Page 225 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Note that the definitions and theorem in this section are true for any ring. We have not restricted
ourselves to commutative rings. But, the case that we are really interested in is when R is a
domain. In the next section we will progress towards this case.

22.2 Some Properties of R[x]

In the previous section you must have realised the intimate relationship between the operations
on a ring R and the operations on R [x]. The next theorem reinforces this fact.
Theorem 2: Let R be a ring.
(a) If R is commutative, so is R [x].

(b) If R has identity, so does R [x].
Proof: (a) Let p (x) = a + a, x + .. . + a,x and
n
0
q(x) = b + b x + ... + b x be in R[x].
0
m m
1
Then (x) . q(x) = c + c x + .. . + c x , where s = m + n and
s
1
0
s
c = a b + a b +. . .+ a b
1
k-1
k 0
k
0 k
= b a + b a 4 ... + b a + b a , since both addition and multiplication are commutative in R.
0 k
k-1 1
1 k-1
k 0
= coefficient of x in q (x) p(x).
k
Thus, for every i  0 the coefficients of x in p(x) q(x) and q(x) p(x) are equal
1
Hence, P(x) q(x) = q(x) p(x).
(b) We know that R has identity 1. We will prove that the constant polynomial 1 is the identity
of R [X]. Take any
p(x) = a + a x + ... + a x  R[x].
n
0
n
1
Then 1.p(x) = c + c x + ... + c x (since deg 1 = 0),
n
1
n
0
where c = a , 1 + ak . 0 + a . 0 + ... + a .0 = a k
-1
k
0
k-2
k
Thus, 1, p(x) = p (x).
Similarly, p(x). 1 = p(x).
This shows that 1 is the identity of R[x]. ,
In the following exercise we ask you to check if the converse of Theorem 2 is true.
Now let us explicitly state a result which will help in showing us that R is a domain iff R [x} is a
domain, This result follows just from the definition of multiplication of polynomials.
Theorem 3: Let R be a ring and f (x) and g (x) be two non-zero elements of R [x]. Then deg (f(x) g
(x))  deg f(x) + deg g (x), with equality if R is an integral domain.
Proof: Let f (x) = a + a x + ... + a x , a  0,
n
n
n
1
0
and g (x) = b + b x + ... + b x , b  0.
m
m
1
m
0
Then deg f (x) = n, deg g (x) = m. We know that
f (x). g (x) = c +c x + ... + c m+n × m+n ,
1
0
where c = a b + a, b + ... + a b ,.
k
1
0 k
k 0
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