Page 44 - Advanced Linear Algebra

28 Advanced Linear Algebra

for some %Á &  9 . Thus,
~ % b &

and since divides both terms on the right, we have “ .
)

To prove 3 , it is clear that if b ~ , then and are relatively prime. For
the converse, consider the ideal º Á » , which must be principal, say
º Á » ~ º%». Then % “ and % “ and so % must be a unit, which implies that
º Á » ~ 9. Hence, there exist Á  9 for which b ~ .
Unique Factorization Domains

Definition An integral domain is said to be a unique factorization domain
9
if it has the following factorization properties:
)
1 Every nonzero nonunit element  9 can be written as a product of a finite
.
number of irreducible elements ~ Ä
2 The factorization into irreducible elements is unique in the sense that if
)
and are two such factorizations, then ~ and
~ Ä ~ Ä
.
after a suitable reindexing of the factors, —
Unique factorization is clearly a desirable property. Fortunately, principal ideal
domains have this property.
Theorem 0.30 Every principal ideal domain 9 is a unique factorization
domain.

Proof. Let  9 be a nonzero nonunit. If is irreducible, then we are done. If
not, then ~ , where neither factor is a unit. If and are irreducible, we

are done. If not, suppose that is not irreducible. Then ~ , where
neither nor is a unit. Continuing in this way, we obtain a factorization of

(
the form after renumbering if necessary)
~ ~ ² ³ ~ ² ³² ³ ~ ² ³² ³ ~ Ä

Each step is a factorization of into a product of nonunits. However, this

process must stop after a finite number of steps, for otherwise it will produce an
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infinite sequence Á Á Ã of nonunits of for which b properly divides .

But this gives the ascending chain of ideals
º »‰º »‰º »‰º »‰Ä

where the inclusions are proper. But this contradicts the fact that a principal
ideal domain satisfies the ascending chain condition. Thus, we conclude that
every nonzero nonunit has a factorization into irreducible elements.
are two such factorizations,
As to uniqueness, if ~ Ä and ~ Ä
then because is an integral domain, we may equate them and cancel like
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factors, so let us assume this has been done. Thus, £ for all Á . If there are
no factors on either side, we are done. If exactly one side has no factors left,