Vol. 11 No 1 - Pi Mu Epsilon

8 CANNON AND STEPHENS
We also note that .IR[x, y] is not a commutative nearring; in other words, composition
of polynomials is not commutative. As an example, consider f(x, y) =
x 2 and g(x, y) = x + y. Thus f(x, y), g(x, y) E .IR[x, y]. Then (f o g)(x, y) =
f(g(x,y),g(x,y)) = f(x+y,x+y) = (x+y) 2 and (gof)(x,y) =g(f(x,y),f(x,y)) =
g(x 2 , x 2 ) = x 2 + x 2 = 2x 2 . Clearly fog f. go f and .lR [x, y] is not commutative.
In addition to lacking commutativity of composition, .JR[x, y] does not have a right
identity. Therefore, there is no two-sided identity either. We verify this in our first
lemma.
LEMMA 1. The nearring (.JR [x, y], +, o) does not contain a right identity.
Proof. Suppose that i(x, y) E .lR [x, y] is a right identity. Then f o i = f for all
f E .IR[x, y]. In particular, let f(x, y) = x and g(x, y) = y. Then f(x, y) and g(x, y)
are elements of .IR[x, y]. Sox = f(x, y) = (f o i)(x, y) = f(i(x, y), i(x, y)) = i(x, y)
and y = g(x, y) = (go i)(x, y) = g(i(x, y), i(x, y)) = i(x, y). Hence i(x, y) = x = y, a
contradiction. So .JR[x,y] does not have a right identity. D
Left identities do exist as shown in the next theorem. We use the notation
f(x, y) = L (aj,k) xiyk to represent an arbitrary polynomial of degree n in
O$j+k$n
.lR [x, y], where aj,k E .lR for all j and k.
THEOREM 2. The following are equivalent:
(i) i(x,y) =
L (aj,k)xiyk is a left identity of.JR[x ,y];
O$j+k$n
(ii) i(x,x)=x;
(iii) ao,o = 0, a1,0 + ao,l = 1, and L aj,k = 0 for all 2 ~ m ~ n.
i+k=m
Proof. Assume condition (i) holds. We show that condition (ii) is true. Since
i(x, y) is a left identity of .lR [x, y], then (i o f)(x, y) = i(f(x, y), f(x, y)) = f(x, y) for
all f(x, y) E .IR[x, y]. In particular, let f(x, y) = x. Then x = f(x, y) = (i o f)(x, y) =
i(f(x, y), f(x, y)) = i(x, x). So condition (i) implies condition (ii).
Now assume that condition (ii) is true. We show that condition (iii) holds. Then
x = i(x, x) = L (aj,k)xixk = L (aj,k)xi+k. But x = L (aj,k)xi+k
O$j+k$n O$j+k$n O$j+k$n
implies that ao,o = 0, a1,0 + ao,1 = 1, and
gives that condition (ii) implies condition (iii).
L aj,k = 0 for all 2 ~ m ~ n. This
j+k=m
It is straightforward to verify that if i(x, y) = L (aj,k)xiyk such that a 0 ,0 =
O$j+k$n
0, a1,0 + ao, 1 = 1, and L aj,k = 0 for all 2 ~ m ~ n, then i(x, x) = x. This gives
j+k=m
(iii) implies (ii).
Finally, assume condition (ii) is true and let f(x, y) E .IR[x, y]. Then (io f)(x , y) =
i(f(x, y) , f(x , y)) = f(x, y). So i(x, y) is a left identity for .lR [x, y]. D
By Theorem 2, the polynomial i 1 (x, y) = x 2 -3y 2 +2xy+3x-2y E .IR[x, y] is a left
identity since the coefficients of the quadratic terms sum to zero, the coefficients of the
linear terms sum to one, and the constant term is zero. Similarly, i2(x, y) = 14x -13y
and i 3 (x, y) = 4x 3 - 3x 2 y- y 3 - 11x + 12y are also left identities in .IR[x, y].
While we have shown that there are no right identities in .JR[x, y], there are some
subnearrings of .lR [x, y] for which a left identity is also a right identity. For each left
identity i(x,y) E .JR[x,y], define the set si = {f(x,y) E .JR[x,y] If oi = f}.
SUBNEARRINGS 9
THEOREM 3. For each left identity i(x, y) E .lR [x, y], Si is a subnearring of
.lR [x, y].
Proof. To show that Si is a subnearring of .IR[x, y], we need to show that (Si, + )
is a subgroup of (.JR[x,y],+) and that Si is closed under composition ([2], p. 5).
Let Sl' S2 E si. So S! oi = Sl and S2 oi = S2· Then (sl -s2)oi = (sl oi)- (s2 oi) =
St - S2· Hence Sl - S2 E si and (Si, +) is a subgroup of (.IR [x, y], + ).
Also, (sl 0 s2) 0 i = S! 0 (s2 0 i) = Sl 0 S2· Hence, S! 0 S2 E si, and si is closed
under composition. Therefore, Si is a subnearring of .lR [x, y]. D
Even though we are not using the operation of standard polynomial multiplication
in our nearring, it is useful to note that the subnearring Si is also closed under this
multiplication.
LEMMA 4. For each left identity i(x, y) E .!R[x, y], Si is closed under standard
polynomial multiplication.
Proof. LetS!, 82 E si. So S! oi = Sl and 82 oi = 82. Therefore (sl (x, y). s2(x, y)) 0
i(x,y) = ((s1 · s2)(x,y)) o i(x,y) = (s1 · s2)(i(x,y),i(x,y)) = s1(i(x,y),i(x,y)) ·
s2(i(x, y), i(x, y)) = [s1(x, y) o i(x, y)] · [s2(x, y) o i(x, y)] = s1 (x, y) · s2(x, y) and
s1 . s2 E si. o
We now completely determine the specific elements in Si. To begin, we first
notice if f(x, y) = c where cis any real number, then f(x, y) E Si since (f o i)(x, y) =
f( i(x, y), i(x, y)) = c = f(x, y). Also, i(x, y) E Si since i(x, y) is a left identity for
.JR[x, y] and (i o i)(x, y) = i(x, y).
Since Si is closed under addition, then f(x, y) = i(x, y) + c is an element of
si for any real number c. Furthermore, since si is closed under standard polynomial
multiplication, then g(x, y) = c[i(x, y)]n is an element of si for any constant
c and any integer n ~ 1. It follows that any polynomial of the form h(x, y) =
n
L (
aj) [i( X' y) ]i is an element of si. In fact, these are all of the elements of si since if
j=O
f(x, y) = L (aj,k)xiyk E si, then f = f 0 i implies that f(x, y) = (f 0 i)(x, y) =
O$j+k$n
n
L (aj,k)[i(x, y)]i[i(x, y)]k = L (am)[i(x, y)]m for appropri-
f(i(x, y), i(x, y)) =
O$j+k$n
m=O
ately chosen values of am. We have just proven the following characterization theorem.
THEOREM 5. For each left identity i(x, y) E .IR[x, y],
S, = {f(x,y) E IR[x, y] f(x,y) = ~(a;)[i(x,y)]; for some 0::; n E &: } .
Now we focus on finding invertible elements in Si. To facilitate this, we first note
that si is more familiar to us than it might appear.
THEOREM 6. For each left identity i(x, y) E .lR [x, y], (Si, +, o) is nearring isomorphic
to (.JR[x], +, o), the nearring of polynomials in one variable.
n
n
Proof. The mapping w: si---+ .JR[x] given by w(:L:(aj)[i(x,y)]i) = L(aj)Xj is
j=O
j=O
a nearring isomorphism. The details of the proof are left to the reader. 0
Thus, to find invertible elements in Si, we only need to look for invertible elements
in .IR[x]. The following known result answers this question.