Galois Field of Order q Exists iff q is Prime Power

== Theorem ==
Let $q \ge 0$ be a positive integer.
Then there exists a Galois field of order $q$ $q$ is a prime power.
== Proof ==
=== Sufficient Condition ===
Let $\struct {F, +, \cdot}$ be a field of order $q$.
By Characteristic of Galois Field is Prime, the characteristic of $F$ is a prime number $p$.
By Field of Prime Characteristic has Unique Prime Subfield the prime subfield of $F$ is $\GF_p := \Z / p \Z$.
By Vector Space on Field Extension is Vector Space, $F$ is an $\GF_p$-vector space.
Since $F$ is finite, $F$ has a finite basis over $\GF_p$.
By Same Dimensional Vector Spaces are Isomorphic, this means that with $k$ equal to the dimension of $F$ there is an isomorphism of vector spaces:
:$F \simeq \GF_p^k$
Finally by the definition of the product of cardinals:
:$\card F = \card {\GF_p}^k = p^k$
So the order of $F$ is a prime power.
=== Necessary Condition ===
Let $\GF_p := \Z / p \Z$ be the field of order $p$.
Consider $x^{p^n} - x \in \GF_p \sqbrk x$ for some $n \in \N$.
By Kronecker’s Theorem, there exists a finite extension $E$ of $\GF_p$ such that $x^{p^n} - x$ splits completely.
Define the following set:
:$\GF_{p^n} = \set {u \in E: u^{p^n} = u}$
It suffices to prove that $\card {\GF_{p^n} } = p^n$ and that $\GF_{p^n}$ is closed under field operations.
$\GF_{p^n}$ contains all the roots of $x^{p^n} - x$ in $E$ by definition, so it contains at most $p^n$ elements.
It suffices to prove $x^{p^n} - x$ has no double roots.
By Double Root of Polynomial is Root of Derivative, and that $\map {\dfrac \d {\d x} } {x^{p^n} - x} = \paren {p^n} x^{p^n - 1} - 1 = -1 \ne 0$ in characteristic $p$, $x^{p^n}-x$ has no double roots and $\card {F_{p^n} } = p^n$.
Clearly $0, 1 \in \GF_{p^n}$, and if $a \neq 0$ and $a^{p^n} = a$ then dividing out by $a^{p^n} \cdot a$ gives $a^{-1} = \paren {a^{-1} }^{p^n}$, so $\GF_{p^n}$ is closed under taking reciprocals.
If $p \neq 2$ and $a^{p^n} = a$ then $p^n$ is odd so $\paren {-a}^{p^n} = -a$. If $p = 2$ then $a = -a$ in any ring of characteristic 2 so still $\paren {-a}^{p^n} = -a$, $\GF_{p^n}$ is closed under negation.
Let:
:$a^{p^n} = a$ 
and:
:$b^{p^n} = b$
Then:
:$\left( ab \right)^{p^n} = \paren {a^{p^n} } \paren {b^{p^n} } = a b$.
So $\GF_{p^n}$ is closed under multiplication.
Consider the map: 
:$x \mapsto x^p$.
We will prove that:
: $\forall a, b \in E$, $\paren {a + b}^p = a^p + b^p$.
By induction, it will follow that:
:$\paren {a + b }^{p^n} = a^{p^n} + b^{p^n}$.
In particular, if $a, b \in \GF_{p^n}$, then:
:$\paren {a + b}^{p^n} = a^{p^n} + b^{p^n} = a + b$.
So $\GF_{p^n}$ will be closed under addition and we will be done.
Suppose $k \in \N$ and $1 \le k \le p - 1$.
Then in characteristic $p$, $\dbinom p k = \dfrac {p!} {k! \paren {p - k}!} = 0$ as the numerator contains a factor of the prime $p$ which the denominator cannot cancel.
By the Binomial Theorem, $\paren {a + b}^p = a^p + \underbrace {\ds \sum_{k \mathop = 1}^{p - 1} \dbinom p k a^k b^{p-k} }_0 + b^p = a^p + b^p$, from the previous statements, $\GF_{p^n}$ is closed under addition. 
Therefore, $\GF_{p^n}$ is a subfield of $E$ with $p^n$ elements.
== Sources ==
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Category:Galois Fields