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300
6. RINGS
Theorem 6.5.19.
Every principal ideal domain is a unique factorization
domain.
Proof.
Let
R
be a principal ideal domain, and let
a
2
R
be a nonzero,
nonunit element. According to Lemma
6.5.17
,
a
has at least one factoriza
tion by irreducibles. We have to show that in any two such factorizations,
the factors are the same, up to order and multiplication by units.
So suppose that
0
±
r
±
s
, and that
p
1
;p
2
;:::;p
r
;q
1
;q
2
;:::;q
s
are
irreducibles, and
p
1
p
2
²²²
p
r
D
q
1
q
2
²²²
q
s
:
(6.5.1)
I claim that
r
D
s
, and (possibly after permuting the
q
i
) there exist units
c
1
;:::;c
r
such that
q
i
D
c
i
p
i
for all
i
.
If
r
D
0
, Equation (
6.5.1
) reads
1
D
q
1
²²²
q
s
. It follows that
s
D
0
, since irreducibles are not invertible. If
r
D
1
, Equation (
6.5.1
) reads
p
1
D
q
1
²²²
q
s
; it follows that
s
D
1
and
p
1
D
q
1
, since otherwise
p
1
has
a proper factorization.
Suppose
r
³
2
. We can assume inductively that the assertion holds
when
r
and
s
are replaced with
r
0
and
s
0
with
r
0
< r
and
s
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
 Fall '08
 EVERAGE
 Algebra

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