5th Cyclotomic Ring is not a Unique Factorization Domain

Theorem
Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.

Then $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is not a unique factorization domain.

The following elements of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are irreducible:


 * $2$
 * $3$
 * $1 + i \sqrt 5$
 * $1 - i \sqrt 5$

Proof
By definition, a unique factorization domain $D$ is an integral domain with the properties that:


 * For all $x \in D$ such that $x$ is non-zero and not a unit of $D$:


 * $(1): \quad x$ possesses a complete factorization in $D$


 * $(2): \quad$ Any two complete factorizations of $x$ are equivalent.

A complete factorization is a tidy factorization
 * $x = u \circ y_1 \circ y_2 \circ \cdots \circ y_n$

such that:
 * $u$ is a unit of $D$
 * all of $y_1, y_2, \ldots, y_n$ are irreducible.

From Units of 5th Cyclotomic Ring, the only units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are $1$ and $-1$.

From Irreducible Elements of 5th Cyclotomic Ring, all of the elements of the set $S$, where:
 * $S := \set {2, 3, 1 + i \sqrt 5, 1 - i \sqrt 5}$

are irreducible in $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$

Then we have:

So there are two tidy factorizations of $6$ in $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ which are not equivalent.

Hence the result.