📂Abstract Algebra

Galois Field

Theorem ¹

For a prime number $p$ and a natural number $n$, a finite field with cardinality $p^{n}$ is defined as the Galois Field of order $p^{n}$, denoted as $\text{GF} \left( p^{n} \right)$. Finite fields are exclusively Galois Fields, and for a given $p$ and $n$, the Galois Field is uniquely determined.

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• The uniqueness implies that even if there are different fields, an isomorphism exists, making them essentially the same field.

Description

Even though no one believed in the existence of finite fields when Gauss first conceived the concept, it is now established that finite fields not only exist but are also unique and their specific structures have been identified. This eliminates the need for redundant research.

For instance, there is no need to ponder the existence of a field with $10$ elements, and since $\text{GF} \left( p \right) = \mathbb{Z}_{p}$ is an integer ring, we already know a lot about it. If more information is desired, one can approach through $\mathbb{Z}_{p}$ without clinging to abstract definitions, and vice versa.

Proof ²

Part 1. All finite fields are Galois Fields.

Let’s denote a finite extension field of field $F$ as $E$ and its degree over $F$ as $n := \left[ E : F \right]$.

If we set $| F | = q$, then $E$ is a $n$-dimensional vector space over $F$, hence $|E| = q^{n}$. A field has a multiplicative identity, and if the characteristic is $0$, an isomorphic subring exists, leading to an infinite field. Therefore, the characteristic of a finite field must be a finite natural number. If the characteristic of finite field $E$ is $p \ne 0$, then since $E$ has a multiplicative identity $1$, it must be $p \cdot 1 = 0$. Being a domain, $$p \cdot 1 = ( p_{1} \cdot 1 ) ( p_{2} \cdot 1 ) = 0$$ cannot have a $p_{1}, p_{2} \in \mathbb{Z}$ satisfying it, thus $p$ must be a prime number. Consequently, $E$ has a subfield isomorphic to the prime field $\mathbb{Z}_{p}$, and since $\left| \mathbb{Z}_{p} \right| = p$, it follows that $|E| = p^{n}$.

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Part 2. Existence of Galois Fields

Part 2-1. Zeros of $x^{p^{n}} - x$

Consider the algebraic closure $\overline{F}$ of a field $F$ with characteristic $p$, $\left( x^{p^{n}} - x \right)$.

Since $\overline{F}$ is algebraically closed, $\left( x^{p^{n}} - x \right) \in \overline{F} [ x ]$ factors into linear terms of $1$. A readily observable fact is $$x^{p^{n}} - x = ( x - 0 ) \left( x^{p^{n}-1} - 1 \right)$$ thus $0$ becomes a zero in $\left( x^{p^{n}} - x \right)$. Considering another zero $\alpha \ne 0$ of $f(x) := x^{p^{n}-1} - 1$, $f \left( \alpha \right) = 0$ hence $$0 = f \left( \alpha \right) = \alpha^{p^{n} - 1} - 1 \implies \alpha^{p^{n} - 1} = 1$$ By this, expressing $f(x)$ as a product of $\left( x - \alpha \right)$, $$\begin{align*} f(x) =& x^{p^{n}-1} - 1 \\ =& x^{p^{n}-1} - \alpha^{p^{n}-1} \\ =& (x - \alpha ) \left( x^{p^{n} - 2 } + \alpha x^{p^{n} - 3 } + \cdots + \alpha^{p^{n} - 3 } x + \alpha^{p^{n} - 2} \right) \end{align*}$$ For convenience, let’s denote the second factor as $$g(x) := \left( x^{p^{n} - 2 } + \alpha x^{p^{n} - 3 } + \cdots + \alpha^{p^{n} - 3 } x + \alpha^{p^{n} - 2} \right)$$ thus, $g(x)$ has $p^{n} - 1$ terms. Substituting $x = \alpha$ gives $$g ( \alpha ) = \alpha^{p^{n} - 2} \cdot \left( p^{n} - 1 \right) = {{\alpha^{p^{n} - 1}} \over { \alpha }} \left( p^{n} - 1 \right)$$ Previously, we stated that $\alpha \ne 0$ is a zero of $f(x)$, hence $\alpha^{p^{n}-1} - 1 = 0$, and since we assumed the characteristic is the prime number $p$, $$g ( \alpha ) = {{1} \over { \alpha }} \cdot (0 - 1) = - {{1} \over { \alpha }} \ne 0$$ Therefore, $\alpha$ is not a repeated root of $f(x) = 0$, and the same applies to other non-$\alpha$ zeros. Consequently, $\left( x^{p^{n}} - x \right)$ has precisely $p^{n}$ distinct zeros.

Part 2-2. Freshman’s Dream

Considering $\alpha , \beta \in F$, computing $\left( \alpha + \beta \right)^{p}$ by the binomial theorem yields $$\begin{align*} \left( \alpha + \beta \right)^{p} =& \sum_{k=1}^{p} \binom{p}{k} \alpha^{k} \beta^{p - k} \\ =& \alpha^{p} + \sum_{k=2}^{p-1} {{p!} \over { ( p - k )! ( k )! }} \alpha^{k} \beta^{p - k} + \beta^{p} \\ =& \alpha^{p} + \beta^{p} + p \sum_{k=2}^{p-1} {{ ( p - 1 )! } \over { ( p - k )! ( k )! }} \alpha^{k} \beta^{p - k} \end{align*}$$ Since the characteristic of $F$ is $p$, the last term becomes $0$, thus $$\left( \alpha + \beta \right)^{p} = \alpha^{p} + \beta^{p}$$ Taking $p$ to the power on both sides once more gives $$\left( \left( \alpha + \beta \right)^{p} \right)^{p} = \left( \alpha^{p} \right)^{p} + \left( \beta^{p} \right)^{p}$$ Simplifying, we get $\left( \alpha + \beta \right)^{p^{2}} =\alpha^{p^2} + \beta^{p^2}$, and repeating this $n$ times results in $$\left( \alpha + \beta \right)^{p^{n}} =\alpha^{p^n} + \beta^{p^n}$$

Now, consider the algebraic closure $\overline{ \mathbb{Z}_{p} }$ of $\mathbb{Z}_{p}$.

Let’s denote the set of all zeros in $\left( x^{p^{n}} - x \right) \in \overline{ \mathbb{Z}_{p} } [ x ]$ as $K \subset \overline{ \mathbb{Z}_{p} }$, and its elements as $\alpha , \beta \in K$.

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 $\left( \alpha + \beta \right)^{p^{n}} =\alpha^{p^n} + \beta^{p^n}$, $$\left( \alpha^{p^{n}} + \beta^{p^{n}} \right) - ( \alpha + \beta ) = \left( \alpha + \beta \right)^{p^{n}} - ( \alpha + \beta ) = 0$$ thus, $( \alpha + \beta ) \in K$.
• (ii) Additive identity: Since $0^{p^{n}} - 0 = 0$, $0 \in K$.
• (iii) Additive inverse: If $\left( - \alpha \right)^{p^{n}} = \left( - 1 \right)^{^{p^{n}}} \left( \alpha \right)^{p^{n}} = \left( - 1 \right)^{^{p^{n}}} \alpha$,
  • If $p=2$, then $-1 = 1$, so $\left( -\alpha \right) = \alpha \in K$.
  • Since $p \ne 2$ is an odd prime, $\left( - \alpha \right)^{p^{n}} - ( - \alpha ) = 0$, i.e., $( - \alpha ) \in K$.
• (iv) Closure under multiplication: Since $\left( \alpha \beta \right)^{p^{n}} = \alpha^{p^{n}} \beta^{p^{n}} = \alpha \beta$, $\left( \alpha \beta \right)^{p^{n}} - \alpha \beta = 0$, thus $\alpha \beta \in K$.
• (v) Multiplicative identity: Since $1^{p^{n}} - 1 = 0$, $1 \in K$.
• (vi) Multiplicative inverse: For $\alpha \ne 0$, taking the inverse of $\displaystyle \left( \alpha \right)^{p^{n}} = \alpha$ gives $\displaystyle {{1} \over {\left( \alpha \right)^{p^{n}} }} = {{1} \over { \alpha }}$, thus $$\left( {{1} \over { \alpha }} \right)^{p^{n}} - {{1} \over { \alpha }} = 0$$ so $\alpha^{-1} \in K$.
• (vii): $| K | = p^{n}$: The characteristic of $\mathbb{Z}_{p}$ is $p$, so by Part 2-1, $\left( x^{p^{n}} - x \right)$ has precisely $p^{n}$ distinct zeros.

Therefore, $K$ is a Galois Field of order

$p^{n}$.

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Part 3. Uniqueness of Galois Fields

From Part 1, the characteristic of $F$ is the prime number $p$, and from Part 2-1, operations in the algebraic closure $\overline{F}$ of $F$, considering the multiplicative identity $1_{F}$ of $F$ as $1_{\mathbb{Z}_{p}}$, are essentially the same as operations in the algebraic closure $\overline{\mathbb{Z}}_{p}$ of $\mathbb{Z}_{p}$.

Part 3-1. The nature of a field with cardinality $p^{n}$, $E \subset \overline{\mathbb{Z}}_{p}$ ³

Lagrange’s Theorem: If $H$ is a subgroup of a finite group $G$, then $|H|$ is a divisor of $|G|$.

Considering the group $\left( E^{\ast} , \times \right)$ under multiplication $\times$ for a field $\left( E , + , \times \right)$ with cardinality $p^{n}$, $E^{\ast}$ consists of $p^{n} - 1$ elements excluding the multiplicative identity $0 \in E$ of $E$ and the identity $1 \in E^{\ast}$. The order of $\alpha \in E^{\ast}$, the cardinality of the cyclic group generated by $\alpha$, is a divisor of $p^{n} - 1$ by Lagrange’s Theorem, hence $$\alpha^{p^{n} - 1} = 1 \implies a^{p^{n}} = \alpha$$ In other words, all elements of $E$ are zeros of $x^{p^{n}} - x$ and, by the Fundamental Theorem of Algebra, the elements of the field $E$ with cardinality $p^{n}$ within the algebraic closure $\overline{\mathbb{Z}}_{p}$ of $\mathbb{Z}_{p}$ are precisely the zeros of $\left( x^{p^{n}} - x \right) \in \mathbb{Z}_{p} [x]$.

Part 3-2. Minimal Polynomials

According to Part 2-1 and Part 3-1, for a given $p$ and $n$, there exists a field $E$ consisting entirely of zeros of $\left( x^{p^{n}} - x \right)$, and considering the characteristic of $F$ as $p$, the operations on its coefficients are the same as those in the prime field $\mathbb{Z}_{p}$. By Part 2-3 and Part 1, $E$ is a Galois Field containing the prime field $\mathbb{Z}_{p}$ as a subfield and satisfies $|E| = p^{n}$, and again by Part 2-1, $E$ is the minimal splitting field of $\left( x^{p^{n}} - x \right)$.

Properties of Minimal Splitting Fields: All minimal splitting fields are isomorphic.

Hence, for a given $p$ and $n$, the Galois Field is unique.

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Corollary: Freshman’s Dream

As an interesting fact, the equation from Part 2-2 $$\left( \alpha + \beta \right)^{p^{n}} =\alpha^{p^n} + \beta^{p^n}$$ is known as the Freshman’s Dream. It’s named because, from the perspective of a newcomer to the subject, the ability to distribute exponentiation inside a bracket would make solving complex problems straightforward without detailed expansions. Notably, in number theory, a similar result can be derived for congruences $\left( \alpha + \beta \right)^{p^{n}} \equiv \alpha^{p^n} + \beta^{p^n} \pmod{ p }$ without mentioning the characteristic.