Finite Fields

We prove each part in turn, building on the previous ones. Throughout, fix an algebraic closure $\overline{\mathbb{F}_p}$ of the prime field $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ (this exists by Algebraic Closure).

Part 1 (Characteristic). Let $F$ be a finite field. The ring homomorphism $\varphi: \mathbb{Z} \to F$, $\varphi(m) = m \cdot 1_F$, has kernel a subgroup $d\mathbb{Z}$ of $\mathbb{Z}$ for some $d \geq 0$. Since $F$ is finite, $\varphi$ cannot be injective (there are infinitely many integers and finitely many elements of $F$), so $d \neq 0$; write $d > 0$. The image $\varphi(\mathbb{Z}) \cong \mathbb{Z}/d\mathbb{Z}$ is a subring of the field $F$, hence an integral domain (no zero divisors, as it embeds in a field). But $\mathbb{Z}/d\mathbb{Z}$ is an integral domain if and only if $d$ is prime. So $d = p$ is prime; this $p$ is by definition the characteristic of $F$, and $\mathbb{Z}/p\mathbb{Z} \hookrightarrow F$ identifies a copy of $\mathbb{F}_p$ inside $F$, called the prime subfield. Now $F$ is a vector space over its prime subfield $\mathbb{F}_p$: addition in $F$ is the vector space addition, and scalar multiplication is multiplication by elements of $\mathbb{F}_p \subset F$; the field axioms for $F$ guarantee the vector space axioms (distributivity, associativity of scalar multiplication with field multiplication) hold. Since $F$ is finite it is finite-dimensional over $\mathbb{F}_p$, say of dimension $n$. Choosing a basis $v_1,\dots,v_n$, every element of $F$ is uniquely $\sum_i c_i v_i$ with $c_i \in \mathbb{F}_p$, giving a bijection $F \leftrightarrow \mathbb{F}_p^n$ as sets. Hence $|F| = p^n$.

Part 2 (Existence). Let $q = p^n$ and consider the polynomial

Let $K \subset \overline{\mathbb{F}_p}$ be the splitting field of $f$ over $\mathbb{F}_p$, i.e. the subfield generated by $\mathbb{F}_p$ and all roots of $f$ in $\overline{\mathbb{F}_p}$; this exists and is unique as a subfield of the fixed algebraic closure. Claim: $f$ has no repeated roots in $\overline{\mathbb{F}_p}$. Compute the formal derivative:

since $q = p^n \equiv 0$ in $\mathbb{F}_p$ (characteristic $p$ kills the coefficient $q$). Since $f'(x) = -1$ is a nonzero constant, $\gcd(f, f') = 1$ in $\overline{\mathbb{F}_p}[x]$ (a nonzero constant is a unit, so the gcd is a unit). A standard fact from the theory of the derivative (used also in the study of Perfect Fields and Inseparability) states that a polynomial has a repeated root $\alpha$ if and only if $f(\alpha) = f'(\alpha) = 0$, i.e. iff $\gcd(f,f') \neq 1$. Since $\gcd(f,f')=1$ here, $f$ has $q$ distinct roots in $\overline{\mathbb{F}_p}$ (counting multiplicity, $\deg f = q$, and no root repeats). Let

be the set of these $q$ distinct roots. We show $K$ is a subfield of $\overline{\mathbb{F}_p}$, hence $K$ is a field of order $q$.

• $0, 1 \in K$: clearly $0^q = 0$ and $1^q = 1$.
• Closed under multiplication: if $\alpha, \beta \in K$ then $(\alpha\beta)^q = \alpha^q \beta^q = \alpha \beta$ (using commutativity), so $\alpha\beta \in K$.
• Closed under multiplicative inverses: if $\alpha \in K$, $\alpha \neq 0$, then from $\alpha^q = \alpha$ we get $\alpha^{q-1} = 1$ (cancel one factor of $\alpha$, valid since $\alpha \neq 0$ in a field), so $\alpha^{-1} = \alpha^{q-2}$, and $(\alpha^{-1})^q = (\alpha^q)^{-1} = \alpha^{-1}$; thus $\alpha^{-1} \in K$.
• Closed under addition: this is the key step and uses characteristic $p$. The Frobenius map $\phi: \overline{\mathbb{F}_p} \to \overline{\mathbb{F}_p}$, $\phi(x) = x^p$, satisfies, for any $\alpha, \beta$ in a commutative ring of characteristic $p$,

because $\binom{p}{k} \equiv 0 \pmod p$ for $0 < k < p$ (the prime $p$ divides $\binom{p}{k} = \frac{p!}{k!(p-k)!}$ since $p$ appears in the numerator but, as $0<k<p$, not in the denominator). So $\phi$ is a ring homomorphism, and hence so is its $n$-fold iterate $\phi^n(x) = x^{p^n} = x^q$. Thus for $\alpha, \beta \in K$: $(\alpha+\beta)^q = \phi^n(\alpha+\beta) = \phi^n(\alpha) + \phi^n(\beta) = \alpha^q + \beta^q = \alpha + \beta$, so $\alpha + \beta \in K$.

• Closed under additive inverses: if $p = 2$ then $-\alpha = \alpha \in K$ automatically. If $p$ is odd, $(-\alpha)^q = (-1)^q \alpha^q = -\alpha^q = -\alpha$ (since $q$ is odd, $(-1)^q = -1$), so $-\alpha \in K$.

Thus $K$ is a subfield of $\overline{\mathbb{F}_p}$ with exactly $q = p^n$ elements (the $q$ distinct roots of $f$). This proves existence: $K$ is a field of order $q$.

Part 4 (Root description), proved alongside uniqueness. We first record: in any field $F$ of order $q = p^n$, every element satisfies $\alpha^q = \alpha$. Indeed $F$ has characteristic $p$ by Part 1 (it is a finite field), so $F$ contains a copy of $\mathbb{F}_p$. The nonzero elements $F^\times$ form a group of order $q - 1$ under multiplication (this is part of the field axioms: nonzero elements are invertible and closed under multiplication, associatively, with identity $1$ — a group of order $|F|-1$). By Lagrange's theorem, every $\alpha \in F^\times$ satisfies $\alpha^{q-1} = 1$, hence $\alpha^q = \alpha$; and $0^q = 0$ trivially. So every element of $F$ is a root of $x^q - x$. Since $\deg(x^q-x) = q = |F|$, and $F$ has $q$ elements each a root, the polynomial $x^q - x$ splits completely in $F$ with $F$ contributing all $q$ of its roots (no repeats, matching the count from Part 2). This confirms Part 4: in an algebraic closure, the elements of $\mathbb{F}_q$ are precisely $\{\alpha : \alpha^q = \alpha\}$, matching the $K$ constructed above.

Part 3 (Uniqueness). Let $F$ be any field of order $q = p^n$. By Part 1, $F$ has characteristic $p$, so $F$ contains (a copy of) the prime field $\mathbb{F}_p$. By the computation just above, every element of $F$ is a root of $x^q - x \in \mathbb{F}_p[x]$, and $F$ has exactly $q = \deg(x^q-x)$ elements. Since $x^q - x$ has at most $q$ roots in any field (a nonzero polynomial of degree $q$ has at most $q$ roots in a field, by unique factorization in $F[x]$), and $F$ already supplies $q$ distinct roots, $x^q - x$ splits completely in $F$ as

and $F$ is generated over $\mathbb{F}_p$ by these roots (indeed $F$ is the set of roots). Hence $F$ is a splitting field of $x^q - x$ over $\mathbb{F}_p$. Splitting fields of a given polynomial over a given field are unique up to isomorphism fixing the base field — this is the standard splitting field uniqueness theorem, proved by extending an isomorphism of base fields step by step through the tower of simple extensions adjoining each root, using that irreducible factors correspond under any field isomorphism extending on coefficients. Since both $F$ and the field $K$ constructed in Part 2 are splitting fields of the same polynomial $x^q-x$ over the same field $\mathbb{F}_p$, we conclude

As $F$ was an arbitrary field of order $q$, all fields of order $q$ are isomorphic to $K$, hence to each other. This proves uniqueness, and justifies the notation $\mathbb{F}_q$ for "the" field of order $q$.

Part 5 (Cyclicity of $\mathbb{F}_q^\times$). Let $G = \mathbb{F}_q^\times$, a finite abelian group of order $m := q - 1$. We use the exponent of a finite abelian group: the exponent $e$ of $G$ is the least common multiple of the orders of all its elements, equivalently the smallest positive integer with $g^e = 1$ for all $g \in G$. We show $e = m$, which forces $G$ to be cyclic, via the following two facts. Fact A: $e \mid m$. By Lagrange's theorem, the order of every element of $G$ divides $|G| = m$; hence the lcm of all element orders, $e$, also divides $m$. (If every order divides $m$, then $\mathrm{lcm}$ of orders divides $m$.) Fact B: $x^e - 1$ has at most $e$ roots in the field $\mathbb{F}_q$, and every element of $G$ satisfies $x^e = 1$ (by definition of exponent as universal exponent), so $|G| \leq e$, i.e. $m \leq e$. This uses that $\mathbb{F}_q$ is a field: a nonzero polynomial of degree $e$ has at most $e$ roots in a field (unique factorization / division algorithm in $F[x]$). Since all $m$ elements of $G$ are roots of $x^e - 1 \in \mathbb{F}_q[x]$, we get $m \leq e$. Combining Fact A ($e \mid m$, so $e \leq m$) and Fact B ($m \leq e$), we conclude $e = m$. Now we invoke the structure theorem for finite abelian groups: a finite abelian group $G$ of order $m$ is cyclic if and only if its exponent equals $m$. (Sketch, for completeness: write $G \cong \mathbb{Z}/d_1 \times \cdots \times \mathbb{Z}/d_k$ with $d_1 \mid d_2 \mid \cdots \mid d_k$ via the classification of finite abelian groups; the exponent of this product is $d_k$, the last invariant factor, and $|G| = d_1 \cdots d_k$; exponent $= |G|$ forces $k=1$, i.e. $G \cong \mathbb{Z}/m\mathbb{Z}$.) Applying this with $e = m$ shows $G = \mathbb{F}_q^\times$ is cyclic of order $q-1$. $\blacksquare$