# 5th Cyclotomic Ring has no Elements with Field Norm of 2 or 3

## Theorem

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

There are no elements of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ whose field norm is either $2$ or $3$.

## Proof

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

Let $z = x + i y$.

Then:

 $\ds \map N z$ $=$ $\ds 2$ $\ds \leadsto \ \$ $\ds x^2 + 5 y^2$ $=$ $\ds 2$ Field Norm on 5th Cyclotomic Ring $\ds \leadsto \ \$ $\ds x^2$ $=$ $\ds 2$ as $y$ is an integer and has to equal $0$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds \sqrt 2$

But Square Root of Prime is Irrational so $x \notin \Z$.

Similarly for $\map N z = 3$.

$\blacksquare$