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 $\map N z = 3$.

$\blacksquare$


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