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

We show that for an integral domainMathworldPlanetmath R, the following are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath.

  1. 1.

    R is a Dedekind domainMathworldPlanetmath.

  2. 2.

    every nonzero proper idealMathworldPlanetmath is a productPlanetmathPlanetmath of maximal idealsMathworldPlanetmath.

  3. 3.

    every nonzero proper ideal is a product of prime idealsMathworldPlanetmathPlanetmath.

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 invertiblePlanetmathPlanetmath ( prime ideal and show that it is maximal. To do this it is enough to show that any aR𝔭 gives 𝔭+(a)=R. First, we have the following inclusions,


Then, consider the prime factorizationsMathworldPlanetmath

𝔭+(a)=𝔭1𝔭m, (1)
𝔭+(a2)=𝔮1𝔮n. (2)

We write a¯ for the image of a under the natural homorphism ( RR/𝔭 and 𝔞¯ for the image of any ideal 𝔞. Equations (1) and (2) give

𝔮¯1𝔮¯n=(a¯)2=(𝔭¯1𝔭¯m)2. (3)

As 𝔭 is strictly contained in 𝔭+(a) and 𝔭+(a2), it must also be strictly contained in 𝔭k and 𝔮k. 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=2m and 𝔭¯k=𝔮¯k=𝔮¯k+m, after reordering of the factors. So 𝔭k=𝔮k=𝔮k+m and,

𝔭+(a2)=𝔮1𝔮n=(𝔭1𝔭m)2=(𝔭+(a))2=𝔭2+a𝔭+(a2). (4)

Then, a𝔭 gives 𝔭(a2)a𝔭 and taking the intersectionMathworldPlanetmath of both sides of (4) with 𝔭,


But 𝔭 was assumed to be invertible, and can be cancelled giving R𝔭+(a), showing that 𝔭 is maximal.

Now let 𝔭 be any prime ideal and a𝔭{0}. Factoring into a product of primes

𝔭1𝔭n=(a)𝔭, (5)

each of the 𝔭k is invertible and, by the above argument, must be maximal. Finally, as 𝔭 is prime, (5) gives 𝔭k𝔭 for some k, so 𝔭=𝔭k 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