---
title: Separable Extensions and Cyclotomic Fields
module: Field Theory
moduleNumber: 10
lessonNumber: 4
order: 1004
summary: >
  A polynomial is separable when its roots are distinct, detected by whether it
  shares a factor with its formal derivative. Over perfect fields — characteristic
  zero and finite fields — every irreducible is separable, and the existence and
  uniqueness of the finite fields follow. Cyclotomic polynomials package
  the roots of unity by order, are irreducible over the rationals, and give the
  cyclotomic field its degree phi(n).
topics: [Field Theory]
sources:
  - book: Dummit & Foote
    ref: "Ch. 13 — Field Theory; §13.5 Separable and Inseparable Extensions, §13.6 Cyclotomic Polynomials and Extensions"
  - book: Judson
    ref: "Ch. 22 — Finite Fields; §22.1 Structure of a Finite Field"
draft: false
---

The Galois correspondence needs its extensions to have distinct roots, so that a
degree-$n$ polynomial contributes $n$ separate roots for automorphisms to
permute. Repeated roots are a defect that appears only in positive
characteristic. A formal derivative detects it without leaving the base field,
and perfect fields never exhibit it. The resulting theory settles the
classification of finite fields and the degree of the
[cyclotomic fields](/abstract-algebra/field-theory/splitting-fields-and-algebraic-closure).

## Multiple roots and separability

Over a splitting field, $f(x) = (x - \alpha_1)^{n_1} \cdots (x - \alpha_k)^{n_k}$
with distinct $\alpha_i$; the exponent $n_i$ is the **multiplicity** of
$\alpha_i$, which is a **simple** root when $n_i = 1$ and a **multiple** root
otherwise.

> **Definition (Separable polynomial).** A polynomial is **separable** if it has
> no multiple roots — all its roots in a splitting field are distinct — and
> **inseparable** otherwise.

Multiplicity does not depend on which splitting field is used, since any two are
isomorphic by a map bijective on roots. So $x^2 - 2$ is separable over
$\mathbb{Q}$ with roots $\pm\sqrt 2$, while $(x^2 - 2)^n$ for $n \ge 2$ is
inseparable.

## The derivative test

Detecting a multiple root looks like it requires the splitting field, but the
formal derivative does it inside $F[x]$.

> **Definition (Formal derivative).** For
> $f(x) = a_n x^n + \cdots + a_1 x + a_0 \in F[x]$, the **derivative** is
> $$
> D_x f(x) = n a_n x^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1 \in F[x].
> $$

This is the calculus formula, but purely algebraic: no limits, valid over any
field. It obeys the sum and product rules, provable directly from the
definition.

> **Proposition (Derivative criterion).** An element $\alpha$ is a multiple root
> of $f$ if and only if $\alpha$ is a root of both $f$ and $D_x f$. Hence $f$ is
> separable if and only if $\gcd(f, D_x f) = 1$.

> **Proof.** If $f = (x - \alpha)^n g$ with $n \ge 2$, the product rule gives
> $D_x f = n(x - \alpha)^{n-1} g + (x - \alpha)^n D_x g$, which has $\alpha$ as a
> root. Conversely if $f = (x - \alpha)h$ and $D_x f(\alpha) = 0$, then
> $D_x f = h + (x - \alpha)D_x h$ forces $h(\alpha) = 0$, so $(x - \alpha)^2$
> divides $f$.[^df-prop33]

The gcd is computed by the Euclidean algorithm in $F[x]$, so separability is
decidable over the base field, no factoring required.

$$
% caption: A double root of f is a shared root of f and its derivative, so it
% survives into the gcd; a separable polynomial has gcd 1 with its derivative.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw, minimum width=26mm, minimum height=11mm, align=center] (f) at (0,1.1) {f};
\node[draw, minimum width=26mm, minimum height=11mm, align=center] (df) at (0,-1.1) {derivative of f};
\node[draw=acc, thick, fill=acc!8, minimum width=30mm, minimum height=11mm, align=center] (g) at (5.2,0) {gcd of f and its derivative};
\draw[->, acc, thick] (f) -- (g);
\draw[->, acc, thick] (df) -- (g);
\node[font=\footnotesize, align=left, text=black] at (9.4,0.6) {nontrivial gcd:\\a repeated root};
\node[font=\footnotesize, align=left, text=black] at (9.4,-0.6) {gcd equals 1:\\separable};
\end{tikzpicture}
$$

In characteristic $0$ this settles the matter. If $p(x)$ is irreducible of degree
$n$, then $D_x p$ has degree $n - 1$, is nonzero, and shares no factor with the
irreducible $p$, so $\gcd(p, D_x p) = 1$.

> **Corollary.** Every irreducible polynomial over a field of characteristic $0$
> is separable, and a polynomial over such a field is separable exactly when it
> is a product of distinct irreducibles.

## Inseparability in characteristic p

The characteristic-$0$ argument used $\deg D_x p = n - 1$. In characteristic $p$
the derivative of $x^{pm}$ is $pm\,x^{pm-1} = 0$, so the derivative can drop by
more than one degree, and $D_x p$ can vanish entirely. When $D_x p = 0$, every
exponent of $p$ is a multiple of the characteristic, so $p(x) = p_1(x^p)$ for
some $p_1$. This is the only way an irreducible can be inseparable.

The standard inseparable example is
$p(x) = x^2 - t$ over the rational function field $F = \mathbb{F}_2(t)$. It is
irreducible (Eisenstein at the prime $t$), but $D_x p = 2x = 0$, and over a
splitting field $x^2 - t = (x - \sqrt t)^2$ since the cross term $2x\sqrt t$
vanishes in characteristic $2$. The single root $\sqrt t$ has multiplicity $2$.

$$
% caption: Over a characteristic-2 function field, x squared minus t is
% irreducible yet its derivative vanishes, so its one root is doubled.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw, minimum width=52mm, minimum height=12mm, align=center] (p) at (0,1.2)
  {p is irreducible over F2(t)\\derivative of p is 0};
\node[draw=acc, thick, fill=acc!8, minimum width=52mm, minimum height=12mm, align=center] (r) at (0,-1.2)
  {one root of multiplicity 2\\p equals (x minus root) squared};
\draw[->, acc, thick] (p) -- (r) node[midway, right=3pt, font=\footnotesize] {char 2};
\node[font=\footnotesize, align=left, text=black] at (5.2,0)
  {the cross term carries\\a factor of 2, which is\\0 in characteristic 2};
\end{tikzpicture}
$$

Much of positive-characteristic algebra follows from the vanishing of the
binomial cross terms.

> **Proposition (Freshman's dream).** In a field of characteristic $p$,
> $(a + b)^p = a^p + b^p$ and $(ab)^p = a^p b^p$, so the map
> $\varphi(a) = a^p$ is an injective field homomorphism.

> **Proof.** Every intermediate binomial coefficient $\binom{p}{i}$ for
> $0 < i < p$ is divisible by $p$, since the $p$ in the numerator survives, so
> those terms vanish; injectivity holds because a field has no nonzero nilpotents,
> so $a^p = 0$ forces $a = 0$.

The map $a \mapsto a^p$ is the **Frobenius endomorphism**. On a finite field it
is injective and therefore surjective, so every element is a $p$th power.

$$
% caption: The Frobenius map raises every element to the p-th power; on a finite
% field of order q it is a bijection, permuting the elements and fixing the prime subfield.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw, circle, minimum size=32mm] (F) at (0,0) {};
\node at (0,0.6) {$\mathbf{F}_q$};
\node[font=\footnotesize] at (0,0.0) {x maps to};
\node[font=\footnotesize] at (0,-0.4) {x to the p};
\draw[->, acc, very thick] (1.9,0.6) arc (60:-60:1.1);
\node[acc, font=\footnotesize, anchor=west] at (2.6,0) {Frobenius,};
\node[acc, font=\footnotesize, anchor=west] at (2.6,-0.45) {a bijection};
\end{tikzpicture}
$$

## Perfect fields and finite fields

> **Definition (Perfect field).** A field $K$ of characteristic $p$ is
> **perfect** if $K = K^p$, i.e. every element is a $p$th power. Every field of
> characteristic $0$ is also called perfect.

The finite-field argument generalizes: over a perfect field the trick that turned
$p(x) = q(x^p)$ into a $p$th power of a polynomial applies, contradicting
irreducibility. So inseparability never arises.

> **Theorem.** Every irreducible polynomial over a perfect field is separable.
> Consequently every finite extension of $\mathbb{Q}$ or of a finite field is
> separable.

This gives the cleanest route to the finite fields themselves.

> **Theorem (Finite fields).** For every prime $p$ and every $n \ge 1$ there is,
> up to isomorphism, exactly one field of order $p^n$, the splitting field
> $\mathbb{F}_{p^n}$ of $x^{p^n} - x$ over $\mathbb{F}_p$.

> **Proof.** The polynomial $x^{p^n} - x$ has derivative $-1$, hence is separable
> with exactly $p^n$ distinct roots. Those roots are closed under addition,
> multiplication, and inverses — the Frobenius identities do the checking — so
> they form a field of order $p^n$, which must be the whole splitting field.
> Conversely any field of order $p^n$ has multiplicative group of order
> $p^n - 1$, so every element satisfies $x^{p^n} = x$ and is a root; it is
> therefore a splitting field of $x^{p^n} - x$, and splitting fields are unique
> up to isomorphism.[^df-finite]

The containment $\mathbb{F}_{p^d} \subseteq \mathbb{F}_{p^n}$ holds exactly when
$d \mid n$, so the subfields of a finite field mirror the divisors of $n$. Their
Galois structure appears in the
[finite fields](/abstract-algebra/galois-theory/finite-fields) lesson.

> **Worked example (building $\mathbb{F}_4$).** Over $\mathbb{F}_2$ the
> polynomial $x^2 + x + 1$ has no root ($0^2 + 0 + 1 = 1$ and
> $1^2 + 1 + 1 = 1$), so it is irreducible and
> $\mathbb{F}_4 = \mathbb{F}_2[x]/(x^2 + x + 1)$ is a field of four elements
> $\{0, 1, \alpha, \alpha + 1\}$ with the single relation
> $\alpha^2 = \alpha + 1$ (read in characteristic $2$). Every product reduces by
> that relation: $\alpha \cdot \alpha = \alpha + 1$, and
> $\alpha(\alpha + 1) = \alpha^2 + \alpha = (\alpha + 1) + \alpha = 1$, so
> $\alpha^{-1} = \alpha + 1$. The three nonzero elements are the successive
> powers $\alpha,\ \alpha^2 = \alpha + 1,\ \alpha^3 = 1$, so
> $\mathbb{F}_4^\times$ is cyclic of order $3$.

Addition is coordinatewise in characteristic $2$, where $x + x = 0$, and
multiplication follows from $\alpha^2 = \alpha + 1$:

| $+$ | $0$ | $1$ | $\alpha$ | $\alpha+1$ |
| --- | --- | --- | --- | --- |
| $0$ | $0$ | $1$ | $\alpha$ | $\alpha+1$ |
| $1$ | $1$ | $0$ | $\alpha+1$ | $\alpha$ |
| $\alpha$ | $\alpha$ | $\alpha+1$ | $0$ | $1$ |
| $\alpha+1$ | $\alpha+1$ | $\alpha$ | $1$ | $0$ |

| $\times$ | $1$ | $\alpha$ | $\alpha+1$ |
| --- | --- | --- | --- |
| $1$ | $1$ | $\alpha$ | $\alpha+1$ |
| $\alpha$ | $\alpha$ | $\alpha+1$ | $1$ |
| $\alpha+1$ | $\alpha+1$ | $1$ | $\alpha$ |

### Separable and inseparable degree

When an irreducible $p(x)$ over a characteristic-$p$ field is inseparable, peel
off powers of $x^p$ until the derivative is nonzero: there is a unique $k$ and a
unique irreducible separable $p_{\mathrm{sep}}$ with
$p(x) = p_{\mathrm{sep}}(x^{p^k})$.

> **Definition (Separable degree).** With $p(x) = p_{\mathrm{sep}}(x^{p^k})$, the
> **separable degree** $\deg_s p$ is $\deg p_{\mathrm{sep}}$ and the
> **inseparable degree** $\deg_i p$ is $p^k$, so that
> $\deg p = \deg_s p \cdot \deg_i p$.

An extension $K/F$ is **separable** if the minimal polynomial of every element of
$K$ is separable. Over perfect fields every finite extension is separable, so the
distinction only matters for function fields and other imperfect fields of
positive characteristic.

## Cyclotomic polynomials

Return to characteristic $0$ and organize the roots of unity by their order. The
$n$th roots of unity form a cyclic group of order $n$; grouping them by which
$d \mid n$ they have as their exact order factors $x^n - 1$.

> **Definition (Cyclotomic polynomial).** The **$n$th cyclotomic polynomial**
> $\Phi_n(x)$ is the monic polynomial whose roots are the primitive $n$th roots
> of unity:
> $$
> \Phi_n(x) = \prod_{\substack{1 \le a \le n \\ \gcd(a,n)=1}} (x - \zeta_n^a),
> \qquad \deg \Phi_n = \varphi(n).
> $$

Since every $n$th root of unity is primitive of exactly one order $d \mid n$, the
linear factors of $x^n - 1$ partition by order:
$$
x^n - 1 = \prod_{d \mid n} \Phi_d(x).
$$
Comparing degrees recovers the identity $n = \sum_{d \mid n} \varphi(d)$. The
factorization is a recursion: $\Phi_1 = x - 1$, and dividing $x^n - 1$ by the
product of the lower $\Phi_d$ gives $\Phi_n$.

> **Worked example (computing $\Phi_{12}$).** The divisors of $12$ are
> $1, 2, 3, 4, 6, 12$, so $x^{12} - 1 = \Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 \Phi_{12}$.
> The five known factors group cleanly:
> $$
> \Phi_1 \Phi_2 \Phi_3 \Phi_6 = (x^3 - 1)(x^3 + 1) = x^6 - 1,
> $$
> using $(x - 1)(x^2 + x + 1) = x^3 - 1$ and $(x + 1)(x^2 - x + 1) = x^3 + 1$.
> Including $\Phi_4 = x^2 + 1$,
> $$
> \prod_{\substack{d \mid 12 \\ d < 12}} \Phi_d = (x^6 - 1)(x^2 + 1)
> = x^8 + x^6 - x^2 - 1.
> $$
> Dividing then gives
> $$
> \Phi_{12}(x) = \frac{x^{12} - 1}{x^8 + x^6 - x^2 - 1} = x^4 - x^2 + 1,
> $$
> of degree $\varphi(12) = 4$, matching $[\mathbb{Q}(\zeta_{12}) : \mathbb{Q}] = 4$.

$$
% caption: The n-th roots of unity split by exact order; the primitive ones
% (order exactly n) are the roots of the n-th cyclotomic polynomial.
\begin{tikzpicture}[scale=1.2, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) circle (1.5);
\draw[black, ->] (-1.8,0) -- (1.8,0);
\draw[black, ->] (0,-1.8) -- (0,1.8);
% 6th roots of unity: primitive are at 60 and 300 degrees
\foreach \k in {0,1,2,3,4,5} \fill[black] (\k*60:1.5) circle (1.6pt);
\fill[acc] (60:1.5) circle (2.6pt);
\fill[acc] (300:1.5) circle (2.6pt);
\node[acc, font=\footnotesize, anchor=south west] at (60:1.5) {primitive};
\node[acc, font=\footnotesize, anchor=north west] at (300:1.5) {primitive};
\node[font=\footnotesize, align=left, text=black] at (2.6,0.9) {n = 6: two roots\\of order exactly 6};
\end{tikzpicture}
$$

Two facts make $\Phi_n$ the minimal polynomial of $\zeta_n$.

> **Lemma.** $\Phi_n(x)$ is a monic polynomial with integer coefficients.

> **Proof.** Induction: $x^n - 1 = f(x)\Phi_n(x)$ where
> $f = \prod_{d \mid n, d < n}\Phi_d$ is monic in $\mathbb{Z}[x]$, and by Gauss's
> lemma the quotient $\Phi_n$ stays in $\mathbb{Z}[x]$.[^df-lemma40]

> **Theorem (Irreducibility).** $\Phi_n(x)$ is irreducible over $\mathbb{Q}$.

> **Proof.** Suppose $\Phi_n = fg$ with $f$ irreducible in $\mathbb{Z}[x]$ and
> $\zeta$ a root of $f$. For a prime $p \nmid n$, the power $\zeta^p$ is again
> primitive, so it is a root of $f$ or of $g$. If $g(\zeta^p) = 0$ then
> $f(x) \mid g(x^p)$, and reducing mod $p$ gives
> $\overline{g}(x)^p = \overline{f}(x)\overline{h}(x)$, forcing $\overline f$ and
> $\overline g$ to share a factor. Then $\overline{\Phi_n}$, and hence
> $\overline{x^n - 1}$, would have a multiple root mod $p$ — impossible for
> $p \nmid n$, since $x^n - 1$ is separable there. So $\zeta^p$ is a root of $f$
> for every such $p$, hence $\zeta^a$ is for every $a$ coprime to $n$, so $f$ has
> all $\varphi(n)$ primitive roots and $f = \Phi_n$.[^df-thm41]

> **Corollary.** $[\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \varphi(n)$.

Since $\Phi_n$ is the minimal polynomial of $\zeta_n$, the degree of the
cyclotomic field is its degree $\varphi(n)$.

| $n$ | $\Phi_n(x)$ | $\varphi(n) = [\mathbb{Q}(\zeta_n) : \mathbb{Q}]$ |
| --- | --- | --- |
| $1$ | $x - 1$ | $1$ |
| $2$ | $x + 1$ | $1$ |
| $3$ | $x^2 + x + 1$ | $2$ |
| $4$ | $x^2 + 1$ | $2$ |
| $5$ | $x^4 + x^3 + x^2 + x + 1$ | $4$ |
| $6$ | $x^2 - x + 1$ | $2$ |
| $8$ | $x^4 + 1$ | $4$ |
| $12$ | $x^4 - x^2 + 1$ | $4$ |

The degrees factor the fields into recognizable pieces. For $n = 8$,
$\varphi(8) = 4$ and $\mathbb{Q}(\zeta_8)$ contains both $i$ and
$\zeta_8 + \zeta_8^{-1} = \sqrt 2$, so
$$
\mathbb{Q}(\zeta_8) = \mathbb{Q}(i, \sqrt 2),
$$
the biquadratic field. This settles the constructibility of regular polygons:
the $n$-gon is constructible when $\tfrac12\varphi(n)$ is a power of
$2$, which is why the $17$-gon works and the $7$-gon and $9$-gon do not, as the
[construction lesson](/abstract-algebra/field-theory/straightedge-and-compass-constructions)
promised.

The cyclotomic fields are the first family of extensions whose automorphisms have
an explicit description: $\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})
\cong (\mathbb{Z}/n\mathbb{Z})^\times$, computed in the
[cyclotomic and abelian extensions](/abstract-algebra/galois-theory/cyclotomic-and-abelian-extensions)
lesson. Separability lets the count of automorphisms match the degree, the
standing hypothesis behind the Galois correspondence.

[^df-prop33]: **Dummit & Foote**, §13.5, Proposition 33 — a multiple root is a common root of $f$ and $D_x f$, so separability is equivalent to $\gcd(f, D_x f) = 1$.
[^df-finite]: **Dummit & Foote**, §13.5, Corollary 34 through the finite-fields example, and Propositions 35, 37 — the Frobenius identities, separability over perfect fields, and existence and uniqueness of $\mathbb{F}_{p^n}$ as the splitting field of $x^{p^n} - x$. See also Judson §22.1.
[^df-lemma40]: **Dummit & Foote**, §13.6, Lemma 40 — $\Phi_n \in \mathbb{Z}[x]$ by induction and Gauss's lemma.
[^df-thm41]: **Dummit & Foote**, §13.6, Theorem 41 and Corollary 42 — irreducibility of $\Phi_n$ over $\mathbb{Q}$ and the resulting degree $\varphi(n)$ of the cyclotomic field.