Finite Fields

Finite fields are the cleanest testing ground for the Galois correspondence. Their Galois groups are cyclic, so every subgroup is normal and the intermediate-field lattice is a divisor lattice. The whole theory of a finite field — how large it can be, which subfields it has, how its elements factor irreducible polynomials — follows from one automorphism, the Frobenius map.

Order and existence

A finite field $\mathbb{F}$ has prime characteristic $p$ (its characteristic cannot be $0$, and a composite characteristic would produce zero divisors), so it contains the prime field ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$. As a vector space over ${\mathbb{F}}_p$ it has some finite dimension $n=[\mathbb{F}:{\mathbb{F}}_p]$, and a vector space of dimension $n$ over a field of $p$ elements has exactly $p^n$ elements.

The multiplicative group ${\mathbb{F}}^{\times }$ has order $p^n-1$, so every nonzero element satisfies ${\alpha }^{p^n-1}=1$ by Lagrange, and hence every element (including $0$) satisfies ${\alpha }^{p^n}=\alpha$. The $p^n$ elements of $\mathbb{F}$ are therefore exactly the $p^n$ roots of $x^{p^n}-x$, which means $\mathbb{F}$ is a splitting field of that polynomial over ${\mathbb{F}}_p$.

Existence is the splitting-field construction: the roots of $x^{p^n}-x$ form a field because the map $\alpha \mapsto {\alpha }^{p^n}$ is a ring homomorphism (the Frobenius, iterated $n$ times), and its fixed set is closed under the field operations. The polynomial is separable — its derivative is $p^n{x}^{p^n-1}-1=-1$, which shares no root with it — so it has $p^n$ distinct roots. Uniqueness holds because any two splitting fields of the same polynomial over ${\mathbb{F}}_p$ are isomorphic.¹

The Frobenius automorphism and the Galois group

Because ${\mathbb{F}}_{p^n}$ is the splitting field of the separable polynomial $x^{p^n}-x$, the extension ${\mathbb{F}}_{p^n}/{\mathbb{F}}_p$ is Galois. Its group is generated by a single map.

The Frobenius fixes ${\mathbb{F}}_p$ pointwise, since every $a\in {\mathbb{F}}_p$ satisfies $a^p=a$ (Fermat's little theorem). Iterating, ${\sigma }_p^k(\alpha )={\alpha }^{p^k}$, and ${\sigma }_p^n=1$ because ${\alpha }^{p^n}=\alpha$ for all $\alpha$. No smaller power is the identity: ${\sigma }_p^k=1$ would force ${\alpha }^{p^k}=\alpha$ for all $p^n$ elements, but $x^{p^k}-x$ has at most $p^k<p^n$ roots. So ${\sigma }_p$ has order exactly $n$.

The Frobenius partitions ${\mathbb{F}}_{p^n}$ into orbits. The orbit of $\alpha$ is $\{\alpha ,{\alpha }^p,{\alpha }^{p^2},\dots \}$, and its length equals the degree of the minimal polynomial of $\alpha$ over ${\mathbb{F}}_p$ — the orbit consists of the Galois conjugates of $\alpha$.

The subfield lattice is the divisor lattice

The Galois group $\mathbb{Z}/n\mathbb{Z}$ has exactly one subgroup of each order dividing $n$: for a divisor $d\mid n$, the subgroup $⟨{\sigma }_p^d⟩$ has order $n/d$. By the Fundamental Theorem, subgroups correspond to intermediate fields, and index corresponds to degree. Since the group is abelian, every subgroup is normal, so every intermediate field is Galois over ${\mathbb{F}}_p$.

The subfield lattice is the divisor lattice of $n$ drawn upside down: containment of fields matches divisibility of exponents. For $n=12$ the divisors $1,2,3,4,6,12$ produce six subfields.

Restricting Frobenius to a subfield recovers its own Frobenius: ${\sigma }_p$ has order $n$ in $\mathrm{Gal}({\mathbb{F}}_{p^n}/{\mathbb{F}}_p)$ but order $d$ in $\mathrm{Gal}({\mathbb{F}}_{p^d}/{\mathbb{F}}_p)$, and $\mathrm{Gal}({\mathbb{F}}_{p^d}/{\mathbb{F}}_p)$ is the quotient $⟨{\sigma }_p⟩/⟨{\sigma }_p^d⟩$.

The multiplicative group is cyclic

Every finite subgroup of the multiplicative group of a field is cyclic, a consequence of a finite field having at most $k$ solutions to $x^k=1$. Applied to ${\mathbb{F}}_{p^n}$ itself:

A primitive element generates the whole field over ${\mathbb{F}}_p$, since it already generates the multiplicative group, so ${\mathbb{F}}_{p^n}={\mathbb{F}}_p(\theta )$ is a simple extension. Its minimal polynomial has degree $n$, which proves that an irreducible polynomial of degree $n$ over ${\mathbb{F}}_p$ exists for every $n\ge 1$.

Worked example: constructing ${\mathbb{F}}_8$

Take $p=2$, $n=3$. The polynomial $x^3+x+1$ has no root in ${\mathbb{F}}_2$ (both $0$ and $1$ evaluate to $1$), so it is irreducible, and

an $8$-element field with basis $1,\theta ,{\theta }^2$. The element $\theta$ is primitive: its powers exhaust ${\mathbb{F}}_8^{\times }$, each computed by reducing with ${\theta }^3=\theta +1$.

The Frobenius orbit of $\theta$ is $\{\theta ,{\theta }^2,{\theta }^4\}$, and indeed

as the table confirms: the coefficient of $x^2$ is $\theta +{\theta }^2+{\theta }^4=\theta +{\theta }^2+({\theta }^2+\theta )=0$ in characteristic $2$, and the constant term is $\theta \cdot {\theta }^2\cdot {\theta }^4={\theta }^7=1$. The remaining orbit $\{{\theta }^3,{\theta }^6,{\theta }^5\}$ multiplies out to the other irreducible cubic $x^3+x^2+1$, and the orbit $\{1\}$ gives $x+1$. Together with the root $0$, the orbits reassemble

the factorization predicted by the theorem below. Since $3$ is prime, the only subfields of ${\mathbb{F}}_8$ are ${\mathbb{F}}_2$ and ${\mathbb{F}}_8$ itself — so ${\mathbb{F}}_4⊈{\mathbb{F}}_8$ even though $4<8$: containment follows exponent divisibility, not size.

Factoring $x^{p^n}-x$ and counting irreducibles

The polynomial $x^{p^n}-x$ vanishes exactly on ${\mathbb{F}}_{p^n}$, whose elements generate subfields ${\mathbb{F}}_{p^d}$ for the various $d\mid n$. Grouping the linear factors by the degree $d$ of the minimal polynomial of their root gives a clean product formula.

This turns the count of irreducibles into a divisor sum. Let $\psi (d)$ be the number of monic irreducibles of degree $d$ over ${\mathbb{F}}_p$. Comparing degrees on both sides of the factorization gives $p^n={\sum }_{d\mid n}d\psi (d)$, and Möbius inversion solves for $\psi$.

The same factorization produces irreducibles recursively. Over ${\mathbb{F}}_2$, the irreducible cubics divide

and the formula agrees: $\psi (3)=\frac{1}{3}(2^3-2)=2$.

A corollary concerns a polynomial irreducible over $\mathbb{Z}$:

For $p=2$, $x^4+1=(x+1{)}^4$. For odd $p$, $8\mid p^2-1$, so $x^8-1\mid x^{p^2-1}-1$; the roots of $x^4+1$ (which divides $x^8-1$) therefore lie in ${\mathbb{F}}_{p^2}$, giving each root degree at most $2$ over ${\mathbb{F}}_p$. A degree-$4$ polynomial with all roots of degree $\le 2$ cannot be irreducible.²

Because ${\mathbb{F}}_{p^m}\subseteq {\mathbb{F}}_{p^n}$ whenever $m\mid n$, the finite fields of characteristic $p$ nest into a directed system, and their union is the algebraic closure:

Application: cyclic codes over finite fields

Error-correcting codes exploit finite-field structure directly. A linear code of length $n$ over ${\mathbb{F}}_q$ is a subspace $C\subseteq {\mathbb{F}}_q^n$; codewords are the vectors it contains, and the receiver corrects transmission errors by mapping a received word back to the nearest codeword. Identifying a vector $(a_0,\dots ,a_{n-1})$ with the polynomial $a_0+a_{1}t+\cdots +a_{n-1}{t}^{n-1}$ realizes ${\mathbb{F}}_q^n$ as the quotient ring

in which multiplication by $t$ cyclically shifts the coefficients. A code closed under cyclic shifts is a cyclic code, and closure under shifts plus linearity amounts to closure under multiplication by every polynomial.³

Choosing a code therefore reduces to factoring $x^n-1$ over ${\mathbb{F}}_q$ and picking a divisor $g(x)$. Over ${\mathbb{F}}_2$ with $n=7$,

so $g(x)=1+x+x^3$ generates a $(7,4)$ code: messages are the $2^4$ polynomials of degree less than $4$, each encoded by multiplying by $g(x)$.

The error-correcting power comes from arranging that $g(x)$ has a run of consecutive powers of a primitive element among its roots. BCH codes choose $g(x)$ as the least common multiple of the minimal polynomials of $\omega ,{\omega }^2,\dots ,{\omega }^{2t}$ for a primitive $(n)$th root of unity $\omega$ in an extension field; the resulting code corrects up to $t$ errors. The construction rests on the Galois theory of finite fields: minimal polynomials over ${\mathbb{F}}_q$ are Frobenius orbits, and the code's guarantees are theorems about roots of unity in ${\mathbb{F}}_{q^m}$.⁴