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

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## 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:

\(\displaystyle \map N z\) | \(=\) | \(\displaystyle 2\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x^2 + 5 y^2\) | \(=\) | \(\displaystyle 2\) | Field Norm on 5th Cyclotomic Ring | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x^2\) | \(=\) | \(\displaystyle 2\) | as $y$ is an integer and has to equal $0$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle \sqrt 2\) |

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

Similarly for $\map N z = 3$.

$\blacksquare$

## Sources

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