Trivial Absolute Value is the only Absolute value for Finite Fields

Theorem

$$|\cdot |=\{\begin{array}{ll}\pm 1& k\text{ is even}\\ 1& k\text{ is odd}\end{array}$$

For all $y^k$ in a Finite Field

Proof

• Let $y\in \mathbb{F}$ be nonzero As $\mathbb{F}$ is a finite field, the set $\{y,y^2,\dots \}$ is finite it follows there exists $0<a<b<\infty$ such that $y^a=y^b$ and so, there exists a $0<k<\infty$ s.t $y^k=1$
• Because $|\cdot |$ is a Norm Function,
• $|y^k|=|y^k|=|1|=1$
• So,

$$|\cdot |=\{\begin{array}{ll}\pm 1& k\text{ is even}\\ 1& k\text{ is odd}\end{array}$$

• So, $|\cdot |$ is an absolute value, it is non-negative, thus $|y|=1$, it follows that $|\cdot |$ is the trivial absolute value