# 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?

• 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 $F$. – 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

## 6 Answers

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 $n$.

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 $|F|$.

• I like this argument better than the vector space one. Surprising that it has so little attention. – R R May 7 '15 at 17:38
• why $(ap+bq)\cdot x = a \cdot (p \cdot x) + b \cdot (q \cdot x)?$ thanks! – scitamehtam Aug 27 '15 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 '15 at 7:06
• After the first paragraph, can’t you just argue: In the equation $x+…+x=0$ ($q$ $x$s) factor out $x$ to get $x(1+…+1)=0$. Since we’re in a field we know that $x^{-1}$ exists so multiply it to both sides to get $1+…+1=0$ ($q$ $1$s). But we know that the order of $1$ in $(F,+)$ is $p$, hence $p|q$. – Coffee_Table Nov 19 '17 at 0:19
• Coffee Table combined with caffeinemachine gives best result :) – Silent Apr 5 at 12:08

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 $x\in F$.

First note that if $$R$$ is a commutative ring with identity then there exists a ring homomorphism from $$\Bbb Z$$ to $$R$$ given by $$1 \mapsto 1_R$$ having kernel $$n \Bbb Z$$ for $$n \ge 0$$. If $$R$$ is a field then the kernel would be $$\{0 \}$$ or $$p \Bbb Z$$ for $$p$$ a prime. Hence the characteristic of a field is either $$0$$ or $$p$$ for $$p$$ a prime. Now let $$R=F$$ be a finite field. Consider a map $$f : \Bbb Z \longrightarrow F$$ as above. Then $$f$$ is a homomorphism. Since $$F$$ is finite $$f$$ cannot be one-one; for otherwise there will be a copy of $$\Bbb Z$$ sitting inside $$F$$. But then $$F$$ will be of infinite order, a contradiction. Hence $$Ker\ (f) \ne \{0 \}$$. Since $$F$$ is a field $$Ker\ (f) = p \Bbb Z$$ for some prime number $$p$$. Then $$\Bbb Z/ p \Bbb Z$$ is embedded in $$F$$. So $$F$$ can be thought of as a vector space over $$\Bbb Z / p \Bbb Z$$. Since $$F$$ is finite, then dimension of $$F$$ as a vector space over $$\Bbb Z / p \Bbb Z$$ is finite. Lets say the dimension to be $$n$$. Then $$|F| = p^n$$. This proves that the order of any field is some power $$n$$ of a prime say $$p$$.

QED

A slight variation on caffeinmachine's answer that I prefer, because I think it shows more of the structure of what's going on:

Let $F$ be a finite field (and thus has characteristic $p$, a prime).

1. Every element of $F$ has order $p$ in the additive group $(F,+)$. So $(F,+)$ is a $p$-group.
2. A group is a $p$-group iff it has order $p^n$ for some positive integer $n$.

The first claim is immediate, by the distributive property of the field. Let $x \in F, \ x \neq 0_F$. We have

\begin{align} p \cdot x &= p \cdot (1_{F} x) = (p \cdot 1_{F}) \ x \\ & = 0 \end{align}

This is the smallest positive integer for which this occurs, by the definition of the characteristic of a field. So $x$ has order $p$.

The part that we need of the second claim is a well-known corollary of Cauchy's theorem (the reverse direction is just an application of Lagrange's theorem).

Let $$n,m$$ be positive integers and $$F$$ be a finite field. Define the operations.

\begin{align} (n \cdot \mathbb{1}_F) (m \cdot \mathbb{1}_F) &= (nm \cdot \mathbb{1_F})\\ (-n)\cdot \mathbb{1}_F &= -(n\cdot \mathbb{1}_F)\\ 0 \cdot \mathbb{1}_F &= 0\\ \end{align}

This then suggests there is a natural homomorphism

\begin{align} \mathbb{Z} &\stackrel{\phi}\to F\\ n &\to n \cdot \mathbb{1}_F \end{align}

Now since $$F$$ is finite, it has prime characteristic $$p$$ and we may see that $$\ker \phi = p\mathbb{Z}$$. Subsequently, we also see

$$\mathbb{Z}/\ker\phi = \mathbb{Z}/p\mathbb{Z} \stackrel{\phi}\to \text{img} \phi \approx F_p \subset F$$

and in fact $$F$$ is a finite extension of $$F_p$$ (say $$[F: F_p] = n$$), that is, $$F_p \leq F$$ or in other words, $$F$$ is a vector space over $$F_p$$. So any element $$z \in F$$ can be written as a linear combination of basis elements $$(z_1, \dots, z_n) \subset F$$ and $$a_i \in F_p$$

$$z = a_1z_1 + \dots a_nz_n$$

Now by counting, each $$a_i$$ has $$p$$ choices. So going through $$n$$ basis elements, there are exactly $$p^n$$ total choices and this counts all the $$z$$s in $$F$$. Therefore $$|F| = p^n$$