Unique Factorization Domain is Integrally Closed

Theorem
Let $A$ be a unique factorization domain (UFD).

Then $A$ is integrally closed.

Proof
Let $K$ be the quotient field of $A$.

Let $x \in K$ be integral over $A$.

Let: $x = a / b$ for $a, b \in A$ with $\gcd \left({a, b}\right) \in A^\times$.

This makes sense because a UFD is GCD Domain.

There is an equation:


 * $\left({\dfrac a b}\right)^n + a_{n-1} \left({\dfrac a b}\right)^{n-1} + \dotsb + a_0$

with $a_i \in A$, $i = 0, \dotsc, n-1$.

Multiplying by $b^n$, we obtain:
 * $a^n + b c = 0$

with $c \in A$.

Therefore:
 * $b \mathrel \backslash a^n$

Suppose $b$ is not a unit.

Then:
 * $\gcd \left({a, b}\right) \notin A^\times$

which is a contradiction.

So $b$ is a unit, and:
 * $a b^{-1} \in A$