Galois Field of Order q Exists iff q is Prime Power

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Theorem

Let $q \ge 0$ be a positive integer.

Then there exists a Galois field of order $q$ if and only if $q$ is a prime power.


Proof


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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.

$\Box$


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.

If $a^{p^n} = a$ 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 for all $a, b \in E$, $\paren {a + b}^p = a^p + b^p$, 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.

$\blacksquare$


Sources

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