proof that a domain is Dedekind if its ideals are products of primes

proof that a domain is Dedekind if its ideals are products of primes

We show that for an

integral domain

R

, the following are

equivalent

.

1.

R

is a

Dedekind domain

.

2.

every nonzero

proper ideal

is a

product

of

maximal ideals

.

3.

every nonzero proper ideal is a product of

prime ideals

.

For the equivalence of

1

and

2

see proof that a domain is Dedekind if its ideals
are products of maximals. Also, as every

maximal ideal is prime

, it is immediate that

2

implies

3

. So, we just need to consider the case where

3

is satisfied and show that

2

follows, for which it enough to show that every nonzero prime ideal is maximal.

We first suppose that

𝔭

is an

invertible

(

http://planetmath.org/FractionalIdeal

) prime ideal and show that it is maximal.
To do this it is enough to show that any

a

∈

R

∖

𝔭

gives

𝔭

+

(

a

)

=

R

. First, we have the following

inclusions

,

𝔭

⊊

𝔭

+

(

a

2

)

⊆

𝔭

+

(

a

)

.

Then, consider the

prime factorizations

𝔭

+

(

a

)

=

𝔭

1

⁢

⋯

⁢

𝔭

m

,

(1)

𝔭

+

(

a

2

)

=

𝔮

1

⁢

⋯

⁢

𝔮

n

.

(2)

We write

a

¯

for the image of

a

under the natural homorphism (

http://planetmath.org/NaturalHomomorphism

)

R

→

R

/

𝔭

and

𝔞

¯

for the image of any ideal

𝔞

. Equations (

1

) and (

2

) give

𝔮

¯

1

⁢

⋯

⁢

𝔮

¯

n

=

(

a

¯

)

2

=

(

𝔭

¯

1

⁢

⋯

⁢

𝔭

¯

m

)

2

.

(3)

As

𝔭

is strictly

contained

in

𝔭

+

(

a

)

and

𝔭

+

(

a

2

)

, it must also be strictly contained in

𝔭

k

and

𝔮

k

. So,

𝔭

¯

k

,

𝔮

¯

k

are nonzero prime ideals, and by uniqueness of prime factorization (see,

prime ideal factorization is unique

) Equation (

3

) gives

n

=

2

⁢

m

and

𝔭

¯

k

=

𝔮

¯

k

=

𝔮

¯

k

+

m

, after reordering of the factors. So

𝔭

k

=

𝔮

k

=

𝔮

k

+

m

and,

𝔭

+

(

a

2

)

=

𝔮

1

⁢

⋯

⁢

𝔮

n

=

(

𝔭

1

⁢

⋯

⁢

𝔭

m

)

2

=

(

𝔭

+

(

a

)

)

2

=

𝔭

2

+

a

⁢

𝔭

+

(

a

2

)

.

(4)

Then,

a

∉

𝔭

gives

𝔭

∩

(

a

2

)

⊆

a

⁢

𝔭

and taking the

intersection

of both sides of (

4

) with

𝔭

,

𝔭

=

𝔭

2

+

a

⁢

𝔭

+

𝔭

∩

(

a

2

)

⊆

𝔭

2

+

a

⁢

𝔭

=

(

𝔭

+

(

a

)

)

⁢

𝔭

.

But

𝔭

was assumed to be invertible, and can be cancelled giving

R

⊆

𝔭

+

(

a

)

, showing that

𝔭

is maximal.

Now let

𝔭

be any prime ideal and

a

∈

𝔭

∖

{

0

}

.

Factoring

into a product of primes

𝔭

1

⁢

⋯

⁢

𝔭

n

=

(

a

)

⊆

𝔭

,

(5)

each of the

𝔭

k

is invertible and, by the above

argument

, must be maximal. Finally, as

𝔭

is prime, (

5

) gives

𝔭

k

⊆

𝔭

for some

k

, so

𝔭

=

𝔭

k

is maximal.

Title

proof that a domain is Dedekind if its ideals are products of primes

Canonical name

ProofThatADomainIsDedekindIfItsIdealsAreProductsOfPrimes

Date of creation

2013-03-22 18:35:07

Last modified on

2013-03-22 18:35:07

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

PrimeIdeal

Generated on Fri Feb 9 17:22:38 2018 by

LaTeXML