Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-IV 101
Then either
(i)
or
n
i
g(x) = b i
x , b n ≠ 0 .
i=
0
deg f(x) < deg g(x)
(ii) deg f(x) ≥ deg g(x)
In the first case we write
f(x) = g(x) 0 + f(x)
so that q(x) = 0 and r(x) = f(x).
In respect of the second case we shall prove the existence of q(x) and r(x) by mathematical induction on
the degree of f(x). If deg f(x) = 1, then the existence of q(x) and r(x) is obvious. Let us suppose that the
result is true when deg f(x) ≤ m−1. If
a
m
h(x) = f(x) −
x m−n g(x)
bn
f(x) = a 0 + a 1 x + … + a m x m
= a m b n −1 x m−n (b 0 + b 1 x + … b n x n )
(iii)
+ (a m−1 − a m b n −1 b n−1 ) x m−1 + (a m−2 − a m b n −1 b n−2 ) x m−2
= a m b n −1 x m−n g(x) + h(x)
then deg h(x) ≤ m−1.
Hence by supposition
h(x) = g(x) q 1 (x) +r(x) ,
where r(x) = 0 or deg r(x) < deg g(x) .
From (iii) and (iv) we have
a
f(x) −
b
That is,
m
n
x m−n g(x) = g(x) q 1 (x) + r(x)
a
f(x) = g(x) [q 1 (x) +
b
= g(x) q(x) + r(x)
m
n
x m−n ] + r(x)
a
m
where q(x) = q 1 (x) +
x m−n
bn
Thus existence of q(x) and r(x) is proved.
Now we shall prove the uniqueness of q(x) and r(x).
Let us suppose that q 1 (x) and r 1 (x) are two polynomials belonging to F[x] such that
f(x) = g(x) q 1 (x) + r 1 (x)
where r 1 (x) = 0 or deg r 1 (x) < deg g(x).
But by the statement of the theorem, q(x) and r(x) are two elements of F(x) such that
f(x) = g(x) q(x) + r(x)
where r(x) = 0 or deg r(x) < deg g(x).
Hence
(iv)