Irreducible Elements of 5th Cyclotomic Ring

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

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


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

Proof
Let $z = x + i y$ be an element of $\Z \sqbrk {i \sqrt 5}$ in the set $S$, where:
 * $S := \set {2, 3, 1 + i \sqrt 5, 1 - i \sqrt 5}$

Let $z$ have a non-trivial factorization:
 * $z = z_1 z_2$

where neither $z_1$ nor $z_2$ are units of $\Z \sqbrk {i \sqrt 5}$.

Let $\map N z$ denote the field norm of $z \in \Z \sqbrk {i \sqrt 5}$.

Then:

Then we have:

From Elements of 5th Cyclotomic Ring with Field Norm 1, the only elements of $\Z \sqbrk {i \sqrt 5}$ whose field norm is $1$ are the units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$: $1$ and $-1$.

From 5th Cyclotomic Ring has no Elements with Field Norm of 2 or 3, none of $4$, $6$ and $9$ have proper divisors which are field norms of elements of $\Z \sqbrk {i \sqrt 5}$.

Thus either $z_1$ or $z_2$ is a unit of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.

So none of the elements of $S$ has a non-trivial factorization in $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.

Hence the result, by definition of irreducible.