proof that a domain is Dedekind if its ideals are invertible

proof that a domain is Dedekind if its ideals are invertible

Let

R

be an

integral domain

with

field of fractions

k

. We show that the following are

equivalent

.

1.

R

is Dedekind. That is, it is

Noetherian

(

http://planetmath.org/Noetherian

),

integrally closed

, and every

prime ideal

is

maximal

(

http://planetmath.org/MaximalIdeal

).

2.

Every nonzero (

integral

) ideal is

invertible

.

3.

Every

fractional ideal

is invertible.

As every fractional ideal is the

product

of an element of

k

and an integral ideal, statements (

2

) and (

3

) are equivalent.
We start by proving that (

3

) implies

R

is Dedekind.

Lemma

.

If every fractional ideal is invertible, then

R

is Dedekind.

Proof.

First, every

invertible ideal is finitely generated

, so

R

is Noetherian.

Now, let

𝔭

be a prime ideal, and

𝔪

be a

maximal ideal

containing

𝔭

. As

𝔪

is invertible, there exists an ideal

𝔞

such that

𝔭

=

𝔪

⁢

𝔞

.
That

𝔭

is a prime ideal implies

𝔞

⊆

𝔭

or

𝔪

⊆

𝔭

. The first case gives

𝔭

⊆

𝔪

⁢

𝔭

and, by cancelling the invertible ideal

𝔭

implies that

𝔪

=

R

, a

contradiction

. So, the second case must be true and, by maximality of

𝔪

,

𝔭

=

𝔪

, showing that all prime ideals are maximal.

Now let

x

be an element of the field of fractions

k

and be integral over

R

.
Then, we can write

x

n

=

c

0

+

c

1

⁢

x

+

⋯

+

c

n

-

1

⁢

x

n

-

1

for coefficients

c

k

∈

R

.
Letting

𝔞

be the fractional ideal

𝔞

=

(

1

,

x

,

x

2

,

…

,

x

n

-

1

)

gives

x

n

∈

𝔞

, so

x

⁢

𝔞

⊆

𝔞

. As

𝔞

is invertible, it can be cancelled to give

x

∈

R

, showing that

R

is integrally closed.
∎

It only remains to show the

converse

, that is if

R

is Dedekind then every nonzero ideal is invertible. We start with the following lemmas.

Lemma

.

Every nonzero ideal

a

contains a product of prime ideals. That is,

p

1

⁢

⋯

⁢

p

n

⊆

a

for some nonzero prime ideals

p

k

.

Proof.

We use

proof by contradiction

, so suppose this is not the case. As

R

is Noetherian, the set of nonzero ideals which do not contain a product of nonzero primes has a

maximal element

(w.r.t. the

partial order

of

set inclusion

) say,

𝔞

.

In particular

𝔞

cannot be prime itself, so there exist

x

,

y

∈

R

such that

x

⁢

y

∈

𝔞

and

x

,

y

∉

𝔞

.
Therefore

𝔞

is strictly

contained

in

𝔞

+

(

x

)

and

𝔞

+

(

y

)

and, by the choice of

𝔞

, these ideals must contain a product of primes. So,

(

𝔞

+

(

x

)

)

⁢

(

𝔞

+

(

y

)

)

=

𝔞

2

+

x

⁢

𝔞

+

y

⁢

𝔞

+

(

x

⁢

y

)

⊆

𝔞

contains a product of primes, which is the required contradiction.
∎

Lemma

.

For any nonzero

proper ideal

a

there is an

element

x

∈

k

∖

R

such that

x

⁢

a

⊆

R

.

Proof.

Let

𝔭

be a maximal ideal containing

𝔞

and

a

be a nonzero element of

𝔞

. By the previous lemma there are prime ideals

𝔭

1

,

…

,

𝔭

n

satisfying

𝔭

1

⁢

⋯

⁢

𝔭

n

⊆

(

a

)

⊆

𝔞

⊆

𝔭

.

We choose

n

as small as possible. As

𝔭

is prime, this gives

𝔭

k

⊆

𝔭

for some

k

and, as every prime ideal is maximal, this is an

equality

. Without loss of generality we may take

𝔭

=

𝔭

n

.
As

n

was assumed to be as small as possible,

𝔭

1

⁢

⋯

⁢

𝔭

n

-

1

is not a subset of

(

a

)

, so there exists

b

∈

𝔭

1

⁢

⋯

⁢

𝔭

n

-

1

∖

(

a

)

. Then,

b

∉

(

a

)

gives

x

≡

a

-

1

⁢

b

∉

R

and

x

⁢

𝔞

⊆

a

-

1

⁢

b

⁢

𝔭

⊆

a

-

1

⁢

𝔭

1

⁢

⋯

⁢

𝔭

n

-

1

⁢

𝔭

⊆

a

-

1

⁢

(

a

)

=

R

as required.
∎

We finally show that every nonzero ideal

𝔞

is invertible. If its

inverse

exists then it should be the largest fractional ideal satisfying

𝔟

⁢

𝔞

⊆

R

, so we set

𝔟

=

{

x

∈

k

:

x

⁢

𝔞

⊆

R

}

.

Choosing any nonzero

a

∈

𝔞

gives

a

⁢

𝔟

⊆

𝔟

⁢

𝔞

⊆

R

so

𝔟

is indeed a fractional ideal. It only remains to be shown that

𝔟

⁢

𝔞

=

R

, for which we use proof by contradiction. If this were not the case then the previous lemma gives an

x

∈

k

∖

R

such that

x

⁢

𝔟

⁢

𝔞

⊆

R

.
By the definition of

𝔟

, this gives

x

⁢

𝔟

⊆

𝔟

and therefore

𝔟

is an

R

⁢

[

x

]

-module. Furthermore, as

R

is Noetherian,

𝔟

will be

finitely generated

as an

R

-module. This implies that

x

is integral over the integrally closed ring

R

, so

x

∈

R

, giving the required contradiction.

Title

proof that a domain is Dedekind if its ideals are invertible

Canonical name

ProofThatADomainIsDedekindIfItsIdealsAreInvertible

Date of creation

2013-03-22 18:34:54

Last modified on

2013-03-22 18:34:54

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Proof

Classification

msc 13A15

Classification

msc 13F05

Related topic

DedekindDomain

Related topic

FractionalIdeal

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