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A finite field is a field with finitely many elements: addition, subtraction, multiplication, and division by a nonzero element all stay inside a finite collection. Write ${\mathbb{F}}_q$, or $\mathrm{GF}(q)$, for a finite field with $q$ elements. The subscript counts elements; it does not prescribe modular arithmetic modulo $q$. The field with four elements is the smallest example where that distinction matters.

Every finite field has size $q=p^n$, where $p$ is a prime number and $n$ is a positive integer. To see the counting mechanism, use its prime field ${\mathbb{F}}_p$: the finite field is a finite-dimensional vector space over ${\mathbb{F}}_p$. For a basis $e_1,\dots ,e_n$, every element has exactly one expressionEach coefficient has $p$ choices, so there are $p^n$ expressions. Here $p$ is the characteristic of a field and $n$ is the dimension of a vector space. Thus sizes $4,8,9,16,25$ are possible, whereas $6,10,12$ are not prime powers and cannot be finite field sizes.

Conversely, every prime power occurs, and any two finite fields of the same size admit a field isomorphism. These are classification statements, beyond the counting argument. See Conrad's Theorems 1.5, 2.2, and 2.7.^[ref] Uniqueness means the arithmetic agrees after relabeling; it does not choose a basis, an irreducible polynomial for a quotient ring representation, or a particular field isomorphism. Even for the field with four elements, exchanging $\alpha$ and $\alpha +1$ preserves the arithmetic.

The appeal for coding theory is exact, finite arithmetic with reliable division: polynomial interpolation and linear algebra work over a finite alphabet. A Reed-Solomon code represents messages by polynomials and adds redundancy by evaluating them at several distinct finite field elements. Forney's MIT notes develop this arithmetic as preparation for such error-correcting codes.^[ref] In cryptography, finite fields likewise supply exact arithmetic on finitely many symbols, with operations that can be implemented using bits.

Historically, Évariste Galois published work on finite fields beyond prime fields in 1830, constructing them by adjoining elements satisfying irreducible polynomials over prime fields. Conrad's historical account also records E. H. Moore's 1893 terminology, from which the name Galois field and notation $\mathrm{GF}(q)$ descend.^[ref]

A K-isomorphism between extensions of $K$ is a field isomorphism fixing $K$ pointwise. Simple algebraic extensions $K(\alpha )$ and $K(\beta )$ admit one sending $\alpha$ to $\beta$ exactly when their minimal polynomials agree: both then identify with the same quotient $K[X]/(m)$.

The polynomial $X^n-2$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion with prime two: all coefficients except the leading one are divisible by two, while the constant coefficient is not divisible by four. Together with Gauss's lemma, this establishes irreducibility in $\mathbb{Q}[X]$. Part (b) shows that its principal ideal is a maximal ideal, soLet $\alpha$ denote the residue class of $X$. The polynomial division algorithm gives every residue class a representative of degree less than $n$, so $1,\alpha ,\dots ,{\alpha }^{n-1}$ span $K_n$ over $\mathbb{Q}$. If a linear combination of them vanished, its degree-less-than-$n$ representative would be a multiple of the degree-$n$ polynomial $X^n-2$, and hence would be zero. They are therefore a basis, givingThe quadratic polynomial $X^2+1$ has no rational root, so it is irreducible over $\mathbb{Q}$. The same maximal-ideal argument proves $L=\mathbb{Q}[X]/(X^2+1)$ is a field.

These two quadratic fields are not isomorphic. In $L$, the class of $X$ squares to $-1$. But $K_2$ embeds in $\mathbb{R}$ by sending its class of $X$ to $\sqrt{2}$: evaluation at $\sqrt{2}$ has kernel $(X^2-2)$, by irreducibility and polynomial division. No real number squares to $-1$, so no element of $K_2$ does. A field isomorphism preserves one, minus one and squaring, so it cannot carry $L$ to $K_2$.

A K-isomorphism $L\to L^{\prime }$ is a field isomorphism fixing every element of $K$. The field automorphism group ${\mathrm{Aut}}_K(L)$ consists of such isomorphisms $L\to L$, with composition.

If an isomorphism sends $\alpha$ to $\beta$, it takes every polynomial relation over $K$ for $\alpha$ to the same relation for $\beta$; the monic minimal polynomials coincide. Conversely, if their common minimal polynomial is $m$, evaluation induces field isomorphisms $K[X]/(m)\cong K(\alpha )$ and $K[X]/(m)\cong K(\beta )$. Their composite sends $\alpha$ to $\beta$ and fixes $K$.

An automorphism of $K(\alpha )$ sends $\alpha$ to a root of $m$ inside $K(\alpha )$ and is determined by this image. More precisely the field automorphism group permutes the set of roots of $m$ lying in this field, a set of size $s\le d$. This action is faithful because fixing $\alpha$ fixes the generated field. Thus ${\mathrm{Aut}}_K(K(\alpha ))$ embeds in $S_s$, and then in $S_d$ by fixing $d-s$ extra labels. In particular it is finite. This argument also covers inseparable minimal polynomials.

For an example not isomorphic even as abstract fields, take $K={\mathbb{F}}_p(t)$, let $\alpha$ generate ${\mathbb{F}}_{p^p}$ over ${\mathbb{F}}_p$, and put $\beta =t^{1/p}$. The constant-field minimal polynomial of $\alpha$ remains irreducible over the rational function field: if it factored over ${\mathbb{F}}_p(t)$, each monic factor's coefficients would be algebraic over ${\mathbb{F}}_p$, being symmetric functions of constant algebraic roots. A rational function algebraic over ${\mathbb{F}}_p$ is constant, so those coefficients would lie in ${\mathbb{F}}_p$, contradicting the original irreducibility. The polynomial $X^p-t$ is irreducible by Eisenstein criterion at $t$ in ${\mathbb{F}}_p[t]$. Both degrees are $p$, butThe first field has $p^p$ roots of $X^{p^p}-X$; the second has only $p$. Indeed a rational function algebraic over a finite constant field must be constant: a nonconstant rational function makes the transcendental variable algebraic over its own rational-function field, and therefore cannot itself be algebraic over the constants. The root counts are isomorphism invariants. These extensions are not isomorphic.

If $K={\mathbb{F}}_q$ is finite, every degree-$d$ extension is a field with $q^d$ elements. Such fields are splitting fields of $X^{q^d}-X$ over $K$, so uniqueness of the splitting field gives a K-isomorphism. There is no finite-base-field example of this kind.