Unique factorization implies normal

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:

$$p(x) = x^n + p_1x^{n-1} + p_2x^{n-2} + \ldots + p_n$$

then the above statement translates to:

$$a^n + p_1a^{n-1}b + \ldots + p_nb^n = 0$$

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.