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Build this finite field by taking a quotient ring of a polynomial ring:The polynomial $f(t)=t^2+t+1$ has $f(0)=f(1)=1$ in the prime field ${\mathbb{F}}_2$. A polynomial of polynomial degree two is an irreducible polynomial exactly when it has no polynomial root in the coefficient field. Thus the quotient by an irreducible polynomial is a field construction applies. Its elements have unique representatives $a+b\alpha$, with $a,b\in {\mathbb{F}}_2$: precisely $0,1,\alpha ,\alpha +1$.

The defining relation is ${\alpha }^2+\alpha +1=0$. Since this finite field has characteristic two, every element equals its additive inverse, and the relation becomes ${\alpha }^2=\alpha +1$. Addition acts separately on the two coefficients:All coefficient calculations use modular arithmetic modulo two. For example, $(\alpha +1)+\alpha =1$ and $(\alpha +1)+(\alpha +1)=0$. Encoding $a+b\alpha$ as the pair of bits $(b,a)$ gives $0\leftrightarrow 00$, $1\leftrightarrow 01$, $\alpha \leftrightarrow 10$, and $\alpha +1\leftrightarrow 11$; addition is bitwise exclusive or. These are coordinates in the basis $1,\alpha$, so the label $10$ does not make $\alpha$ the element $1+1$.

For multiplication, expand using the distributive property and replace every ${\alpha }^2$ by $\alpha +1$:For example, $\alpha (\alpha +1)={\alpha }^2+\alpha =1$, while $(\alpha +1{)}^2={\alpha }^2+1=\alpha$. The following complete tables are computed from the coefficient rules. Colors identify results, rather than their numerical sizes. Every row of the addition table lists all four elements, and every nonzero row of the multiplication table does too: adding an element, or multiplying by a nonzero element, can be undone.

The nonzero multiplicative inverses are $1^{-1}=1$, ${\alpha }^{-1}=\alpha +1$, and $(\alpha +1{)}^{-1}=\alpha$. Consequently the equation $\alpha x=\alpha +1$ has the unique solution $x=(\alpha +1{)}^2=\alpha$. Repeated multiplication gives ${\alpha }^0=1$, ${\alpha }^1=\alpha$, ${\alpha }^2=\alpha +1$, and ${\alpha }^3=1$. They cycle through every element of the multiplicative group of a finite field.

Compare the ring $\mathbb{Z}/4\mathbb{Z}$, also with four elements. Its addition satisfies $1+1=2\ne 0$, and its multiplication satisfies $2^2=0$. Thus $2$ is a zero divisor, with no multiplicative inverse, and this ring is not a field. Multiplying by $2$ collapses different inputs to the same output: both $0$ and $2$ give $0$, while both $1$ and $3$ give $2$. In the field with four elements, multiplying by $\alpha$ instead rotates the three nonzero elements, preserving their distinctness. The figure makes visible why counting four elements alone does not specify their arithmetic.

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]

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. Zero is excluded because it has no multiplicative inverse. In the field with four elements, these elements are $1,\alpha ,\alpha +1$, obtained by repeated multiplication by $\alpha$: they form a cyclic group of group order three. Its addition behaves differently: every nonzero element added to itself gives zero. Keeping the two operations distinct prevents confusing the size of the finite field with the group order of its multiplicative group of a finite field.

Let $D$ be a finite integral domain. For any $a\ne 0$, multiplication $m_a:D\to D$ is injective: $ax=ay$ implies $a(x-y)=0$, hence $x=y$ because there are no zero divisors. An injective map of a finite set to itself is surjective, so $ab=1$ for some $b$. Every nonzero element is invertible; therefore a finite integral domain is a field.

For an irreducible polynomial $f\in F[X]$, consider a nonzero class $[g]$ in the quotient ring $F[X]/(f)$. Since $f$ does not divide $g$, irreducibility gives $\gcd (f,g)=1$. The Euclidean algorithm in the polynomial ring gives polynomials $u,v$ with $uf+vg=1$. Modulo $(f)$, $[v][g]=1$. Thus the quotient by an irreducible polynomial is a field.

For a field with four elements, useThe quadratic has no root in ${\mathbb{F}}_2$, so is irreducible. Write $\alpha =[X]$. Polynomial division shows the elements are $0,1,\alpha ,\alpha +1$, with ${\alpha }^2=\alpha +1$. Its multiplication table is

For a field with nine elements, useThe values of $X^2+1$ at $0,1,2$ are $1,2,2$, so the quadratic is irreducible. Every class is uniquely $a+b\beta$, $a,b\in {\mathbb{F}}_3$, with ${\beta }^2=-1=2$. These nine classes form a finite field; multiplication is $(a+b\beta )(c+d\beta )=(ac-bd)+(ad+bc)\beta$, with coefficients modulo three.

A suitable field with four elements isWrite $\alpha$ for the class of $t$. Monic polynomial division over a ring gives a unique representative $a+b\alpha$, with $a,b\in {\mathbb{F}}_2$: existence follows by division, and uniqueness follows because a nonzero multiple of the degree-two modulus cannot have degree less than two. The four elements are therefore $0,1,\alpha ,\alpha +1$.

The defining relation gives ${\alpha }^2=\alpha +1$, and hence $\alpha (\alpha +1)=1$. Thus the nonzero elements have inversesThe quotient ring is commutative and every nonzero element is a unit, so it is a field. This verifies the field property directly, without quoting a theorem about irreducible-polynomial quotients.

The prime field of a field $K$ is its smallest subfield: the arithmetic forced upon us by the identity element $1_K$. Every subfield must contain $1_K$, its repeated sums, their additive inverses, and the multiplicative inverses of nonzero results. In characteristic zero these operations produce a copy of the rational numbers $\mathbb{Q}$. In positive characteristic of a field $p$, they produce a copy of ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$, with addition and multiplication given by modular arithmetic.

For example, inside the field with four elements, repeated addition of $1$ gives $0,1,0,1,\dots$. Thus its prime field is $\{0,1\}$, and the other two elements cannot be reached by repeatedly adding $1$. This is why a finite field need not consist entirely of ordinary integer residues: a larger finite field extends the arithmetic of its prime field with additional elements.

The positive characteristic of a field must be a prime number. If its least vanishing multiple of $1_K$ were $rs$ with $1<r,s<rs$, then $(r{1}_K)(s{1}_K)=0$ would exhibit zero divisors. The absence of zero divisors rules this out. The prime field supplies the scalars when viewing $K$ as a vector space; its basis then explains the possible sizes of a finite field.