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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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    $\begingroup$ What is the cardinality of any finite-dimensional vector space over the field with $p$ elements? $\endgroup$
    – NKS
    Oct 15 '11 at 17:50
  • $\begingroup$ What exactly is the scalar field and the multiplication law? $\endgroup$
    – Mohan
    Oct 15 '11 at 18:14
  • $\begingroup$ 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. $\endgroup$ Oct 15 '11 at 18:37
  • $\begingroup$ First prove that if $charF=p $ then $F_p$ is a subfield of $F$. $\endgroup$
    – Math.mx
    Oct 15 '11 at 19:27
  • $\begingroup$ The answers to this question contain all the information that you need. IOW this is almost an exact duplicate. $\endgroup$ Oct 16 '11 at 6:17
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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 $p\le n$.

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    $\begingroup$ I like this argument better than the vector space one. Surprising that it has so little attention. $\endgroup$
    – R R
    May 7 '15 at 17:38
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    $\begingroup$ @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 $m\cdot x=0$. $\endgroup$ Aug 27 '15 at 7:06
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    $\begingroup$ 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 $p|q$. $\endgroup$ Nov 19 '17 at 0:19
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    $\begingroup$ Coffee Table combined with caffeinemachine gives best result :) $\endgroup$
    – Silent
    Apr 5 '19 at 12:08
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    $\begingroup$ 0 has positive order, namely 1, in any group. So I think you should say $|x| = q > 1$, not merely that it's positive. $\endgroup$
    – Junglemath
    Sep 6 '20 at 7:00
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  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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    $\begingroup$ Denoting $\underbrace{1+1+\dots+1}_{n \text{ times} }$ by $ n \cdot 1$, where $1$ is the identity, we can prove that the set of the elements in $F$ of the form $(n \cdot {} 1)^{-1} (m \cdot 1)$ is the subfield of $F$, say $S$. Here the vector space have a scalar multiplication being a function of $S \times F$ to $F$, and the element of the vector space being in $F$. Kinda weird, but it works beautifully. $\endgroup$ Jul 5 at 19:48
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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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  • $\begingroup$ I'm having trouble seeing why the map you defined is a homomorphism. $\endgroup$
    – Junglemath
    Sep 6 '20 at 7:11
  • $\begingroup$ It's an additive homomorphism: $\phi(x+y)=\phi(x)+\phi(y)$. $\endgroup$
    – Ehsaan
    Oct 11 '20 at 3:30
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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

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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 $|F| = p^n$

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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).

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