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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?

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What is the cardinality of any finite-dimensional vector space over the field with $p$ elements? –  NKS Oct 15 '11 at 17:50
What exactly is the scalar field and the multiplication law? –  Mohan Oct 15 '11 at 18:14
The scalar field is the subfield consisting of the elements $0$, $1$, $1+1$, $1+1+1$, $1+1+1+1$ and so forth. (You first have to prove this is in fact a subfield, of course). Multiplication is whatever passes for multiplication in the finite field. –  Henning Makholm Oct 15 '11 at 18:37
First prove that if $charF=p $ then $F_p$ is a subfield of $F$. –  Frank Murphy Oct 15 '11 at 19:27
The answers to this question contain all the information that you need. IOW this is almost an exact duplicate. –  Jyrki Lahtonen Oct 16 '11 at 6:17

3 Answers 3

up vote 22 down vote accepted
  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 $n$.

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Let $F$ be a finite field. Then the underlying additive group of the field (let's denote this by $F^+$) has this interesting property:

For every two non-identity (i.e. non-zero) elements $a$ and $b\in F^+$, there is an automorphism $\phi$ of the additive group such that $\phi(a)=b$.

This can be seen by examining the map $(x\mapsto ba^{-1}x)$.

This means the set of automorphisms of $F^+$ act transitively on $F^+$. Since automorphisms permute elements of the same order, we can conclude that every element in $F$ has the same order.

But a finite group in which all non-identity elements have the same order is necessarily a $p$-group such that every element has prime order. This can be shown by Cauchy's Theorem.

Suppose the order of $F^+$ had two distinct prime factors $p$ and $q$. Then $F^+$ would contain an element of order $p$ and another element of order $q$ by Cauchy's Theorem. This contradicts that every element has the same order. So the order of $F$ is indeed a prime-power. Cauchy's Theorem implies that $F$ has an element of order $p$, thus all elements have order $p$ by the hypothesis.

So, $F$ must be of prime-power order $p^n$, and we have that $px=0$ for all $x\in F$.

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Let $p$ be the characteristic of a finite field $F$. 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 positive order in $(F,+)$.

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

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I like this argument better than the vector space one. Surprising that it has so little attention. –  R R May 7 at 17:38
why $(ap+bq)\cdot x = a \cdot (p \cdot x) + b \cdot (q \cdot x)?$ thanks! –  scitamehtam Aug 27 at 5:26
@scitamehtam An inductive argument should easily settle that $(m+n)\cdot x=m\cdot x+n\cdot x$ for all integers $m$ and $n$ and any $x\in F$. Further, an inductive argument can be used to settle $(mn)\cdot x= m\cdot(n\cdot x)$ for all integers $m$ and $n$ and $x\in F$. I guess the source of the confusion is probably the fact that here the '$\cdot$' does not denote the multiplication in $F$. It is simply a notation that if $m$ is a positive integer, then $m\cdot x$ is $x$ summed $m$ times. If $m$ is negative then $m\cdot x:=(-m)\cdot x$. If $m=0$ then $m\cdot x=0$. –  caffeinemachine Aug 27 at 7:06

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