# 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?

• $a^{q-1} = 1$.. – reuns Aug 30 '17 at 0:45
• What are the requirements of absolute value? – fleablood Aug 30 '17 at 1:00
• Why are you reading Gouvea's book without any background in abstract algebra? – Xam Aug 30 '17 at 2:14
• For some reason p-adics are a topic in a pure maths course thats taken before a first course in abstract algebra (or real analysis, but I've been trying to learn a bit in my own time) at my uni. It's about a weeks worth of content and as a result not a lot of it is built up very well, so I wanted to use a textbook to understand the concept behind them better. I couldn't find much besides Gouvea's book and I thought it was really good so far up to the stuff on valuation theory. Do you have any other recommendations? – paulw Aug 30 '17 at 9:21

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.