Page 223 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes c = a b + a b + a b +a b + a b = 10 (since a = 0 = b ).
4 4 0 3 1 2 2 1 3 0 4 4 4
c = a b +a b + a b + a b +a b + a b, = 14 (since a = 0 = b ).
5
5 0
5
5
2 3
4 1
3 2
0
1 4
So p(x). q(x) = 2 + 3x +2x - 3x + 10x + 14x . 5
2
4
3
Note that p(x). q(x)  Z[X], and deg (p(x) q(x)) = 5 = deg p (x) + deg q (x).
As another example, consider
2
6
2

p(x)  1 2x,q(x)   3x  Z [x].
Then, p(x). q(x) = 2 + 4x + 3x + 6x = 2 + 4x + 3x .
2
2
3
Here, deg (p(x). q(x)) = 2 < deg p (x) + deg q (x) (since deg p (x) = 1, deg q (x) = 2).
In the next section we will show you that
deg (f(x) g(x))  deg f(x) + deg g(x)
By now you must have got used to addition and multiplication of polynomials. We would like
to prove that for any ring R, R[x] is a ring with respect to these operations. For this we must note
that by definition, + and . are binary operations over R [x].
Now let us prove the following theorem. It is true for any ring, commutative or not,
Theorem 1: If R is a ring, then so is R[x], where x is an indeterminate.
Proof: We need to establish the axioms R1 – R6 of Unit 14 for (R[x], + , .).
(i) Addition is Commutative: We need to show that
p(x) + q(x) = q(x) + p(x) for any p(x) , q(x)  R [x].

Let p (x) = a + a x + ... + a,x , and
n
1
0
q(x) = b + b x + ... + b x be in R[x].
m
1
m
0
Then, p (x) + q(x) = c + c x + ... + c x , t
1
0
1
where c = a + b and t = max (m,n).
i
i
i
Similarly,
q(x) + p(x) = d + d x + ... + d x , s
1
s
0
where d = b + a , s = max (n, m) = t.
i
i
i
Since addition is commutative in R, c, =d  i  0.
i
So we have
p(x) + q(x) = q(x) + p(x).
(ii) Addition is Associative: Again, by using the associativity of addition in R, we can show
that if p(x), q(x), s(x)  R[x], then
{p(x) + q (x)} + s(x) = p(x) + {q(x) + s(x)}.
(iii) Additive Identity: The zero polynomial is the additive identity in R [x]. This is because,
for any p(x) = a + a, x+ ... + a x  R[x],
n
0

n
0 + p(x) = (0 + a,) + (0 +a )x + ... +(0 + a )x n
n
1
= a + a x + ... +a x n
0 l n
= p(x)
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