Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

98
ADVANCED ABSTRACT ALGEBRA
(a 0 , a 1 , a 2 ,….) [(b 0 , b 1 , b 2 ,….) (c 0 , c 1 , c 2 ,….)] = (f 0 , f 1 , f 2 ,…)
where
Hence
f m =
j
j+ k+
q=
m
a b c .
k
q
[(a 0 , a 1 , a 2 ,…) (b 0 , b 1 ,..) ] (c 0 , c 1 , c 2 ,….)
= (a 0 , a 1 , a 2 ,…) [(b 0 , b 1 , b 2 ,..) (c 0 , c 1 , c 2 ,..)]
(vi)
where
(a 0 , a 1 , a 2 ,…) [(b 0 , b 1 , b 2 ,…) + (c 0 , c 1 , c 2 ,…)]
= (a 0 , a 1 , a 2 ,…) (b 0 + c 0 , b 1 + c 1 , b 2 + c 2 ,….)
= (d 0 , d 1 , d 2 ,…)
d m = a (b c )
j k
+
j+
k=
m
= a
jbk
+ a
jc
j+
k=
m
k
j+
k=
m
k
= f m + g m , say .
Also
(a 0 , a 1 , a 2 ,…) (b 0 , b 1 , b n ) = (f 0 , f 1 , f 2 ,….) ,
(a 0 , a 1 , a 2 ,…) (c 0 , c 1 , c 2 ,…) = (g 0 , g 1 , g 2 ,…)
Hence
(a 0 , a 1 , a 2 ,…) [(b 0 , b 1 , b 2 ,…) + (c 0 , c 1 , c 2 ,…)]
= (a 0 , a 1 , a 2 ,…) (b 0 , b 1 , b 2 ,…) + (a 0 , a 1 , a 2 ,….) (c 0 , c 1 ,…)
Hence P is a ring. We call this ring of polynomials as polynomial ring over R and it is denoted by R[x].
Let
Q = {(a, 0, 0,…) | a ∈ R}
Then a mapping f : R → Q defined by f(a) = (a, 0, 0,…) is an isomorphism. In fact,
f(a+b) = (a + b, 0, 0,…)
= (a, 0, 0, …) + (b, 0, 0,…)
= f(a) + f(b),
f(ab) = (ab, 0, 0, ….)
= (a, 0, 0, …) (b, 0, 0, …)
= f(a) f(b)
and
f(a)
= f(b) (a, 0, 0,…) = (b, 0, 0, ….)
a = b .
Hence
R ~ Q
(i)
So we can identify the polynomial (a, 0, 0, …) with a.
If we represent (0, 1, 0, …) by x then we can see that
x 2 = (0, 0, 1, 0, …)