proof that a Noetherian domain is Dedekind if it is locally a PID

We show that for a NoetherianPlanetmathPlanetmathPlanetmath ( domain R with field of fractionsMathworldPlanetmath k, the following are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath.

  1. 1.

    R is Dedekind. That is, it is integrally closedMathworldPlanetmath and every nonzero prime idealMathworldPlanetmathPlanetmath is maximal.

  2. 2.

    for every maximal idealMathworldPlanetmath 𝔪,


    is a principal ideal domainMathworldPlanetmath.

For a given maximal ideal 𝔪 and ideal 𝔞 of R, we shall write 𝔞¯ for the ideal generated by 𝔞 in R𝔪, which consists of the elements of the form s-1a for a𝔞 and sR𝔪. It is then easily seen that 𝔭𝔭¯ gives a bijection between the prime ideals of R contained in 𝔪 and the prime ideals of R𝔪, with inversePlanetmathPlanetmathPlanetmath 𝔭R𝔭. In particular 𝔪¯ is the unique maximal ideal of R𝔪, which is therefore a local ringMathworldPlanetmath.

Now suppose that R is Dedekind, then the localizationMathworldPlanetmath R𝔪 will be a Dedekind domainMathworldPlanetmath (localizations of Dedekind domains are Dedekind) with a unique maximal ideal, so it is a principal ideal domain (Dedekind domains with finitely many primes are PIDs).

Only the converseMathworldPlanetmath remains to be shown, so suppose that R is a Noetherian domain such that R𝔪 is a principal ideal domain for every maximal ideal 𝔪. In particular, R𝔪 is integrally closed and every nonzero prime ideal is maximal, so it contains a unique nonzero prime ideal 𝔪¯.

We start by showing that every nonzero prime ideal 𝔭 of R is maximal. Choose a maximal ideal 𝔪 containing 𝔭. Then, 𝔭¯ is a nonzero prime ideal, so 𝔭¯=𝔪¯ and therefore 𝔭=𝔪 is maximal.

We finally show that R is integrally closed. So, choose any x integral over R and lying in its field of fractions. Let 𝔞 be the ideal


We use proof by contradictionMathworldPlanetmathPlanetmath to show that 𝔞 is the whole of R. So, supposing that this is not the case, there exists a maximal ideal 𝔪 containing 𝔞. Then x will be integral over the integrally closed ring R𝔪 and therefore xR𝔪. So, x=s-1y for some sR𝔪 and yR. Then, sx=yR so s𝔞𝔪, which is the required contradictionMathworldPlanetmathPlanetmath. Therefore, 𝔞=R and, in particular, 1𝔞 and x=1xR, showing that R is integrally closed.

Title proof that a Noetherian domain is Dedekind if it is locally a PID
Canonical name ProofThatANoetherianDomainIsDedekindIfItIsLocallyAPID
Date of creation 2013-03-22 18:35:27
Last modified on 2013-03-22 18:35:27
Owner gel (22282)
Last modified by gel (22282)
Numerical id 4
Author gel (22282)
Entry type Proof
Classification msc 13F05
Classification msc 13A15