205
Internat. J. Math. & Math. Sci.
VOL. 12 NO. 1 (1989) 205-207
ON ANTI-COMMUTATIVE SEMIRINGS
J.S. RATTI and Y.F. LIN
Department of Mathematics
University of South Florida
Tampa, Florida 33620
(Received March 28, 1988)
An anticommutative semiring is completely characterized
by the types of multiplications that are permitted.
It is shown
ABSTRACT.
that a semiring is anticommutative if and only if it is a product
of two semirings R
R2
and R
2
such that R
is
left multiplicative and
is right multiplicative.
KEY WORDS AND PHRASES. Semiring, anticommutative, isomorphism.
1980 AMS SUBJECT CLASSIFICATION CODES.
16A78
A
semiring is a non-empty set
R equipped with two binary
called addition + and multiplication (denoted by
juxtaposition), such that R is multiplicatively a semigroup.
operations,
additively a commutative semigroup and multiplication is
distributive across the addition both from the left and the right.
A semiring R is called anti-commutative if and only if for
arbitrary x, y e R the relation x
y always implies xy
yx.
Let R and R be semirings, then
is the semiring with
2
RlXR 2
the following operations:
(Xl’ x2)
(Xl’ x2)
+
(YI’ Y2
(Yl’ Y2
(Xl + YI’ x2 + Y2 )
(XlYl’ x2Y2)"
J.S. PATTI AND Y.F. LIN
206
is a commutative semigroup under +,
Suppose R
R of type
define multiplication in
(T1)
xy
and if we
x
for all x,y E R
or
(T2)
for all x,y E R,
y
xy
then it is easily seen that R is an anti-commutative semiring.
A natural
Suppose R
Does the multiplication in R
question that arises is the following:
is an anti-commutative semiring.
(T1)
have to be of type
or
(T2)?
to answer
this question, we
prove the following:
THEOREM 1.
is isomorphic to
A
R
semiring R is anti-commutative if and only if
x R
multiplication of type
multiplication of type
in
(T1) and
(T2).
R2
y, z,
is a seniring with
2
xyz
PROOF OF THEOREM 1.
aR.
2
xz
Since
R
is non empty,
By using the lemma,
it
multiplication in aR is of type
f"
f(x)
R
f(xy)
Thus,
type
Set R
(T1)
and
(T2).
Ra x aR, such that for each xER.
f(y)
(ya,ay).
(x+y)a, a(x+y)
(xa + ya, ax + ay)
(xa,ax) + {ya,ay)
f(x) + f(y).
(xya. axy)
(xaya, axay) [By part (ii) of the Lemma]
f(x)f(y).
f is a homomorphism.
Ra
that Ra and aR are
(xa,ax}.
then for yR,
f(x+y)
let aER.
is obvious
semirings and multiplication in Ra is of
Let
is contained
Let R be an anti-commutative semiring, then for arbitrary
R we have
(i)
(ii}
and R
R
is a semirlng with
We shall need the following lemma, whose proof
[1,p.75]. to prove Theorem 1.
LEMMA.
x,
2.
where R
ANTI-COMMUTATIVE SEMIRINGS
To show f
is an isomorphism,
such that g(xa.ay)
207
let us define g" Ra x aR
R.
xy.
Then
2
g(xa,ax)
g(f(x))
(gof}(x}
xa x
x
2
x,
and
(fog)(xa,ay}
f[g(xa,ay}]
f(xy}
(xya,axy}
{xa,ay).
This shows that f" is an isomorphism.
The proof for the converse is left to the reader.
THEOREM 2.
arbitrary
PROOF"
Let R be an anti-commutative semiring.
xER. x + x
Then for an
x.
As in the proof of Theorem 1, we have
x
g(xa,ax).
Thus,
x + x
g(xa + xa. ax + ax)
+ x2a, ax 2 + ax
g(x2a
2)
(x(x + x)a. a(x + x)x)
(xa. ax)
X.
REFERENCES
1.
LJAPIN, E.S.
SemiKroups,
American Ma.th
Providence,
Rhode Island (1963}
Society Translation