How to show the only absolute value on a finite field is the trivial one.

Define the trivial absolute value $|\cdot|$ by $|x| = 1$ if $x \neq 0$ or $|x| = 0$ if $x=0$. The textbook I'm currently reading (Gouvêa - P-adic Numbers An Introduction) asked me to show that for a finite field $\mathbb{K}$ the only possible absolute value is the trivial one. I haven't done any abstract algebra yet, so I can't think of any properties of finite fields that I could use to help me show this. Any hints?

Solution 1:

Let $\mathbb K^*$ be the multiplicative group of non-zero elements in $\mathbb K$. Then, suppose that $|x| = L \neq 1,0$.

Claim : Then, $x$ has infinite order in the group $\mathbb K^*$.

This is because $|x|^n = |x^n|$, so if $x$ had some finite order, say $x = x^m$, then $|x| = |x|^m \implies L = L^m$. Note that $L$ is a real number, so this becomes $L(L^{m-1} - 1) = 0$, so that $L = 0$ or $L = 1$, neither of which is allowed.

Finally, of course as $K$ is finite, the above cannot happen, so there is no element with absolute value not equal to zero or one. From here, it follows that the trivial absolute value is the only possible absolute value on a finite field.