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LINEAR RIGHT IDEAL NEARRINGS
KENNETH D. MAGILL JR.
(Received 16 February 2001)
Abstract.We determine, up to isomorphism, all those topological nearringsᏺn whose
additive groups are then-dimensional Euclidean groups,n >1, and which containn one-dimensional linear subspaces{Ji}n
i=1which are also right ideals of the nearring satisfying several additional properties. Specifically, for each w∈ᏺn, we require that there exist wi∈Ji, 1≤i≤n, such thatw=w1+w2+ ··· +wnand multiplication on the left ofw
yields the same result as multiplication by the same element on the left ofwn. That is, vw=vwnfor eachv∈ᏺ
n.
2000 Mathematics Subject Classification. 16Y30, 54H13.
1. Introduction. A topological nearringᏺhere is a triplet(N,+,·)where(N,+)is a (not necessarily commutative) topological group,(N,·)is a topological semigroup and the following right distributive law holds:
(u+v)w=uw+vw ∀u, v, w∈N. (1.1)
In particular, our nearrings are right nearrings. For information about abstract near-rings, one should consult [2,4,5]. Ann-dimensional Euclidean nearring is any topolog-ical nearring whose additive group is then-dimensional Euclidean groupRn. The main result of this paper is inSection 2where we determine, up to isomorphism, forn >1, alln-dimensional Euclidean nearringsᏺn which containndistinct one-dimensional linear subspaces{Ji}n
i=1which are also right ideals ofᏺnsuch that for eachw∈ᏺn, there existwi∈J
i, 1≤i≤n, such thatw=w1+w2+···+wnandvw=vwnfor all
v∈ᏺn. As a corollary to the latter result, we obtain the fact that a two-dimensional Euclidean nearringᏺ2has two distinct nonzero proper closed, connected right ideals J1 and J2 such that for everyw∈ᏺ2, there existw1∈J1 and w2∈J2 such that w=w1+w2andvw=vw2for allv∈ᏺ
2if and only ifᏺ2is isomorphic to one of the seven two-dimensional Euclidean nearrings whose multiplications follow:
(1) vw=vfor allv, w∈ᏺ2, (2) vw=0 for allv, w∈ᏺ2, (3) vw=(v1,0)for allv, w∈ᏺ2, (4) vw=(0, v2)for allv, w∈ᏺ2, (5) vw=(v1|w2|r, v2w2)
(6) vw=
v1w2r, v2w2
forw2≥0,
(7) vw=
v1aw2r, av2w2 forw2≥0,
v1bw2r, bv2w2 forw2>0, wherer >0 in each case,a≤0,b≥0, anda2+b2≠0.
2. The main theorem. An elementw∈Rnwill be represented as a row vector, that is,w=(w1, w2, . . . , wn)and it will be assumed throughout the remainder of the paper thatn >1. Forwi∈J
iwhereJiis any subset ofᏺn, we havewi=(w1i, w2i, . . . , wni). Furthermore, for allv, w∈ᏺn, theith coordinate of the productvwwill be denoted by(vw)i. In general, the function which maps all of a spaceXinto a single pointx will be denoted byx. Its domain will be apparent from context. Finally, forr >0, we define 0r=0.
Definition2.1. Alinear right ideal nearringis anyn-dimensional Euclidean
near-ring which containsndistinct right ideals{Ji}ni=1, each of which is a one-dimensional linear subspace ofRn, such that for allw∈ᏺ
nthere existwi∈Ji, 1≤i≤n, such thatw=w1+w2+···+wnandvw=vwnfor allv∈ᏺ
n.
Theorem2.2. Ann-dimensional Euclidean nearringᏺnis a linear right ideal
near-ring if and only ifᏺnis isomorphic to one of the four types of nearrings whose multi-plications follow:
(vw)i=0 or (vw)i=vi for1≤i≤n, (2.1)
(vw)i=viwnri fori≠n, (vw)n=vnwn, (2.2)
whereri>0for1≤i < n,
(vw)i=
vi
wn ri
forwn≥0, −viwnri forwn<0,
(2.3)
fori≠nand(vw)n=vnwnwhereri>0,
(vw)i=
vi
awn
ri forw
n≤0, i≠n,
vi
bwn ri
forwn>0, i≠n,
(vw)n=
avnwn forwn≤0,
bvnwn forwn>0,
(2.4)
whereri>0,a≤0,b≥0, anda2+b2≠0.
It is convenient to prove Theorem 2.2via a sequence of lemmas. Recall that the constant function which maps all ofRninto the real numberr is denoted byrand letKi= {v∈ᏺn:vj=0 forj≠i}.
Lemma2.3. The nearringᏺnis ann-dimensional Euclidean nearring such that each
Ki,1≤i≤n, is a right ideal ofᏺnandvw=v(0,0, . . . ,0, wn)for allv, w∈ᏺnif and only if there existncontinuous selfmaps{fi}ni=1of the real numbersRsuch that
and the multiplication onᏺnis given by
(vw)i=vifi
wn
for1≤i≤n. (2.6)
Proof. Suppose first that there existncontinuous selfmaps ofRsatisfying (2.5)
and the multiplication in ᏺn is given by (2.6). We must show that multiplication is associative and that eachKiis a right ideal ofᏺn. For 1≤i≤n, we have
(uv)wi=(uv)ifi
wn
=uifi vn fi wn
=uifi
vnfi
wn
=uifi
(vw)n
=u(vw)i (2.7)
and we see that multiplication is associative. It follows immediately from (2.6) that eachKiis a right ideal ofᏺn.
Suppose, conversely, thatᏺnis ann-dimensional Euclidean nearring such that each
Ki, 1≤i≤n, is a right ideal ofᏺn andvw=v(0,0, . . . ,0, wn)for allv, w∈ᏺn. We must show that there exist n continuous selfmaps {fi}n
i=1 of the real numbersR such that (2.5) and (2.6) are satisfied. According to [3, Theorem 2.10] there existn2 continuous functions{gij}, 1≤i, j≤n, fromRntoRsuch that
(vw)i= n
j=1
vjgij(w) ∀v, w∈Rn. (2.8)
Leteibe a vector such thatei
j=0 forj≠iandeii=1. From (2.8), we get
ekw i=
n
j=1
ekjgij(w)=gik(w). (2.9)
SinceJkis a right ideal ofᏺn, we haveekw∈Jkfor allw∈ᏺn. It follows from this and (2.9) thatgik= 0fork≠iand this, together with (2.8), implies that
(vw)i=vigii(w) ∀v, w∈Rn,1≤i≤n. (2.10)
For 1≤i≤n, define a continuous selfmapfibyfi(x)=gii(0,0, . . . ,0, x). Sincevw=
v(0,0, . . . ,0, wn), it follows from (2.10) that
(vw)i=v
0,0, . . . ,0, wn
=vigii
0,0, . . . ,0, wn
=vifi
wn
(2.11)
for 1≤i≤nand we see that the multiplication onᏺn is, indeed, given by (2.6). It follows from (2.11) that
u(vw)i=uifi
(vw)n
=uifi
vnfn
wn
,
(uv)wi=(uv)ifi
wn
=uifi vn fi wn , (2.12)
for allu, v, w ∈Rn. Letu
i=1,vn=x, andwn=y in (2.12) and, sinceu(vw)=
(uv)w, conclude that
fi
xfn(y)
=fi(x)fi(y) ∀x, y∈R,1≤i≤n, (2.13)
Our next task is to characterize those functions which satisfy (2.5). Takei=nin (2.5) and getfn(x, fn(y))=fn(x)fn(y)for allx, y∈Rwhich implies that
fn(r x)=r fn(x) ∀x∈R, ∀r∈Ran
fn
, (2.14)
where Ran(fn)denotes the range offn. Such a function was referred to in [3] as a semilinear map and it was shown in [3, Theorem 3.3] that eitherfn= 0,fn= 1, or Ran(fn)=R, or Ran(fn)=R+whereR+denotes the nonnegative reals.
Lemma2.4. Supposefn= 0. Then (2.5) is satisfied if and only if eitherfi= 0or
fi= 1for1≤i < n.
Proof. Suppose (2.5) is satisfied. Thenfi(x)fi(y)=fi(0)for allx, y∈R. Thus
(fi(0))2=fi(0)which means that eitherfi(0)=0 orfi(0)=1. Since(fi(x))2=fi(0) for allx∈R, it readily follows that eitherfi= 0orfi= 1for 1≤i≤n.
One easily verifies that if eitherfi= 0 or fi= 1 for 1≤i < n, then (2.5) is satisfied and, with this observation, the proof is complete.
The proof ofLemma 2.5is similar to the proof ofLemma 2.4and, for that reason, is omitted.
Lemma2.5. Supposefn= 1. Then (2.5) is satisfied if and only if eitherfi= 0or
fi= 1for1≤i < n.
Lemma2.6. Lethbe a continuous selfmap ofRand suppose that
h(x+y+b)=h(x)+h(y) ∀x, y∈R, (2.15)
wherebis any fixed real number. Then there exists a real numberrsuch thath(x)=
r x+r bfor allx∈R.
Proof. We haveh(x+z+b)=h(x)+h(z)for allx, z∈R. Lets= −bandz=y+s
and the latter transforms into
h(x+y)=h(x)+h(y+s) ∀x, y∈R. (2.16)
Leth(2s)=tand it then follows from (2.16) thath(x+s)=h(x)+tor, equivalently, h(y+s)=h(y)+tfor ally∈R. We can now rewrite (2.16) as
h(x+y)=h(x)+h(y)+t ∀x, y∈R. (2.17)
One uses induction on (2.17) to show that
h(mx)=mh(x)+(m−1)t. (2.18)
For any nonzero integern, letx=1/nin (2.18) and get
h m n
=mh 1
n
+(m−1)t. (2.19)
Letm=1 andh(1)=ain (2.19) and get
h 1 n
From (2.19) and (2.20), we get
h m
n
=m
n
a+(1−n)t=m
n(a+t)−t. (2.21)
Letr =a+t and conclude from (2.21) thath(x)=r x−tfor all rational numbers. Sincehis continuous, we conclude that
h(x)=r x−t ∀x∈R. (2.22)
We use the latter fact and (2.15) to get
r x+r y+r b−t=h(x+y+b)=h(x)+h(y)=r x−t+r y−t (2.23)
for allx, y∈Rwhich implies thatr b= −t. It follows from this and (2.22) thath(x)=
r x+r bfor allx∈Rand the proof is complete.
Recall that for any positive numberr, we define 0r=0.
Lemma2.7. Letabe a nonzero real number and lethbe a nonconstant continuous
selfmap ofR. Ifa <0and
h(axy)=h(x)h(y) ∀x∈R, ∀y≤0, (2.24)
then there exists a positive real numberrsuch that
h(x)=(ax)r ∀x≤0. (2.25)
Ifa >0and
h(axy)=h(x)h(y) ∀x∈R, ∀y≥0, (2.26)
then there exists a positive real numberrsuch that
h(x)=(ax)r, ∀x≥0. (2.27)
Proof. Suppose first thata <0 andhsatisfies (2.24). Suppose further thath(y)=
0 for somey <0. Then (2.24) implies thath(axy)=0 for allx∈Rwhich results in the contradiction thath= 0. Consequently,h(y)≠0 whenever y <0. It also follows from (2.24) thath(ax2)=(h(x))2for allx <0. Since every negative number is of the formax2 for appropriatex, we conclude thath(−∞,0)⊆(0,∞). Define a homeomorphismϕfromRonto(−∞,0)byϕ(x)= −exand define a homeomorphism
ψfrom(0,∞)ontoRbyψ(x)=lnx and letα=ψ◦h◦ϕ. In view of our previous observations,αis a continuous selfmap ofR. Letb=ln(−a). Then
α(x+y+b)=ψh−ex+y+b=ψh−eb−ex−ey
=ψha−ex−ey=ψh−exh−ey
=ψh−ex+ψh−ey=α(x)+α(y)
(2.28)
for allx, y∈Rand it follows fromLemma 2.5thatα(x)=r x+r bfor somer∈R. Now,h=ψ−1◦αϕ−1andψ−1(x)=ex whileϕ−1(x)=ln(−x). Lets=er b and for
x <0, we have
h(x)=ψ−1◦αln(−x)=ψ−1rln(−x)+r b
=ψ−1rln|x|+r b=er beln|x|r
Evidentlys >0. Supposer <0. Then
lim
x→0−h(x)=xlim→0−s|x|
r= ∞, (2.30)
which contradicts the continuity of h at zero. Thus, r ≥ 0. Suppose r =0. Then h(x)=sfor allx≤0. Letx=y= −1 in (2.24) and gets=ha(−1)(−1)=h(−1)2= s2. This means thats=1 sinces >0 and we now haveh(x)=1 for allx≤0. Next, letx=1 and lety <0 in (2.24). Thenh(ay)=h(a·1·y)=h(1)h(y)=h(1). Since every positive number is of the formay for some negative numbery, we conclude thath(x)=h(1)for allx >0. Thus, we haveh(x)=1 for allx∈Rbut this contra-dicts the fact thathis a nonconstant map. Hence, we must conclude thatr≠0 which meansr >0. Takex=y= −1 in (2.24) and use (2.29) to get
s|a|r=h(a)=ha(−1)(−1)=h(−1)2
=s2, (2.31)
which implies thats= |a|r. It follows from this fact and (2.29) that (2.25) holds.
The case wherea >0 is quite similar to the previous case so we will be content with making a few remarks and omitting most of the details. Supposeh(x)=0 for somex >0. Then, for ally >0, we haveh(axy)=h(x)h(y)=0 and we conclude that h(x)=0 for allx≥0. Next, letx <0 and we have 0=h(ax2)=h(x)h(x). This means thath(x)=0 for allxwhich contradicts the fact thath is nonconstant. Therefore, h(x)≠0 forx >0. Moreover, for anyx >0, (2.26) assures thath(ax2)=(h(x))2and it follows thath(0,∞)⊆h(0,∞). This time, define a homeomorphismϕfromRonto (0,∞)byϕ(x)=ex and a homeomorphismψ from(0,∞)ontoR byψ(x)=lnx. Again letα=ψ◦h◦ϕand one easily verifies thatα(x+y+b)=α(x)+α(y)for allx, y∈Rwhereb=lna. As before, it follows fromLemma 2.6thatψ◦h◦ϕ(x)=
α(x)=r x+r b for somer∈R. With a few minor modifications of the arguments used in the previous case, one shows that the latter implies that (2.27) holds.
Lemma2.8. Leta≠0and lethbe a nonconstant continuous selfmap ofR. Thenh
satisfies the functional equation
h(axy)=h(x)h(y) ∀x, y∈R (2.32)
if and only if there exists a positive numberr such that either
h(x)= |ax|r ∀x∈R, (2.33)
or
h(x)=
(ax)r forax≥0,
−|ax|r forax <0. (2.34)
Proof. Consider first the case wherea <0. It follows fromLemma 2.7that
h(x)=(ax)r ∀x≤0. (2.35)
Next, letx >0 andy=1 in (2.32). Thenax <0 and we appeal to (2.32) and (2.35) to get
Nowh(1)≠0 since(a2x)r ≠0 wheneverx≠0 so we letp=(a2)r/h(1)and from (2.36), we conclude thath(x)=pxr forx >0. Takex=y=1 in (2.32) and since
a <0, we get
|a|2r=a2r
=h(a)=h(a·1·1)=h(1)2. (2.37)
It follows from (2.37) that eitherh(1)= |a|r orh(1)= −|a|r.
Case1(h(1)= |a|r). Thenp=(a2)r /h(1)=(|a|r)2/|a|r= |a|r and we conclude
thath(x)=pxr = |a|rxr = |ax|r for allx >0. It follows from this and (2.35) that
h(x)= |ax|rfor allx∈R. That is, the functionhis given by (2.33).
Case2(h(1)= −|a|r). Thenp=(a2)r/h(1)=(|a|r)2/− |a|r= −|a|r and we see
thath(x)=pxr= −|a|rxr= −|ax|r forx >0. Sincea <0, it follows from this fact and (2.35) that, in this case, the functionh is given by (2.34) and this verifies the lemma for the case wherea <0. The case wherea >0 is quite similar to the previous case and for that reason we omit the details.
Lemma2.9. Letabe a nonpositive real number andba nonnegative real number
such thata2+b2≠0and lethbe a nonconstant continuous selfmap ofR. Then
h(axy)=h(x)h(y) ∀x∈R, y≤0, (2.38)
h(bxy)=h(x)h(y) ∀x∈R, y≥0, (2.39)
if and only if there exists anr >0such that
h(x)=
(ax)r forx≤0,
(bx)r forx≥0. (2.40)
Proof. Suppose that (2.38) and (2.39) are both satisfied. We first consider the case
wherea≠0≠b. It then follows immediately fromLemma 2.7that there exist positive real numbersrandssuch that
h(x)=
(ax)r forx≤0,
(bx)s forx≥0. (2.41)
Takex=1 andy= −1 and it follows from (2.38) and (2.41) that
(−a)sbs=(−ab)s=f (−a)=fa·1·(−1)=f (1)f (−1)=bs(−a)r (2.42)
which readily implies that(−a)s=(−a)r which, in turn, implies thatr=s. Thus,h is given by (2.40). It is a routine matter to check that ifhis given by (2.40), then it satisfies both (2.38) and (2.39) and we omit the details.
We next consider the case wherehsatisfies (2.38) and (2.39) anda=0. Thenb≠0 and it follows from (2.39) andLemma 2.7that there exists a positive numberr such thath(x)=(bx)r for allx≥0. It follows from (2.38) thath(0)=h(x)h(y)for all
Consequently, we must haveh(0)=0 which meansh(x)h(y)=0 for allx∈Rand ally≤0. Letx=1 and getbrh(y)=0 for ally≤0. Thus,h(y)=0 for ally≤0 and we have shown thatf is given by (2.40) in this case also. It is immediate that ifh satisfies (2.40) anda=0, thenhsatisfies both (2.38) and (2.39). The case whereb=0 is very similar to the preceding case and we say no more about it.
Theorem2.10. The functionsfi(1≤i≤n)satisfy (2.5) if and only if
fi= 0 or fi= 1 for1≤i≤n, (2.43)
or there exist a nonzero numbera and positive numbers ri (1≤ i < n)such that
fn(x)=axand eitherfi(x)= |ax|ri or
fi(x)=
(ax)ri forax≥0,
−|ax|ri forax <0, (2.44)
or there exist ana≤0, ab≥0witha2+b2≠0andr
i>0(1≤i < n)such that
fn(x)=
ax forx≤0,
bx forx >0, (2.45)
fi(x)=
(ax)ri forx≤0,
(bx)ri forx >0, (2.46)
for1≤i < n.
Proof. It is immediate that the functions{fi}n
i=1satisfy (2.5) if (2.43) holds and it is a routine matter to show that if fn(x)=ax for some a≠0 and there exist
n−1 positive numbers{ri}n−1
i=1 such that eitherfi(x)= |ax|rifor 1≤i < norfi(x)is given by (2.44), then the functions{fi}n
i=1satisfy (2.5). Similarly, it follows in a routine manner that iffnandfi(1≤i < n)are given by (2.45) and (2.46), respectively, then the functions{fi}n
i=1again satisfy (2.5).
Suppose, conversely, that the functions{fi}n
i=1 satisfy (2.5). We must show that they must be one of the families described in the statement of the theorem. First of all, we observed following the proof ofLemma 2.3that by taking i=nin (2.5), we getfn(r x)=r fn(x)for all x∈Rand allr∈Ran(fn)and it was shown in [3, Theorem 3.3] that eitherfn= 0,fn= 1, or Ran(fn)=Ror Ran(fn)=R+.
Case3(fn= 0orfn= 1). It follows from Lemmas2.4and2.5thatfi= 0or
fi= 1for 1≤i≤nand thus, (2.43) is satisfied.
Case4(Ran(fn)=R). In this case, we havefn(xy)=xfn(y)for allx, y∈R. Let
y=1, letf (1)=aand conclude thatfn(x)=axfor allx∈R. Evidently,a≠0 since Ran(fn)=R. Since (2.5) is satisfied, we havefi(axy)=fi(x)fi(y)for allx, y∈R and it follows fromLemma 2.8that there exists a positive numberrisuch that either
fi(x)= |ax|ri for allx∈Ror
fi(x)=
(ax)ri forax≥0,
Case5(Ran(fn)=R+). In this case, we havefn(xy)=xfn(y)for allx≥0 and
ally∈R. Takey=1 andb=fn(1)and conclude thatfn(x)=bxforx≥0. Next, letx <0, takey= −1, leta= −fn(−1)and conclude thatfn(x)=fn((−x)(−1))= −xfn(−1)=ax. Evidently,a≤0,b≥0, anda2+b2≠0 since Ran(fn)=R+. In this case,fnsatisfies (2.45) and it follows from (2.5) thatfi(axy)=fi(x)fi(y)fory≤0 andfi(bxy)=fi(x)fi(y)fory≤0. This, in view ofLemma 2.9, implies that eachfi, 1≤i < n, is given by (2.46). This completes the proof.
Our next result is an immediate consequence of Lemma 2.3and Theorem 2.10. Recall thatKi= {v∈Rn:vj=0 forj≠i}.
Theorem2.11. Letᏺnbe ann-dimensional Euclidean nearring. Then eachKi,1≤
i≤n, is a right ideal ofᏺnandvw=v(0,0, . . . ,0, wn)if and only if the multiplication inᏺnis given by one of the following:
(vw)i=0 or (vw)i=vi, for1≤i≤n, (2.48)
(vw)i=viawnri fori≠n, (vw)n=avnwn, (2.49)
wherea≠0andri>0for eachi,
(vw)i=
vi
awn
ri foraw
n≥0, −viawn
ri foraw n<0,
(2.50)
fori≠nand(vw)n=avnwnwherea≠0andri>0,
(vw)i=
viawnri forwn≤0, i≠n,
vi
bwn ri
forwn>0, i≠n,
(vw)n=
avnwn forwn≤0,
bvnwn forwn>0,
(2.51)
whereri>0,a≤0,b≥0, anda2+b2≠0.
Proof ofTheorem2.2. Suppose first thatᏺn is a Euclidean nearring which is
isomorphic to a nearringᏺn1whose multiplication is given by any one of (2.1), (2.2), (2.3), and (2.4) inclusive. It is immediate that eachKi, 1≤i≤nis a right ideal ofᏺn1. For eachw∈ᏺn1, letwibe the vector such thatwii=wiandwji=0 forj≠i. Then
wi∈K
i,w=w1+w2+ ··· +wn, andvw=vwnfor allv∈ᏺn1. That is,ᏺn1is a linear right ideal nearring and so isᏺnsince any isomorphism fromᏺn1ontoᏺnis a linear automorphism ofRn.
Suppose, conversely, thatᏺnis a linear right ideal nearring with linear right ideals
ϕ(v)ϕ(w)=ϕ(wn). Sincew=ϕ−1(w1)+ϕ−1(w2)+···+ϕ−1(wn)andϕ−1(wi)∈
Kifor 1≤i≤n, it readily follows that(ϕ−1(wi))i=wifor eachiand(ϕ−1(wi))j=0 forj≠i. Consequently, all this implies that
v∗w=ϕ−1ϕ(v)ϕ(w)=v∗ϕ−1wn=v∗0,0, . . . ,0, w n
(2.52)
and it follows fromTheorem 2.11that the multiplication in ᏺn1 is given by one of (2.48) to (2.51). The proof will be complete once we show that the nearring whose multiplication is given by (2.49) is isomorphic to the one whose multiplication is given by (2.2) and the nearring whose multiplication is given by (2.50) is isomorphic to the nearring whose multiplication is given by (2.3). Denote by ᏺn1 the nearring whose multiplication is given by (2.2) and denote byᏺn2the nearring whose multiplication is given by (2.49). The multiplication in both cases will be denoted by juxtaposition. Sincea≠0, we are able to define a bijectionϕfromᏺn1ontoᏺn2by
ϕ(v)i= vi
|a|ri for 1≤i < n,
ϕ(v)n=vn
a . (2.53)
It is immediate thatϕis an additive isomorphism. As for multiplication, for 1≤i < n we have
ϕ(vw)i=(vw)i |a|ri =
viwnri
|a|ri , (2.54)
ϕ(v)ϕ(w)i=ϕ(v)iaϕ(w)nri= vi |a|ri
awn
a
ri
=viwnri
|a|ri . (2.55) In addition to this, we have
ϕ(vw)n=vnwn
a =a
ϕ(v)nϕ(w)n=ϕ(v)ϕ(w)n (2.56)
and it follows from (2.54), (2.55), and (2.56) thatϕ is an isomorphism from ᏺn1 ontoᏺn2.
This time, letᏺn1be a nearring whose multiplication is given by (2.3) and letᏺn2 be a nearring whose multiplication is given by (2.50). Again, define a bijection from ᏺn1ontoᏺn2 just as in (2.53). Here, there are a number of cases to consider for a productvwdepending upon whethera <0,a >0,wn<0, orwn≥0. Consider the case wherea <0 andwn<0. It then follows that for 1≤i < n, we have
ϕ(vw)i=
(vw)i |a|ri = −
viwn ri
|a|ri . (2.57)
Sincea(ϕ(w))n=wn<0, we also have
ϕ(v)ϕ(w)i= −
ϕ(v)ia
ϕ(w)n
ri= −viwn ri
|a|ri . (2.58) In addition, we have
ϕ(vw)n=
(vw)n
a =
vnwn
a =a
ϕ(v)n
ϕ(w)n=
ϕ(v)ϕ(w)n, (2.59)
Corollary2.12. A two-dimensional Euclidean nearringᏺ2has two distinct nonzero
proper closed, connected right idealsJ1andJ2such that for everyw∈ᏺ2there exist w1∈J1andw2∈J2such thatw=w1+w2andvw=vw2 for allv∈ᏺ
2 if and only ifᏺ2is isomorphic to one of the seven two-dimensional Euclidean nearrings whose multiplications follow:
vw=v ∀v, w∈ᏺ2, vw=(0,0) ∀v, w∈ᏺ2, vw=v1,0 ∀v, w∈ᏺ2, vw=0, v2 ∀v, w∈ᏺ2,
vw=v1w2r, v2w2 ∀v, w∈ᏺ2,
vw=
v1w2r, v2w2
forw2≥0,
−v1w2r, v2w2 forw2<0,
vw=
v1aw2r, av2w2 forw2≥0,
v1bw2r, bv2w2 forw2>0,
(2.60)
wherer >0in each case,a≤0,b≥0, anda2+b2≠0.
Proof. Suppose that the multiplication on a two-dimensional Euclidean nearringᏺ
is given by any one of (2.60) inclusive. ThenK1= {v∈ᏺn:v2=0}andK2= {v∈ᏺn:
v1=0}are two distinct proper closed, connected right ideals ofᏺ. For anyw∈ᏺn, letw1=(w1,0)andw2=(0, w2). Thenwi∈K
i,w=w1+w2, andvw=vw2for all
v∈ᏺ. Thus, ifᏺ2is isomorphic toᏺ, thenᏺ2contains two distinct nonzero proper closed, connected right idealsJ1andJ2such that for everyw∈ᏺ2there existw1∈J1 andw2∈J2such thatw=w1+w2andvw=vw2for allv∈ᏺ
2.
Suppose, conversely, that a two-dimensional Euclidean nearringᏺ2has two distinct nonzero proper closed, connected right idealsJ1andJ2such that for everyw∈ᏺ2 there exist w1∈J1and w2∈J2 such thatw=w1+w2 and v∗w=v∗w2 (it is convenient to denote the multiplication inᏺ2by∗) for allv∈ᏺ2. Since the additive subgroupJ1+of the idealJ1is a closed, connected nonzero subgroup ofR2, it follows from [1, Proposition 3, page 71] thatJ1+contains a one-dimensional linear subspace L1ofR2. According to one of the statements preceding [1, Proposition 3, page 71],J+ 1 is isomorphic toRp×ZqwhereZis the group of integers and 0≤p+q≤2. Butq=0 sinceJ+1 is connected andp=1 sinceJ1+is a proper nonzero subgroup ofR2. That is, J1is a linear subspace ofR2and, of course, the same is true ofJ2. Thus,ᏺ
2is a linear right ideal nearring and it now follows fromTheorem 2.2thatᏺ2is isomorphic to one of the seven two-dimensional Euclidean nearrings whose multiplications are given by (2.60).
References
[1] N. Bourbaki,Elements of Mathematics. General Topology. Part 2, Hermann and Addison-Wesley, Paris and Massachusetts, 1966.MR 34#5044b. Zbl 301.54002.
[3] K. D. Magill Jr.,Topological nearrings whose additive groups are Euclidean, Monatsh. Math. 119(1995), no. 4, 281–301.MR 96c:16058. Zbl 830.16032.
[4] J. D. P. Meldrum,Near-rings and Their Links with Groups, Research Notes in Mathematics, vol. 134, Pitman, Massachusetts, 1985.MR 88a:16068. Zbl 658.16029.
[5] G. Pilz,Near-rings. The Theory and Its Applications, 2nd ed., North-Holland Math. Studies, vol. 23, North-Holland Publishing, Amsterdam, 1983.MR 85h:16046. Zbl 521.16028.
Kenneth D. Magill Jr.: Mathematics Building, Rm.244, SUNY at Buffalo, Buffalo, NY14260-2900, USA