Proposition 1   Every gcd domain is integrally closed.

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

1. $\GCD(ma,mb)=m\GCD(a,b)$
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 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 $\gcd(a,b)=1$

Write $$f(x)=x^n+c_{n-1}x^{n-1}+\cdots+c_0.$$ So we have $$0=(a/b)^n+c_{n-1}(a/b)^{n-1}+\cdots+c_0.$$ Multiply the equation by $b^n$ then rearrange, and we get $$-a^n=c_{n-1}ba^{n-1}+\cdots+c_0b^n=b(c_{n-1}a^{n-1}+\cdots+c_0b^{n-1}).$$ 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 $a/b\in D$

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 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.