Finite fields are rigid: their size determines them up to isomorphism. Their structure combines field theory, cyclic groups, and Frobenius automorphisms.
Existence and uniqueness
Theorem. A finite field has \(p^n\) elements for some prime \(p\) and integer \(n\ge1\). Conversely, for every \(p^n\) there is a finite field with \(p^n\) elements, unique up to isomorphism.
Proof. The prime subfield is \(\mathbb F_p\), and the field is a finite-dimensional vector space over it, so its size is \(p^n\). Existence comes from the splitting field of \(x^{p^n}-x\), whose roots form a field. Uniqueness follows because any field of size \(p^n\) is exactly the set of roots of this polynomial.
Construction by irreducible polynomials
Proposition. If \(f\in\mathbb F_p[x]\) is irreducible of degree \(n\), then \(\mathbb F_p[x]/(f)\) is a field with \(p^n\) elements.
Example. Since \(x^2+x+1\) is irreducible over \(\mathbb F_2\), \[ \mathbb F_4\cong\mathbb F_2[x]/(x^2+x+1). \] If \(\alpha=x+(x^2+x+1)\), then \(\alpha^2=\alpha+1\).
Multiplicative group
Theorem. The multiplicative group \(\mathbb F_q^\times\) of a finite field is cyclic of order \(q-1\).
Remark. A generator of \(\mathbb F_q^\times\) is called a primitive element. This theorem is one of the most useful places where finite field theory uses finite abelian group structure.
Frobenius automorphism
Theorem. The Frobenius map \(\operatorname{Frob}_p:x\mapsto x^p\) is an automorphism of \(\mathbb F_{p^n}\). The Galois group \(\operatorname{Gal}(\mathbb F_{p^n}/\mathbb F_p)\) is cyclic of order \(n\), generated by Frobenius.
Proof. In characteristic \(p\), \((x+y)^p=x^p+y^p\), so Frobenius is a homomorphism. It is injective on a field, hence bijective because the field is finite. Its fixed field inside \(\mathbb F_{p^n}\) is \(\mathbb F_p\), and \(\operatorname{Frob}_p^n=1\).
Subfields
Proposition. The subfields of \(\mathbb F_{p^n}\) are exactly \(\mathbb F_{p^d}\) for \(d\mid n\).
Calculation. The field \(\mathbb F_{2^6}\) has subfields \(\mathbb F_2\), \(\mathbb F_4\), \(\mathbb F_8\), and \(\mathbb F_{64}\), corresponding to divisors \(1,2,3,6\) of \(6\).
Irreducible polynomials and roots
Proposition. An irreducible polynomial \(f\in\mathbb F_q[x]\) of degree \(d\) divides \(x^{q^n}-x\) iff \(d\mid n\). Equivalently, its roots lie in \(\mathbb F_{q^n}\) exactly when \(d\mid n\).
Existence and uniqueness
Theorem. For every prime power \(q=p^n\), there exists a field \(\mathbb F_q\), and any two fields with \(q\) elements are isomorphic. It is the splitting field over \(\mathbb F_p\) of \[ x^q-x. \]
Proof. The derivative of \(x^q-x\) is \(-1\), so its roots in a splitting field are distinct. They form a field: in characteristic \(p\), \[ (a+b)^q=a^q+b^q,\qquad (ab)^q=a^qb^q, \] and inverses are preserved. Hence there are exactly \(q\) roots. Conversely, in any field of order \(q\), every nonzero element satisfies \(a^{q-1}=1\), so all elements are roots. Thus every such field is the same splitting-field construction up to isomorphism.
Frobenius and subfields
The Galois group is cyclic: \[ \operatorname{Gal}(\mathbb F_{p^n}/\mathbb F_p) =\langle x\mapsto x^p\rangle \cong\mathbb Z/n\mathbb Z. \] The fixed field of the \(d\)-th Frobenius power has \(p^{\gcd(n,d)}\) elements. Consequently \(\mathbb F_{p^m}\) embeds in \(\mathbb F_{p^n}\) exactly when \(m\mid n\), and then it is the unique subfield of that size.
Irreducible polynomials
Proposition. If \(N_q(n)\) denotes the number of monic irreducible polynomials of degree \(n\) over \(\mathbb F_q\), then \[ q^n=\sum_{d\mid n}dN_q(d),\qquad N_q(n)=\frac1n\sum_{d\mid n}\mu(d)q^{n/d}. \]
Proof. The polynomial \(x^{q^n}-x\) is the product of all monic irreducibles whose degrees divide \(n\), each once. Compare degrees and apply M?bius inversion.