Elements of 5th Cyclotomic Ring with Field Norm 1

Theorem

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

The only elements of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ whose field norm equals $1$ are the units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$: $1$ and $-1$.

Proof

From Units of 5th Cyclotomic Ring, $1$ and $-1$ are the only units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.

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

Let $z \in \Z \sqbrk {i \sqrt 5}$ such that $\map N z = 1$.

Let $z = x + i y$.

Then:

\(\ds \map N z\)

\(=\)

\(\ds 1\)

Field Norm on 5th Cyclotomic Ring

\(\ds \leadsto \ \ \)

\(\ds x^2 + 5 y^2\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds x^2\)

\(=\)

\(\ds 1\)

\(\, \ds \land \, \)

\(\ds y\)

\(=\)

\(\ds 0\)

The result follows.

$\blacksquare$

Sources

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