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 $\gcd (a,b)$ 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 $\gcd (a,b)=1$.

Write

$$f(x)=x^n+c_{n-1}{x}^{n-1}+\mathrm{\cdots }+c_0.$$

So we have

$$0={(a/b)}^n+c_{n-1}{(a/b)}^{n-1}+\mathrm{\cdots }+c_0.$$

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

$$-a^n=c_{n-1}b{a}^{n-1}+\mathrm{\cdots }+c_0{b}^n=b(c_{n-1}{a}^{n-1}+\mathrm{\cdots }+c_0{b}^{n-1}).$$

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 $a/b\in D$.

∎

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 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.

∎