Intereting Posts

Partial fractions for $\pi \cot(\pi z)$
Sums of two probability density functions
Limit $\lim_{x\to 0} \frac{\tan ^3 x – \sin ^3 x}{x^5}$ without l'Hôpital's rule.
Limit of factorial function: $\lim\limits_{n\to\infty}\frac{n^n}{n!}.$
If $x \equiv 1 \pmod 3$ and $x \equiv 0 \pmod 2$, what is $x \pmod 6$?
The area of the superellipse
How to define countability of $\omega^{\omega}$ and $\omega_1$? in set theory?
solution for integral $\int_0^{\infty} \frac{k}{k^3-a}J_0\left(k \, r\right) dk $ involving Bessel function (Hankel transform)
Locus of vertex of parabolas through three points
Does $(\mathbf A+\epsilon \mathbf I)^{-1}$ always exist? Why?
What is $n!$ when $n=0$?
Closed set in $\ell^1$
Show that if $ R $ is an integral domain then a polynomial in $ R $ of degree $ d $ can have at most $ d $ roots
What is the “opposite” of the Axiom of Choice?
A sum including binomial coefficients

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?

- What happens if we remove the requirement that $\langle R, + \rangle$ is abelian from the definition of a ring?
- Rings with $a^5=a$ are commutative
- Are projective modules “graded projective”?
- Algorithms for symbolic definite integration?
- Given $f, g \in k$ coprime, why can we find $u,v \in k$ such that $uf + vg \in k\setminus\{0\}$?
- Group presentation for semidirect products
- Projective but not free (exercise from Adkins - Weintraub)
- Where does the word “torsion” in algebra come from?
- Automorphism group of the elliptic curve $y^2 + y = x^3$
- Differences between infinite-dimensional and finite-dimensional vector spaces

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.

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….

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