UFD is Integrally Closed

Theorem

Let $A$ be a unique factorisation domain.

Then $A$ is integrally closed.

Proof

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

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

Write $x = a/b$, $a,b \in A$ with $\operatorname{gcd}(a,b) \in A^\times$ (this makes sense because a UFD is GCD Domain).

There is an equation:

$\displaystyle \left(\frac ab\right)^n + a_{n-1}\left(\frac ab\right)^{n-1} + \cdots + a_0$

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

Multiplying by $b^n$, we obtain $a^n + bc = 0$, with $c \in A$.

Therefore $b | a^n$, so if $b$ is not a unit, then $\operatorname{gcd}(a,b) \notin A^\times$, a contradiction.

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

$\blacksquare$