Separable Extensions

We use the algebra of the formal derivative, the The Minimal Polynomial and elementary properties of gcd's, computed either in $F[x]$ or in $\overline{F}[x]$ (a gcd computed by the Euclidean algorithm in $F[x]$ does not change when we extend scalars to $\overline{F}[x]$, since the Euclidean algorithm only uses field operations in $F$, and the gcd is determined up to a unit by the sequence of remainders it produces).

Step 1: repeated roots detect common factors with $f'$. Work in $\overline{F}[x]$, where $f$ splits completely:

Apply the product rule (valid formally in any commutative ring, and it is straightforward to check by induction on the number of factors that the usual product and sum rules for polynomial derivatives hold over any field):

Fix an index $i_0$ with $e_{i_0} \geq 2$ (such an index exists iff $f$ has a repeated root). Every term of the sum for $f'$ with $j \neq i_0$ contains the factor $(x-\alpha_{i_0})^{e_{i_0}}$ (since that whole factor survives from the product $\prod_{i\ne j}(x-\alpha_i)^{e_i}$, as $e_{i_0}\ge 2 \ge 1$), and the term with $j = i_0$ contains $(x-\alpha_{i_0})^{e_{i_0}-1}$ with $e_{i_0}-1 \geq 1$. So every term is divisible by $(x - \alpha_{i_0})$, hence $(x-\alpha_{i_0}) \mid f'(x)$ in $\overline{F}[x]$. Since also $(x-\alpha_{i_0}) \mid f(x)$, the two polynomials share the common factor $(x-\alpha_{i_0})$ in $\overline{F}[x]$, so $\gcd(f,f') \ne 1$ (up to units) when computed in $\overline{F}[x]$. Since the gcd computed in $F[x]$ generates the same ideal after tensoring to $\overline F[x]$ (Euclidean algorithm is field-operation-only, hence unaffected by field extension), $\gcd(f,f')\ne1$ already in $F[x]$. Conversely, suppose all $e_i = 1$, i.e. $f$ has no repeated root. Fix any root $\alpha_{i_0}$. In the derivative sum above, every term with $j \ne i_0$ still contains the factor $(x - \alpha_{i_0})^{e_{i_0}} = (x-\alpha_{i_0})$ (since $e_{i_0}=1$ means the untouched product $\prod_{i \ne j} (x-\alpha_i)^{e_i}$ for $j \ne i_0$ still includes the $i=i_0$ factor once), while the $j = i_0$ term is $c\cdot 1 \cdot \prod_{i \ne i_0}(x - \alpha_i)$, which does not vanish at $x = \alpha_{i_0}$ because all other roots are distinct from $\alpha_{i_0}$. Hence $f'(\alpha_{i_0}) \ne 0$ for every root $\alpha_{i_0}$ of $f$, so $f$ and $f'$ share no common root in $\overline F$, so (since $\overline F$ is algebraically closed and a nonunit common factor of $f,f'$ in $\overline F[x]$ would force a common root) $\gcd(f,f')=1$ in $\overline F[x]$, hence also in $F[x]$. This proves the equivalence: $f$ has a repeated root in $\overline{F}$ $\iff$ $\gcd(f,f') \neq 1$ in $F[x]$. Statement (1) is the negation of this.

Step 2: the irreducible case. Assume now $f$ is irreducible in $F[x]$. If $f' = 0$, then trivially $\gcd(f,f') = f \ne 1$ (as $f$ is nonconstant), so by Step 1, $f$ has a repeated root. Conversely suppose $f' \neq 0$. Since $\deg f' < \deg f$ (as $\deg f \geq 1$ and differentiation strictly lowers degree for a nonzero derivative) and $f' \ne 0$, and since $f$ is irreducible, the only divisors of $f$ in $F[x]$ are units and unit multiples of $f$ itself. Because $\deg f' < \deg f$, the polynomial $f$ itself cannot divide $f'$ unless $f'=0$; since $f' \ne 0$, $f$ does not divide $f'$, so $f$ is not an associate of any common divisor built from $f$, i.e. $\gcd(f,f')$, being a divisor of $f$, must be $1$ (the only proper divisors of an irreducible $f$ are units). Hence $\gcd(f,f')=1$, and by Step 1, $f$ has no repeated root. This proves (2).

Step 3: characteristic $0$. Suppose $\operatorname{char} F = 0$ and $f$ is irreducible of degree $n \geq 1$, say $f(x) = a_n x^n + \cdots + a_0$ with $a_n \neq 0$. Then $f'(x) = n a_n x^{n-1} + \cdots + a_1$. Since $\operatorname{char} F = 0$, the element $n \in F$ (meaning $n \cdot 1_F$) is nonzero for every positive integer $n$, so $n a_n \ne 0$, so $f' \ne 0$ and has degree exactly $n - 1 \geq 0$. If $n = 1$ then $f$ is linear, trivially separable (one root, multiplicity one) — consistent since a linear polynomial always has a single simple root. If $n \geq 2$, $f' \ne 0$, so by Step 2, $f$ has no repeated root, i.e. $f$ is separable. This proves (3).

Step 4: characteristic $p$. Suppose $\operatorname{char} F = p > 0$ and $f(x) = \sum_{i=0}^n a_i x^i$ is irreducible with $f' = 0$. Then $f'(x) = \sum_{i=1}^n i a_i x^{i-1} = 0$ means $i a_i = 0$ in $F$ for every $i = 1, \dots, n$. If $a_i \ne 0$ for some $i$ not divisible by $p$, then $i \ne 0$ in $F$ (since $\operatorname{char} F = p$ and $p \nmid i$ implies $i \cdot 1_F \ne 0$; this is because the additive order of $1_F$ is exactly $p$, so $i\cdot 1_F = 0$ iff $p \mid i$), forcing $i a_i \ne 0$, a contradiction. So $a_i = 0$ whenever $p \nmid i$. Hence every nonzero coefficient of $f$ occurs at an exponent divisible by $p$, i.e.

Conversely, if $f(x) = g(x^p)$ for some $g \in F[x]$, then writing $g(y) = \sum_k b_k y^k$,

since $pk \equiv 0 \pmod p$ makes each coefficient $b_k(pk) = 0$ in $F$ (as $\operatorname{char} F = p$). By Step 2 (using irreducibility of $f$), $f' = 0$ is equivalent to $f$ having a repeated root, so $f$ is inseparable. This proves (4). $\blacksquare$

Characteristic $0$ is Theorem part (3). For $F$ finite of characteristic $p$ with $|F| = p^m$: the Frobenius map $\varphi: F \to F$, $\varphi(a) = a^p$, is a ring homomorphism (it is additive because $(a+b)^p = a^p+b^p$ in characteristic $p$, by the binomial theorem and the fact that $\binom{p}{i} \equiv 0 \pmod p$ for $0<i<p$; it is multiplicative trivially), and it is injective since $F$ is a field (its kernel is an ideal, hence $0$ or $F$, and $\varphi(1)=1\ne0$ so the kernel is not all of $F$). An injective map from a finite set to itself is surjective, so $\varphi$ is an automorphism of $F$; in particular every element of $F$ is a $p$-th power of some (unique) element of $F$. Now suppose, for contradiction, $f \in F[x]$ is irreducible and inseparable. By Theorem part (4), $f(x) = g(x^p) = \sum_k b_k x^{pk}$ with $b_k \in F$. Since $\varphi$ is surjective, write $b_k = c_k^p$ for some $c_k \in F$. Then, using that the Frobenius is a ring homomorphism on $F$ so that $c^p x^{p} = (cx)^p$ termwise, and that in characteristic $p$ we have $(\sum_k c_k x^k)^p = \sum_k c_k^p x^{pk}$ (again by the Frobenius property applied to the ring $F[x]$, since the "freshman's dream" identity holds in any commutative ring of characteristic $p$):

So $f = h^p$ with $\deg h = \deg f / p \geq 1$ (as $f$ nonconstant), contradicting the irreducibility of $f$ (an irreducible polynomial cannot be a nontrivial power of a lower-degree polynomial, i.e. cannot factor as $h \cdot h \cdots h$ with $p \geq 2$ factors each of positive degree). This contradiction shows no such inseparable irreducible $f$ exists. $\blacksquare$

We argue by induction on $[K:F]$, using the tower law and the standard extension theorem for embeddings (that a simple extension $F(\alpha)/F$ admits exactly as many $F$-embeddings into $\overline F$ as the number of distinct roots of $m_\alpha$ in $\overline F$, a fact itself proved from the isomorphism $F(\alpha) \cong F[x]/(m_\alpha(x))$ and the universal property of quotient rings).

Base case. If $[K:F] = 1$ then $K = F$ and there is exactly one embedding, the inclusion, so equality holds trivially and $K/F$ is (vacuously, or directly, since the only element is in $F$ itself with linear, hence separable, minimal polynomial) separable.

Simple extension step. Suppose $K = F(\alpha)$ with minimal polynomial $m_\alpha \in F[x]$ of degree $d = [K:F]$. Any $F$-embedding $\sigma: K \to \overline{F}$ is determined by $\sigma(\alpha)$, which must be a root of $m_\alpha$ in $\overline{F}$ (apply $\sigma$ to the relation $m_\alpha(\alpha) = 0$, using that $\sigma$ fixes coefficients of $m_\alpha \in F[x]$). Conversely, for any root $\beta \in \overline{F}$ of $m_\alpha$, the map $F[x]/(m_\alpha(x)) \to \overline{F}$ sending $x \mapsto \beta$ is a well-defined $F$-algebra homomorphism (since $m_\alpha(\beta)=0$), and composing with the isomorphism $F(\alpha) \cong F[x]/(m_\alpha(x))$ (which sends $\alpha \mapsto x$) gives an embedding $\sigma$ with $\sigma(\alpha) = \beta$. Distinct roots give distinct embeddings. Hence

with equality iff $m_\alpha$ has $d$ distinct roots, i.e. iff $m_\alpha$ is separable, i.e. iff $\alpha$ is separable over $F$, i.e. (since $K = F(\alpha)$, every element of $K$ is an $F$-polynomial in $\alpha$) iff $K/F$ is a separable extension — we verify this last equivalence fully in the general inductive step below, since a priori separability of $K/F$ requires every element, not just $\alpha$, to be separable; this is subsumed by the induction.

Inductive step (general tower). Suppose $[K:F] = n > 1$ and the result holds for all extensions of degree less than $n$. Choose $\alpha \in K \setminus F$ and set $E = F(\alpha)$, so $1 < [E:F] = d \le n$ and $[K:E] = n/d < n$. By the tower law, $[K:F] = [K:E][E:F]$. Every $F$-embedding $\tau: K \to \overline{F}$ restricts to an $F$-embedding $\tau|_E : E \to \overline F$. By the simple extension step, there are at most $d$ possible restrictions $\sigma_1,\dots,\sigma_r$ ($r \le d$) that occur as restrictions of some embedding of $E$. For each such $\sigma_i : E \to \overline F$, the number of embeddings $\tau : K \to \overline F$ extending $\sigma_i$ equals the number of $\sigma_i$-embeddings of $K$ into $\overline F$; this is bounded, by the same simple-extension counting argument applied iteratively to a chain of simple extensions from $E$ to $K$ (or directly by induction, viewing $K/E$ as a finite extension and $\overline F$ still algebraically closed so it receives $\sigma_i$ and remains a valid target), by $[K:E]$. (This uses that $\overline F$ is algebraically closed, hence also serves as an algebraic closure for $\sigma_i(E)$, so the same embedding-extension machinery applies over the field $\sigma_i(E)$.) Therefore

Equality analysis. Equality $|\operatorname{Emb}_F(K,\overline{F})| = [K:F]$ forces, tracing the inequalities above, both $r = d$ (equality in the simple-extension step for $\alpha$, i.e. $\alpha$ separable over $F$) and, for each $i$, exactly $[K:E]$ extensions of $\sigma_i$ to $K$ (equality in the induction hypothesis applied to $K/E$, transported via $\sigma_i$, i.e. $K/E$ is a separable extension in the sense that every element of $K$ has separable minimal polynomial over $\sigma_i(E)$, hence — since minimal polynomials and separability are intrinsic and $\sigma_i$ is an $F$-isomorphism onto its image fixing $F$ — every element of $K$ has separable minimal polynomial over $E$). We must now show: [$\alpha$ separable over $F$] and [every element of $K$ separable over $E = F(\alpha)$] together imply every element of $K$ is separable over $F$, i.e. $K/F$ is separable, and conversely. ($\Leftarrow$, $K/F$ separable $\Rightarrow$ equality): if $K/F$ is separable, every $\beta \in K$, in particular $\alpha$, is separable over $F$; moreover for $\beta \in K$, the minimal polynomial of $\beta$ over $E \supseteq F$ divides the minimal polynomial of $\beta$ over $F$ (since the latter has $\beta$ as a root and lies in $E[x]$, being separable it has no repeated roots in $\overline F$; a divisor of a separable polynomial is separable, since repeated roots of the divisor would be repeated roots of the whole product). So every $\beta \in K$ is separable over $E$ too, giving $K/E$ separable, and by the induction hypothesis (applicable to $K/E$, a smaller-degree extension) equality holds for $K/E$; combined with equality for $E/F$ (since $\alpha$ separable over $F$ gives $r = d$), we get the full equality chain above, hence $|\operatorname{Emb}_F(K,\overline F)| = [K:F]$. ($\Rightarrow$, equality $\Rightarrow K/F$ separable): assume equality holds. As shown, this forces $\alpha$ separable over $F$ and, via the induction hypothesis (now used in its "equality $\Rightarrow$ separable" direction for the smaller extension $K/E$), every element of $K$ separable over $E$. Take any $\beta \in K$; we must show $\beta$ is separable over $F$. Let $m^F_\beta, m^E_\beta$ denote its minimal polynomials over $F, E$ respectively; $m^E_\beta \mid m^F_\beta$ in $E[x]$ (minimal polynomial over the bigger field divides minimal polynomial over the smaller one, since $m^F_\beta(\beta)=0$ and $m^F_\beta \in F[x]\subseteq E[x]$). We know $m^E_\beta$ is separable (has distinct roots in $\overline F$). We want $m_\beta^F$ separable too. This requires one more ingredient, proved next as a standalone lemma, since it is not immediate that a polynomial dividing a separable one forces the whole polynomial ($m^F_\beta$, not just $m^E_\beta$) to be separable — we need the converse direction via degrees, using the multiplicativity established by the Tower/Embedding count itself applied to the tower $F \subseteq F(\beta) \subseteq K$, which by definition of $\alpha, E, \beta$ falls under the same inductive machinery: apply the entire argument above with $\beta$ playing the role of $\alpha$. Concretely, redo the simple/tower decomposition using the chain $F \subseteq F(\beta) \subseteq K$ instead of $F \subseteq F(\alpha) \subseteq K$: since $[K:F]$ is fixed and equality $|\operatorname{Emb}_F(K,\overline F)|=[K:F]$ is a property of $K/F$ not depending on the chosen intermediate field, the same equality-forces-separability argument (with $\beta$ in place of $\alpha$) shows $\beta$ is separable over $F$. Since $\beta \in K$ was arbitrary, $K/F$ is separable. $\blacksquare$