the construction of Theorem 4.5 gives all such near-rings. Theorem 6.7.
Let ( N , + , ∗ ) be an f-near-ring of order p which is not a near-field.Then N is also an a-near-ring . Moreover, there exists a group ( V , +) isomorphic to ( N , +) and a map a : U → U with a ( ) = and a an automorphism of the multi-plicative group ( U , ∗ ) such that ( N , + , ∗ ) is isomorphic to the near-ring ( V , + , ∗ ) ,where the near-ring operations are defined as in Theorem 4.5.Proof. That U is a subfield of the near-ring of order p follows from Theorem6.5. According to Theorem 6.3, J ( N ) = { } and since 1 ∈ N , J ( N ) = N by [6,Proposition 5.43]. Let n ∈ N \ U . Then, Nn = N is an N -group of size p . Since N has an identity element, for any non-zero element mn ∈ Nn we have { } 6 = Nmn ⊆ Nn . But Nn is a group of order p , so we must have Nmn = Nn and consequentlythe N -group Nn is of type 2. Since J ( N ) is the intersection of the annihilators ofall N -groups of type 2 of a near-ring N (see [6, Definition 5.1]), J ( N ) Nn = { } .Since 1 ∈ N we see that J ( N ) n = { } . Also, we have that N = J ( N ) + U and J ( N ) ∩ U =
0. Consequently, for each n ∈ N there is a unique j ∈ J ( N ) and aunique u ∈ U such that n = j + u . Let u ∈ U . Since U is a group of order p , u isadditively generated by the identity element, so u = + . . . +
1, with k summands,say. We will use the notation u = k · k and an element n ∈ N , then k · n will mean n + . . . + n with k summands.The product of two natural numbers m , m will be also denoted by m · m .Since J is a group of order p , the groups J and U are isomorphic by a groupisomorphism y . We let y ( J ) = U . Note that the near-ring multiplication ∗ when NITS IN NEAR-RINGS 15 restricted to the units of the near-ring is a field multiplication because ( U , + , ∗ ) isa field.Let u ∈ U and 0 = y − ( ) ∈ J . Then, 0 = y − ( ) ∗ u because u is a unit. Also, J ( N ) is an ideal of the near-ring and therefore, y − ( ) ∗ u ∈ J ( N ) . Since J ( N ) is a group of order p , any element in J ( N ) is additively generated by y − ( ) .So, there is a smallest natural number d u ∈ { , . . . , p − } such that y − ( ) ∗ u = y − ( ) + . . . + y − ( ) = d u · y − ( ) . Now consider that map a : U → U , , = u d u · a is well defined. Note that for a unit u we cannot have a ( u ) = u , u ∈ U and suppose that a ( u ) = a ( u ) . Then, y − ( ) ∗ u = y − ( ) ∗ u and by fixedpointfreeness of the action of the units we get u = u . So, a is injectiveand consequently by finiteness bijective. We now show that a is a multiplicativehomomorphism of ( U , ∗ ) . To this end let u , u ∈ U . Then, by associativity of thenear-ring multiplication we have ( y − ( ) ∗ u ) ∗ u = y − ( ) ∗ ( u ∗ u ) . ( y − ( ) ∗ u ) ∗ u = ( d u · y − ( )) ∗ u = ( y − ( ) + . . . + y − ( )) ∗ u with d u summands. y − ( ) ∗ u = d u · y − ( ) (here we have d u many summands of y − ( ) ). Usingright distributivity and the fact that J ( N ) has order p we get ( y − ( ) ∗ u ) ∗ u =( d u · d u ( mod p )) · y − ( ) . On the other hand, y − ( ) ∗ ( u ∗ u ) = d u ∗ u · y − ( ) and from that we see that d u ∗ u = d u · d u ( mod p ) . But this precisely means that a is a multiplicative homomorphism. So, a is a zero preserving automorphism of themultiplicative group ( U , ∗ ) , as required in the construction of Theorem 4.5.We now will show that w.r.t. the field multiplication ( U , ∗ ) , N is isomorphic to anear-ring as constructed in Theorem 4.5. To this end let ( V , + , ∗ ) be a near-ring asconstructed in Theorem 4.5, where ( V , +) = ( U × U , +) , + the field addition in ( U , + , ∗ ) and multiplication ∗ defined as ( x , x ) ∗ ( , a ) : = ( x ∗ a ( a ) , x ∗ a ) and if a = ( x , x ) ∗ ( a , a ) : = ( x ∗ a , x ∗ a ) .We now claim that the map b : N → V , j + u ( y ( j ) , u ) is a near-ring iso-morphism. Since y ( J ) = U and ( N , +) is the direct sum of the groups ( J , +) and ( U , +) it is clear that b is an isomorphism of groups, so it remains to showthat b is a multiplicative map. We distinguish two cases. First we calculate b (( j + u ) ∗ ( j + u )) with j =
0. Let n = j + u an arbitrary element in N and n = j + u a non-unit in N , thus, j =
0. Then, using right distributivity ofthe near-ring, n n = j ( j + u ) + u ( j + u ) . j ( j + u ) = n is a non-unit. Again by right distributivity and using that 1 is the identity of the near-ringand u = k · u ( j + u ) = ( + . . . + )( j + u ) = ( j + u ) + . . . + ( j + u ) = k · ( j + u ) . Since ( N , +) is an abelian group (by Theorem 5.3, anyhow any groupof order p is abelian), this is k · j + k · u = ( + . . . + ) j + ( + . . . + ) u = u j + u u . Thus we have that the units distribute from the left hand side and n n = ( j + u )( j + u ) = u n = u j + u u in case n is a non-zero non-unit.Thus, b (( j + u ) ∗ ( j + u )) = b ( u ∗ j + u ∗ u ) . Since b is an additive ho-momorphism, this gives b ( u ∗ j ) + b ( u ∗ u ) = ( y ( u ∗ j ) , ) + ( , u ∗ u ) . Let u = k ·
1. Then the last expression equals ( y ( k · j ) , ) + ( , u ∗ u ) . Now y is anadditive homomorphism and k · j = j + . . . + j with k many summands. So wehave b (( j + u ) ∗ ( j + u )) = ( k · y ( j ) , ) + ( , u ∗ u ) .On the other hand, b ( j + u ) ∗ b ( j + u ) = ( y ( j ) , u ) ∗ ( y ( j ) , u ) = ( u ∗ y ( j ) , u ∗ u ) = (( k · ) ∗ y ( j ) , u ∗ u ) = ( k · y ( j ) , u ∗ u ) . So, in case j = b is a multiplicative map.Now consider the second case, namely b (( j + u ) ∗ u ) = b ( j ∗ u + u ∗ u ) =( y ( j ∗ u ) , u ∗ u ) . On the other hand, b ( j + u ) ∗ b ( u ) = ( y ( j ) , u ) ∗ ( , u ) = ( y ( j ) ∗ a ( u ) , u ∗ u ) . In order to show that these expressions are the same, wehave to compare the elements y ( j ∗ u ) and y ( j ) ∗ a ( u ) and show that they areequal. Since J is a cyclic group of order p , j is a sum of the elements y − ( ) , j = l · y − ( ) , say with l ∈ { , . . . , p − } . So, j ∗ u = ( l · y − ( )) ∗ u = d u · ( l · y − ( )) = ( l · d u ) · y − ( ) . Since y is a homomorphism we get that y ( j ∗ u ) = y (( l · d u ) · y − ( )) = ( l · d u ) · y ( y − ( )) = ( l · d u ) ·
1. By definitionof a , a ( u ) = d u ·
1. So, y ( j ) ∗ a ( u ) = y ( l · y − ( )) ∗ ( d u · ) = ( l · y ( y − ( )) ∗ ( d u · ) , using that y is an additive homomorphism. This gives ( l · d u ) ·
1. So, in-deed y ( j ∗ u ) = y ( j ) ∗ a ( u ) and this proves that b is an isomorphism. (cid:3) Hence we see that the construction method of Theorem 4.5 gives us all f-near-rings of order p , p a prime number.7. C ONCLUSION
In this paper we have looked at the situation for near-rings with their units obey-ing one or the other of two special properties: acting fixed point freely (f-near-rings) and being additively closed (including the zero of the near-ring) (a-near-rings). For rings with DCCL the situation is clear: all such rings are fields. Itremains open whether infinite a-rings or f-rings exist that are not fields. For propernear-rings, we have seen that more complex and interesting situations can arise.We have investigated a-near-rings only in the case that the group of units isnontrivial. We have seen that a-rings with trivial group of units are a direct productof copies of Z . It remains open what a-near-rings exist with trivial unit group andwhat can be said about them more than that their additive groups are elementaryabelian 2-groups.We have seen that the additive group of a-near-rings are of prime exponent,while f-near-ring additive groups are elementary abelian. It is shown that in manycases, for instance when the order of the nearring is p q for primes p , q , the unitsform a prime field. Examples have been constructed to show that all near-fieldscan arise as the units of an af-near-ring. On the other hand, it is unclear whether allfixedpointfree automorphism groups can appear as the units in an f-near-ring.It has been shown that f-near-rings which are not near-fields are nonsimple andnot J -semisimple. It remains open what special structures the J radical possesses.We have seen that the radical has an important role in determining the structure of f-nearrings of order p . It is an open question to the authors if the Jacobson radical oftype 2 in the type of near-rings we study is always nilpotent and if the constructionmethod of Theorem 4.5 also gives all af-near-rings of some higher prime powerorder. The question of deciding when f-near-rings constructed by Theorems 4.1,4.5 are isomorphic would be an interesting problem to consider.8. A CKNOWLEDGEMENTS
We would like to thank Prof. G ¨unter Pilz and our colleagues Wen-Fong Ke andPeter Mayr for input during the process of preparing this paper.R
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NSTITUT F ¨ UR A LGEBRA , J
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