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

1. 1.

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

2. 2.

   If $\operatorname{GCD}(a,b)=[1]$ and $\operatorname{GCD}(a,c)=[1]$, then $\operatorname{GCD}(a,bc)=[1]$.

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 $\operatorname{GCD}(a,b)$.

Let $K$ be the field of fraction of $D$, and $a/b\in K$ ($a,b\in D$ and $b\neq 0$) is a root of a monic polynomial $p(x)\in D[x]$. We may, from property (1) above, assume that $\gcd(a,b)=1$.

Write

So we have

Multiply the equation by $b^{n}$ then rearrange, and we get

Therefore, $b\mid a^{n}$. Since $\gcd(a,b)=1$, $1=\gcd(a^{n},b)=b$, 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 $a/b\in D$.

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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 $c\mid ab$ with $a,b\in D$. Let $r=\gcd(a,c)$ and $rt=a$, $rs=c$. Then $1=\gcd(s,t)$ 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 $\gcd(s,t)=1$ 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.

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