proof that a gcd domain is integrally closed

Proposition 1.

Let D be a gcd domain. For any a,bD, let GCD(a,b) be the collection of all gcd’s of a and b. For this proof, we need two facts:

  1. 1.


  2. 2.

    If GCD(a,b)=[1] and GCD(a,c)=[1], then GCD(a,bc)=[1].

The proof of the two properties above can be found here ( For convenience, we let gcd(a,b) be any one of the representatives in GCD(a,b).

Let K be the field of fractionMathworldPlanetmath of D, and a/bK (a,bD and b0) is a root of a monic polynomial p(x)D[x]. We may, from property (1) above, assume that gcd(a,b)=1.



So we have


Multiply the equation by bn then rearrange, and we get


Therefore, ban. Since gcd(a,b)=1, 1=gcd(an,b)=b, by repeated applications of property (2), and one application of property (1) above. Therefore b is an associateMathworldPlanetmath of 1, hence a unit and we have a/bD.

Together with the additional property (call it property 3)

if GCD(a,b)=[1] and abc, then ac (proof found here (,

we have the following

Proposition 2.

Every gcd domain is a Schreier domain.


That a gcd domain is integrally closed is clear from the previous paragraph. We need to show that D is pre-Schreier, that is, every non-zero element is primal. Suppose c is non-zero in D, and cab with a,bD. Let r=gcd(a,c) and rt=a, rs=c. Then 1=gcd(s,t) by property (1) above. Next, since cab, write cd=ab so that rsd=rtb. This implies that sd=tb. So stb together with gcd(s,t)=1 show that sb by property (3). So we have just shown the existence of r,sD with c=rs, ra and sb. Therefore, c is primal and D is a Schreier domain.

Title proof that a gcd domain is integrally closed
Canonical name ProofThatAGcdDomainIsIntegrallyClosed
Date of creation 2013-03-22 18:19:27
Last modified on 2013-03-22 18:19:27
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 6
Author CWoo (3771)
Entry type Derivation
Classification msc 13G05