The Number of Elements in a Finite Field is a Power of a Prime Number

Problem 726

Let $\F$ be a finite field of characteristic $p$.

Prove that the number of elements of $\F$ is $p^n$ for some positive integer $n$.

Proof.

First note that since $\F$ is a finite field, the characteristic of $\F$ must be a prime number $p$. Then $\F$ contains the prime field $\F_p$ and $\F$ is a finite extension of $\F_p$, say, of degree $n$.

This means that we have a basis $\{v_1, \dots, v_n\}$ of $\F$ as a vector space over $\F_p$. Hence any element $x\in \F$ can be uniquely written as
$x=a_1v_1+\cdots a_n v_n,$ where $a_i \in \F_p$ for $i=1, \dots, n$.

It follows that the fields $\F$ has $p^n$ elements.

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• Yu says:

Dear Dickens,

Prove that $\F_3[x]/(x^2+1)$ is a Field and Find the Inverse Elements
Let $\F_3=\Zmod{3}$ be the finite field of order $3$. Consider the ring $\F_3[x]$ of polynomial over $\F_3$ and its ideal...