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 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:

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

This article, or a section of it, needs explaining. In particular: This is not actually an equation, it has no equals sign in it. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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:

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

By definition of integrally closed integral domain:

$A$ is integrally closed if and only if it is integrally closed in its field of fractions $K$.

Since $x$ is an arbitrary element in the integral closure of $A$ in $K$:

$A$ is integrally closed.

$\blacksquare$

Sources

• 2011: J.S. Milne: Algebraic Number Theory: Chapter $2$ Ring of Integers: Integral elements: Proposition $2.9$