04 - Principal Ideal Domain are Unique Factorisation Domains

Principal Ideal Domain are Unique Factorisation Domains

Let $R$ be a PID then

$R$ is a UFD

Proof

As irreducibles are prime in a PID then by

Suppose for contradiction $a=a_1\in R$ and $a$ is not a product of irreducible elements
As $a$ is not irreducible then

$$a=b.c\text{ where }b\text{ nor }c\text{ is a unit}$$

If both $b$ and $c$ can be written as a product of prime elements then so is $a$
Hence one of $b$ or $c$ cannot be written as a product of prime element

WLOG suppose it is $b$ and define $a_2=b$
Let

$$I_1=⟨a_2⟩\text{ and }{I}_2=⟨a_2⟩$$

Then

$$I_1⊈I_2$$

As $a_2$ cannot be irreducible then exists $a_3$ such that

$$I_2=⟨a_2⟩⊈⟨a_3⟩=I_3$$

Continuing to get nested sequence of ideals $I_k$ with each strictly bigger than previous
By Ascending chain condition on ideals in a PID, then it cannot happen if $R$ is a pid
Thus $a$ does not exist

By Equivalent statements for an unique factorisation domain then $R$ is a UFD