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?
Finite Fields – Proving the Trivial Absolute Value
abstract-algebrafinite-fieldsvaluation-theory
Related Solutions
$F_q((x)):=\left\{\sum_{n=k}^{\infty}a_nz^n:\ \text{for } n\in\mathbb{N},\text{ and }a_n\in F_q,\ n=k,k+1,\ldots\right\}$
For $F:=\sum a_nz^n, G:=\sum b_nz^n\in F_q((z))$ and $a\in F_q$ we have
- $F+G:=\sum (a_n+b_n)z^n$
- $aF:=\sum aa_nz^n$
- $FG:=\sum c_nz^n$, with $c_n:=\sum_k a_nb_{n-k}$
- For $F:=\sum_{n=k}^{\infty}a_nz^n$, with $a_k\neq0$ we define $|F|:=q^{-k}$
The absolute value of $3$ is non-Archimedean. In fact, in general, we have
$$|F+G|\leq\max\left(|F|,|G|\right).$$
In particular $|z|=q^{-1}$ and therefore $|n\cdot z|=|z+z+\ldots+z|\leq\max(|z|,\ldots,|z|)=q^{-1}$. On the other hand we have $|1|=q^0=1$. Therefore we can never make $|n\cdot z|>1$ for any $n\in\mathbb{N}$.
The short answer is that you do not need to view the elements of finite fields as polynomials, but it simply is the most convenient presentation for many a purpose.
The slightly longer answer is that those elements really aren't polynomials, but instead they are cosets in a ring of polynomials, and we simply select the lowest degree polynomial to represent the entire coset. This version of the answer is pedagogically possibly the worst I'm gonna give, but it does come in handy when you do calculations in a finite field in a computer program. Namely, with a bit of care you can use either those low degree polynomials, or you can use monomials to represent the same elements. The former are useful for addition, the latter for multiplication. This is because we can use discrete logarithm tables. I'm not gonna discuss this aspect now, but if you are interested you can take a peek at this Q&A I prepared for referrals like this in mind.
Then the long answer - building up on the explanation given by Mathmo123 (+1).
When working with algebraic extensions of number fields, say $\Bbb{Q}$, students often take comfort in thinking of those extensions as consisting of some explicit numbers. Been there, done that. This works because we can always think of a copy of that extension field inside the field of complex numbers $\Bbb{C}$. This does have some pitfalls (there are often several distinct ways of view the given field as such a subset), but we can do this because A) $\Bbb{C}$ is algebraically closed, B) at that point in our studies the concept of a number is still closely tied to $\Bbb{C}$ and its subsets.
Consider the following. It is certainly more natural to work with $\sqrt2$ rather than the coset $x+\langle x^2-2\rangle$ in the quotient ring $\Bbb{Q}[x]/\langle x^2-2\rangle$. We easily do arithmetic as follows $$ (3+\sqrt2)(5+4\sqrt2)=15+(5+12)\sqrt2+4(\sqrt2)^2=23+17\sqrt2. $$ But notice that we multiplied those numbers involving $\sqrt2$ exactly like we would multiply the polynomials $$ (3+x)(5+4x)=15+17x+4x^2. $$ Only in the end we replaced $x^2$ with $2$. All because we think of $x$ having "value" $x=\sqrt2$. Saying that we here actually did arithmetic with polynomials modulo $x^2-2$ is just a fancy way of saying that we use $\sqrt2$ exactly as we use $x$ except in the end we replace all occurrences of $(\sqrt2)^2$ with $2$.
Building up on this we can similarly do arithmetic with $\root3\of2$. Only this time we can only simplify third powers and higher. So compare (I write $\root3\of 4$ in place of $(\root3\of2)^2$ and similarly for higher powers) $$ \begin{aligned} (1+2\root3\of2+3\root3\of4)(1+\root3\of4)&=1+2\root3\of2+(3+1)\root3\of4+2\root3\of8+3\root3\of{16}\\ &=(1+2\cdot2)+(2+3\cdot2)\root3\of2+4\root3\of4\\ &=5+8\root3\of2+4\root3\of4 \end{aligned} $$ replacing $\root3\of 8$ with $2$ and $\root3\of{16}$ with $2\root3\of2$ (and $\root3\of{32}$ with $2\root3\of4$ should the need arise). This is, again, exactly like working with polynomials $$ (1+2x+3x^2)(1+x^2)=1+2x+(3+1)x^2+2x^3+3x^4, $$ where this time we replaced $x^3$ with $2$ and $x^4$ with $2x$.
What's the deal you may ask? In these two example we could have simply used the real numbers $\sqrt2$ and $\root3\of2$. We might even have tried to use their decimal approximations to do the arithmetic with a pocket calculator. Sure, that would introduce rounding errors whereas doing it as above is precise, because it is easier to identify the numbers exactly. By this I mean, it is in this sense more informative to write $$ (\sqrt2-1)^{100}=94741125149636933417873079920900017937-66992092050551637663438906713182313 772 \sqrt{2} $$ instead of $$ (\sqrt2-1)^{100}=5.2775391806914391296141\cdot10^{-39}, $$ which is what a calculator might give you. For example, from that decimal expansion it is very hard to realize that it actually is a number of the form $a-b\sqrt2$ for some integers $a,b$.
Forward we go. What is you want to do arithmetic with the largest real zero of $x^5-2x^3-2x+2$. Plotting that polynomial shows that this number is slightly bigger than $1.5$. Well, this time we DO NOT HAVE A NICE EXPRESSION FOR THAT NUMBER IN TERMS OF RADICALS. So what to do? Just call the number $\alpha$, do arithmetic with its polynomials as above, and keep in mind that by the defining equation $$ \alpha^5=2\alpha^3+2\alpha-2. $$ Therefore we can calculate for example that $$ \begin{aligned} \alpha^7&=\alpha^2\alpha^5\\ &=\alpha^2(2\alpha^3+2\alpha-2)\\ &=2\alpha^5+2\alpha^3-2\alpha^2\\ &=2(2\alpha^3+2\alpha-2)+2\alpha^3-2\alpha^2\\ &=6\alpha^3-2\alpha^2+4\alpha-4. \end{aligned} $$
What has this got to do with doing arithmetic in finite fields? There the situation is much like in that last case. It is not convenient to use a numerical value for $\alpha$, when we need to do exact arithmetic and avoid rounding errors and such. All we have to work with is that polynomial equation satisfied by $\alpha$. We use that equation to do our arithmetic. The price we need to pay is that $\alpha$ does not have a precise identity. In the above example any of the five zeros of that polynomial - two of them complex numbers - would lead to the same arithmetic. This is because the related fields are isomorphic.
Enough of why polynomials (modulo an equation). You asked for alternatives. Let's look at the field of four elements, denote it $\Bbb{F}_4$ or $GF(4)$, whichever you are more familiar with. The field looks like $$ \Bbb{F}_4=\{0,1,\alpha,\alpha+1\}, $$ and its arithmetic follows from it having characteristic two (so $1+1=0=\alpha+\alpha$) and the special equation $\alpha^2=\alpha+1$.
We can produce $\Bbb{F}_4$ as a quotient ring much the same way that we produce the field of seven elements as $\Bbb{Z}/7\Bbb{Z}$ - residue classes of integers modulo seven. Here's one way. Let $\omega=(-1+i\sqrt3)/2$ be the usual complex primitive third root of unity. Let us look at the ring $$ \Bbb{Z}[\omega]=\{a+b\omega\mid a,b\in\Bbb{Z}\}. $$ We can then reduce modulo two in the ring $\Bbb{Z}[\omega]$. In other words, we can do arithmetic as usual, but do operations with $a,b$ modulo two, so effectively both $a$ and $b$ will have two choices, $0,1$, and we have four combinations of them altogether. What do we get? Because $$ 0=\omega^3-1=(\omega-1)(\omega^2+\omega+1) $$ we can deduce that $$ \omega^2=-\omega-1. $$ Now, if we reduce this modulo two, we see that, because $1\equiv-1\pmod2$, $\omega^2=\omega+1$. In other words $\omega$, or more precisely, its residue class modulo two, takes the role of $\alpha$ in $\Bbb{F}_4$
We can produce any finite field in this way. In infinitely many different ways. But using the resulting representations of complicated numbers with coefficients modulo a prime number does not really help us to do any arithmetic operations. So we just usually won't bother.
This became even longer than I anticipated. Hope it helps :-)
Best Answer
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.