# Unique Factorization Domain is Integrally Closed

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## Theorem

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

Then $A$ is integrally closed.

## Proof

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

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

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

This makes sense because a UFD is GCD Domain.

There is an equation:

- $\paren {\dfrac a b}^n + a_{n - 1} \paren {\dfrac a b}^{n - 1} + \dotsb + a_0$

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:

- $b \divides a^n$

Aiming for a contradiction, suppose $b$ is not a unit.

Then:

- $\gcd \set {a, b} \notin A^\times$

which is a contradiction.

So $b$ is a unit, and:

- $a b^{-1} \in A$

$\blacksquare$