Unique factorization implies normal

This article gives the statement and possibly, proof, of an implication relation between two commutative unital ring properties. That is, it states that every commutative unital ring satisfying the first commutative unital ring property must also satisfy the second commutative unital ring property
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Statement

Verbal statement

Any unique factorization domain is a normal domain.

Proof

Let $R$ be a unique factorization domain. We want to prove that $R$ is inetgrally closed in its field of fractions. In other words, suppose $a,b\in R$ and $a/b$ satisfies a monic polynomial with coefficients in $R$. Then we should prove that $a/b\in R$.

Since $R$ is a UFD, we can, without loss of generality, assume that $a/b$ is in reduced form, viz $gcd(a,b)=1$. Now consider a monic polynomial $p(x)$ such that $p(a/b)=0$. Suppose:

$p(x)=x^{n}+p_{1}x^{n-1}+p_{2}x^{n-2}+\ldots +p_{n}$

then the above statement translates to:

$a^{n}+p_{1}a^{n-1}b+\ldots +p_{n}b^{n}=0$

Suppose there exists a prime $q$ dividing $b$. Then $q$ divides all terms with a $b$ in them, and hence $q|a^{n}$. Since $q$ is prime, this forces $q|a$, contradicting our assumption of $a/b$ being in reduced form.

Thus, there is no prime dividing $b$, so $b$ is a unit. Thus $a/b\in R$ and we are done.