Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-IV 109
Lemma 4. If f(x) is an irreducible polynomial of positive degree in D[x], it is also irreducible in K[x]
where K is the quotient field of D.
Proof. Let if possible, f(x) be reducible in K[x] so that we have a relation of the form
f(x) = g(x) h(x)
where g(x), h(x) are in K[x] and are of positive degree.
Now
g(x) =
h(x) =
a
b
1
1
a
b
2
2
g 1 (x)
h 1 (x)
where a 1 , b 1 , a 2 , b 2 ∈ D and g 1 (x) and h 1 (x) are primitive in D[x].
Thus we have
a1a
2
f(x) = g 1 (x) h 1 (x)
b b
1
2
(b 1 b 2 ) f(x) = (a 1 a 2 ) g 1 (x) h 1 (x)
But by Lemma 1, g 1 (x) h 1 (x) is primitive. The constant of right hand side in a 1 a 2 . Also f(x) being
irreducible in D[x] is primitive and the constant of the left hand side is b 1 b 2 . Therefore, a 1 a 2 = b 1 b 2 .
Therefore
f(x) = g 1 (x) h 1 (x)
This contradicts the fact that f(x) is irreducible in D[x].
Therefore f(x) is irreducible in K[x].
Theorem. The polynomial ring D[x] over a unique factorisation domain D is itself a unique factorisation
domain.
Proof. Let a(x) be any non-zero non-unit member of D[x]. We have
a(x) = ga 0 (x)
where g ∈ D and a 0 (x) is a primitive polynomial belonging to D[x].
Since D is a U.F.D. we have
g = p 1 p 2 …. p r
where p i ’s are prime elements of D.
If now a 0 (x) is reducible, we have
a 0 (x) = a 01 (x) a 02 (x)
where a 01 (x) and a 02 (x) are both primitive of positive degree.
Proceeding in this manner, we shall after a finite number of steps, arrive at a relation of the form
a(x) = p 1 p 2 …. p r a 1 (x) …. a s (x)
where each factor on the right is irreducible.
This shows that D[x] is a f.d.
To show uniqueness, let us suppose that
a(x) = p 1 p 2 …. p r a 1 (x) …. a s (x) = p 1 ′ p 2 ′ …. p l ′ a 1 ′(x) a 2 ′ (x) … a m ′(x)