ON THE ALGEBRA OF ORTHOGONAL LATIN SQUARES
by
JACK BRUCE GOEBEL
A THESIS
submitted to
OREGON STATE UNIVERSITY
in partial fulfillment of
the requirements for the
degree of
DOCTOR OF PHILOSOPHY
June 1962
APPROVED
Redacted for privacy
Professor/hf Mathematic
Charge of Major
Redacted for privacy
Chairman of Department of Mathematics
Redacted for privacy
Chairman of Shoo1 of Science Graduate Committee
Redacted for privacy
Dean of Graduate School
Date thesis is presented
Typed by JolAn Erttss
Ju y 26, 1961.
TABLE OF CONTENTS
Page
INTRODUCTION
1
CHAPTER I.
MANN-RINGS
3
CHAPTER II.
BOSE-RINGS
32
CHAPTER III.
SOME EXAMPLES
40
CHAPTER IV.
APPLICATIONS OF MANN-RINGS TO
ORTHOGONAL LATIN SQUARES
53
APPENDIX
59
BIBLIOGRAPHY
61
ON THE ALGEBRA OF ORTHOGONAL LATIN SQUARES
INTRODUCTION
In the book Design and Analysis of Experiments by
H. B. Mann, (7), a method is illustrated by which orthogonal Latin squares may be obtained from Galois fields.
If we inspect some of his proofs, we find he does not
use all of the properties of a field. If we delete from
the axioms of a field many of the properties not used,
the remaining system is what we call a Mann-ring. Its
formal definition is given below.
In this paper we demonstrate that a Mann-ring determines at least two mutually orthogonal Latin squares.
This suggests a method for attacking problems arising from
the design of experiments (7), and from various combinatorial problems dating back to the Kirkman schoolgirl
problem (6). In recent years much progress has been made
toward the construction of mutually orthogonal Latin
squares (2,4,7,9). Some of these have used finite geomet-
ries and some have used finite fields. In 1959 a variety
of new methods appeared in papers by E. T. Parker, R. C.
Bose, S. S. Shrikande, et al. Of paramount interest to
Parker was the 10X10 Latin square. Our attempt to find a
triple of 10X10 mutually orthogonal Latin squares leads to
the first theorems of Chapter I.
Since the construction of special types of Mannrings will lead to a set of mutually orthogonal Latin
squares, we proceed through the remainder of Chapter I
and all of Chapter II to study the algebraic structure
of such systems. Some of the statements demonstrated
there were suggested by (1).
At the end of the paper will be found an Appendix
listing some of the unsolved problems in this area,
3
CHAPTER I
MANN-RINGS
Definition:
M
Let
binary operation.
if
all
in
y
H.14
Indeed
groupoid.
A
Definition:
on
M,
for all
x
in
S
and for
say
be a left-S-
(M, 5,')
and
11
is associative then we
is a left-S-semi-group.
left-S-groupoid in which
and
is a groupoid.
Let two compositions be defined
M.
in such a manner that
.,
(M, s,.)
(M,S,+,$)
S=M
be a non-empty set of elements and
M
Let
be a subset of
a group and
system
is a left-S- roupoid
may even be undefined.
If the operator
(M, s,.)
S
be a
may be non-associative or non-
y. x
S C M
Let
Definition:
let
M
"*"
M.
commutative.
THEOREM:
M
We say that
We note that
say
be two sets and
S
is an element of
y
x
and
is a left-S-groupoid.
(MO.)
Then the
is called a Mann-ring if the system
satisfies the following postulates:
(1)
There exists an element
e.m.m
for all
m
in
M.
e
in
S
such that
is
4
There
exists an element
dsa=dsb
implies
d-e is in
for all
S
and
(d-e)-m=dsm-m
Definition:
in
for all
such that
S
and
a
(d-e)sa=(d-e)eb
b
and
a
a=b
d
in
for all
in
b
implies
M.
a=b
M.
m
in
M.
A Mann-ring in which (M,S*)
is a Ieft-S-
semi-group will be called an associative Mann-ring.
Note:
A Mann-ring is not a Veblen-Wedderburn system.
In
Chapter III we put some additional restrictions on our
system and get a Bose-ring which is a Veblen-Wedderburn
system with two postulates missing.
Thus a Veblen-
Wedderburn system is a special type of Mann-ring.
THEOREM 0:
A Mann-ring determines at least two orthogonal
Latin squares.
Proof:
The addition table of the group
square with the
(i,j) element being m
a new Latin square whose
(i j)
This is a Latin square because if
then
tinct.
mj = mk
so
j = k
(M,+)
+mj
.
element is
ci.mmi
is a Latin
We construct
dsm +m
= d-m
js
m
k
as the group elements are dis-
We conclude that the row elements are distinct.
We can check to see if each column had distinct elem nts
by supposing
dsmi + mj = dsmk + mj
and therefore
d.m. = drnk and so
mi
ink
by postulate (ii).
To see orthogonality we suppose that
(rai+raj, thmil-mj) = (ms+mt, dims-lint). That is, suppose
the (i,j) element of the superposed squares is equal to
the (s,t) element. Then
mi +mj =ms +mt
and
d.m.+mj=d-m s
t
By using the associativity of
M
and adding
on the left side of both members of (a) we get:
mj=-mi+(ms+mt).
get d.mi
.+m +
Substitute this in the left of (b) and
.m +mt
by associativity.
Whence we
now add
to the right side of each member and get
d.mi- i=d.ms-ms
again by associativity. Therefore
(d-e).mi=(d-e).ms by (iii) and hence mimes by (ii ).
i=s as the group elements are distinct.
Hence
mj=mt
from (a) and therefore 3=t and thus we have orthogonality
and the theorem is proved.
It can be shown that as a consequence of (ii),
(ii'), and (iii), S must contain at least three elements
if M/f01. If M={0} and M is a group then 060=0 so
S=f01=M and e=d=d-e=0 and all four postulates are
clearly satisfied if we define 00=O. We will call [03
Note:
6
the trivial Mann-ring and in the rest of the paper we will
mean
M / [01
when we say Mann-ring.
contains at least three
distinct elements and if d/e+e then M contains at least
four distinct elements.
THEOREM:
A Mann-ring
(M,S,+,*)
Call it 0.
Then postulates (i), (ii) and (ii') require S to contain
e, d and d-e. We note that e cannot be 0, for if
Proof: M contains an additive identity.
e=0
then d-e=d so postulate (iii) becomes
for all
dm=dm-m
or
This is a contradiction so e is not
O.
Furthermore e is not d for if d=e then postulate
(iii) implies 0em=0 for all m, and (ii') tells us
0.a=0b implies a=b for all a and b. Hence 0=0
m=0
implies a=b
contradiction.
m.
for all
a
and
b.
This implies M=(O, a
is distinct from 0 and d.
We claim that d is not O. If d=0 then by (iii),
(0-e)bm=0m-m, or 0em=0 for all m. Now combine this
with (ii) and we get that 0=0 implies a=b for all a
and b, i.e., that M=[01, a contradiction. We now have
that 0,e, and d are distinct. It may be that d=e+e
as in GF(3). Otherwise 0,11,d, and d e are all distinct since d-e=e would mean d=e+e, the case we are
ruling out, and d-e=d means d-d=e or e=0 and we know
e is not 0 so d-e is not d. Furthermore d-e cannot
Hence
e
or d and e would not be distinct, a fact we
have already proved. Therefore if Oei-e, then we have
be
0
four distinct elements in the Mann-ring.
Now we attempt to construct a Mann-ring of order
10 using the ring of integers modulo 10 as the group M.
We would like a Mann-ring (M S,+.) where S=M and
"." obeys the following rules:
M
is closed under the operation ".".
There exists a left identity e for ft.".
The left cancellation law holds for two elements d and d-e.
There is a right distributive law for d-e.
We let a and b be any two elements from M and define
the multiplication by:
1)
a.b=xaP+ybr-za bt where
x,y,z,p,r,s,
arbitrary but fixed constants from
and
t
are
M.
Fixed constants x.y.z,p,r s, and t cannot
be chosen in M such that the multiplication defined by
THEOREM:
(I) will have properties 1), 2), 3), and 4).
Proof: The proof is divided into several short steps labeled as Lemmas 1 through 11.
Lemma 1:
9
9 =9.
If
g
is an element of
M
then
=g
and
8
Clearly 25=2
Proof:
and
6=24
Also 4=22
29=2.
and
and
so the Lemma is true for 2.4,6, and 8.
also clear for
g=3
and since
9=32
7=33
and
8=2 3
It is
it is
true for 3, 7, and 9. We know that 5=5 modulo 10 for
all t>0 so g9=g5=g modulo 10 for all g in M.
In view of Lemma 1, the statement "let s=1",
[or let p, r, or t, etc. be 11, will mean the same as
"let s=1, or 5, or 9" wherever these symbols are used
as exponents in this theorem.
If a and b are in M then (a+b)n -an bn
is an even number, for any integer n, 0<n<10.
Lemma 2:
If n=1,5, or 9 then (a+b)n-an-bn=0 by Lemma 1.
If n=2, this expression reduces to 2ab. Otherwise
Proof:
(a+b)n-an.bn contains a factor of the form ab(a+b),
ab(a2 +b2 )*
a
or
b
or
etc. Such factors are always even for if either
is even then so is ab and if both are odd then
the second factor, [i.e., a+b,
a2 +b2
etc.J, is always
1
even.
Lemma 3:
all
b
If there is an
e
in
M
such that eb=b for
then either:
(1)
x=5
and
e
(ii)
e=5
and
x
is even or
is even or
9
or
x=0
e=0.
Proof: If e'b=b
fore xeP=0. If
x
for all
b
then e0=0 and thereIf
we have case (iii).
x=0
or eP is a divisor of zero,
[e.g.,
x=2
xp00
and
then
eP=5).
This implies either eP=5 and therefore e=5 and x is
even which is (ii) or else e is even and x=5 which is
(i) or else
If
Lemma 4:
Proof:
xeP=0).
e=0.
If
and
x=5
e
e
is even then either
t=1
and
r=1
or
t=1
and
y=5
or
r=1
and
y=1.
is even and e.b=b then ybr-z sb b (as
Now let b=1
is even. Put
and we see y=ze5+1
y=ze5+1
is odd since
in ybr-ze sbt.b and got
(br-bt)=b-br. Call this equation (1). This must hold
for b=2.3,4,6,7,8,9. It is clearly true for b=0,1, and
ze
5.
Now recall 4=22. 8=23, 6=24, 9=32,
10 so put these in equation (1) for
ing seven equations:
b
and
7=33 modulo
and get the follow-
10
zes (2r.2t)=2.2r
zes (22r_22t).4.22r
ze5(23r-23t) 8-23r
zes(24r_24t) 6.24r
zes( 3r.3t)=3_3r
ze (3
3 2t )=9-32r
zes(3
..33t)=7...33r.
We observe that equations 2,,4 ,6
and
7
contain fac-
tors in their left hand members of the form u2 v2
u3 -v3 ,
and
u4 v4.
Hence by proper multiplication of
equations 1 and 5 by factors such as
u+v, u2 +uv+v2 ,
etc., we may use 1 and 5 to eliminate
by equating the right hand members. This done we get:
u3+u2v+uv2+v3,
2')
3,)
(2-2r)(2r+2t)
(2-2r)(2+2r)
(2_2r)(22r4.2r2 22t) = (2-2r)(2 +2.2r+22r)
2r22t 23t
4') (2-2r)(23r+22r
(3-3r)(3r+3t)
6_24r,
32r
=
(3 3)(32 3 3
) =7-3
If we factor these in the obvious way we get:
2')
(2-2r)(2t 2)
0
or
11
3')
4,)
(2-2r)(2 2)(2r+2t+2) = 0
6")
(3-3r)(3t-3) = 0
(2_2r)(2t_2)(22r4.212t.0.2r4.22t4,2 e2t +22 )
7') (3-3r)(3t-3)(3r+3 3)
Clearly the only solutions are r=1, or t=1, Ei.e.,
r=1,5,9 or t=1,5,91. We check these solutions. If
r=1 in 1) through 7) then either t=1 or ze5=0 or
ze5=5 (since b'-b is even for all b).
Now
e
was even
so zes is not 5 and therefore ze =0 and since y=ze5+1
this implies y=1. The solutions are:
r=1 and t=1 or
r=1
if
t=1
above.
and
y=1
or
in 1) through 7) then r=1 is a solution as
If t=1 then y=5 is also a solution, where
y=ze5+1.
Hence
Lemma 5:
If
e=5
t=1
and
Proof: We want eb=b
and
x
y=5
Is a solution.
is even then r=1.
for all
b
so
e.b=5.b=x5P+ybr-z5sbt=-.
Since x is even we know
b
x5P=0. Hence we have ybr_b_z5sbt=0 and
br_b
y=ze5+1=z55+1=5z+1. Therefore z55(br-bt)
and
br -bt
is always even so we have
implies that r=1 (or 5 or 9).
b
0
0
which
12
is even then the circle operation fails to have the required right distributive property.
If
Lemma 6:
Proof:
and
e=5
By Lemma 5,
x
r 1. Therefore a.b=xalp (5z+1).b.zas t
since y=ze +1. We want
Since r=1
and
e=5
(d-Oc=d.c-e.c=d.c-c for all c.
this reduces to:
xL(d-5)13-dP-5P1-yc ze [(d-5)s-d6-55]=0
Now
y=5z4.1
for all
c
modulo 10.
is even so let x=2k. This is true
and therefore is true for c=2. So let c=2
and
x
and we get:
5)P-dP-5/21-(5z4-1)2-z2t[(d-5)5-d5-5s1
0 modulo 10.
Hence k[(d-5)P-dP 5P]-(5z+1)-z2t-1[(d-5)9-d5-55] = 0
However since 5 E 0 modulo 5 we have:
modulo 5.
kte-d] ,1_z2t-l[d5.d5]
0 modulo 5,
or -1 E 0 modulo 5
and this is clearly an impossibility so we have no right
distributive law for case (ii) of Lemma 3.
We now prove a similar result for case ( ) of
Lemma 3.
is even, the circle operation
If x=5 and
does not have the right distributive property.
Lemma 7:
Proof: By Lemma 4 either:
t=1
and
r=1
or
t=1
and
y=5
or
13
(c)
Suppose (a).
r=1
and
y1.
Then the right distributive property is:
5[(d-e)P-dP-eP3-zcE(d-e)s-d
for all c.
sl=yc modulo 10
is even and hence
Now (d-e)P-dP-eP
c(-z[(d- )s-da-081 yi 5 0 modulo 10.
We know
y
is odd
is even and so the expression in brackets is
odd by Lemma 2. We then take c to be odd and get an
because
e
odd number congruent to 0 malujo 10, an impossibility.
Suppose (b) is true. Then the distributive law would
imply:
5[(d-e)P-0-eP1-acE(d-05-d5-
5c
modulo 10.
Since
the expression in brackets is even, let c-5 and get
0=5 modulo 10 which is impossible.
Suppose (c).
Then
a.b=5aP+b-za5b
implies -zct[(d-
5[(d-e)P-e-e] = 0
055 modulo 10.
and (d-e).c=dic-e.c
$] E c modulo 10, since
by Lemma 2.
Now let
c=5
and get
This is again an impossibility so the
Lemma is proved.
Lemma 8:
Suppose
e=0
as in (iv) of Lemma 3.
Then
a.b
does not have the right distributive property.
eb=b implies that 0b=b and hence that yb b.
This is to hold for all b so we try b=1 and get y=1.
Proof:
14
Hence
br=b.
So if
e=0
).c
(d
This means
r=1 (or 5 or 9).
then a.b=xab +b-zas bt .
We want
doc-e.c. Therefore
x[(d )1)-e-eP]c-zc [(d-e)s-ds
si =0 modulo 10.
Since 0=0 this says
x[dP-ej-c-zet[ds-ds] = 0
modulo 10. Hence -c=0 modulo 10 for all c. This is
impossible so case (iv) of Lemma 3 has no right distributive law.
If
Lemma 9:
x=0,
as in case (iii) of Lemma 3, then
either
Proof:
If
x=0
then
and
r=1
and
zes
a.blaybr- a
=
or
.
bt. We want an identity
such that e.b=b and hence ybr-ze bt=b for all b.
If we let b=1 we see that Y=ze5+1.
Our right distributive property requires that
-zet((d-e)s-ds-esj=yer modulo 10 for all c. If we let
c=5 we get 0 on the left in virtue of Lemma 2. Hence
5y=0 modulo 10 and this implies that y is even. Therefore
ze5+1
is even. Furthermore putting y zes+1 in
e'b=b reduces e.b to zes(br-bt)+br-b which is the
very same equation we solved in Lemma 4. The volutions
15
were:
or
and
t=1
and
ze6=0
r=1
and
ze =5 or
t=1
and
y=5.
or
even so (d) and (b) are
not possible and therefore either:
In the present case we have
y
or
r=1
and
t=1
r1
and
zes=5.
The second case of Lemma 9 is impossible be-
Lemma 10:
cause it lacks the right distributive property.
Proof:
Let
e =5 and so
z
and therefore both z and
see that z=1,3,507, or 9
e
is odd.
Also e5
is odd
are odd numbers. Hence we
and e=1,3070 or 9. We know
is not 5 by Lemma 6.
If z=1 then e5=5 and e=5 but this contradicts Lemma 6.
e
If
z=3
then
3
and so
1,8=5
and so e5 but this
contradicts Lemma 6. Similarly if z=7 or 9 then ze55
implies e=5, a contradiction of Lemma 6. So z=5 is the
only possible case. Let
e5=10347, or 9. Then
and
e=1 307. or 9.
a.b=yb-5 sbt and y=ze5+1=6
a.b=6b-5 sbt. Now suppose
Then
z=5
Then
so
(d-e)ic=dc-e*c for all c.
esi = 6c modulo 10 and letting c=2
16
gives 0=2 modulo 10,
a contradiction.
The first case of Lemma 9 does not have a left
cancellation law for any two "consecutive" elements a and
Lemma 11:
a -e.
Proof:
or 9,
In Lemma 9 we sdw y=ze5+1
and
is even so z=1,3,7,
e=1,3,7, or 9,[e is not 5 by Lemma 6 and
is not 5 by Lemma 10].
So
a.b=yb-zasbk
z
If we want
a.b=a.c to imply b=c then (y-za5)(b.c)=0 implies
b=c,
so we need
(y-za5,10)=1.
This means that
y-za5=1,3,7, or 9. Since y is even we need a to be
odd. Hence we cannot cancel a on the left unless a is
odd. But e was odd so if a is odd then a-e is even
and cannot be cancelled. This completes the proof of the
theorem.
Since definition I of the circle operation fails
we could try some new definition, such as a.b=p(a)b or
a.b=ap(b) etc., where p(a) is some polynomial in a
with coefficients in M. This suggssts the following
theorem.
There do not exist polynomials p and q with
coefficients in M such that a.b=p(a)q(b) has properties
THEOREM:
(1), (2), (3), and (4).
17
Proof: We want the following to hold:
(d-e).c=d.c-c;
d.a=d.b implies a=b;
implies a=b.
If we take
then
or
c=1
(d
and
)-a= d-e .b
in (1) such that (d-e).1=del-1
p(d-e)q(1)=p(d)q(1)-1.
So
p(d-e)q(1)-p(d)q(1)=-1
p(d)q(1)-p(d-e)q(1)]..
Thence
q(1)[p(d)-p(d-e))=1.
since
p(d-e)
is not equal to
p(d)
This also implies that
p(d)-p(d-e)
is odd, for an even number could not be congruent to 1
modulo 10.
parity.
Thus
p(d)
and
Now (2) requires
p(d-e)
p(d)q(a)=p(d)q(b)
a=b,
and also
a=b.
These are equivalent to:
p(d-e)q(a)=p(d-e)q(b)
p(d)[q(a)-q(b)]=0
implies
a=b.
q(a)=q(b)
implies
a=b
and
is to imply that
p(d-e)[q(a)-q(b)]=0
Thus each of these conditions implies that
p(d)[q(a)-q(b)]=0
p(d-e)[q(a)-q(b))=0
and p(d-e)
implies
implies
q(a)=q(b)
q(a)=q(b)
a=b.
and
we know that
cannot be divisors of zero or else the
above implications would be impossible.
This is because
we can always obtain a factor of 2 or of 5 in
by choosing
a=2
since
and p(d-e)
p(d)
to imply
since this is certainly a consequence of
Since
p(d)
are of opposite
and
b=4
or
a=5
and
q(a)-q(b)
b=0, etc.
Now
are not zero-divisors, they both
must be mdd numbers, a contradiction since they were of
18
opposite parity.
THEOREM:
If
M
is a Mann-ring such that:
M=S;
is associative;
both distributive laws hold;
H.11
then
is a ring with left identity.
M
Proof: We need to see that
(e+e)(a+b) for any a and
M
b
is abelian. Consider
in
M
and
e
the left
identity.
(e+e) a-i-b)=t4 a+b)+e( a +b)a+b+a+b
and
(e+e)(a+b)=(e+e)a+(e+e)b=a+a+b+b so
b+a=a+b.
be an additive group and S be a
set of left operators on M with the following properties:
There exists an e in S such that ea=a for all
Definition:
a
(ii')
in
Let (Mo+)
M;
There exists an element d in S such that
da=db implies a=b for all a and b in M;
d-e is in $ and (d-e)a=(d-e)b implies a b,
all a and b in M;
(d-e)a=da-a for all a in Ms
We call the system 6=(M,S0-,) a left-Mann-module.
If
is a left module with the set S from a ring with
left cancellation and a left identity then
is a
Mann-module.
19
Definition: Let(M,S + e)
Mann-rings.
(M.,S'
and
We call a mapping 0 of
M
into
if:
(m1+m2)0=m1oft20 for all ml' m2 in M.
m0
is in S.
is in S' if
(sm)0=(80)(m0) for all s in S, and
morphism of
u
into
s)
be two
M'
a homo-
3,
m
in M.
We note that a homomorphism is defined in terms of
m
the addition and multiplication of the left-S-groupoid
only and is independent of the four additional postulates
defining a Mann-ring. Therefore we would expect that a
homomorphic image of a Mann-ring might or might not be a
Mann-ring as the homomorphism might not preserve the left
cancellation properties of postulates (ii) and (ii').
If
THEOREM:
a Mann ring
d
is a homomorphism of a Mann-ring g into
then the image of
M
is a group in
M'.
Proof: 0 is a row" homomorphism and this is a known
theorem for groups.
Definition: Let tr(M,S,+,.) be a Mann-ring. We say
that h=(N,C,+,*) is a Mann-sub-ring if h is a Mannring and (N,+) is a subgroup of (M,+) and C is a sub-
set of
S.
We note that if
ring
into a Mann-ring
is a homomorphism of a Mann-
then the kernel
K
d
of 0
20
is a normal subgroup of
(M,+)
since
o
is a group homo-
morphism.
In standard ring theory it is natural to define a
subset of a ring called an ideal. However Mann-rings not
only lack both distributive laws but multiplication may
be undefined for some elements. Since an ideal is a subring closed under multiplication by ring elements, the
definition of ideal is not applicable in our case.
Further difficulties arise from the fact that a
Mann-ring must contain a left identity and thus the subring must contain a left identity. We shall see that in
many cases of Mann-rings there is only one non-zero idem-
potent and hence only one left identity.
Another approach to the theory of ideals is by way
of the kernels of homomorphisms.
We can show that if
there is a homomorphism of a Mann-ring, its kernel is indeed a normal subgroup of (M,+). Actually this is all
that can be expected since the homomorphism is defined in
terms of the left-S-groupoid and not in terms of the postulates for a Mann-ring. Thus a generalization of the
concept of ideal appears not to be fruitful.
However one is tempted to try to prove theorems con-
Just as in
the case of the ideal concept, difficulties arise because
of the lack of total cancellation and distributive laws.
cerning the homomorphic images of Mann-rings.
21
Since a homomorphism does not depend upon the Mann-ring
postulates, there is no reason for them to be preserved.
Indeed, one type of Mann-ring would be that for which S=M
and we have a groupoid. In that case we could add inverses and get a quasi-group and we know there exist quasigroups possessing homomorphism upon systems which are not
quasi-groups -- and similarly for loops (see reference
(1), page 92, and page 1101 section 1). It is still an
open question as to what loops G have the property that
every image of
under a multiplicative homomorphism is
also a loop. The same problem for Mann-rings appears
even more formidable.
G
Definitions Let a=(M,S,+,-) be a Mann-ring. If xx=x
in 3, we call x an idempotent of S. If the Mann-ring
is associative then x=xx=xex, an element of xSx since
e is in S. So we define x as regular in S" if x
is in S and x=xsx for some s in So i.e., x is
in xSx.
All idempotent elements are regular.
If x=x(six), we call s1 the left relative inverse of
x.
If s =s (xs
)
then
x ansi are relative inverses
of each other.
THEOREM:
If
is an associative Mann-ring and if
x
is
22
with right inverse then
potent if and only if xe=e.
an element of
C
x
is an idem-
If xx=x then xe=x(xx-1)=(xx)x 1=xx-i=e. So
xe=e. Conversely if xe=e then (xe)x=ex, so
xx=x(ex)=(xe)x=ex=x and therefore x is idempotent.
Proof:
Corollary: In an associative Mann-ring, an idempotent
with right inverse has the property that
m,
and hence is a left identity.
xm=m
x
for all
Definition: Let
i:=(M,S,+.°) be a Mann-ring. By the
MacNeish kernel S* of Es we mean the maximal subset of
M
such that every element of
(ii), (ii'), and (iii) and also
except e
S*
S*
satisfies
has left cancella-
tion, right distributive law, and is closed under subtraction.
We form S* by starting with all suitable elements
in S and then extending our definition of multiplication
from SXM into M to a multiplication of S*XM into M.
So S* depends upon S, upon the original multiplication
and upon
(M0-).
be a Mann-ring and let
i*..(m,s*,+,.) be the "MacNeishft-ring formed by replacing
Now let
S
go.(m,s0.,..)
by the MacNeish kernel S*.
23
Every element in
THEOREM:
has a unique right in-
S*
verse.
Proof:
We have left cancellation for all
in
s
S*
so
M= sM.
Corollary 1:
In an associative MacNeish-ring, the only
idempotent is the left identity
Proof:
So
xx=x
If
by the two previous theorems.
xe=e
then
e.
is the unique right inverse of
e
of all idempotents and
xm=m
so
x
and therefore
x
is a left identity.
(x-e)m=xm.m=0
But then the distributive law tells us that
for all
So we choose a different from
m.
that
(x-e)a=0
a=b,
unless
and
x-e=0.
So
x-e=0
The left identity
Corollary 2:
and
e
and get
(x-e) =(x-e)b
hence
(x-e)b=0
b
hence
x=e.
is unique in an asso-
ciative MacNeish -ring.
Proof:
fm=m
If
for all
th n
m
so fe by
ff=f
corollary 1.
Let
6
M S*,+..)
be an idempotent such that
for
m
in
So define
M we see
xM=(xm:m
be a MatNeish-ring and let
x
and
e x
are in
m=xm-1-(m-xm)=xm+(e-x)m,
is in
M1
(e-x)M*f(e-x)m:m
and define
is in
M.
S*.
x
Then
24
Then we note that
b
is in
xMn(e-x)Me0
So
(e-x)M.
is in
b
if
xbreb
the left Pierce decomposition of
decomposes a MacNeish -ring
if
if
xbe0
and
MexM 0 (e-x)M
and
So an idempotent
M.
and
x
xM
are in
e-x
x
S*.
We now proceed to define and discuss hyper-Mannrings and hypo-Mann-rings.
ring and let
to be in
y.m
we have
be in
M
for
M
y.m
choices for
n
m
all
could let
y.me0
is not
and let
0
rm
such that
for all
in
Then define
if the group
M
ym=m for all
or we could let
m,
0m=0,
=sufyi
S.
for each given
M.
m
is of order
m
or we
y.mem
if
or any other multiplication
was an element of
Now define
be a Mann-
3e(M,Sice,y)
but not in
For example we could let
n.
y
y
Let
M.
We then see that
(M,51,+,')
*
Is a Mann-ring where the dot operation is the old dot multiplication on
into
SXM
into
M
and is the new dot on laM
M.
So any Mann-ring is a subring (i.e., "is isomorphic
to a subring") of "larger" Mann-rings formed by increasing
the size of
S
by any "suitable" definition of left mul-
tiplication for the elements of
original
n.
S.
We are not
M
that were not in the
These "larger" Mann-rings are still of order
adding elements to
ing elements into
S
M,
rather we are shift-
that were not originally in
S.
Since the postulates of a Mann r ng still hold for
25
any such extension of S, we see a "maximal" Mann-ring
of the same order as
would be (M,M,+,*). Call any
such ring a hyper-Mann-ring containing 3 and label it
by b7(M,M,+,.). There will be many such rings depending
law*
upon the different multiplications or defines.
Similarly we see that S really need contain only
the elements e,d, and d-e to qualify as a Mann-ring
and it may even be that d-e=e. So define 3=(M,S,4-,*),
where S=[,,d,d -el, as the hypo-Mann-ring contained in
The above discussion demonstrates the following theorem.
is embeddable in a hyperMann-ring T. We note that a Mann-ring will be embeddable
In as many hyper-Mann-rings as there are binary operations
definable from (M-S)XM into M. By embeddable of course
we mean the multiplication agrees with the original mulTHEOREM:
Every Mann-ring
tiplication on
5
SXM.
Now choose some fixed multiplication from MXM into
M
and thus a fixed 3 containing 3 (isomorph cally).
we have the folThen considering the hypo-Mann-ring
lowing theorem.
THEOREM:
Every hypo-Mann-ring
of order 3 and
in a fixed
M
is of order
3-0.(M,S0-,e)
n
where
has 2n-3-1
$
is
extensions
26
Proof: SC SiC S2C.C S =M
is any element not in
a
where
S.
where
S3+1=S5U [al
The number of distinct
S 's then depends upon the ways of choosing the distinct
This can be done in
a.
(
n-3)
(n-3
(n-3 )
n-3
2
1
different ways.
Q.E.D.
We see that the Mann-rings formed by using these
subsets of M as the S of the left-S-groupoid can all
be regarded as subrings of each other as
of
M.
Corollary:
n-2
3C EiZ
Hence,
If
S
is a subset
=
C
is of order
M
2
then there will be
extensions for each fixed multiplication.
is isomorWe have noted that every Mann-ring
phic to a subring of a hyper-Mann-ring 5 and contains a
hypo-Mann-ring Z. Suppose S is of order t, where
t=2 or 3. Then there is a chain of sub-Mann-rings Ek
every order k, t<k<n, such that for each fixed multiplication,
2
1
00-eitc at+1C
For a given
C
ft
fx(M,S,+ .)
is or order
+ IC
where
C 141
is of order
n
and
we can determine the number of distinct
binary operations ft 0 " from MXM into M such that
" 0 " agrees with the original multiplication on SXM.
S
a
27
For
in
M-S
in
yl
we want
For a chosen
M.
of the
elements of
n
There are
0
in
yi
n-elements by
possible mappings of
M-S
yip m
(M-S)Xm
nn2-an
S
and
if
there are
into
Now m
M.
n
n(n -a)
If
is of order
S
(2n
no
a
So associ-
with
.1)(nn(n-al+
.)
S
Mann
is of order 3 then the family of Mann-rings
S
(2n -3 -1)(nn(n-a))
Definition:
such that
Mann-rings between
distinct multiplications.
we have a total of
a,
If
Mann-rings.
Two Mann-rings are equivalent, (we write
if they have the same group
groupoid
was any
multiplication.
"."
2n-2-1
was of order 2.
S
0
ated with each "parent" Mann-ring of order
contains
n-a
so there are
is the original
Recall that there were
rings.
M -S.
and each can be
possible multiplications
restricted to
of order
in
y2
elements so we get a total of
n
(nn-a)n
Similarly for
M.
m
y10 m beEll one
we can let
elements
n -a
mapped onto
one of
m
defined for each
yl 0 m
(M,S,)
if and only if
(M.+)
and the same left-S-
in the sense that
s.m=s0m
for all
),
$
(M,S,+,°)=(M,S,+,G)
in
S
and all
m
in
M.
We will now investigate specific methods of getting
28
new Mann-rings from old ones.
Suppose given a Mann-ring
= (M,S,+,.). If M is of order n we can use the
original multiplication tt.11, to define n multiplications say "(a)ft .
For each element
from
SXM
in M we define the operation (a)
a
into M by:
d(a)m=d.m+a
and
(d-e)(a)m=d(a)m-m,
if
is different from
and
s(a)m=s.m
d-e.
So the new ring 3(a) is the old system E with
only the multiplication for d and d-e altered. We
verify that Zi(a) is a Mann ring:
s
d
and
d(a)a=d(a)b means da+a=d.b+a so d.a=d.b and so
(d-e)(a)a=(d-e)(a)b is the same as d.a+a a=d.b+a-b.
So
d.a-a-d.b-b or (d-e)a=(d-e)b
so
a
Therefore
a=b.
postulates (ii) and (iA') are verified and (iii) is valid
by the definition of (a). To see that these rings are all
distinct, we suppose that
d(a)m = d(rk)m for
a,PeM
and some
m
in
We note also that if a=0 we
get (a)=".
and therefore we get 3 back again.
We can get n more distinct Mann-rings if e+e .
In that case we define ral to be a mapping f SXM into
Then
dm+a=d.m+p
so
a=P
.
29
M
b
dCalm = (d-e)-m + a and
(d-e)[alm . Calm -m
and all other elements map as before under the original
multiplication. It should be noted that the multiplication for d-e is completely determined by the addition
table and the multiplication for d because of postu-
late (iii). Thus it is not really necessary that the
multiplication of d-e be specified. $o if our original group (M,+) had a multiplicative left identity e
of additive order 2, (i.e.,
e+e=0),
then we do have
distinct Mann-rings generated by this device of redefining multiplication for d and d-e. This demon2n
strates that for every
pn
GF(pn)
there exist at least
Mann-rings and for every GF(2n)
least
there exist at
2n+1.
We end this chapter on Mann-rings with some models
of order four to illustrate the independence of our pos-
tulates. In all four models we choose the fours group
for the additive group. So let (M,+) be the fours
group (i,e,d,d-eI, with addition defined by x+x=0 for
all x. Let S=1.(e,d d-el. We will use d+e in place of
d-e
since +e=-e.
30
Example (1):
We define multiplication by the fol-
lowing table:
Here we see that (ii), (ii'), and (iii) are valid but
(i) fails so (i) is independent of the other postulates.
Example(ii): We let (MO.) be the fours group
again (where x+x=0) and Si.4e,d,d-el. We define mul-
tiplication by the following table:
Here we see (i), (ii'), and (iii) are satisfied but (ii)
is not.
Therefore (ii) is an independent postulate.
Example (lit): For the same
define multiplication by the table:
M
and
S
as before,
31
Here (i), (ii), and (iii) hold and (ii') fails so (ii')
is an independent postulate.
Example (iii): Use the same
define multiplication by:
ON+
and
S
and
Clearly (/). (ii). and (ii') hold but (d+e)d=d while
dd+d=0 and d is not 0 so (iii) fails and so postulate
(iii) is independent of the rest of the system.
32
CHAPTER II
BOSE-RINGS
In the previous chapter we discussed how to generate new Mann-rings. However each new Mann-ring used the
d
and
d-e of the original ring. We saw in Theorem 0
that such a Mann-ring will generate at least two mutually
orthogonal Latin squares.
In order that we may assert that our system will
indeed generatz more than two such squares we want more
elements to satisfy the postulates. We begin by requir-
ing that S=M-[01, and all elements to satisfy postulates
(ii) and (iiT) as per the following definition.
Definition: A Bose-ring is a Mann-ring ij=(M,S.+,.) with
S=M-10/ and such that:
ma=mb implies a=b for all a.b in M and
all
m
not
(a+b)m=amtbm
O.
for all
a,b,m in
M.
So a Bose-ring has a left cancellation and a right distributive law.
By an associative Bose-ring we man a Bose-
ring in which the multiplication is associative.
THEOREM:
If a Bose-ring Z=(M,M-0,4-,.) is of order
then it determines a set of n-1 pairwise orthogonal
n
33
Latin squares.
Each of the n-1 elements of S is a suitable
"d" for the proof of Theorem 0 in Chapter I. So we get
Proof:
squares in that manner and these include the original
square. Their mutual orthogonality follows from the proof
of Theorem 0 in Chapter I.
n-1
Example:
The Veblen-Wedd rburn systems are examples of
Bose-rings.
THEOREM:
In an associative Bose-ring,
Proof:
If
a
Ox=(x-x)x=xx-xx=0.
is not
0
neither equal to
Then
x0=a.
Now suppose
then choose
O.
Ox=x0=0.
b
and
c
not equal and
x0=x(0b)=(x0)b=ab
and similar-
ly, x0=x(0c)=(x0)c=ac. So ab=ac and if a is not 0
then b=c, a contradiction to the assumption on b and
c. Thus it must be that a=0, i.e., x0=0. Q.E.D.
Note:
The above theorem proves that an associative Bose-
ring is a y-ring as defined and discussed in the following
page. The question of when x0=0 is partly answered in
the following proposition.
Definition: Let
a
be a Mann-ring with S=M. We call
op-ring if x0=0 for all x in B.
We note that if x0 is not 0 for some x then
B
34
we can let
x0=p
where
following phenomenon. if
pz=(x0)z=x(0z)=x0=p.
that
pz=p
is not
p
and we observe the
0
is associative:
B
p
So there exists a non-zero
for all
in
z
B.
such
This characteristic of a
non-T-ring leads to the following theorem.
An associative Mann-ring B is a T -ring if and
only if zx=z for all x in B implies z=0.
THEOREM:
Proof:
B.
Then
fore
in
Suppose
z0=z.
z=0.
B
is a trring and let
B
B
But
is a
T-ring so
Conversely suppose that
implies
(x0)b=x(0b)=x0
Consider
z=0.
so
(x0)b=x0
therefore by hypothesis
x0
z0=0
for some
b
in
x
in
and there-
for all
zx=z,
for all
all
zx=z,
x
in
B
and
x
B.
x0=0.
Q.E.D.
This suggests areas where not to look for
T-rings.
For example, consider the set of transformations of a ring
into itself.
Let
Tb
be that transformation taking every
element of the ring into the element
for all
Tx
b.
Then
in this set of transformations.
would not form a T-ring.
T T =T
b x b
Thus they
Example 3 of Chapter III shows
that they can be formed into a Mann-ring.
THEOREM:
In an associative Bose-ring, the right
35
cancellation law holds.
Proof: By a previous Theorem,
left cancellation, if
xa=x0
a=0.
x0=0,
for all
x.
By
is not 0 and xa=0 we have
so a=0. Thus if x is not 0, xa=0 implies
Now suppose c is not 0 and that ac=bc. We
x
must see that a=b. We note ac-bc=0 and so (a-b)c=0.
a-b is not
But we chose c not
Now if
0
0
then
c=0
so
a-b=0
by left cancellation.
and therefore
a=b
and we have right cancellation for all non-zero elements
of
M.
Q.E.D.
Corollary 1: (proved later as an independent theorem)
In a Bose-ring in which the associative law holds, the
only idempotent is e or 0.
Proof:
If
xx=x
is not
so that x-e is not 0.
then (x-e)x=0. If x is non-zero then x-e=0
Suppose
x
e
since we have right cancellation from the Theorem. But
x-e is not 0 and therefore e is the only non-zero
idempotent.
Corollary 2:
In an associative Bose-ring,
unique left identity.
Proof: Use corollary 1.
e
is the
36
Later in this paper we will remove the restriction
of associativity from both corollaries 1 and 2.
Now a previous theorem tells us that right cancellation holds. Thus xa=ya implies x=y if a is nonzero. Hence Ma=M and there exists a11
in M such
Note:
that
a e. So right cancellation implies the existence
a1
of left inverses. Also the left cancellation implies
-1
rightinversesasuch
that
orem will show that if
x
such that
xi =xr Imx
aa -1
is not
xx
x
e. Our next thethen
0
x=e.
In an associative Bose-ring, the left inverse
equals the right inverse for any non-zero element.
THEOREM:
If
Proof:
that
is not
x
lx=e
0
let the left inverse be
and the right inverse be
r
1
such that
such
xr=e.
We note that:
rx=(er) =((lx)r)x 1(xr)x=1(e)x=lx.
r=1
Thus
rx=lx
and so
by the associative law and the right cancellation.
Q.E.D.
THEOREM:
e
In an associative Bose-ring the left identity
is also a right identity.
Proof:
For
x
in
M
and
x
not
0
let
xe=a.
Now
37
choose an element
and write (xe)b=ab or
xb=ab. Then by right cancellation we see x=a and thus
xe=x. We already knew that 0e=0 so xe=x for all x
in M.
not
b
0
Q.E.D.
In a Bose-ring in which
THEOREM:
for all
x0=0
the
x,
right cancellation law holds.
If
Proof:
But
c
(x-y)0=0
is not zero let
so
xc=ye
so that (x-y)c=0.
(x-y)0=(x-y)c and therefore c=0 by
left cancellation, if (x-y) is not zero, and
contradiction so x-y=0
right cancellation law.
and
x=y.
c=0
is a
This demonstrates the
Q.E.D.
Corollary:
exists a
c
if
is in M and therefore there
such that c-1c=e and so c has a left
Mc=M
c
inverse.
Definition: By a subring of a Bose-ring
mean a Bose-ring *n=(N,E,+,) such that
group of
Note:
NI,S,+,)
(N 4.)
we
is a sub-
(M,1).
is the only idempotent in a Bose-ring so the same
left identity e must be used for all subrings of a Bosering.
e
38
In a Bose-ring, the left identity is unique.
THEOREM:
Proof:
Choose two elements
different from b. If
ties then
Also
and
and
e2
(e1 e2 )a=eI a -e2 a=0
(el -e2)b=elb-e2b=0
That is,
e
a
is
are two left identi-
b
such that
since they are identities.
and so the left sides are equal.
(el -e2)a=(el -e2)b.
But we have left cancella-
tion for every non-zero element of the Bose-ring so
el e2 non-zero would imply a=b, a contradiction.
01=02
a
Hence
and the left identity is unique.
Q.E.D.
THEOREM:
In an associative Bose -ring, there is only one
non-zero idempotent.
Proof: Choose any element
in M. Suppose x is a
non-zero idempotent. So xx=x. Let xa=b. Then
x(xa) xb or xa=xb and since x is not 0, we have
left cancellation and so a=b. That is, xa=a. But a
was arbitrary so x is a left identity. Yet e was the
unique left identity so x=e.
Q.E.D.
We can remove the restriction of associativity if
we change to a different type of proof. We do that in the
following theorem.
39
In a Bose-ring with right cancellation, the only
THEOREM:
non-zero idempotent is e.
If
Proof:
is not
x
and
0
xx=x
then xx=ex
x=e.
so
Q.E.D.
If
THEOREM:
is a homomorphism of a Bose-ring
o
into a Bose-ring Eil=(Ms1M"-(01,4-..)
3=(M,M-(010-1-)
the kernel of
Proof:
If
exists an
x
is zero.
o
is in
M
and
x
is not
such that xx- =e.
x-1
then
= (xo)(x -1 o) = eo.
Also
So
0
(xx
then there
)cy
(em)o = (eo)(mo) = inc so 00
Is the left identity in
Now if
x
is not
0
and
x0=0'
we have
)0 (x0)(x-10)=0"(x 0)=0'0 since 0/m1=0, in a
Bose-ring P. Therefore mo=0' for 11 m. This is impossible so xo is not 0' unless x=0. Then
eo=(xx
Oa = (m-m)0 = mo-mo = 0'.
So the kernel of
o
is
(01
Q.E.D.
If we considered ideals as kernels of homomorphisms
then the above shows that a Bose-ring has no proper ideals
Note:
and no proper homomorphic images.
40
CHAPTER III
SOME EXAMPLES
If
Example 1.
and
S=M
M
then
is a field with the usual operations
is a Bose-ring.
a field,
such that d-e is not
Example 2.
(M,S,+,)
M
is a Mann-ring.
example 3.
GF(3)
Rg=-g
all
e
[e,d,d-el for any element
and d-e is not 0 then
Let
M
be the set of all transformations of
into GF(3). M contains 27 elements. Let
be defined by (Ti+Ti)g=Tig+Tjg. We define
R
for all
-T=RT
for
on
GF(3).
T, (M,+)
Now let
g
in GF(3).
Then letting
is a group.
S
be the symmetric group
33
is contained in M and (M,S,+,*) is a Mann-ring
with multiplication defined by the usual operator notation.
It can be verified that we can take d to be
S
/012)
'210
and
by
d_e . 1012 ) or we can take At7)121
`201r
' ='1021
d-e = (Tg).
In this example each element
T1
is a
'
r1(1
41
transformation of GF(3) 10,1,21 and there are 27 ways
of mapping (0,1,21 into a triple (a,b,c) and these
are the ternary representation of the integers
For example (1,0,1)
10 and (2,1,2) is equivalent to
23 to the base 3. Therefore we will arrange and label the
27 transformations such that Ti
base 3.
Then
1
T15 T19'T21
012%
and
012"
T21 the only choices for
andare
11
and
M
For example
T1+Ti=T(141)mod 27.
S = [T5TTl
T
'T19
are the respective d-e's.
(T-u'
S = (T5
T26
I
T
j
=
012%
1abc )t
S
and
T
1
Hence
abc.j in base 3
and
=83
although we could
be any subset containing T51T 1,T15 or else
T5,T21,T19.
1-1
d,
1 = 53.
TI ,T15,T19
It seems natural to take
let
abc=i
since 17 base 3= 122. Thus
(T.1)
T17
where
(°a)
At least we need both
and therefore in
53
d
d-e
and
in order that
(II,
to be
S#4. ')
be
a Mann-ring.
The similar example of all transformations on the
field
GF(22)
does not work as there are no 1-1
42
transformations
is not in
d-e. For all
and
d
in
d
-e
4'
so that example fails.
S4
To continue with example 3 we attempt to form a
homomorphism of
into itself.
3
the additive group
Since T +TfT(.3.4.nmod 27
is the abelian group of integers modulo 27 generated by Tl and so it is cyclic.
(M,+)
Hence the only subgroups are the cyclic subgroups
c)(
generated by T9 and the subgroup
3 =IT0' T9' T18 I
3(9=1T0,T3,T6,T9,T12,T150T18 'T21* T241
generated by T3.
These are normal of course and are the only candidates
for the kernels of proper homomorphisms of M. We find
that we can not get a homomorphism of 3. If we take
t.v3
as the kernel then T a=Tj a if ii mod 9.
Therefore
MI
-4 (0*3'
.A.4(
Scr = (T/ T/ T1 T/ T'T/1
for d,
namely
Tli
... # T8
T1
and
3
We see also our two choices
and
T21
map onto
T
and
Ti
respectively.
We must define multiplication for the left -Sgroupoid (MI,S*,+.8). Let S'=S. In order that we
insure the homomorphic properties of a we define multiplication by (TiTi)a = (Tiv)(Tia) for T in S.
Since each element in
Ss
has a pre-image in 5,
this
43
determines the multiplication if we agree that for
S
a=T'
Tsi
and
For example
T
in
ij
then T'T' = (Ts Tj )a.
TIT.;
(T19Tj)a
and
T'
T'i
2
(T1 TJ )a' etc.
The multiplication table for S' would then be:
/
T"
IT
/
t
t
1
/
T'4
I
I
T
I
T3
'
T1
T
T,6
T°
1
/
8
$
that this multiplication gives da=db implies
for any choice of d. We also see that T; is our
We observe
a=b
new identity. However no choice of
d
can satisfy pos-
tulate (iii). Thus while (i), (ii), and (ii') are satisfied, postulate (iii) fails.
as a
The only other possibility was to use
kernel and to have M'=Ma = (T'TIT'l. This is an addi-
tive group with S=S'
so
Se=M°.
However
M'
has no
44
left identity e and under our usual multiplication
(sm)a is not (s )(ma) so a is not a homomorphism.
Any left vector space (left module) is a
Example 4.
Mann-module.
A Mann-ring need not be finite. Consider a
Example 5.
of the set of analytic functions of a complex
variable. Let M be the set of simple (schlicht) functions on a region D. Then f(z) is analytic, onevalued, and does not take on one value more than once in
D. So D of the z-plane is represented as D' of
w=f(z)-plane in such a way that there exists a 1-1 corsubset
M
respondence between them. We know a simple function of
a simple function is simple (Titchmarsh, Theory of Functions, 2nd edition, p. 198). Thus fog=f(g) is defined
and M is closed under 4.o. We may take S to be any
set consisting of at least e and one other element d
such that d-e is not 0 or e. We know d -1 is
simple. Then if d is in S there exists d- in M
such that
d -1
and
Example 6.
d-1 d
= e. We therefore take
S
to include
(d
Any Veblen-Wedderbu n system is a Mann-ring
and indeed is a Bose-ring.
45
Example 7.
ments d,
The rings of integers having consecutive ele-
d-1 relatively prime to their characteristics are Mann-rings. For example 115 with d=8,
and
d-1=7, etc.
The Mann-rings of order 3.
If (M,S,+,-) has three elements then M=(0,e,d)
only possible + and
tables are:
Example 8.
and the
where we get the multiplication table by using postulate
(iii) and the addition table. We see it is not necessary
to specify left multiplication for zero, i.e., we can
table is that of GF(3) and the
multiplication table is a subset of that of GF(3). So
if we define 0.m=0 then this is of necessity GF(3). If
take S=iedl.
The
we define
to be not
+
at all then this is not
0
or if we do not define
0.m
GF(3).
The Mann-rings of order 4.
We have noted that given one Mann-ring we can devise
ExaTple 9.
46
more by defining our multiplication for dem in terms of
one of the elements a. For example d(a)m = d.m + a.
We saw this would give rise to as many as 2n rings if
el* = 0. We know GF(22) is a Mann-ring of order 4 and
characteristic 2 so these devices give us 8 Mann-rings of
order 4. We prove these are the only ones and then display these 8. First we make some observations, The group
(M.+)
group.
.
is either the cyclic group of order 4 or the fours
The two addition tables are displayed.
The cyclic group.
We note that ab=c always.
Now the additive group of GF(22)
2.
The fours group.
is the fours group and
hence the 8 Mann-rings already mentioned have table 2 as
their group. We will show that no Mann-ring of order 4
has the cyclic group 1 as its group.
Suppose first we consider the case where e+e is
not equal to d. S contains at least 3 distinct elements
47
e, d,
d-e. If e+e is not d and e+e is not
then we have either:
Case (a): e+e=0 or
Case (b): e+e=d-e.
In Case (a) where e+ =00 e corresponds to b
in table 1 and again we have two choices, namely letting
d correspond to a or to c in table 1. Call these
Cases a 1 and a2. We first let d correspond to a
and
and check out
caseal . A somewhat tedious but routine
calculation shows that it is impossible to define a multiplication from SXM into M satisfying (i)-(iii).
The calculation is simplified by noting that postulates (iii), (i) and the multiplication for d will
determine the table of multiplication. An example is as
follows: If we let d*O=d-e, ded=d0 and d.e=e then
(iii) implies that (d-e)s0=d-e, (d-e)sd=0,
and
(d-e)e=0, and the last two equalities are a contradic-
tion of (ii').
All other choices lead to similar contradictions so
case
al fails. Next we try case a2 which is symmet-
rical and the same simple calculations show that it fails.
Next we try Case b where e+e=d-e so d=e+e+e.
Since we are letting e+e=d-e, we cannot have e+e=0 and
therefore e cannot equal b in table 1. Thus there are
two cases: Case bl where we let e correspond to a,
48
d
to c,
where e is
Now in case
d-e to
and
b
is a,
in table 1; and Case
d-e
is
b2
in table 1.
c,
d
b
we get the addition table displayed:
1
and
b
Again we see that postulate (iii) and this table determines the multiplication of (d-e)..m if d.m is known.
There are only 24 choices for dem and all lead to a
contradiction of postulate (ii'). Similarly Case b2
fails to produce a multiplication table that yields a
Mann-ring.
The only choice left would be to let e+extd.
e=d -e. Then M contains elements (0,e,d,x1 and
e+e=d so d.m=m+m, all m. Now d is non-zero so
is not b in table 1. Therefore we get:
Case (a) e is a in table 1, or
So
Case (b) e is c in table 1.
In both cases d must correspond to b in table 1 so
49
We see then that dm=m+m, all m.
Hence dee=d and dx=d, a contradiction of (ii) so no
Mann-ring can exist whose additive group is the cyclic
group of order 4.
Thus any Mann ring of order 4 must have (Ms+)
equivalent to the fours group (table 2). We note that
e+e=d
and
x+x=d.
so that e+e is not d and so S contains 3 distinct elements and M=[0,e,d,d- 1 which we can write as
e+e=0
10 e d d+e1
since -e=+e.
Now our postulates cannot distinguish between d
and d e where e+e=0 so we get the same + table if
we let e be a, or e be b, or e be c. In fact
these correspondences are all just permutations of the
rows and columns of our additive group since x+x=0 and
0+m=m so we can write down the only possible addition
table:
And as before we use
(d+e m=dm+m,
i.e. postulate (i ),
50
and this addition table to determine the multiplication
table. There are then only 41 cases for d.m and 8 of
these we know give Mann-rings and a simple check show the
rest fail.
We write down the 8 possible cases now. We get
them as follows: Start with GF(22). We know that its
multiplication gives a Mann-ring. We define four new mul-
tiplications as follows:
d(e)m=dm+e
and
d(0)m=dm
d(d)m=dm+d
d(d+e)m=dm+d+e.
This gives 4 Mann-rings whose multiplication tables are:
(e)
e
(
e
d
IMIN
0 111111 d
1111 d+e
d+e 1111
d
d+e
d+e
d
0
0
d+e
The opera ion associated th the element y is indicated
by (Y).
Now since e+e 0, we can get 4 more Mann-rings.
We can use the fact that e+e=0, e e, and hence
to verify that the following definitions actually
yield Mann-rings. Let "." be the multiplication of
d-e=d+e
GF(4).
Define th operation [y]
as follows:
associated with the element
y
d[O]m = (d+e).m + 0;
d[ejm = (d+e).m. + a;
dfdlm = (d+e).m + d:
d(d+e]m = (d+e).m + d+e.
These four multiplications give 4 Mann-rings with the mul-
tiplication tables as follows:
53
CHAPTER Iv
APPLICATIONS OF MANN RINGS TO ORTHOGONAL LATIN SQUARES
Orthogonal Latin squares were first considered in
a paper by Euler in 1782 (5). Early papers on the subject include those of MacNeish and Bose.
ni
n
If pl'p2'p3 *0. P W is the prime power decomposition of an integer
function n(v)
by
v
der v.
n(v)
n
n
n(v) =min(p110,24',
it is known [MacNeish-Mann]
at least
and we define the arithmetic
-1
that there exists a set of
mutually orthogonal Latin squares of or-
be the maximum number of mutually
orthogonal Latin squares of order v. Then the MannLet N(v)
MacNeish theorem is N(v)n(v). MacNeish conjectured that
N(v)=n(v). Since n(v)=1 in this case, this implied the
correctness of Euler% conjecture about the non-existence
of two orthogonal Latin squares of order v=4t+2.
MacNeish's conjecture was disproved by Parker.
Bose and Shrikhande (3) obtained counter examples to
Euler's conjecture.
We saw in Chapter I, (Theorem 0), that Mann-rings
lead to mutually orthogonal Latin squares. In this
54
chapter we investigate further the relation between Mannrings and mutually orthogonal Latin squares. We know the
addition table of (M,+) in a Mann-ring [=(M,$,+,.)
forms a -Latin square whose i,j th element is mi+mj.
also know that another Latin square can be formed whose
.th element is dmi+mj and that these two squares are
J..)
orthogonal if d satisfies postulates (ii), (ii'), and
(iii). Now if there are two such des in S, say d
and d2
and
then they determine two Latin squares, say
M
M2.
To make the discussion easier we introduce some
notation. Let the Mann-ring postulates (i), (ii), (ii'),
and (iii) be referred to as postulate set I. Let Postulate set II be the followings
di-dj is in S.
(di -dj)a=(di-dj)b implies a=b.
(di-dj) =d
THEOREM 4.0.
If
Ml
and
m-dj m.
M2
are the Latin squares de-
fined by d1 and d2 respectively and if d1 and d2 sa -
tisfy postulate sets I and II then
M2
M1
is orthogonal to
55
Proof:
Suppose the
(I j) element equals the (s,t) ele-
ment in the superposition of
(d 1m
.1'
d2 i+mi) = (di s+mt,
d 1m1
- dims
I .e.,
with
Mi
s +mt ).
Then
and
m
d2 m1 + mi = d2m5
So
dimi -d2, i
or
(di -d2)
dims -d2m
= (di -d2)ms by 3) of postulate set II
and therefore
by 2)
mi=ms
and so
mi
Q.E.D.
If di
THEOREM 4.1.
and
they determine Latin squares
Proof:
In
the
Mi
(
ms = m
and
M
,t) element is
Now suppose that dims + m
so
satisfy postulate set
di
M
d1m5
mt.
dimp + mt. Then dims= imp
and therefore s p and each column element
is unique. Similarly, row elements are unique.
If
THEOREM 4.2.
and II then
Proof:
4
d
and
d
satisfy postulate sets I
Is orthogonal to
M4.
The proof is the same as for Theorem 4.0.
56
Notation:
ing
e
Let
S*
be the minimum subset of
such that every element of
S*
except
contain-
S
e
satis-
fies postulate sets I and II. Then we have the following
corollary.
Corollary 4.2: If
is of order
a
then the Mann-ring
3 . (m,s +,.) determines a set of
s
pairwise orthogo-
S*
nal Latin squares.
Proof:
Theorems 4.0, 4.1,
and
4.2.
Now it may be possible to adjoin elements of M-S
to the set S in such a manner that the order of S* is
increased and
is still a Mann-ring. For example, if d is in S but d-e is not in S we can
join d-e to S with the stipulation that the left multiplication for d-e be determined by postulate (iii).
If d-e then failed to satisfy (ii') we would reject it
as a new element of S. If d.m is to satisfy postulate
(ii), there exist only n2 choices for d.m for fixed
d and m so it can be determined if a suitable multiplication is definable from dXM into M such that postulate sets I and II are satisfied. In view of the foregoing discussion, it seems natural to make the following
(M,S,+,.)
57
definition.
Definition: The MacNeish kernel
maximal subset of M containing
is the
S*
of
S*
such that every
either satisfies postulate sets I and II
or can be made to satisfy sets I and II by a suitable
element of
S*
extension of the multiplication of SXM onto
If
M.
is of order k then the Mann-ring
(M,S*,+,0 determines a set of k pairwise orthogonal
THEOREM 4.3.
S*
Latin squares.
Proof: Theorems 4.0, 4.1, 4.2, Corollary 4.2, and the
above definition.
Q.E.D.
THEOREM 4.4.
Let
S*
be of order a. Then there is no
set R in M such that there exists a suitable extension of multiplication from S to R such that
= (M,R,+,.) determines a set of a+1 pairwise orthogonal Latin squares.
Proof:
is defined to be maximal with respect to postulate sets I and II and thus with respect to the number
of pairwise orthogonal Latin squares determined by 3.
S*
No proper subset of S* could determine a
pairwise orthogonal Latin squares.
THEOREM 4.5.
set of
a
58
Proof:
Every element of
satisfies postulate sets I
and not containing deS* is
S*
and II. If a set
C S*
such that (M,k,+,.) determines a pairwise orthogonal
Latin squares then the set i U fdl is in S* and will
determine a+l pairwise orthogonal Latin squares, a contradiction of Theorem 4.4 above.
Q.E.D.
If
THEOREM 4.6.
S* = M-(61
is
then te =
a Bose-ring.
Proof: Routine verification.
We note that
S*
is contained in
S*
so the
MacNeish kernel of 3 will depend upon the original set
S in 3 . (moso.,.) and upon the multiplication from
S*XM
into
M.
So
S*
really is dependent upon all of
We note that a consequence of theorems proved in
the sections on Mann and Bose-rings is that if the multiplication is associative then e is unique in S* and
is the only idempotent element in S.
59
APPENDIX
Some obvious questions may have arisen in the
mind of the reader, some of which should not be asked
and some of which are not new to this paper.
It has already been pointed out that the concept
of "ideal" really does not apply here as the "ring"
structure is too weak to support it.
usual arithmetic of Mann-rings allows
bizarre behavior.
Furthermore the unx.0=0
and other
For this reason the name of "ring"
should not lure one into too serious consideration of
ideals, quotient-Mann-rings, or kernels of homomorphisms.
It would seem more fruitful to investigate unions
inter-
sections, central elements, isomorphisms. etc.
In this paper we mentioned a direct sum decomposi-
tion which may suggest further exploitation.
Some questions are
In a Bose-ring, is
e
the only non-zero idempotent?
In a finite Bose-ring, is
not
0
x.0=0?
We know
in some infinite Bose-rings and
x.0
zero in some Mann-rings so we expect perhaps
not
0
in some finite Bose-rings.
is
is nonx*0
is
The answer has
significance in view of theorems concerning
in Chapter II.
x.0
T-rings
It is postulated in the definition of
Veblen-Wedderburn systems and therefore probably i
60
an independent axiom.
Can we define some sort of "strong homomorphism" that
will preserve postulates (ii), (110), and (iii)?
Then
the image of such a homomorphism would of course be a
Mann-ring.
Can the restriction of associativity be removed from
the theorems concerning idempotents in Bose-rings?
Which of these theorems carry over from Bose-rings
Mann-rings?
What can be said about the set of idempotents in
a
Mann-ring?
We notice that in the proof of Theorem 0 in Chapter I we used the associative law of (M,+) to establish
orthogonality of the two Latin squares. Since we did not
need this associative property of (M.+) to show that 3
actually determined a Latin square, it would appear as if
the group M might be replaced by a loop and still have
Theorem 0 hold true. This would mean a new proof of orthogonality would have to be devised and a new algebraic
structure different from the present Mann-ring would
emerge.
61
BIBLIOGRAPHY
I. Bruck, R. H. A survey of binary systems. BerlinGottingen-Heidelberg, Springer-Verlag, 1958.
185 p.
Bose, R. C.
On the application of the properties of
Galois fields to the problem of construction of
hyper-Graeco Latin squares. Sankhya 3:323-338. 1938.
Bose, R. C. and S. S. Shrikhande. On the falsity of
Euler's conjecture concerning 10X10 orthogonal Latin
squares. Proceedings of the National Academy of
Sciences 45:734-737.
1959.
Bose, R. C., S. S. Shrikhande and E. T. Parker.
Further results on the construction of mutually orthogonal Latin squares and the falsity of Euler's
conjecture. Canadian Journal of Mathematics
12:189-203.
1960.
Euler, Leonhard. Recherches sur une nouvelle espece
de quarres magique. In: Leonhardi Euleri Opera amnia,.
ser. 1, vol. 7.
Leipzig, B. G. Teubner, 1923.
p. 291-392.
Gardner, Martin. Mathematical games.
American, November 19480 p. 181-188.
Scientific
Mann, H. B.
Analysis and design of experiments.
New York, Dover, 1949. 195 p.
MacNeish, Harris F. Euler squares.
matics 23.221-227. 1922.
Annals of Mathe-
Parker E. T. Orthogonal Latin squares.
Proceedings
of the National Academy of Sciences 45:359-862. 1959.