an integral domain is lcm iff it is gcd

an integral domain is lcm iff it is gcd

Proposition 1

.

Let

D

be an

integral domain

. Then

D

is a

lcm domain

iff it is a gcd domain.

This is an immediate

consequence

of the following

Proposition 2

.

Let

D

be an integral domain and

a

,

b

∈

D

. Then the following are

equivalent

:

1.

a

,

b

have an lcm,

2.

for any

r

∈

D

,

r

⁢

a

,

r

⁢

b

have a gcd.

Proof.

For arbitrary

x

,

y

∈

D

, denote

LCM

⁡

(

x

,

y

)

and

GCD

⁡

(

x

,

y

)

the sets of all lcm’s and all gcd’s of

x

and

y

, respectively.

(

1

⇒

2

)

. Let

c

∈

LCM

⁡

(

a

,

b

)

. Then

c

=

a

⁢

x

=

b

⁢

y

, for some

x

,

y

∈

D

. For any

r

∈

D

, since

r

⁢

a

⁢

b

is a multiple of

a

and

b

, there is a

d

∈

D

such that

r

⁢

a

⁢

b

=

c

⁢

d

. We claim that

d

∈

GCD

⁡

(

r

⁢

a

,

r

⁢

b

)

. There are two steps: showing that

d

is a common

divisor

of

r

⁢

a

and

r

⁢

b

, and that any common divisor of

r

⁢

a

and

r

⁢

b

is a divisor of

d

.

1.

Since

c

=

a

⁢

x

, the equation

r

⁢

a

⁢

b

=

c

⁢

d

=

a

⁢

x

⁢

d

reduces to

r

⁢

b

=

x

⁢

d

, so

d

divides

r

⁢

b

. Similarly,

r

⁢

a

=

y

⁢

d

, so

d

is a common divisor of

r

⁢

a

and

r

⁢

b

.

2.

Next, let

t

be any common divisor of

r

⁢

a

and

r

⁢

b

, say

r

⁢

a

=

u

⁢

t

and

r

⁢

b

=

v

⁢

t

for some

u

,

v

∈

D

. Then

u

⁢

v

⁢

t

=

r

⁢

a

⁢

v

=

r

⁢

b

⁢

u

, so that

z

:=

a

⁢

v

=

b

⁢

u

is a multiple of both

a

and

b

, and hence is a multiple of

c

, say

z

=

c

⁢

w

for some

w

∈

D

. Then the equation

a

⁢

x

⁢

w

=

c

⁢

w

=

z

=

a

⁢

v

reduces to

x

⁢

w

=

v

. Multiplying both sides by

t

gives

x

⁢

w

⁢

t

=

v

⁢

t

. Since

v

⁢

t

=

r

⁢

b

=

x

⁢

d

, we have

x

⁢

d

=

x

⁢

w

⁢

t

, or

d

=

w

⁢

t

, so that

d

is a multiple of

t

.

As a result,

d

∈

G

⁢

C

⁢

D

⁢

(

r

⁢

a

,

r

⁢

b

)

.

(

2

⇒

1

)

. Suppose

k

∈

GCD

⁡

(

a

,

b

)

. Write

k

⁢

i

=

a

,

k

⁢

j

=

b

for some

i

,

j

∈

D

. Set

ℓ

=

k

⁢

i

⁢

j

, so that

a

⁢

b

=

k

⁢

ℓ

. We want to show that

ℓ

∈

LCM

⁡

(

a

,

b

)

. First, notice that

ℓ

=

a

⁢

j

=

b

⁢

i

, so that

a

∣

ℓ

and

b

∣

ℓ

. Now, suppose

a

∣

t

and

b

∣

t

, we want to show that

ℓ

∣

t

as well. Write

t

=

a

⁢

x

=

b

⁢

y

. Then

t

⁢

a

=

a

⁢

b

⁢

y

and

t

⁢

b

=

a

⁢

b

⁢

x

, so that

a

b

∣

t

a

and

a

b

∣

t

b

. Since

GCD

⁡

(

t

⁢

a

,

t

⁢

b

)

≠

∅

, we have

t

⁢

k

∈

GCD

⁡

(

t

⁢

a

,

t

⁢

b

)

(see proof of this here (

http://planetmath.org/PropertiesOfAGCDDomain

)), implying

a

b

∣

t

k

. In other words

t

⁢

k

=

a

⁢

b

⁢

z

for some

z

∈

D

. As a result,

t

⁢

k

=

a

⁢

b

⁢

z

=

k

⁢

ℓ

⁢

z

, or

t

=

ℓ

⁢

z

. In other words,

ℓ

∣

t

, as desired.
∎

Since the first statement is equivalent to

D

being an lcm domain, and the second statement is equivalent to

D

being a gcd domain,

Proposition

1 follows.

Another way of stating Proposition 1 is the following: let

L

be the set of

equivalence classes

on the integral domain

D

, where

a

∼

b

iff

a

and

b

are

associates

.

Partial order

L

so that

[

a

]

≤

[

b

]

iff

a

⁢

c

=

b

for some

c

∈

D

. Then

L

is a

semilattice

(upper or lower) implies that

L

is a

lattice

.

Title

an

integral domain is lcm iff it is gcd

Canonical name

AnIntegralDomainIsLcmIffItIsGcd

Date of creation

2013-03-22 18:19:38

Last modified on

2013-03-22 18:19:38

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

10

Author

CWoo (3771)

Entry type

Derivation

Classification

msc 13G05

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