Galois Field
Theorem1
For a prime $p$ and a natural number $n$, a finite field whose cardinality is $p^{n}$ is defined as a Galois field of order $p^{n}$, denoted by $\text{GF} \left( p^{n} \right)$. The only finite fields are Galois fields, and for given $p$ and $n$, the Galois field exists uniquely.
- Here, uniqueness means that even if two such fields are distinct, an isomorphism exists between them, so they are essentially the same field.
Explanation
It is said that when Gauss first conceived the notion of a finite field, nobody believed in its actual existence, but nowadays it is known not only that finite fields exist, but that they are unique and that even their concrete form has been revealed. Since the form of every finite field has been characterized, there is no need for pointless research.
For example, there is no need to even wonder whether a field with $10$ elements exists, and since $\text{GF} \left( p \right) = \mathbb{Z}_{p}$ is an integer ring, we already know a great deal about it. If there is anything more one is curious about, there is no need to cling to the abstract definition; one can approach it through $\mathbb{Z}_{p}$, and vice versa.
Proof2
Part 1. Every finite field is a Galois field.
Let $E$ be a finite extension field of a field $F$, and let the degree over $F$ be $n := \left[ E : F \right]$.
If $| F | = q$, then $E$ is an $n$-dimensional vector space over $F$, so $|E| = q^{n}$. A field has a unity, and if the characteristic is $0$, there exists a subring isomorphic to $\mathbb{Z}$, which makes the field infinite. Therefore, the characteristic of a finite field must be a finite natural number. If the characteristic of the finite field $E$ is $p \ne 0$, then since $E$ has a unity $1$, it must hold that $p \cdot 1 = 0$. Since a field is an integral domain, there cannot exist $p_{1}, p_{2} \in \mathbb{Z}$ satisfying $$ p \cdot 1 = ( p_{1} \cdot 1 ) ( p_{2} \cdot 1 ) = 0 $$ and $p$ must necessarily be a prime. Therefore $E$ has a subfield isomorphic to the prime field $\mathbb{Z}_{p}$, and since $\left| \mathbb{Z}_{p} \right| = p$, it follows that $|E| = p^{n}$.
Part 2. Existence of Galois Fields
Part 2-1. Zeros of $x^{p^{n}} - x$
For $\left( x^{p^{n}} - x \right)$, consider the algebraic closure $\overline{F}$ of a field $F$ with characteristic $p$.
Since $\overline{F}$ is algebraically closed, $\left( x^{p^{n}} - x \right) \in \overline{F} [ x ]$ factors into factors of degree $1$. What can be seen immediately is that $$ x^{p^{n}} - x = ( x - 0 ) \left( x^{p^{n}-1} - 1 \right) $$ so $0$ is a zero of $\left( x^{p^{n}} - x \right)$. Considering another zero $\alpha \ne 0$ of $f(x) := x^{p^{n}-1} - 1$, since $f \left( \alpha \right) = 0$, $$ 0 = f \left( \alpha \right) = \alpha^{p^{n} - 1} - 1 \implies \alpha^{p^{n} - 1} = 1 $$ and accordingly, expressing $f(x)$ as a product with $\left( x - \alpha \right)$ gives $$ \begin{align*} f(x) =& x^{p^{n}-1} - 1 \\ =& x^{p^{n}-1} - \alpha^{p^{n}-1} \\ =& (x - \alpha ) \left( x^{p^{n} - 2 } + \alpha x^{p^{n} - 3 } + \cdots + \alpha^{p^{n} - 3 } x + \alpha^{p^{n} - 2} \right) \end{align*} $$ Meanwhile, if for convenience we denote the second factor by $$ g(x) := \left( x^{p^{n} - 2 } + \alpha x^{p^{n} - 3 } + \cdots + \alpha^{p^{n} - 3 } x + \alpha^{p^{n} - 2} \right) $$ then the number of terms of $g(x)$ is $p^{n} - 1$. Therefore, substituting $x = \alpha$ yields $$ g ( \alpha ) = \alpha^{p^{n} - 2} \cdot \left( p^{n} - 1 \right) = {{\alpha^{p^{n} - 1}} \over { \alpha }} \left( p^{n} - 1 \right) $$ Above, we said that $\alpha \ne 0$ is a zero of $f(x)$ so that $\alpha^{p^{n}-1} - 1 = 0$, and since the characteristic was assumed to be a prime $p$, $$ g ( \alpha ) = {{1} \over { \alpha }} \cdot (0 - 1) = - {{1} \over { \alpha }} \ne 0 $$ Therefore $\alpha$ is not a repeated root of $f(x) = 0$, and the same holds for the other zeros besides $\alpha$. In the end, $\left( x^{p^{n}} - x \right)$ has exactly $p^{n}$ distinct zeros.
Part 2-2. Freshman’s Dream
Meanwhile, for $\alpha , \beta \in F$, computing $\left( \alpha + \beta \right)^{p}$ by the binomial theorem gives $$ \begin{align*} \left( \alpha + \beta \right)^{p} =& \sum_{k=1}^{p} \binom{p}{k} \alpha^{k} \beta^{p - k} \\ =& \alpha^{p} + \sum_{k=2}^{p-1} {{p!} \over { ( p - k )! ( k )! }} \alpha^{k} \beta^{p - k} + \beta^{p} \\ =& \alpha^{p} + \beta^{p} + p \sum_{k=2}^{p-1} {{ ( p - 1 )! } \over { ( p - k )! ( k )! }} \alpha^{k} \beta^{p - k} \end{align*} $$ Since the characteristic of $F$ is $p$, the last term becomes $0$, and therefore $$ \left( \alpha + \beta \right)^{p} = \alpha^{p} + \beta^{p} $$ Raising both sides to the $p$-th power once more gives $$ \left( \left( \alpha + \beta \right)^{p} \right)^{p} = \left( \alpha^{p} \right)^{p} + \left( \beta^{p} \right)^{p} $$ Rearranging gives $\left( \alpha + \beta \right)^{p^{2}} =\alpha^{p^2} + \beta^{p^2}$, and repeating this $n$ times yields the following. $$ \left( \alpha + \beta \right)^{p^{n}} =\alpha^{p^n} + \beta^{p^n} $$
Now consider the algebraic closure $\overline{ \mathbb{Z}_{p} }$ of $\mathbb{Z}_{p}$.
Let $K \subset \overline{ \mathbb{Z}_{p} } $ be the set collecting all the zeros of $\left( x^{p^{n}} - x \right) \in \overline{ \mathbb{Z}_{p} } [ x ]$, and let its elements be $\alpha , \beta \in K$.
Part 2-3. $K$ is a Galois field.
- (i) Closure under addition: $$ \begin{cases} \alpha^{p^{n}} - \alpha = 0 \\ \beta^{p^{n}} - \beta = 0 \end{cases} $$ Adding both sides, by Part 2-2 $\left( \alpha + \beta \right)^{p^{n}} =\alpha^{p^n} + \beta^{p^n}$, $$ \left( \alpha^{p^{n}} + \beta^{p^{n}} \right) - ( \alpha + \beta ) = \left( \alpha + \beta \right)^{p^{n}} - ( \alpha + \beta ) = 0 $$ so $( \alpha + \beta ) \in K$.
- (ii) Additive identity: Since $0^{p^{n}} - 0 = 0$, we have $0 \in K$.
- (iii) Additive inverse: We have $\left( - \alpha \right)^{p^{n}} = \left( - 1 \right)^{^{p^{n}}} \left( \alpha \right)^{p^{n}} = \left( - 1 \right)^{^{p^{n}}} \alpha$.
- If $p=2$, then $-1 = 1$, so $\left( -\alpha \right) = \alpha \in K$.
- If $p \ne 2$, it is an odd prime, so $\left( - \alpha \right)^{p^{n}} - ( - \alpha ) = 0$, that is, $( - \alpha ) \in K$.
- (iv) Closure under multiplication : Since $\left( \alpha \beta \right)^{p^{n}} = \alpha^{p^{n}} \beta^{p^{n}} = \alpha \beta$, we get $\left( \alpha \beta \right)^{p^{n}} - \alpha \beta = 0$, that is, $\alpha \beta \in K$.
- (v) Multiplicative identity: Since $1^{p^{n}} - 1 = 0$, we have $1 \in K$.
- (vi) Multiplicative inverse: For $\alpha \ne 0$, taking the reciprocal of $\displaystyle \left( \alpha \right)^{p^{n}} = \alpha$ gives $\displaystyle {{1} \over {\left( \alpha \right)^{p^{n}} }} = {{1} \over { \alpha }}$, that is, $$ \left( {{1} \over { \alpha }} \right)^{p^{n}} - {{1} \over { \alpha }} = 0 $$ so $\alpha^{-1} \in K$.
- (vii): $| K | = p^{n}$ : Since the characteristic of $\mathbb{Z}_{p}$ is $p$, by Part 2-1, $\left( x^{p^{n}} - x \right)$ has exactly $p^{n}$ distinct zeros.
Therefore, $K$ is a Galois field of order $p^{n}$.
Part 3. Uniqueness of Galois Fields
Let us point out that in Part 1 the characteristic of $F$ was a prime $p$, and that in Part 2-1 the operations in the algebraic closure $\overline{F}$ of $F$ were, when the unity $1_{F}$ of $F$ is regarded as $1_{\mathbb{Z}_{p}}$, in fact no different from the operations in the algebraic closure $\overline{\mathbb{Z}}_{p}$ of $\mathbb{Z}_{p}$.
Part 3-1. The Identity of a Field $E \subset \overline{\mathbb{Z}}_{p}$ with Cardinality $p^{n}$3
Lagrange’s theorem: If $H$ is a subgroup of a finite group $G$, then $|H|$ is a divisor of $|G|$.
For a field $\left( E , + , \times \right)$ with cardinality $p^{n}$, consider the group $\left( E^{\ast} , \times \right)$ under multiplication $\times$; then $E^{\ast}$ has the $p^{n} - 1$ elements of $E$ excluding the identity $0 \in E$ for $+$, together with the identity $1 \in E^{\ast}$. The order of $\alpha \in E^{\ast}$, that is, the cardinality of the cyclic group generated by $\alpha$, namely $\left| \alpha \right| = \left| \left< \alpha \right> \right|$, is a divisor of $p^{n} - 1$ by Lagrange’s theorem, and therefore we obtain $$ \alpha^{p^{n} - 1} = 1 \implies a^{p^{n}} = \alpha $$ In other words, every element of $E$ is a zero of $x^{p^{n}} - x$, and by the fundamental theorem of algebra, the elements of a field $E$ with cardinality $p^{n}$ contained in the algebraic closure $\overline{\mathbb{Z}}_{p}$ of $\mathbb{Z}_{p}$ are exactly the zeros of $\left( x^{p^{n}} - x \right) \in \mathbb{Z}_{p} [x]$.
Part 3-2. Minimal Splitting Property
By Part 2-1 and Part 3-1, for given $p$ and $n$ there exists a field $E$ all of whose elements consist exactly of the zeros of $\left( x^{p^{n}} - x \right)$; note that, since the characteristic of $F$ is $p$, the operations on those coefficients were also the same as the operations in the prime field $\mathbb{Z}_{p}$. By Part 2-3 and Part 1, $E$ is a Galois field that has the prime field $\mathbb{Z}_{p}$ as its prime field and must satisfy $|E| = p^{n}$, and once again by Part 2-1, we could see that $E$ is the minimal splitting field of $\left( x^{p^{n}} - x \right)$.
Property of minimal splitting fields: The minimal splitting fields of $f(x) \in F [ x ]$ are all isomorphic.
By the property of minimal splitting fields, for given $p$ and $n$ the Galois field is unique.
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Lemma: Freshman’s Dream
As a fact that is simply fun to know, the identity that appeared in Part 2-2, $$ \left( \alpha + \beta \right)^{p^{n}} =\alpha^{p^n} + \beta^{p^n} $$ is called the Freshman’s Dream. This is because, from the standpoint of a freshman who has just entered school, if the exponent could go inside the parentheses, difficult problems could be solved without complicated expansions. For reference, in number theory the congruence $\left( \alpha + \beta \right)^{p^{n}} \equiv \alpha^{p^n} + \beta^{p^n} \pmod{ p }$ can be derived in the same way even without any mention of the characteristic.