Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

100
ADVANCED ABSTRACT ALGEBRA
If m > n, then
f(x) + g(x) = (a 0 + b 0 ) + (a 1 +b 1 )x + … + (a n +b n )x n
+ a n+1 x n+1 + … + a m x m
Therefore in this situation
deg [f(x) + g(x)] = m
Similarly it can be seen that if m < n, then
deg [f(x) + g(x)] = n
It follows therefore that if m ≠ n, then
deg [f(x) + g(x)] = max (m, n)
Also ,
f(x) g(x) = a 0 b 0 + (a 0 b 1 + a 1 b 0 )x + … + a m b n x m+n
Therefore
deg[f(x) g(x)] = m + n if a
mbn
≠ 0
.
< m + n, where a b = 0
If R is without zero divisor, then
a m b n ≠ 0 since a m ≠ 0, b n ≠ 0 .
Hence for such a ring R we have
deg [f(x) g(x)] = m+n = deg f(x) + deg g(x)
If R is without zero divisor and f(x) and g(x) are non-zero polynomial of R[x], then
deg f(x) ≤ deg [f(x) g(x)] (Θ deg g(x) ≥ 0).
Theorem. If R is an integral domain, then so is also polynomial ring R[x].
Proof. R is a commutative ring with unity. Therefore R[x] is commutative with unit element. It suffices
to prove that R[x] is without zero divisor. Let
and
m
i
f(x) = a x , a i m ≠ 0
i=
0
n
i
g(x) = b x , b i n ≠ 0 ,
i=
0
be two non-zero polynomials of R[x] and let m and n be their degrees respectively.
Since R is an integral domain and a m ≠ 0 , b n ≠ 0 , therefore a m b n ≠ 0 . Hence f(x) g(x) ≠ 0. Hence R[x]
is without zero divisor and therefore an integral domain.
Division Algorithm for polynomials over a field.
Theorem. Corresponding to any two polynomials f(x) and g(x) ≠ 0 belonging to F[x] there exist
uniquely two polynomials q(x) and r(x) also belonging to F[x] such that
f(x) = g(x) q(x) + r(x)
where
r(x) = 0 or deg r (x) < deg g(x) .
m
n
Proof. Let
m
i
f(x) = a i
x , a m ≠ 0
i=
0