Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

108
ADVANCED ABSTRACT ALGEBRA
We have k = gh −1
f 1 (x) = kf 2 (x) where 0 ≠ k ∈ K
∴ hf 1 (x) = gf 2 (x)
where g ∈ D , h ∈ D
∴ f 1 (x) ~ f 2 (x) in D[x]
(Application of Lemma III).
Lemma 3. Every non-zero member f(x) of D[x] is expressible as a product cg (x) of c ∈ D and of a
primitive member g(x) of D[x] and this expression is unique apart from the differences in associateness.
Proof. Let c be the H.C.F. of the set
{a 0 , a 1 ,…, a i ,…, a n }
of the coefficients of f(x).
Let
a i = cb i , 0 ≤ i ≤ n
Consider the set
{b 0 ,…., b i ,….,b n }
This set has no common factor other than units. Thus
n
i
g(x) = b ix
i=
0
is a primitive polynomial member of D[x] and we have f(x) = cg(x)
which expresses f(x) as required.
We now attend to the proof of the uniqueness part of the theorem.
If possible let
f(x) = c g (x)
f(x) = d h(x)
where g(x) and h(x) are primitive members of D[x].
We have therefore
cg (x) = dh(x)
cb i = dc i
This implies that each prime factor of c is a factor of dc i for all 0 ≤ i ≤ n. This prime factor of c must
not, however be a factor of some c i .
It follows that each prime factor of c is a factor of d that c is a factor of d.
Similarly, it follows that d is a factor of c. Thus c and d are associates. Let c = ed where e is a unit.
Also since
cg(x) = dh(x)
it follows that
eg(x) = h(x)
implying that g(x) and h(x) are associates.
Hence the lemma.
Definition. A polynomial p(x) in F[x] is said to be irreducible over F if whenever p(x) = a(x) b(x), with
a(x), b(x) ∈ F[x], then one of a(x) or b(x) has degree zero (i.e. is constant).