proof that a gcd domain is integrally closed
Proposition 1.
Every gcd domain is integrally closed^{}.
Proof.
Let $D$ be a gcd domain. For any $a,b\in D$, let $\mathrm{GCD}(a,b)$ be the collection of all gcd’s of $a$ and $b$. For this proof, we need two facts:

1.
$\mathrm{GCD}(ma,mb)=m\mathrm{GCD}(a,b)$.

2.
If $\mathrm{GCD}(a,b)=[1]$ and $\mathrm{GCD}(a,c)=[1]$, then $\mathrm{GCD}(a,bc)=[1]$.
The proof of the two properties above can be found here (http://planetmath.org/PropertiesOfAGcdDomain). For convenience, we let $\mathrm{gcd}(a,b)$ be any one of the representatives in $\mathrm{GCD}(a,b)$.
Let $K$ be the field of fraction^{} of $D$, and $a/b\in K$ ($a,b\in D$ and $b\ne 0$) is a root of a monic polynomial $p(x)\in D[x]$. We may, from property (1) above, assume that $\mathrm{gcd}(a,b)=1$.
Write
$$f(x)={x}^{n}+{c}_{n1}{x}^{n1}+\mathrm{\cdots}+{c}_{0}.$$ 
So we have
$$0={(a/b)}^{n}+{c}_{n1}{(a/b)}^{n1}+\mathrm{\cdots}+{c}_{0}.$$ 
Multiply the equation by ${b}^{n}$ then rearrange, and we get
$${a}^{n}={c}_{n1}b{a}^{n1}+\mathrm{\cdots}+{c}_{0}{b}^{n}=b({c}_{n1}{a}^{n1}+\mathrm{\cdots}+{c}_{0}{b}^{n1}).$$ 
Therefore, $b\mid {a}^{n}$. Since $\mathrm{gcd}(a,b)=1$, $1=\mathrm{gcd}({a}^{n},b)=b$, by repeated applications of property (2), and one application of property (1) above. Therefore $b$ is an associate^{} of 1, hence a unit and we have $a/b\in D$.
∎
Together with the additional property (call it property 3)
if $\mathrm{GCD}(a,b)=[1]$ and $a\mid bc$, then $a\mid c$ (proof found here (http://planetmath.org/PropertiesOfAGcdDomain)),
we have the following
Proposition 2.
Every gcd domain is a Schreier domain.
Proof.
That a gcd domain is integrally closed is clear from the previous paragraph. We need to show that $D$ is preSchreier, that is, every nonzero element is primal. Suppose $c$ is nonzero in $D$, and $c\mid ab$ with $a,b\in D$. Let $r=\mathrm{gcd}(a,c)$ and $rt=a$, $rs=c$. Then $1=\mathrm{gcd}(s,t)$ by property (1) above. Next, since $c\mid ab$, write $cd=ab$ so that $rsd=rtb$. This implies that $sd=tb$. So $s\mid tb$ together with $\mathrm{gcd}(s,t)=1$ show that $s\mid b$ by property (3). So we have just shown the existence of $r,s\in D$ with $c=rs$, $r\mid a$ and $s\mid b$. 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  20130322 18:19:27 
Last modified on  20130322 18:19:27 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  6 
Author  CWoo (3771) 
Entry type  Derivation 
Classification  msc 13G05 