[proofplan] We verify the two conditions for a Galois extension separately: separability follows from the general fact that every algebraic extension of a finite field is separable, and normality follows from the characterisation of $\mathbb{F}_{q^n}$ as the splitting field of $t^{q^n} - t$ over $\mathbb{F}_q$. Once we know the extension is Galois, the Galois group has order $[\mathbb{F}_{q^n} : \mathbb{F}_q] = n$. We then show the Frobenius automorphism $\operatorname{Fr}_q: \alpha \mapsto \alpha^q$ has exact order $n$ in this group: the iterate $\operatorname{Fr}_q^k$ acts by $\alpha \mapsto \alpha^{q^k}$, so $\operatorname{Fr}_q^n = \operatorname{id}$ because every element of $\mathbb{F}_{q^n}$ is a root of $t^{q^n} - t$, while $\operatorname{Fr}_q^k \ne \operatorname{id}$ for $1 \le k < n$ because the polynomial $t^{q^k} - t$ has only $q^k < q^n$ roots. Since $\operatorname{Fr}_q$ has order $n$ in a group of order $n$, it generates the entire group, giving $\operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q) = \langle \operatorname{Fr}_q \rangle \cong \mathbb{Z}/n\mathbb{Z}$. [/proofplan] [step:Verify that $\mathbb{F}_{q^n}/\mathbb{F}_q$ is separable] Write $q = p^d$ for a prime $p$ and integer $d \ge 1$. By [Finite Field Extensions Are Separable](/theorems/1268), every algebraic extension of a finite field is separable. Since $\mathbb{F}_{q^n}/\mathbb{F}_q$ is a finite (hence algebraic) extension, it is separable. [guided] Separability is the first of two conditions needed for a Galois extension. Recall that an algebraic extension $L/K$ is separable if for every $\alpha \in L$, the minimal polynomial $\operatorname{min}_K(\alpha) \in K[t]$ has no repeated roots. By [Finite Field Extensions Are Separable](/theorems/1268), every algebraic extension of a finite field is separable. The key idea behind that result is: if an irreducible polynomial $f \in \mathbb{F}_q[t]$ were inseparable, then $f(t) = g(t^p)$ for some $g \in \mathbb{F}_q[t]$. Writing $g(t) = \sum_k a_k t^k$, each coefficient $a_k \in \mathbb{F}_q$ has a $p$-th root in $\mathbb{F}_q$ (since the Frobenius endomorphism $x \mapsto x^p$ is injective on $\mathbb{F}_q$, and an injective map on a finite set is surjective). Writing $a_k = b_k^p$, we get $f(t) = \sum_k b_k^p t^{pk} = (\sum_k b_k t^k)^p$, contradicting the irreducibility of $f$. Hence every irreducible polynomial over $\mathbb{F}_q$ is separable. Since $[\mathbb{F}_{q^n} : \mathbb{F}_q] = n < \infty$, the extension $\mathbb{F}_{q^n}/\mathbb{F}_q$ is algebraic, so the theorem applies and $\mathbb{F}_{q^n}/\mathbb{F}_q$ is separable. [/guided] [/step] [step:Verify that $\mathbb{F}_{q^n}/\mathbb{F}_q$ is normal by identifying $\mathbb{F}_{q^n}$ as a splitting field] By the [Existence and Uniqueness of Finite Fields](/theorems/1275), $\mathbb{F}_{q^n}$ is the splitting field of the polynomial $t^{q^n} - t \in \mathbb{F}_q[t]$ over $\mathbb{F}_q$. (The elements of $\mathbb{F}_{q^n}$ are precisely the $q^n$ distinct roots of $t^{q^n} - t$: every $\alpha \in \mathbb{F}_{q^n}$ satisfies $\alpha^{q^n} = \alpha$ by Lagrange's theorem applied to $\mathbb{F}_{q^n}^\times$, and the derivative of $t^{q^n} - t$ is $-1$ in characteristic $p$, so all $q^n$ roots are distinct.) Since $\mathbb{F}_{q^n}$ is the splitting field of a polynomial over $\mathbb{F}_q$, the extension $\mathbb{F}_{q^n}/\mathbb{F}_q$ is normal. [guided] Normality is the second condition for a Galois extension. An algebraic extension $L/K$ is normal if every irreducible polynomial in $K[t]$ that has at least one root in $L$ splits completely over $L$. Equivalently, $L/K$ is normal if and only if $L$ is a splitting field of some family of polynomials over $K$. We use the characterisation of $\mathbb{F}_{q^n}$ from the [Existence and Uniqueness of Finite Fields](/theorems/1275). That theorem establishes that $\mathbb{F}_{q^n}$ consists of precisely the roots of $f(t) := t^{q^n} - t \in \mathbb{F}_p[t]$, viewed inside a splitting field of $f$ over $\mathbb{F}_p$. We verify this claim directly. Every element $\alpha \in \mathbb{F}_{q^n}^\times$ has multiplicative order dividing $|\mathbb{F}_{q^n}^\times| = q^n - 1$ (by Lagrange's theorem applied to the finite group $\mathbb{F}_{q^n}^\times$), so $\alpha^{q^n - 1} = 1$, hence $\alpha^{q^n} = \alpha$. For $\alpha = 0$, we have $0^{q^n} = 0$. Therefore every element of $\mathbb{F}_{q^n}$ is a root of $t^{q^n} - t$. The polynomial $t^{q^n} - t$ has degree $q^n$ and $|\mathbb{F}_{q^n}| = q^n$, so $\mathbb{F}_{q^n}$ accounts for all roots. Moreover, the formal derivative of $t^{q^n} - t$ is $q^n t^{q^n - 1} - 1 = -1$ (since $q^n = p^{dn}$ and $\operatorname{char}(\mathbb{F}_q) = p$), which is a nonzero constant, so $\gcd(t^{q^n} - t, \, (t^{q^n} - t)') = 1$, confirming that all $q^n$ roots are distinct. Now, $t^{q^n} - t \in \mathbb{F}_p[t] \subset \mathbb{F}_q[t]$ (since $\mathbb{F}_p \subset \mathbb{F}_q$), and $\mathbb{F}_{q^n}$ is generated over $\mathbb{F}_q$ by the roots of this polynomial, with all roots lying in $\mathbb{F}_{q^n}$. Therefore $\mathbb{F}_{q^n}$ is the splitting field of $t^{q^n} - t$ over $\mathbb{F}_q$. By the equivalence between splitting fields and normal extensions, $\mathbb{F}_{q^n}/\mathbb{F}_q$ is normal. [/guided] [/step] [step:Conclude that $\mathbb{F}_{q^n}/\mathbb{F}_q$ is Galois of degree $n$] Since $\mathbb{F}_{q^n}/\mathbb{F}_q$ is both separable and normal, it is a Galois extension. By the definition of a Galois extension, the Galois group satisfies \begin{align*} |\operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q)| = [\mathbb{F}_{q^n} : \mathbb{F}_q] = n. \end{align*} [/step] [step:Define the Frobenius automorphism and verify it lies in $\operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q)$] Define the Frobenius endomorphism \begin{align*} \operatorname{Fr}_q: \mathbb{F}_{q^n} &\to \mathbb{F}_{q^n} \\ \alpha &\mapsto \alpha^q. \end{align*} We verify that $\operatorname{Fr}_q$ is a field automorphism of $\mathbb{F}_{q^n}$ fixing $\mathbb{F}_q$ pointwise. **Ring homomorphism.** For all $\alpha, \beta \in \mathbb{F}_{q^n}$: the identity $(\alpha + \beta)^q = \alpha^q + \beta^q$ holds because $\operatorname{char}(\mathbb{F}_{q^n}) = p$ and $q = p^d$, so the binomial coefficients $\binom{q}{k}$ for $1 \le k \le q - 1$ are divisible by $p$ (since $q$ is a power of $p$) and hence vanish in $\mathbb{F}_{q^n}$. Furthermore, $(\alpha \beta)^q = \alpha^q \beta^q$ and $1^q = 1$ hold in any commutative ring. Therefore $\operatorname{Fr}_q$ is a ring homomorphism. **Injectivity.** A ring homomorphism from a field to itself is injective (its kernel is an ideal of a field, hence $\{0\}$ or the whole field; since $\operatorname{Fr}_q(1) = 1 \ne 0$, the kernel is $\{0\}$). **Surjectivity.** An injective map from a finite set to itself is surjective. **Fixes $\mathbb{F}_q$.** For $\alpha \in \mathbb{F}_q$, we have $\alpha^q = \alpha$ (as shown in the previous step: every element of $\mathbb{F}_q$ is a root of $t^q - t$). Therefore $\operatorname{Fr}_q(\alpha) = \alpha$ for all $\alpha \in \mathbb{F}_q$. Hence $\operatorname{Fr}_q \in \operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q)$. [guided] The Frobenius automorphism is the map "raise to the $q$-th power." The fact that this is a field homomorphism relies crucially on the characteristic-$p$ identity $(\alpha + \beta)^p = \alpha^p + \beta^p$, which fails in characteristic zero. We verify each property explicitly. **Additivity.** We show $(\alpha + \beta)^q = \alpha^q + \beta^q$ for all $\alpha, \beta \in \mathbb{F}_{q^n}$. Since $q = p^d$, it suffices to show $(\alpha + \beta)^p = \alpha^p + \beta^p$ (and then iterate $d$ times). By the binomial theorem, \begin{align*} (\alpha + \beta)^p = \sum_{k=0}^{p} \binom{p}{k} \alpha^k \beta^{p-k}. \end{align*} For $1 \le k \le p - 1$, the binomial coefficient $\binom{p}{k} = \frac{p!}{k!(p-k)!}$ is divisible by $p$ (since $p$ is prime and the denominator $k!(p-k)!$ is a product of integers smaller than $p$, so $p$ does not cancel). Since $\operatorname{char}(\mathbb{F}_{q^n}) = p$, these terms vanish, leaving $(\alpha + \beta)^p = \alpha^p + \beta^p$. Iterating $d$ times gives $(\alpha + \beta)^{p^d} = \alpha^{p^d} + \beta^{p^d}$, i.e., $(\alpha + \beta)^q = \alpha^q + \beta^q$. **Multiplicativity** is immediate: $(\alpha\beta)^q = \alpha^q \beta^q$ holds in any commutative ring. **Preserves identity:** $1^q = 1$. These three properties establish that $\operatorname{Fr}_q$ is a ring homomorphism $\mathbb{F}_{q^n} \to \mathbb{F}_{q^n}$. **Injectivity.** The kernel $\ker(\operatorname{Fr}_q) = \{\alpha \in \mathbb{F}_{q^n} : \alpha^q = 0\}$. If $\alpha^q = 0$, then $\alpha = 0$ (since $\mathbb{F}_{q^n}$ is a field and has no nonzero nilpotent elements). So $\ker(\operatorname{Fr}_q) = \{0\}$, and $\operatorname{Fr}_q$ is injective. **Surjectivity** follows from the fact that an injective function from a finite set to itself is a bijection: $|\mathbb{F}_{q^n}| = q^n < \infty$, so $\operatorname{Fr}_q$ is surjective. **Fixes $\mathbb{F}_q$ pointwise.** An element $\alpha \in \mathbb{F}_q$ satisfies $\alpha^q = \alpha$ (since $\mathbb{F}_q$ consists of the roots of $t^q - t$, as established in the previous step). Therefore $\operatorname{Fr}_q(\alpha) = \alpha^q = \alpha$ for all $\alpha \in \mathbb{F}_q$, confirming that $\operatorname{Fr}_q$ is an $\mathbb{F}_q$-automorphism of $\mathbb{F}_{q^n}$. Why is the Frobenius the natural automorphism to consider? In characteristic zero, "raise to the $q$-th power" is not additive, so it cannot be a field homomorphism. The Frobenius exists precisely because of the "freshman's dream" identity $(\alpha + \beta)^p = \alpha^p + \beta^p$ in characteristic $p$, which makes exponentiation by $p$ (and hence by $q = p^d$) a ring endomorphism. [/guided] [/step] [step:Show that $\operatorname{Fr}_q$ has exact order $n$ in $\operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q)$] For any positive integer $k$, the $k$-fold iterate of $\operatorname{Fr}_q$ acts by \begin{align*} \operatorname{Fr}_q^k(\alpha) = \alpha^{q^k} \quad \text{for all } \alpha \in \mathbb{F}_{q^n}. \end{align*} This follows by induction: $\operatorname{Fr}_q^1(\alpha) = \alpha^q$, and if $\operatorname{Fr}_q^k(\alpha) = \alpha^{q^k}$, then $\operatorname{Fr}_q^{k+1}(\alpha) = (\alpha^{q^k})^q = \alpha^{q^{k+1}}$. The condition $\operatorname{Fr}_q^k = \operatorname{id}_{\mathbb{F}_{q^n}}$ means $\alpha^{q^k} = \alpha$ for all $\alpha \in \mathbb{F}_{q^n}$, i.e., every element of $\mathbb{F}_{q^n}$ is a root of $t^{q^k} - t$. **$\operatorname{Fr}_q^n = \operatorname{id}$:** Every $\alpha \in \mathbb{F}_{q^n}$ satisfies $\alpha^{q^n} = \alpha$ (as every element of $\mathbb{F}_{q^n}$ is a root of $t^{q^n} - t$, shown in the normality step). Hence $\operatorname{Fr}_q^n = \operatorname{id}_{\mathbb{F}_{q^n}}$. **$\operatorname{Fr}_q^k \ne \operatorname{id}$ for $1 \le k < n$:** If $\operatorname{Fr}_q^k = \operatorname{id}_{\mathbb{F}_{q^n}}$, then every element of $\mathbb{F}_{q^n}$ is a root of $t^{q^k} - t \in \mathbb{F}_q[t]$. This polynomial has degree $q^k$, so it has at most $q^k$ roots in $\mathbb{F}_{q^n}$. Since $|\mathbb{F}_{q^n}| = q^n$ and $k < n$ gives $q^k < q^n$, the polynomial $t^{q^k} - t$ cannot have $q^n$ roots. This is a contradiction, so $\operatorname{Fr}_q^k \ne \operatorname{id}_{\mathbb{F}_{q^n}}$. Therefore $\operatorname{Fr}_q$ has exact order $n$ in $\operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q)$. [guided] The order of an automorphism $\sigma$ in the Galois group is the smallest positive integer $m$ such that $\sigma^m = \operatorname{id}$. We need to compute the order of $\operatorname{Fr}_q$ in $\operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q)$. First, we establish a formula for iterates. The $k$-th iterate $\operatorname{Fr}_q^k$ acts as $\alpha \mapsto \alpha^{q^k}$. The base case $k = 1$ is the definition. For the inductive step, assuming $\operatorname{Fr}_q^k(\alpha) = \alpha^{q^k}$, we compute \begin{align*} \operatorname{Fr}_q^{k+1}(\alpha) = \operatorname{Fr}_q(\operatorname{Fr}_q^k(\alpha)) = \operatorname{Fr}_q(\alpha^{q^k}) = (\alpha^{q^k})^q = \alpha^{q^{k+1}}. \end{align*} The condition $\operatorname{Fr}_q^k = \operatorname{id}_{\mathbb{F}_{q^n}}$ is therefore equivalent to: $\alpha^{q^k} = \alpha$ for every $\alpha \in \mathbb{F}_{q^n}$, which in turn means that every element of $\mathbb{F}_{q^n}$ is a root of the polynomial $t^{q^k} - t \in \mathbb{F}_q[t]$. **Upper bound: $\operatorname{Fr}_q^n = \operatorname{id}$.** From the normality step, every $\alpha \in \mathbb{F}_{q^n}$ satisfies $\alpha^{q^n} = \alpha$. Thus $\operatorname{Fr}_q^n(\alpha) = \alpha^{q^n} = \alpha$ for all $\alpha$, giving $\operatorname{Fr}_q^n = \operatorname{id}_{\mathbb{F}_{q^n}}$. The order of $\operatorname{Fr}_q$ divides $n$. **Lower bound: $\operatorname{Fr}_q^k \ne \operatorname{id}$ for $1 \le k \le n - 1$.** Suppose for contradiction that $\operatorname{Fr}_q^k = \operatorname{id}_{\mathbb{F}_{q^n}}$ for some $k$ with $1 \le k \le n - 1$. Then every element of $\mathbb{F}_{q^n}$ is a root of $t^{q^k} - t$. Since $t^{q^k} - t$ has degree $q^k$ and is a polynomial over a field, it has at most $q^k$ roots in $\mathbb{F}_{q^n}$. But $|\mathbb{F}_{q^n}| = q^n$, and since $k < n$ and $q \ge 2$, we have $q^k < q^n$. This is a contradiction: $q^n$ elements cannot all be roots of a polynomial of degree $q^k < q^n$. Combining: the order of $\operatorname{Fr}_q$ divides $n$ (since $\operatorname{Fr}_q^n = \operatorname{id}$), and the order is not any positive integer less than $n$ (since $\operatorname{Fr}_q^k \ne \operatorname{id}$ for $1 \le k \le n-1$). Therefore $\operatorname{ord}(\operatorname{Fr}_q) = n$. Note that this argument uses nothing about divisors of $n$ — we prove directly that no power strictly less than $n$ gives the identity, which is stronger than just checking divisors. [/guided] [/step] [step:Conclude that $\operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q) = \langle \operatorname{Fr}_q \rangle \cong \mathbb{Z}/n\mathbb{Z}$] The Galois group $\operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q)$ has order $n$, and $\operatorname{Fr}_q$ is an element of order $n$ in this group. Therefore the cyclic subgroup $\langle \operatorname{Fr}_q \rangle$ has order $n$ and is contained in $\operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q)$. Since both sets are finite with the same cardinality, \begin{align*} \operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q) = \langle \operatorname{Fr}_q \rangle = \{\operatorname{id}, \operatorname{Fr}_q, \operatorname{Fr}_q^2, \ldots, \operatorname{Fr}_q^{n-1}\}. \end{align*} Since $\langle \operatorname{Fr}_q \rangle$ is a cyclic group of order $n$, the map \begin{align*} \mathbb{Z}/n\mathbb{Z} &\to \operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q) \\ \bar{k} &\mapsto \operatorname{Fr}_q^k \end{align*} is a well-defined group isomorphism (it is a surjective homomorphism between groups of the same finite order). Therefore $\operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q) \cong \mathbb{Z}/n\mathbb{Z}$. [/step]