Finite field (${\mathbb{F}}_q$)

A finite field is a field with finitely many elements; its size is a prime power.

The finite field ${\mathbb{F}}_9$ can be represented as ${\mathbb{F}}_3[X]/(X^2+1)$, since the quadratic has no root modulo three. Its elements are $a+b\beta$ for $a,b\in {\mathbb{F}}_3$, with ${\beta }^2=-1$. Multiplication reduces to $(a+b\beta )(c+d\beta )=(ac-bd)+(ad+bc)\beta$ modulo three.

The quotient by an irreducible polynomial is a field construction gives ${\mathbb{F}}_4={\mathbb{F}}_2[X]/(X^2+X+1)$. The quadratic has no root in ${\mathbb{F}}_2$. With $\alpha =[X]$, the four elements are $0,1,\alpha ,\alpha +1$, and ${\alpha }^2=\alpha +1$, $\alpha (\alpha +1)=1$, $(\alpha +1{)}^2=\alpha$. These relations determine the full multiplication table.

In a splitting field over ${\mathbb{F}}_q$, the polynomial $X^{q^n}-X$ has $q^n$ distinct roots because its derivative is $-1$. Its roots are closed under addition, multiplication, additive inverses, and nonzero inverses, so they form the field ${\mathbb{F}}_{q^n}$. The splitting field consequently has degree $n$ over ${\mathbb{F}}_q$.

Every monic irreducible polynomial over ${\mathbb{F}}_q$ whose degree divides $n$ divides $X^{q^n}-X$. In particular, if $f$ has degree $n$ and $\alpha$ is one of its roots, then ${\mathbb{F}}_q(\alpha )$ has $q^n$ elements, so ${\alpha }^{q^n}=\alpha$ and the minimal polynomial $f$ divides $X^{q^n}-X$.

Let $N_q(n)$ be the number of monic irreducible polynomials of polynomial degree $n$ over ${\mathbb{F}}_q$. Factoring $X^{q^n}-X$ into all such polynomials whose degrees divide $n$ givesThe Möbius inversion formula therefore gives

Every degree-$n$ extension of ${\mathbb{F}}_q$ is isomorphic to ${\mathbb{F}}_{q^n}$.

The map $x\mapsto x^q$ is an automorphism of ${\mathbb{F}}_{q^n}$ fixing ${\mathbb{F}}_q$. Its powers are all the ${\mathbb{F}}_q$-automorphisms, and its order is $n$.

Over ${\mathbb{F}}_q$, the roots of the minimal polynomial of $\alpha \in {\mathbb{F}}_{q^n}$ are its distinct iterates $\alpha ,{\alpha }^q,{\alpha }^{q^2},\dots$ under the finite-field Frobenius automorphism.

Every finite field extension is a Galois extension, and

The multiplicative group of a finite field ${\mathbb{F}}_q$ is the group ${\mathbb{F}}_q^{\times }$ of its $q-1$ nonzero elements under multiplication.

A nonzero element $a$ of a finite field is a quartic residue when $a=b^4$ for some nonzero $b$. If $q\equiv 1(\mathrm{mod}4)$, the multiplicative group of a finite field is cyclic of order divisible by four, so the quartic residues form a subgroup of index four; equivalently $a^{(q-1)/4}=1$.

The nonzero elements of every finite field form a cyclic group. For a field with $q$ elements, combine elements realizing each prime-power factor of the group exponent to obtain an element of exponent order; the root bound for $X^m-1$ then forces that order to be $q-1$.

Every finite subgroup $H$ of the multiplicative group of a field is cyclic. If $m$ is the exponent of $H$, commutativity lets one combine elements whose orders realize the prime-power factors of $m$, producing an element of order $m$. Every element of $H$ is a root of $X^m-1$, so the Lagrange root bound over a field gives $|H|\le m$. Since $m\le |H|$, equality holds and that element generates $H$.

In a finite field with $q$ elements, the equation $x^k=1$ has exactly $\gcd (k,q-1)$ solutions. Indeed, after choosing a generator $g$ of the multiplicative group, $x=g^j$ is a solution exactly when $(q-1)\mid kj$.

For odd $p$, the nonzero squares form an index-two subgroup of ${\mathbb{F}}_p^{\times }$. Given nonzero $a,b$, either $a$ or $b$ is a square, or both occupy the nontrivial coset and $ab$ is a square. Taking $a=2$ and $b=3$ shows that one of $2,3,6$ is a square, apart from the immediate zero case $p=3$.