Page 224 - DMTH403_ABSTRACT_ALGEBRA

Unit 22: Polynomial Rings

(iv) Additive Inverse: For p (x) = a, + a x +... + a x  R[x], consider the polynomial –p(x) = Notes
n
n
1
–a, –a x – ... –a x , – a being the additive inverse of ai in R. Then
n
1 n i
p(x) + (–p(x)) = (a,, –a,,) + (a – a ) x + ... + (a – a )x n
n
1
1
1
= 0 + 0.x + 0.x + ... + 0.x n
2
= 0.
Therefore, – p(x) is the additive inverse of p(x).
(v) Multiplication is Associative:
Let p(x) =a, +a x + ... +a x ,
n
1
n
q(x) = b + b x+ ... +b x ,
m
m
1
0
and t (x) = d +d x + ... + d x , be in R [x]
r
0
r
1
Then
p(x) . q(x) = c + c x + .. . + c x where s = m + n and
s
,
s
1
0
ck = a b + a b + ... + a b  k = 0,l, ...,s .
k 0
k–1 1
0 k
Therefore,
{p(x) . q(x)} t (x) = e + e x + ... +e x , t
1
1
0
where t = s + r = m+n+r and
e = c d + c d + ... + c d k
k
1
0
k
0
k-1
= (a b + ... + a b )d + (a ,b + ... + a b ) d + ... + a b d .
0
0 0
k
0 k
k 0
k-1
0 k-1
0
1
Similarly, we can show that the coefficient of x (for any k  0) in p(x) (q (x) t(x))
k
is a b d + a , (b d + b d ) + ... + a (b d + b , d + ... + b d )
1
0
0
k
k-1
k
0
k 0
0
0
1
1
0
k-1
= e , by using the properties of + and . in R.
k
Hence, {p(x).q(x)} . t(x) = p(x) . {q (x). t (x)}
(vi) Multiplication Distributes over Addition:
Let p(x) = a +a x + ... + a x ,
n n
0
1
q(x) = b + b x + ...+ b x m
l
m
0
and t(x) = d + d, x + . . . + d x be in R[x],
r
0
r
The coefficient of xk in p (x). (q(x) + t (x)) is
c = a (b + d ) + a(b + d ) + (b + d ) + ... + a, (b + d ).
1
0
k
k
k
k
1
1
1
0
And the coefficient of xk in p (x) q (x) + p (x) t(x) is
(a b + a b + ... + a b ) + (a d + a d + ... +a d ),
0 k
k 0
0
k
k
k-1
1
0
k-1 1
= a (b + d ) +a (b + d ) + ... +a (b + d ) = c k
k
k-1
0
k
k
1
0
0
This is true  k  0.
Hence, p (x) . (q(x) + t (x) ] = p (x) . q(x) + p (x). t.(x).
Similarly, we can prove that
{q(x)+t(x).p(x)=q(x).p(x)+t(x).p(x)
Thus, R [x] is a ring.
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