[proofplan] We prove each of the four parts in turn. Part 1 constructs $\mathbb{F}_{p^n}$ as the splitting field of $f(x) = x^{p^n} - x \in \mathbb{F}_p[x]$ and identifies it with the root set of $f$, using Frobenius to show $f$ is separable. Uniqueness follows from the [Existence and Uniqueness of Splitting Fields](/theorems/1258). Part 2 shows every finite field $F$ of characteristic $p$ has order $p^n$ for some $n$ and its elements are exactly the roots of $x^{p^n} - x$. Part 3 uses the separability of $\mathbb{F}_{p^n}/\mathbb{F}_p$ and the [Equivalent Characterisations of Galois Extensions](/theorems/3310) to establish the Galois property, then identifies the Galois group as cyclic generated by Frobenius. Part 4 relates subfields to divisors of $n$ via the [Tower Law](/theorems/1248) and unique root sets. [/proofplan]

[step:Construct $\mathbb{F}_{p^n}$ as the root set of $x^{p^n} - x$ and prove existence and uniqueness]Set $q = p^n$ and define the polynomial \begin{align*} f: \mathbb{F}_p[x] \to \mathbb{F}_p[x], \quad f(x) = x^q - x. \end{align*} By the [Existence and Uniqueness of Splitting Fields](/theorems/1258), a splitting field $K$ of $f$ over $\mathbb{F}_p$ exists and is unique up to $\mathbb{F}_p$-isomorphism. Let $R = \{\alpha \in K : \alpha^q = \alpha\}$ be the set of roots of $f$ in $K$. We show $|R| = q$ and $R$ is a field. **$f$ is separable.** The formal derivative of $f$ is $f'(x) = qx^{q-1} - 1 = -1$ (since $q = p^n \equiv 0$ in $\mathbb{F}_p$). Since $\gcd(f, f') = \gcd(f, -1) = 1$, the polynomial $f$ has no repeated roots. Therefore $|R| = \deg f = q$. **$R$ is a subfield of $K$.** We verify the subfield criteria. The element $0$ satisfies $0^q = 0$, so $0 \in R$. The element $1$ satisfies $1^q = 1$, so $1 \in R$. For $\alpha, \beta \in R$: by the Frobenius endomorphism, $(\alpha - \beta)^q = \alpha^q - \beta^q = \alpha - \beta$ (using the binomial expansion in characteristic $p$, where $\binom{q}{k} \equiv 0 \pmod{p}$ for $0 < k < q$), so $\alpha - \beta \in R$. For $\beta \neq 0$: $(\alpha \beta^{-1})^q = \alpha^q (\beta^q)^{-1} = \alpha \beta^{-1}$, so $\alpha \beta^{-1} \in R$. Hence $R$ is a subfield of $K$. Since $R$ is a subfield of $K$ containing all roots of $f$, and $K$ is generated over $\mathbb{F}_p$ by the roots of $f$ (as $K$ is the splitting field), we have $K = R$. Therefore $K$ is a field of order $q = p^n$. Uniqueness: any two fields of order $p^n$ are splitting fields of $f(x) = x^{p^n} - x$ over $\mathbb{F}_p$ (as we show in Part 2), so they are isomorphic by the uniqueness clause of the [Existence and Uniqueness of Splitting Fields](/theorems/1258).[/step]

[guided]We construct $\mathbb{F}_{p^n}$ explicitly. Set $q = p^n$ and consider $f(x) = x^q - x \in \mathbb{F}_p[x]$. **Why this polynomial?** The idea is that a field of order $q$ must have multiplicative group of order $q - 1$, so every nonzero element satisfies $\alpha^{q-1} = 1$, hence every element (including $0$) satisfies $\alpha^q = \alpha$. This means the elements of a field of order $q$ are precisely the roots of $x^q - x$. By the [Existence and Uniqueness of Splitting Fields](/theorems/1258), a splitting field $K$ of $f$ over $\mathbb{F}_p$ exists and is unique up to $\mathbb{F}_p$-isomorphism. The existence clause requires $f$ to have degree $\geq 1$, which holds since $\deg f = q \geq p \geq 2$. Define $R = \{\alpha \in K : \alpha^q = \alpha\}$. **Separability of $f$.** We compute the formal derivative: $f'(x) = qx^{q-1} - 1$. Since we are in characteristic $p$ and $q = p^n$, we have $q \equiv 0 \pmod{p}$, so $f'(x) = -1$. A polynomial and its formal derivative share a common root if and only if they share a common nonconstant factor. Since $f'(x) = -1$ is a nonzero constant, $\gcd(f, f') = 1$, so $f$ has no repeated roots. Therefore $f$ has exactly $\deg f = q$ distinct roots in $K$, and $|R| = q$. **$R$ is a subfield of $K$.** We verify the subfield axioms. Containment of $0$ and $1$ is immediate: $0^q = 0$ and $1^q = 1$. For closure under subtraction: let $\alpha, \beta \in R$. We use the Frobenius endomorphism in characteristic $p$. For any elements $a, b$ in a field of characteristic $p$, the identity $(a + b)^p = a^p + b^p$ holds because all binomial coefficients $\binom{p}{k}$ for $0 < k < p$ are divisible by $p$. Iterating $n$ times, $(a + b)^{p^n} = a^{p^n} + b^{p^n}$, i.e., $(a + b)^q = a^q + b^q$. (The same applies with $-b$ in place of $b$.) Therefore $(\alpha - \beta)^q = \alpha^q - \beta^q = \alpha - \beta$, so $\alpha - \beta \in R$. For closure under division: if $\beta \neq 0$, then $(\alpha \beta^{-1})^q = \alpha^q (\beta^{-1})^q = \alpha^q (\beta^q)^{-1} = \alpha \beta^{-1}$, so $\alpha \beta^{-1} \in R$. Hence $R \leq K$ is a subfield of cardinality $q$. **$K = R$.** The splitting field $K$ is generated over $\mathbb{F}_p$ by the roots of $f$, and all roots of $f$ lie in $R$. Hence $K \subseteq R$. But $R \subseteq K$ by definition. So $K = R$ is a field of order $q = p^n$. **Uniqueness.** Suppose $F$ is any field of order $q$. Then every element of $F$ satisfies $\alpha^q = \alpha$ (Part 2 below), so $F$ is a splitting field of $f(x) = x^q - x$ over its prime subfield $\mathbb{F}_p$. By the uniqueness clause of the [Existence and Uniqueness of Splitting Fields](/theorems/1258), $F \cong K = \mathbb{F}_{p^n}$.[/guided]

[step:Show every finite field has prime-power order and is isomorphic to some $\mathbb{F}_{p^n}$]Let $F$ be a finite field. Since $F$ has positive characteristic, $\operatorname{char}(F) = p$ for some prime $p$ (by [Characteristic of an Integral Domain Is Zero or Prime](/theorems/2578), the characteristic of the integral domain $F$ is $0$ or prime; it cannot be $0$ since $F$ is finite). The prime subfield of $F$ is $\mathbb{F}_p$. Since $F$ is a finite-dimensional vector space over $\mathbb{F}_p$, set $n = [F : \mathbb{F}_p]$. Then $|F| = p^n$. The multiplicative group $F^\times = F \setminus \{0\}$ has order $p^n - 1$. By [Lagrange's Theorem](/theorems/782) applied to the group $F^\times$, every $\alpha \in F^\times$ satisfies $\alpha^{p^n - 1} = 1$, hence $\alpha^{p^n} = \alpha$. This also holds for $\alpha = 0$. Therefore every element of $F$ is a root of $x^{p^n} - x$. Since $|F| = p^n$ and $x^{p^n} - x$ has at most $p^n$ roots, the elements of $F$ are precisely the roots of $x^{p^n} - x$. Hence $F$ is a splitting field of $x^{p^n} - x$ over $\mathbb{F}_p$, and by the uniqueness of splitting fields ([Existence and Uniqueness of Splitting Fields](/theorems/1258)), $F \cong \mathbb{F}_{p^n}$.[/step]

[guided]Let $F$ be a finite field. We must show $|F| = p^n$ for some prime $p$ and positive integer $n$, and that $F \cong \mathbb{F}_{p^n}$. **The characteristic is a prime $p$.** The characteristic $\operatorname{char}(F)$ is well-defined: it is the smallest positive integer $m$ such that $\underbrace{1 + \cdots + 1}_{m} = 0$ in $F$, or $0$ if no such $m$ exists. Since $F$ is finite, $\operatorname{char}(F) \neq 0$ (otherwise the elements $1, 1+1, 1+1+1, \ldots$ would be infinitely many distinct elements). By [Characteristic of an Integral Domain Is Zero or Prime](/theorems/2578) (a field is an integral domain), $\operatorname{char}(F)$ is prime. Call it $p$. **The order is a prime power.** The prime subfield (the subfield generated by $1$) is isomorphic to $\mathbb{Z}/p\mathbb{Z} = \mathbb{F}_p$. The field $F$ is a vector space over $\mathbb{F}_p$, and since $F$ is finite, this vector space is finite-dimensional. Let $n = \dim_{\mathbb{F}_p} F = [F : \mathbb{F}_p]$. Choosing a basis $\{e_1, \ldots, e_n\}$, every element of $F$ is uniquely written as $\sum_{i=1}^n a_i e_i$ with $a_i \in \mathbb{F}_p$. There are $p$ choices for each $a_i$, giving $|F| = p^n$. **$F$ is a splitting field of $x^{p^n} - x$.** Set $q = p^n = |F|$. The multiplicative group $F^\times$ has order $q - 1$. By [Lagrange's Theorem](/theorems/782), every $\alpha \in F^\times$ satisfies $\alpha^{q-1} = 1$, so $\alpha^q = \alpha$. For $\alpha = 0$, $0^q = 0$. Hence every element of $F$ is a root of $g(x) = x^q - x$. Since $\deg g = q$ and $g$ has at most $q$ roots in any field, and $F$ provides exactly $q$ roots, the polynomial $g$ splits completely in $F$ and $F$ is generated over $\mathbb{F}_p$ by the roots. Therefore $F$ is a splitting field of $x^q - x$ over $\mathbb{F}_p$. By the uniqueness of splitting fields ([Existence and Uniqueness of Splitting Fields](/theorems/1258)), $F \cong \mathbb{F}_{p^n}$.[/guided]

[step:Prove $\mathbb{F}_{p^n}/\mathbb{F}_p$ is Galois with cyclic Galois group generated by Frobenius]We verify that $\mathbb{F}_{p^n}/\mathbb{F}_p$ satisfies the conditions of the [Equivalent Characterisations of Galois Extensions](/theorems/3310). **Separability.** By [Finite Field Extensions Are Separable](/theorems/1268), every algebraic extension of a finite field is separable. The extension $\mathbb{F}_{p^n}/\mathbb{F}_p$ is finite (hence algebraic), so it is separable. **Normality.** The field $\mathbb{F}_{p^n}$ is the splitting field of $x^{p^n} - x$ over $\mathbb{F}_p$ (as shown in Part 1). By [Normal Equals Splitting Field](/theorems/1269), a finite extension is normal if and only if it is a splitting field of some polynomial. Hence $\mathbb{F}_{p^n}/\mathbb{F}_p$ is normal. Since $\mathbb{F}_{p^n}/\mathbb{F}_p$ is both normal and separable, it is Galois by the [Equivalent Characterisations of Galois Extensions](/theorems/3310), and $|\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)| = [\mathbb{F}_{p^n} : \mathbb{F}_p] = n$. **The Frobenius automorphism generates the Galois group.** Define \begin{align*} \varphi_p: \mathbb{F}_{p^n} &\to \mathbb{F}_{p^n}, \\ x &\mapsto x^p. \end{align*} We verify $\varphi_p \in \operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)$. The map $\varphi_p$ is a ring homomorphism: $\varphi_p(xy) = (xy)^p = x^p y^p = \varphi_p(x)\varphi_p(y)$ and $\varphi_p(x + y) = (x+y)^p = x^p + y^p = \varphi_p(x) + \varphi_p(y)$ (using the Frobenius identity in characteristic $p$). It is injective since $\ker \varphi_p = \{0\}$ (if $x^p = 0$, then $x = 0$). Since $\mathbb{F}_{p^n}$ is finite, an injective map from $\mathbb{F}_{p^n}$ to itself is surjective. Hence $\varphi_p$ is a field automorphism. For $a \in \mathbb{F}_p$, $\varphi_p(a) = a^p = a$ (since $\mathbb{F}_p^\times$ has order $p - 1$, so $a^{p-1} = 1$ for $a \neq 0$, and $0^p = 0$). Thus $\varphi_p$ fixes $\mathbb{F}_p$ and $\varphi_p \in \operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)$. **Order of $\varphi_p$.** The $k$-th iterate $\varphi_p^k$ sends $x \mapsto x^{p^k}$. We have $\varphi_p^k = \operatorname{id}$ if and only if $x^{p^k} = x$ for all $x \in \mathbb{F}_{p^n}$, i.e., every element of $\mathbb{F}_{p^n}$ is a root of $x^{p^k} - x$. Since $\mathbb{F}_{p^n}$ has $p^n$ elements and $x^{p^k} - x$ has at most $p^k$ roots, $\varphi_p^k = \operatorname{id}$ requires $p^n \leq p^k$, hence $k \geq n$. And $\varphi_p^n(x) = x^{p^n} = x$ for all $x \in \mathbb{F}_{p^n}$ (since $\mathbb{F}_{p^n}$ consists of the roots of $x^{p^n} - x$). Therefore $\operatorname{ord}(\varphi_p) = n$. Since $\varphi_p$ has order $n$ in a group of order $n$, $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) = \langle \varphi_p \rangle \cong \mathbb{Z}/n\mathbb{Z}$.[/step]

[guided]We must show $\mathbb{F}_{p^n}/\mathbb{F}_p$ is Galois and identify its Galois group. **Establishing the Galois property.** By the [Equivalent Characterisations of Galois Extensions](/theorems/3310), it suffices to show $\mathbb{F}_{p^n}/\mathbb{F}_p$ is both normal and separable. Separability: the result [Finite Field Extensions Are Separable](/theorems/1268) states that every algebraic extension of a finite field is separable. The hypothesis is that the base field is finite, which $\mathbb{F}_p$ certainly is. The extension $\mathbb{F}_{p^n}/\mathbb{F}_p$ is finite, hence algebraic, so the theorem applies and $\mathbb{F}_{p^n}/\mathbb{F}_p$ is separable. Normality: in Part 1 we showed $\mathbb{F}_{p^n}$ is the splitting field of $x^{p^n} - x$ over $\mathbb{F}_p$. By [Normal Equals Splitting Field](/theorems/1269), a finite extension is normal if and only if it is a splitting field of some polynomial. The condition is satisfied, so $\mathbb{F}_{p^n}/\mathbb{F}_p$ is normal. The [Equivalent Characterisations of Galois Extensions](/theorems/3310) now gives $|\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)| = [\mathbb{F}_{p^n} : \mathbb{F}_p] = n$. **Identifying the Galois group.** We claim $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)$ is cyclic of order $n$, generated by the Frobenius map $\varphi_p: x \mapsto x^p$. First, is $\varphi_p$ actually a field automorphism of $\mathbb{F}_{p^n}$? We need three properties: 1. **Ring homomorphism**: $\varphi_p(xy) = (xy)^p = x^p y^p$ (exponentiation distributes over multiplication). For addition, $\varphi_p(x + y) = (x + y)^p = x^p + y^p$. Why? Expanding by the binomial theorem, $(x+y)^p = \sum_{k=0}^p \binom{p}{k} x^k y^{p-k}$. For $0 < k < p$, $\binom{p}{k} = \frac{p!}{k!(p-k)!}$, and since $p$ is prime and $0 < k < p$, $p$ divides the numerator but not the denominator, so $p \mid \binom{p}{k}$. In characteristic $p$, these terms vanish. 2. **Injective**: if $\varphi_p(x) = 0$, then $x^p = 0$, so $x = 0$ (a field has no zero divisors). Hence $\ker \varphi_p = \{0\}$. 3. **Surjective**: an injective map from a finite set to itself is surjective. Hence $\varphi_p$ is a field automorphism. Does it fix $\mathbb{F}_p$? For $a \in \mathbb{F}_p^\times$, $a^{p-1} = 1$ by [Lagrange's Theorem](/theorems/782) applied to $\mathbb{F}_p^\times$ (which has order $p - 1$), so $a^p = a$. And $0^p = 0$. Thus $\varphi_p$ fixes $\mathbb{F}_p$ pointwise, confirming $\varphi_p \in \operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)$. **Computing the order of $\varphi_p$.** The $k$-th iterate is $\varphi_p^k: x \mapsto x^{p^k}$. The condition $\varphi_p^k = \operatorname{id}$ means $x^{p^k} = x$ for every $x \in \mathbb{F}_{p^n}$. The polynomial $x^{p^k} - x$ has degree $p^k$ and therefore at most $p^k$ roots in any field. But $\mathbb{F}_{p^n}$ has $p^n$ elements that would all need to be roots. So we need $p^k \geq p^n$, i.e., $k \geq n$. Conversely, when $k = n$: every element of $\mathbb{F}_{p^n}$ satisfies $x^{p^n} = x$ (since $\mathbb{F}_{p^n}$ is the root set of $x^{p^n} - x$), so $\varphi_p^n = \operatorname{id}$. Therefore $\operatorname{ord}(\varphi_p) = n = |\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)|$. A cyclic subgroup $\langle \varphi_p \rangle$ of order $n$ inside a group of order $n$ must be the whole group: \begin{align*} \operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) = \langle \varphi_p \rangle \cong \mathbb{Z}/n\mathbb{Z}. \end{align*}[/guided]

[step:Characterise subfields of $\mathbb{F}_{p^n}$ by divisors of $n$]We show: $\mathbb{F}_{p^n}$ contains a subfield isomorphic to $\mathbb{F}_{p^m}$ if and only if $m \mid n$, and in that case the subfield is unique. **($\Leftarrow$) If $m \mid n$, then $\mathbb{F}_{p^m}$ embeds in $\mathbb{F}_{p^n}$.** Suppose $m \mid n$. Then $p^m - 1 \mid p^n - 1$ (since $p^n - 1 = (p^m)^{n/m} - 1$ and $a - 1 \mid a^k - 1$ for any integer $a$ and positive integer $k$). Therefore $x^{p^m - 1} - 1 \mid x^{p^n - 1} - 1$ in $\mathbb{F}_p[x]$, and multiplying by $x$, $x^{p^m} - x \mid x^{p^n} - x$. Every root of $x^{p^m} - x$ in $\mathbb{F}_{p^n}$ is also a root of $x^{p^n} - x$, hence lies in $\mathbb{F}_{p^n}$. The polynomial $x^{p^m} - x$ is separable (its derivative is $-1$), so it has exactly $p^m$ roots in $\mathbb{F}_{p^n}$. The set $S = \{\alpha \in \mathbb{F}_{p^n} : \alpha^{p^m} = \alpha\}$ is a subfield of $\mathbb{F}_{p^n}$ (by the same Frobenius argument as in Part 1) of cardinality $p^m$, hence $S \cong \mathbb{F}_{p^m}$. **($\Rightarrow$) If $\mathbb{F}_{p^n}$ contains a subfield $F \cong \mathbb{F}_{p^m}$, then $m \mid n$.** By the [Tower Law](/theorems/1248), $[\mathbb{F}_{p^n} : \mathbb{F}_p] = [\mathbb{F}_{p^n} : F] \cdot [F : \mathbb{F}_p]$, i.e., $n = [\mathbb{F}_{p^n} : F] \cdot m$. Hence $m \mid n$. **Uniqueness.** The subfield isomorphic to $\mathbb{F}_{p^m}$ is the root set $\{\alpha \in \mathbb{F}_{p^n} : \alpha^{p^m} = \alpha\}$. Since this is determined by the equation $x^{p^m} - x = 0$, there is exactly one such subfield.[/step]

[guided]We must show that the subfield lattice of $\mathbb{F}_{p^n}$ corresponds exactly to the divisors of $n$. **($\Leftarrow$) Constructing the subfield when $m \mid n$.** Suppose $m \mid n$, say $n = mq$. We claim the root set $S = \{\alpha \in \mathbb{F}_{p^n} : \alpha^{p^m} = \alpha\}$ is a subfield of $\mathbb{F}_{p^n}$ isomorphic to $\mathbb{F}_{p^m}$. Why do all $p^m$ roots of $x^{p^m} - x$ lie in $\mathbb{F}_{p^n}$? Because $x^{p^m} - x \mid x^{p^n} - x$ in $\mathbb{F}_p[x]$. To see the divisibility: write $a = p^m$ and note $a^q - 1 = (a - 1)(a^{q-1} + a^{q-2} + \cdots + 1)$, so $(p^m - 1) \mid (p^n - 1)$. This means $x^{p^m - 1} - 1 \mid x^{p^n - 1} - 1$ (since $x^{p^m-1} - 1$ divides $x^{k(p^m-1)} - 1$ for any positive integer $k$, and $p^n - 1 = k(p^m - 1)$ for $k = (p^n - 1)/(p^m - 1)$). Multiplying both sides by $x$: $x^{p^m} - x \mid x^{p^n} - x$. Since every element of $\mathbb{F}_{p^n}$ is a root of $x^{p^n} - x$, and $x^{p^m} - x \mid x^{p^n} - x$, every root of $x^{p^m} - x$ that lies in any extension of $\mathbb{F}_p$ inside $\mathbb{F}_{p^n}$ is indeed in $\mathbb{F}_{p^n}$. The polynomial $x^{p^m} - x$ has formal derivative $-1$, so it is separable and has exactly $p^m$ distinct roots. All $p^m$ roots lie in $\mathbb{F}_{p^n}$. The set $S$ is a subfield by the same Frobenius argument used in Part 1: $S$ is closed under subtraction (since $(\alpha - \beta)^{p^m} = \alpha^{p^m} - \beta^{p^m} = \alpha - \beta$) and under division (since $(\alpha\beta^{-1})^{p^m} = \alpha^{p^m}(\beta^{p^m})^{-1} = \alpha\beta^{-1}$). Since $|S| = p^m$, we have $S \cong \mathbb{F}_{p^m}$. **($\Rightarrow$) A subfield forces $m \mid n$.** Suppose $F$ is a subfield of $\mathbb{F}_{p^n}$ with $|F| = p^m$. Then $\mathbb{F}_p \subseteq F \subseteq \mathbb{F}_{p^n}$, and by the [Tower Law](/theorems/1248): \begin{align*} n = [\mathbb{F}_{p^n} : \mathbb{F}_p] = [\mathbb{F}_{p^n} : F] \cdot [F : \mathbb{F}_p] = [\mathbb{F}_{p^n} : F] \cdot m. \end{align*} Hence $m \mid n$. **Uniqueness of the subfield.** If $F$ and $F'$ are two subfields of $\mathbb{F}_{p^n}$ both isomorphic to $\mathbb{F}_{p^m}$, then every element of $F$ satisfies $x^{p^m} = x$ (since $F$ consists of roots of $x^{p^m} - x$), and similarly for $F'$. Hence $F$ and $F'$ are both subsets of $S = \{\alpha \in \mathbb{F}_{p^n} : \alpha^{p^m} = \alpha\}$. Since $|F| = |F'| = |S| = p^m$, we conclude $F = S = F'$. There is exactly one subfield of each admissible order.[/guided]