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How to prove that for every finite field its cardinality is $p^n$ where $p$ is prime and $n\in\mathbb{N}$?

Thanks in advance!

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Have you seen these proofs at Wikipedia? –  joriki Nov 30 '12 at 7:13

2 Answers 2

up vote 4 down vote accepted

The prime field of $F$, which is the smallest subfield of $F$ or the field obtained by taking the elements $0, 1, 1+1, 1+1+1, \ldots$ must be some $\mathbb Z/p\mathbb Z$. To make this a field, $p$ must be prime. Then $F$ is a vector space over $\mathbb Z/p\mathbb Z$, which makes its cardinality $|\mathbb Z/p\mathbb Z|^{\operatorname{dim}F}$.

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The characteristic of a finite field $F$ is some prime number $p$. Then for any $a\in F$ holds that $p\cdot a=0$ (here $p\cdot a$ means $a$ plus itself $p$ times). So, in $F$, considered simply as an additive group, every element has order $p$. Thus it is a $p$-group and since any $p$-group (by Cauchy's Theorem) has order a power of $p$ the proof is complete.

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