Geometric reasons finite fields have prime power orders?

Question

All variations of proofs that finite fields have prime power orders have a very algebraic feel to them. I was wondering - is there a more geometric way to see why this is true?

Answer 1

A finite field is a vector space over some $\mathbb{F}_p$, so it has a basis and its elements are in bijection with some $n$-uples of coordinates in $\mathbb{F}_p$, which tells you it has cardinal $p^n$. I don't know if you consider that a geometric or algebraic argument though.

Answer 2

If $k$ is a field (finite or not) you can associate to it the one point scheme $X=\{x\}=\operatorname {Spec}(k)$.
Like all schemes $X$ comes with a canonical morphism $can: X\to \operatorname {Spec}(\mathbb Z)$ whose image is the point $can(x)=z\in \operatorname {Spec}(\mathbb Z)$.
We then get the induced extension of residual fields $can^\ast :\kappa(z)\to \kappa (x)=k$ and since $k$ is finite so must be $\kappa (z)$, which implies that $z=p\mathbb Z\in \operatorname {Spec}(\mathbb Z)$ for some prime $p$:
Indeed $\operatorname {Spec}(\mathbb Z) $ consists of the prime ideals $p\mathbb Z$ and of the ideal $0.\mathbb Z$, the generic point of $\operatorname {Spec}(\mathbb Z)$, but $z_0= 0.\mathbb Z$ is excluded because $\kappa(z_0)=\mathbb Q$ is infinite.
Thus $\kappa(z)=\mathbb F_p $ and from the set-finiteness of $k$ follows that the field extension $can^\ast: \kappa(z)=\mathbb F_p\to k$ makes of $k$ a finite-dimensional $\mathbb F_p$-vector space of cardinality $p^n$ for some positive $n$, just as required.

Needless to say all that can be said purely algebraically (and more simply!) but Grothendieck and his school have shown how such a vision of field theory can lead to an astonishingly general theory, linking for example Galois theory and the topological theory of covering spaces.