# 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

A particular theorem is missing. In particular: For the concept of irreducibility to be defined, it needs to be demonstrated that $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is actually an integral domain. |

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:

\(\ds \map N z\) | \(=\) | \(\ds \map N {z_1 z_2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map N {z_1} \map N {z_2}\) | Definition of Field Norm of Complex Number |

Then we have:

\(\ds \map N 2\) | \(=\) | \(\ds 2^2 + 5 \times 0^2\) | Field Norm on 5th Cyclotomic Ring | |||||||||||

\(\ds \) | \(=\) | \(\ds 4\) | ||||||||||||

\(\ds \map N 3\) | \(=\) | \(\ds 3^2 + 5 \times 0^2\) | Field Norm on 5th Cyclotomic Ring | |||||||||||

\(\ds \) | \(=\) | \(\ds 9\) | ||||||||||||

\(\ds \map N {1 + i \sqrt 5}\) | \(=\) | \(\ds 1^2 + 5 \times 1^2\) | Field Norm on 5th Cyclotomic Ring | |||||||||||

\(\ds \) | \(=\) | \(\ds 6\) | ||||||||||||

\(\ds \map N {1 - i \sqrt 5}\) | \(=\) | \(\ds 1^2 + 5 \times 1^2\) | Field Norm on 5th Cyclotomic Ring | |||||||||||

\(\ds \) | \(=\) | \(\ds 6\) |

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.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $9$: Rings: Exercise $19 \ \text {(iii)}$