Page 210 - DMTH403_ABSTRACT_ALGEBRA

Unit 20: Principal Ideal Domains

Thus, <a> is a maximal ideal of R. Notes
What Theorem 9 says is that the prime ideals and maximal Ideals coincide in a PID.

Now, take any integer n. Then we can have n = 0, or n = ± 1, or n has a prime factor. This property
of integers is true for the elements of any PID, as you will see now.
Theorem 10: Let R be a PID and a be a non-zero non-invertible element of R. Then there is some
prime element p in R such that a.
Proof: If a is prime, take p = a. Otherwise, we can write a =albl, where neither a, nor b is an
1
associate of a. Then < a >  < a >. If a is prime, take p = a . Otherwise, we can write a = a b ,
1
1
1
1
2 2
where neither a nor b is an associate of a,. Then <a >  < a >. Continuing in this way we get
2
1
2
2
an increasing chain
<a>  <a >  <a >  ...
1
2
By Theorem 8, this chain stops with some < a, >. Then a, will be prime, since it doesn’t have any
non-trivial factors. Take p = a,, and the theorem is proved.
And now we are in a position to prove that any non-zero non-invertible element of a PID can be
uniquely written as a finite product of prime elements (i.e., irreducible elements).
Theorem 11: Let Rt be a PID. Let a  R such that a  0 and a is not a unit. Then a = p ,p ....p , where
1
2
r
p ,p .... p , are prime elements of R.
r
2
1
Proof: If a is a prime element, there is nothing to prove. If not, then P | a, for some prime p in
1
1
R, by Theorem 10. Let a = p a . If p a . If a is a prime, we are through. Otherwise P | a, for some
l l
1 1
1
2
prime p in R. Let a , = p a . Then a = p p a . If a is a prime, we are through. Otherwise we
2
2 2
1
2 2
1
2
continue the process. Note that since al is a non-trivial factor of a, <a>  <a >. Similarly, <a > 
1
1
< a >. So, as the process continues we get an increasing chain of ideals,
2
<a>  <a >  <a >  ...
1
2
in the PID R. Just as in the proof of Theorem 10, this chain ends at < a, > for some m  N, and a,
is irreducible.
Hence, the process stops after m steps, i.e., we can write a = p p ... p , where p is a prime element
m
2
i
1
of R  i = 1, .... m.
Thus, any non-zero non-invertible element in a PID can be factorised into a product of primes.
What is interesting about this factorisation is the following result that you have already proved
for Z in Unit 1.
Theorem 12: Let R be a PID and a  0 be non-invertible in R. Let a = p p ....p = q q ....q , where
n
2
m
1 2
1
p and q are prime elements of R. Then n = m and each p is an associate of some q for 1  i | n,
j
i
i
j
1  j | m.
Before going into the proof of this result, we ask you to prove a property of prime elements that
you will need in the proof.
Task Use induction on n to prove that if p is a prime element in an integral domain
R and if p | a a ... a, (where a , a ,.. .., a,  R), then p | a , for some i = 1, 2. .... n.
2
l
1 2
i
Now let us start the proof of Theorem 12.
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