How to prove that UFD implies normal? or is there is a reference containing this proof?
a ring $R$ is a normal ring if $R$ is an integral domain that is integrally closed in $\operatorname{frac}(R).$
And $R$ is integrally closed in a commutative ring $S$ iff the integral closure of $R$ in $S$ (int($R$)) equals $R$. And int($R$) consists of all $x \in S$ that is integral over R. And note that an element is integral over $R$ if there exists a monic polynomial $f(x) \in R[x]$ such that $f(x) = 0.$
The proof is the same as in the case of integers and rationals.
Let $R$ be a UFD and $Q$ its field of fractions. Let $q = \frac{a}{b} \in Q$ be such that $q$ satisfies a monic polynomial over $R$.
WLOG, we may assume that $a$ and $b$ have no common prime factors. We show that $b$ is a unit and hence, $q \in R$.
Let $f(x) = x^n + a_1x^{n-1} + \cdots + a_n \in R[x]$ be a polynomial that $q$ satisfies. In other words, $f(q) = 0$. Multiplying this equation with $b^n$ gives $$a^n + a_1ba^{n-1} + \cdots + a_nb^n = 0$$ or $$a^n = -b(\underbrace{\cdots}_{\in R}).$$ Thus, every prime factor of $b$ is also one of $a$. By hypothesis, there is no such prime. Thus, $b$ must be a unit.
