Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-IV 105
If x, y ∈ A , then there exist integers i and j such that
x ∈ (a i ) , y ∈ (a j )
Without loss of generality suppose that i ≥ j . Then x, y ∈ (a i ) . This implies that x − y ∈ (a i ) and α ∈
(a i ) where α ∈ D. Hence x−y, αx ∈ A.
Since D is a principal ideal domain, therefore ∃ an element β of D such that
A = (β) .
There exists, therefore, an ideal member (a m ) of the system such that
β ∈ (a m )
and accordingly
β ∈ (a n ) for all n ≥ m
(β) ⊂ (a n ) for all n ≥ m
Also since (β) is the union of the ideals, we have
(β) ⊃ (a n ) for all n
Thus from (II) and (III)
(β) = (a n ) for all n ≥ m,
(a m ) = (a m+1 ) = (a m+2 ) = ….
which is a contradiction to (I). Hence case (ii) cannot arise.
Thus we have proved that every non-zero non-unit element of a principal ideal domain is expressible as a
product of prime element. Hence D is a F.d.
To prove the uniqueness, let
a = p 1 p 2 … p r = q 1 q 2 …q s
(IV)
where each p and q is prime. We shall prove the result by induction on r. The result is obvious if r = 1.
Suppose now that the result is true for each natural number < r. Since D is a principal ideal domain,
every prime element generates a prime ideal. Therefore p 1 p 2 … p n ∈ (p 1 )
which implies q 1 q 2 …. q s ∈ (p 1 )
Therefore one of the factors q 1 q 2 …. q s should belong to (p 1 ). Without loss of generality say, q 1 ∈ (p 1 ).
Then p 1 |q 1 . As q 1 is prime, this implies q 1 /p 1 and therefore p 1 and q 1 are associates. Let
q 1 = e 1 p 1
(V)
where e 1 is a unit.
From (IV) and (V) we have
p 2 p 3 … p 2 = (e 1 q 2 ) q 3 … q s
(VI)
By the assumed hypothesis
r−1 = s−1
and each factor, on the right of (VI) is an associate of some factor on the left and vice-versa. This proves
the theorem.
(II)
(III)
Theorem. In a principal ideal domain a prime element generates a maximal ideal.
Proof. Let p be a prime element in a principal ideal domain R and let
p = (p)
be an ideal of R generated by p.