Page 45 - Advanced Linear Algebra

Preliminaries 29

then we have expressed as a product of irreducible elements, which is not

possible since irreducible elements are nonunits.
Suppose that both sides have factors left, that is,

Ä ~ Ä

for some . We can
where £ . Then “ Ä , which implies that “

assume by reindexing if necessary that ~ . Since is irreducible
must be a unit. Replacing by and canceling gives

Ä c ~ Ä c
This process can be repeated until we run out of 's or 's. If we run out of 's

"
first, then we have an equation of the form " Ä ~ where is a unit,

which is not possible since the 's are not units. By the same reasoning, we

cannot run out of 's first and so ~ and the 's and 's can be paired off as

associates.
Fields
For the record, let us give the definition of a field a concept that we have been
(
using .
)
Definition A field is a set , containing at least two elements, together with two
-
(
binary operations, called addition denoted by b ) and multiplication
(denoted by juxtaposition ), for which the following hold:
)
1 - is an abelian group under addition.
2 The set - ) i of all nonzero elements in - is an abelian group under
multiplication.
3 )(Distributivity ) For all Á Á  - ,
² b ³ ~ b and ² b ³ ~ b
We require that have at least two elements to avoid the pathological case in
-
which ~ .
Example 0.17 The sets , and , of all rational, real and complex numbers,
d
rs
respectively, are fields, under the usual operations of addition and multiplication
of numbers.
is a field if and only if is a prime number. We

Example 0.18 The ring {
is not a field if is not prime, since a field is also an

have already seen that {
integral domain. Now suppose that ~ is a prime.
is an integral domain and so it remains to show that every
We have seen that {
nonzero element in { has a multiplicative inverse. Let £  { . Since
 , we know that and are relatively prime. It follows that there exist
integers and for which
"
#