For each prime power \(q=p^n\) there is a field \(\FF_q\) with \(q\) elements, unique up to isomorphism, and \(\FF_{p^n}/\FF_p\) is Galois with cyclic Galois group generated by the Frobenius automorphism \(x\mapsto x^p\).
A finite field of characteristic \(p\) has prime subfield \(\FF_p\). If \(f\in\FF_p[x]\) is irreducible of degree \(n\), then \(\FF_p[x]/\gens f\) is a field with \(p^n\) elements, and every field with \(p^n\) elements is a splitting field of \(x^{p^n}-x\) over \(\FF_p\).
Tower law for field degrees (Tag PR-ZO73V)
Let \(k\subseteq F\subseteq L\) be field extensions with \([L:F]\) and \([F:k]\) finite. Then \([L:k]\) is finite and \[ [L : k] = [L: F]\,[F: k]. \]
Let \(\alpha_1,\ldots,\alpha_m\) be a basis of \(F\) over \(k\) and \(\beta_1,\ldots,\beta_n\) a basis of \(L\) over \(F\). Every \(x\in L\) is \(\sum_j c_j\beta_j\) with \(c_j\in F\), and each \(c_j=\sum_i a_{ij}\alpha_i\) with \(a_{ij}\in k\), so the \(\alpha_i\beta_j\) span \(L\) over \(k\). If \(\sum_{i,j}a_{ij}\alpha_i\beta_j=0\) with \(a_{ij}\in k\), then linear independence of the \(\beta_j\) over \(F\) gives \(\sum_ia_{ij}\alpha_i=0\) for each \(j\), and linear independence of the \(\alpha_i\) over \(k\) gives \(a_{ij}=0\). So the \(mn\) products \(\alpha_i\beta_j\) form a basis of \(L\) over \(k\).
Construction of $\GF(p^n)$ as $\FF_p[x]/(f)$ (Tag T-QBQLM)
Let \(p\) be a prime, \(n\geq 1\), and \(f \in \FF_p[x]\) an irreducible polynomial of degree \(n\), and let \(\alpha\) be a root of \(f\) in an extension field of \(\FF_p\). Then \[ \GF(p^n)\cong \frac{\FF_p[x]}{(f)} \cong \FF_p[\alpha] = \spanof_{\FF_p}\theset{1, \alpha, \ldots, \alpha^{n-1}}. \]
Prime subfield and isomorphism type of a finite field (Tag PR-ADLQN)
Let \(F\) be a finite field. Then \(\ch F = p\) for a prime \(p\), the prime subfield of \(F\) is isomorphic to \(\FF_p\), and \(F \cong \FF_{p^n}\) for the unique integer \(n = [F:\FF_p] \geq 1\), so that \(\abs{F} = p^n\).
Subfields of the finite field $\FF_{p^k}$ (Tag PR-Q2IFX)
Let \(p\) be a prime and \(k,\ell\geq 1\). The field \(\FF_{p^k}\) contains a subfield with \(p^\ell\) elements if and only if \(\ell\mid k\). In that case the subfield is unique, equal to \(\theset{x\in\FF_{p^k}\suchthat x^{p^\ell}=x}\).
If \(F\subseteq\FF_{p^k}\) is a subfield with \(p^\ell\) elements, then \(\FF_{p^k}\) is an \(F\)-vector space of some dimension \(d\), so \(p^k=(p^\ell)^d\) and \(\ell\mid k\).
Conversely, suppose \(\ell\mid k\). Every element of \(\FF_{p^k}\) is a root of \(x^{p^k}-x\), which has degree \(p^k\), so \(x^{p^k}-x=\prod_{a\in\FF_{p^k}}(x-a)\). Since \(\ell\mid k\), \(p^\ell-1\) divides \(p^k-1\), so \(x^{p^\ell-1}-1\) divides \(x^{p^k-1}-1\) and \(x^{p^\ell}-x\) divides \(x^{p^k}-x\). Hence \(x^{p^\ell}-x\) has \(p^\ell\) distinct roots in \(\FF_{p^k}\). They form the fixed set of the field automorphism \(x\mapsto x^{p^\ell}\), which is a subfield with \(p^\ell\) elements. Every element of a subfield with \(p^\ell\) elements satisfies \(x^{p^\ell}=x\), so such a subfield is contained in, and hence equal to, this set.
The subfields of \(\FF_{p^n}\) are the fields \(\FF_{p^m}\) with \(m\divides n\), one for each such \(m\), and \(x^{p^n}-x\) is the product of the monic irreducible polynomials over \(\FF_p\) of degree dividing \(n\). Hence a monic irreducible \(f\in\FF_p[x]\) of degree \(d\) divides \(x^{p^n}-x\) if and only if \(d\divides n\), if and only if \(f\) splits over \(\FF_{p^n}\).
A field with $p^n$ elements is a splitting field of $x^{p^n} - x$ (Tag PR-32R3E)
Let \(p\) be a prime, let \(n \geq 1\), and let \(F\) be a field with \(p^n\) elements, whose prime subfield is \(\FF_p\). Then \(F\) is a splitting field of \(\rho(x) = x^{p^n} - x\) over \(\FF_p\), and the elements of \(F\) are exactly the \(p^n\) roots of \(\rho\) in \(F\).
$x^{p^n}-x$ is the product of the monic irreducibles over $\FF_p$ of degree dividing $n$ (Tag PR-73MCN)
Let \(p\) be prime and \(n\geq 1\), and set \(\rho_n(x) \coloneqq x^{p^n} - x\). Then \[ \rho_n(x) = \prod_{\substack{f\in\FF_p[x]\text{ monic irreducible}\\ \deg f\mid n}} f(x). \] Equivalently, for a monic irreducible \(f\in\FF_p[x]\), \[ f(x)\mid \rho_n(x) \iff \deg f\mid n. \]
The roots of \(\rho_n(x)=x^{p^n}-x\) in an algebraic closure of \(\FF_p\) are exactly the elements of \(\FF_{p^n}\). If \(f\) is monic irreducible of degree \(d\) and \(\alpha\) is one of its roots, then \(\FF_p(\alpha)\cong\FF_{p^d}\). Hence \[ f\mid \rho_n \iff \alpha\in\FF_{p^n} \iff \FF_{p^d}\subseteq\FF_{p^n} \iff d\mid n. \] Finally, \(\rho_n'(x)=-1\), so \(\rho_n\) is squarefree; therefore each of these irreducible factors occurs exactly once, giving the displayed product.
Finite fields are not algebraically closed (Tag PR-DEG36)
A finite field \(F\) is not algebraically closed.
Write \(F = \theset{a_1, \ldots, a_q}\) and let \(f(x) \coloneqq 1 + \prod_{i=1}^q (x - a_i) \in F[x]\). Then \(f\) is nonconstant, and \(f(a_j) = 1 \neq 0\) for every \(j\), so \(f\) has no root in \(F\).
(Structure of finite fields)
\(\FF_{p^n}\) exists and is unique up to isomorphism, as the splitting field of \(x^{p^n} - x\) over \(\FF_p\).
\(\FF_{p^m}\) embeds in \(\FF_{p^n}\) if and only if \(m \divides n\), so the lattice of subfields of \(\FF_{p^n}\) is the lattice of divisors of \(n\).
\(\FF_{p^n}/\FF_p\) is Galois with cyclic Galois group of order \(n\), generated by \(x\mapsto x^p\).
The multiplicative group \(\FF_{q}^\times\) is cyclic of order \(q-1\).
Hence every extension \(\FF_{p^n}/\FF_{p^m}\) is Galois with cyclic Galois group of order \(n/m\), generated by \(x\mapsto x^{p^m}\).