Every finite field has prime power order

Let F be a finite field of characteristic p. Prove that |F| = p^n for some n.

Suppose F has a nonzero element \alpha of additive order q. That is, q is minimal such that q\alpha = 0. Note that q \leq p, since p annihilates every element of F (being the characteristic of F). If q < p, then by the division algorithm in \mathbb{Z} we have p = qb + r for some b and r with 0 \leq r < q. If r \neq 0, then r \alpha = 0, violating the minimalness of q. If r = 0, then p = qb. Since p is prime, either q = 1 (so \alpha = 0) or b = 1 (so q = p), which both give a contradiction. Thus q = p.

So every element of F has additive order p, and thus by Cauchy’s theorem |F| is a power of p.

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