Order of finite fields is $p^n$

Let $F$ be a finite field. How do I prove that the order of $F$ is always of order $p^n$ where $p$ is prime?

Solution 1:

Let $p$ be the characteristic of a finite field $F$.${}^{\text{Note 1}}$ Then since $1$ has order $p$ in $(F,+)$, we know that $p$ divides $|F|$. Now let $q\neq p$ be any other prime dividing $|F|$. Then by Cauchy's Theorem, there is an element $x\in F$ whose order in $(F,+)$ is $q$.

Then $q\cdot x=0$. But we also have $p\cdot x=0$. Now since $p$ and $q$ are relatively prime, we can find integers $a$ and $b$ such that $ap+bq=1$.

Thus $(ap+bq)\cdot x=x$. But $(ap+bq)\cdot x=a\cdot(p\cdot x)+b\cdot(q\cdot x)=0$, giving $x=0$, which is not possible since $x$ has order at least $2$ in $(F,+)$.

So there is no prime other than $p$ which divides $|F|$.

Note 1: Every finite field has a characteristic $p\in\mathbb N$ since, by the pigeonhole principle, there must exist distinct $n_1< n_2$ both in the set $\{1, 2, \dots, \lvert F\rvert +1\}$ such that $$\underbrace{1+1+\dots+1}_{n_1}=\underbrace{1+1+\dots+1}_{n_2},$$ so that $\underbrace{1+1+\dots+1}_{n_2-n_1}=0$. In fact, this argument also implies $p\le n$.

Solution 2:

1. Prove that the smallest multiple $m$ of 1 that gives zero has to be a prime. (Otherwise there are divisors of $m$ which are then divisors of zero.)

2. Prove that a field is a vector space over a subfield.

3. Count the elements of the field if the dimension of this vector space is $n$.