every PID is a UFD

every PID is a UFD

Theorem 1

.

Every

Principal Ideal Domain

(PID) is a

Unique Factorization Domain

(UFD).

The first step of the proof shows that any PID is a

Noetherian ring

in which every

irreducible

is prime. The second step is to show that any Noetherian ring in which every irreducible is prime is a UFD.

We will need the following

Lemma 2

.

Every PID

R

is a

gcd domain

. Any two gcd’s of a pair of elements

a

,

b

are

associates

of each other.

Proof.

Suppose

a

,

b

∈

R

. Consider the

ideal generated by

a

and

b

,

(

a

,

b

)

. Since

R

is a PID, there is an element

d

∈

R

such that

(

a

,

b

)

=

(

d

)

. But

a

,

b

∈

(

a

,

b

)

, so

d

∣

a

,

d

∣

b

. So

d

is a common

divisor

of

a

and

b

. Now suppose

c

∣

a

,

c

∣

b

. Then

(

d

)

=

(

a

,

b

)

⊂

(

c

)

and hence

c

∣

d

.

The second part of the lemma follows since if

c

,

d

are two such gcd’s, then

(

c

)

=

(

a

,

b

)

=

(

d

)

, so

c

∣

d

and

d

∣

c

so that

c

,

d

are associates.
∎

Theorem 3

.

If

R

is a PID, then

R

is

Noetherian

and every

irreducible element

of

R

is prime.

Proof.

Let

I

1

⊂

I

2

⊂

I

3

⊂

…

be a chain of (principal) ideals
in

R

. Then

I

∞

=

∪

k

I

k

is also an ideal. Since

R

is a PID, there is

a

∈

R

such that

I

∞

=

(

a

)

, and thus

a

∈

I

n

for some

n

. Then for each

m

>

n

,

I

m

=

I

n

. So

R

satisfies the

ascending chain condition

and thus is Noetherian.

To show that each irreducible in

R

is prime, choose some irreducible

a

∈

R

, and suppose

a

=

b

⁢

c

. Let

d

=

gcd

⁡

(

a

,

b

)

. Now,

d

∣

a

, but

a

is irreducible. Thus either

d

is a unit, or

d

is an associate of

a

. If

d

is an associate of

a

, then

a

⁢

∣

d

∣

⁢

b

so that

a

∣

b

and

c

is a unit. If

d

is itself a unit, then we can assume by the lemma that

d

=

1

. Then

1

∈

(

a

,

b

)

so that there are

x

,

y

∈

R

such that

x

⁢

a

+

y

⁢

b

=

1

. Multiplying through by

c

, we see that

x

⁢

a

⁢

c

+

y

⁢

b

⁢

c

=

c

. But

a

∣

x

a

c

and

a

∣

y

b

c

=

y

a

. Thus

a

∣

c

so that

b

is a unit. In either case,

a

is prime.
∎

Theorem 4

.

If

R

is Noetherian, and if every irreducible element of

R

is prime, then

R

is a UFD.

Proof.

We show that any nonzero nonunit is

R

is expressible as a

product

of irreducibles (and hence as a product of primes), and then show that the factorization is unique.

Let

𝒰

⊂

R

be the set of ideals generated by each element of

R

that cannot be written as a product of irreducible elements of

R

. If

𝒰

≠

∅

, then

𝒰

has a

maximal element

(

r

)

since

R

is Noetherian.

r

is not irreducible by construction and thus not prime, so

(

r

)

is not prime and thus not

maximal

. So there is a proper

maximal ideal

(

s

)

with

(

r

)

⊊

(

s

)

, and

s

∣

r

.

Since

(

r

)

is maximal in

𝒰

, it follows that

(

s

)

∉

𝒰

and thus that

s

is a product of irreducibles. Choose some irreducible

a

∣

s

; then

a

∣

r

and

r

=

a

⁢

b

for some

b

∈

R

. If

(

b

)

∉

𝒰

(note that this includes the case where

b

is a unit), then

b

and hence

r

is a product of irreducibles, a contradiction. If

(

b

)

∈

𝒰

then

(

r

)

⊂

(

b

)

(since

b

∣

r

).

(

r

)

≠

(

b

)

since

a

is not a unit, and thus

(

r

)

⊊

(

b

)

. This contradicts the presumed maximality of

(

r

)

in

𝒰

. Thus

𝒰

=

∅

and each element of

R

can be written as a product of irreducibles (primes).

The proof of uniqueness is identical to the standard proof for the

integers

. Suppose

a

=

p

1

⋅

…

⋅

p

n

=

q

1

⋅

…

⋅

q

m

where the

p

i

and

q

j

are primes. Then

p

1

∣

q

1

⋅

…

⋅

q

m

; since

p

1

is prime, it must divide some

q

j

. Reordering if necessary, assume

j

=

1

. Then

p

1

=

u

⋅

q

1

where

u

is a unit.

Factoring

out these terms since

R

is a domain, we get

p

2

⋅

…

⋅

p

n

=

u

⋅

q

2

⋅

…

⋅

q

m

We may continue the process, matching

prime factors

from the two sides.
∎

Title

every PID is a UFD

Canonical name

EveryPIDIsAUFD

Date of creation

2013-03-22 16:55:51

Last modified on

2013-03-22 16:55:51

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

9

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 13F07

Classification

msc 16D25

Classification

msc 11N80

Classification

msc 13G05

Classification

msc 13A15

Related topic

UFD

Related topic

UniqueFactorizationAndIdealsInRingOfIntegers

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