The sum of the elements in a field of at least three elements is 0 This statement seems so simple yet I don't quite know how to start with this proof in a substantial way.  Can anybody help me here?  
 A: To even speak of a sum, I suppose the field $F$ is finite. If $f\colon F\to F$ is any bijection, then we can conclude that $\sum_{x\in F}x=\sum_{x\in F}f(x)$. Let $\alpha\in F$ such that $\alpha\ne 0$. Then $x\mapsto \alpha x$ is such a bijection. We conclude that $\sum_{x\in F} x=\sum_{x\in F}\alpha x$, i.e. $(1-\alpha)\sum x=0$. If $F$ has more than two elements, we can pick $\alpha\in F\setminus\{0,1\}$ and are done.

To elaborate:
Why does $\sum_{x\in F}x=\sum_{x\in F}f(x)$ hold if $f\colon F\to F$ is a bijection? More generally, if $I$ is any finite set and for each $i\in I$, $a_i$ is a number (or element of an additive abelian group such as $F$), and $f\colon I\to I$ is a bijection, then $\sum_{i\in I} a_i = \sum_{i\in I}a_{f(i)}$. This is because the summands on the right are precisely the same summnds as on the left and hence equality is guaranteed by the laws of associativity and commutativity. (I suppose this rough reference to merely changed order of addition suffices; otherwise a formal proof is possible by induction based on the fact that the group of permutations of $I$ is generated by transpositions ("swappng two adjacent elements"))
After this, if $F$ has more than two elements, it certainly has more elements than only $0$ and $1$. Therefore there exixts $\alpha$ that is both $\ne0$ and $\ne 1$. With such $\alpha$ the key steps in the proof above work.
The first condition, $\alpha\ne 0$,  guarantees that the map $F\to F$, $x\mapsto \alpha x$ is a bijection (namely with $x\mapsto \alpha^{-1}x$ as inverse map) and so allows us to conclude than $(1-\alpha)\sum_{x\in F} x=0$.
The second condition $\alpha\ne 1$, allows us to conclude that the first factor $(1-\alpha)$ is nonzero, hence the second factor must be zero, i.e. $\sum_{x\in F} x=0$ as was to be shown.
A: Another approach: 
If $\;\Bbb F\;$ is a finite field, then there exists a prime $\;p\;$ such that $\;\Bbb F_p=\Bbb Z/p\Bbb Z\;$ is the prime subfield of $\;\Bbb F\;$, so that $\;|\Bbb F|=p^n\;$ , for some $\;n\in\Bbb N\;$. 
But then we know, if we denote by $\;\overline{\Bbb F_p}\;$ an algebraic closure of the prime field, that:
$$\Bbb F=\left\{a\in\overline{\Bbb F_p}\;:\;\;a^{p^n}-a=0\right\}=\left\{a\in\overline{\Bbb F_p}\;;\;a\;\;\text{is a root of}\;\;x^{p^n}-x\in\Bbb F_p[x]\right\}$$
and then the sum of all elements in $\;\Bbb F\;$ is the coefficient of$\;x^{p^n-1}\;$ in the above polynomial....
A: Here's a different proof which does not use the field structure of a general $\mathbb F_q$ at all. Instead I will show with just basic linear algebra and tiny combinatorics that, if $V$ is any $n$-dimensional $\mathbb F_p$-vector space, then
$$\sum_{v \in V} v = 0$$
unless $p=2$ and $n=1$.
Namely, say $e_1, ..., e_n$ is a basis of $V$, then an arbitrary element $v \in V$ is written as $\sum_{i=1}^n a_i e_i$ with unique $a_i \in \mathbb F_p$. Then a moment's consideration shows that for each $i \in \{1, ...,n\}$ and $j= \bar 0, \bar 1, ..., \overline{p-1} \in \mathbb F_p$, there are exactly $p^{n-1}$ elements $v \in V$ such that  $a_i = j$. So the coefficient of $e_i$ in $\sum_{v \in V} v$ is the residue mod $p$ of
$$p^{n-1} \cdot (1+2+...(p-1)) = p^{n-1}\cdot \frac{(p-1)p}{2} = p^n \cdot \frac{p-1}{2}$$
which obviously is divisible by $p$, except in the case that simultaneously $n=1$ and $p=2$.
