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Order of finite fields is $p^n$

If $F$ is a finite field then the number of elements of $F$ is of the form $p^n$ for some $n\in\mathbb N$.

I know that characteristic of $F$ is a prime number $p$ but I don't know how to show the number of elements of $F$ is of the form $p^n$. Can anyone help me?

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marked as duplicate by Jyrki Lahtonen, t.b., Gerry Myerson, Zhen Lin, Asaf Karagila Apr 16 '12 at 6:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Wikipedia may be helpful –  Raymond Manzoni Apr 15 '12 at 11:44
    
I hate to spoil your fun, but this question is essentially a duplicate of this and also this question asked and answered earlier. –  Jyrki Lahtonen Apr 15 '12 at 16:36

6 Answers 6

up vote 4 down vote accepted

Since the characteristic of $F$ is $p$, $F$ will contain a subfield $K$(say) isomorphic to the field $(Z_{p},+_{p},._{p})$ ($\because$ If the characteristic of a field $F$ is prime $p$, then $F$ contains a subfield isomorphic to $(Z_{p},+_{p},._{p})$.) $\therefore$ The field $F$ will contain $p$ elements. Now the field $F$ can be considered as an extension of the field $K$. i.e. $F$ is a linear space over field $K$. As $F$ is a finite field, its dimension will also finite. If the dimension of $F$ is $n$, then it has a basis $\{x_{1},x_{2},....,x_{n}\}$ conataining $n$ elements. Now we know that if $B=\{x_{1},x_{2},....,x_{n}\}$ is basis of an $n$ dimensional linear space $(V,+,.,F)$, then each vector of $V$ has a unique representation as a linear combination of vectors $x_{1},x_{2},...,x_{n}$.

$\therefore$ for each $x\in F$ has a unique representation $$x=\alpha_{1}x_{1}+\alpha_{2}x_{2}+...\alpha_{n}x_{n}, \alpha_{i}\in K (i=1,..n)$$ $\alpha_{i}\in K$ implies that we have $p$ choices for each $\alpha_{1},\alpha_{2},....,\alpha_{n}$. Thus the total possible choices for $x$ will be $p^{n}$ and these choices will give the entire field $F$, i.e. the number of elements in $F$ is $p^{n}$. Note: From this result you can also derive following important result. If a finite field $F$ contains $p^{n}$ elements, then each element of $F$ satisfies the equation $x^{p^{n}}-x=0.$

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I hope my answer is very useful to you and others. –  Kns Apr 16 '12 at 5:53

Ok so the fact that the field $F$ is finite tells you a lot. First the characteristic must be $p$ for some prime since fields of characteristic $0$ are infinite. Secondly, by that fact, $F$ must be finitely generated as a vector space over $\mathbb{Z}/p\mathbb{Z}$.

So let's give $F$ a basis $x_1, ..., x_n$. Then the elements of $F$ can all be written uniquely as $\alpha_1 x_1 + \alpha_2 x_2 + ... + \alpha_n x_n$ for $\alpha_1, \alpha_2, ..., \alpha_n\in\mathbb{Z}/p\mathbb{Z}$. Also each such linear combination gives an element of $F$.

How many choices are there for the alphas? Well there are $p$ choices for each alpha so we have that $|F| = p^n$.

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Ok i understand.......thanks a lot –  Siddhant Trivedi Apr 15 '12 at 15:13

If the characteristic of $F$ is $p$, you can see $F$ as a $\mathbb{Z}/p\mathbb{Z}$ vector space. While $F$ is finite, the cardinal of $F$ is a power of $p$.

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Here are two hints:

1) It is suffiecient to show that the characteristic is a prime.

2) If the characteristic is not a prime, then there exists a nonzero element, which is not invertible.

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A finite field of characteristic $p$ is a vector space of finite dimension $n$ over $F_p = \mathbb Z/p\mathbb Z$.

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Let $p$ be the characteristic of the field $F$, which is the additive order of the multiplicative identity $1$. If $q$ is prime and divides the order of $F$, then there is an element $x$ of order $q$ in $(F, +)$. Thus $qx = 0$ and $(q1)x = 0$, which implies $q1 = 0$. Thus $q$ divides the additive order of $1$, which means that $p = q$ since $p$ is prime.

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