Page 209 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Now, why do you think we have said that Theorem, 7 is true for a PID only? You can see that one
way is true for any domain. Is the other way true for any domain? That is, is every irreducible
element of a domain prime? You will get an answer to this question.

Example: Just now we will look at some uses of Theorem 7.
Theorem 7 allows us to give a lot of examples of prime elements of F[x]. For example, any linear
polynomial over F is irreducible, and hence prime. In the next unit we will particularly consider
irreducibility (and hence primeness) over Q[x].
Now we would like to prove a further analogy between prime elements in a PID and prime
numbers, namely, a result analogous. For this we will first show g very interesting property of
the ideals of a PID. This property called the ascending chain condition, says that any increasing
chain of ideals in a PID must stop after a finite number of steps.
Theorem 8: Let R be a PID and I ,I ,.. .. .. be an infinite sequence of ideals of R satisfying.
1
2
1  1 ....
2
1
Then 3 m  N such that I, = I m+1 = I m+2 = ....

Proof: Consider the set I = I,  I  ... =  I n . We will prove that I is an ideal of R.
2
m 1

Firstly, I  , since I   and I  I.
1
1
Secondly, if a,b  I, then a  I, and b  I for some r,s  N.
s
Assume r  s. Then I  I,. Therefore, a, b  I,. Since I, is an ideal of R, a–b  I,  I. Thus,
s
a–b  I  a, b  I.
Finally, let x  R and a  I. Then a  I, for some r  N.
 xa  I,  I. Thus, whenever x  R and a  I, xa  I.

Thus, I is an ideal of R. Since R is a PID, I = <a> for some a  R. Since a  I, a  I, for some m  N.
Then I  I,. But I,  I. So we. see that I = I .
m
Now, I,  I m+1  Therefore, I, = I m+1
m
Similarly, I, = I m+2 and so on. Thus, Im = I m+1 = I m+2 = ...
Now, for a moment let us go back, where we discussed prime ideals. Over there we said that an
element p  R is prime iff < p > is a prime ideal of R. If R is a PID, we shall use Theorem 7 to make
a stronger statement.
Theorem 9: Let R be a PID. An ideal < a > is a maximal ideal of R iff a is a prime element of R.
Proof: If < a > is a maximal ideal of R, then it is a prime ideal of R. Therefore, a is a prime element
of R.
Conversely, let a be prime and let I be an ideal of R such that < a >  I. Since R is a PID, I = < b
> for some b  R. We will show that b is a unit in R.
< b > = R, i.e., I = R.
Now, < a >  < b >  a = bc for some c  R. Since a is irreducible, either b is an associate of a or
b is a unit in R. But if b is an associate of a, then <b> = <a>, a contradiction. Therefore, b is a unit
in R. Therefore, I = R.

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