Category: Abstract Algebra

Cyclotomic Polynomials

Let $\mathbb F$ be a field.

Definition 1. Define the n-th cyclotomic extension of $\mathbb F$ by $\mathcal Z(f_n) \subseteq \bar{\mathbb F}$ , where $f_n(x) =x^n - 1$ . We call $\alpha \in \mathcal Z(f_n)$ a primitive n-th root of unity if $\alpha^r \neq 1$ for any $1 \leq r < n$ .

Problem 1. Show that $\mathcal Z(f_n)$ contains a primitive $n$ -th root of unity if and only if $\mathrm{char}(\mathbb F) \nmid n$ . In this case, there are $\varphi(n)$ primitive n-th roots of unity, where $\varphi$ denotes the Euler totient function.

(Click for Solution)

Solution. If $\mathrm{char}(\mathbb F) = 0$ , then $\mathbb F$ contains $\mathbb Q$ as a subfield, so that $\bar{\mathbb F} \supseteq \bar{\mathbb Q}$ contains the usual primitive $n$ -th root of unity $\zeta_n := e^{2\pi i/n}$ . Otherwise, suppose $\mathrm{char}(\mathbb F) = p$ for some prime $p$ .

If $p \mid n$ , then $n = kp$ for some integer $k$ . Then for any $\alpha \in \mathcal Z(f_n)$ ,

$(\alpha^k-1)^p = \alpha^n - 1 = 0,$

so that $\alpha^k = 1$ , and $\alpha$ is not a primitive $n$ -th root of unity.

Suppose $p \nmid n$ . By Euler’s totient theorem,

$[p^{\varphi(n)}]_n = [p]_n^{\varphi(n)} = [1]_n.$

Therefore, $n \mid (p^{\varphi(n)} - 1) = |\mathbb F_{p^{\varphi(n)}}^*|$ , so that there exists an integer $k$ such that

$p^{\varphi(n)} - 1 = k \cdot n.$

Recall that $\mathbb F_{p^{\varphi(n)}}^*$ is cyclic, so that there exists $\alpha \in \mathbb F_{p^{\varphi(n)}}^*$ such that $\mathbb F_{p^{\varphi(n)}}^* = \langle \alpha \rangle$ . In particular,

$(\alpha^k)^n = \alpha^{k \cdot n} = \alpha^{p^{\varphi(n)}-1} = 1.$

In particular, $\alpha^k$ is a primitive $n$ -th root of unity.

Problem 2. Let $\zeta_n$ be a primitive $n$ -th root of unity. Show that $\mathbb F(\zeta_n) \supseteq \mathbb F$ is Galois, and $\mathrm{Gal}(\mathbb F(\zeta_n)/\mathbb F)$ is isomorphic to some subgroup of $\mathbb Z_n^*$ .

(Click for Solution)

Solution. Let $\zeta_n$ be a primitive $n$ -th root of unity and write $\mathcal Z(f_n) = \langle \zeta_n \rangle$ . Then $\mathbb F(\zeta_n)$ is the splitting field of the separable polynomial $f_n$ , and thus $\mathbb F(\zeta_n) \supseteq \mathbb F$ is Galois. For any $\sigma \in \mathrm{Gal}(\mathbb F(\zeta_n)/\mathbb F)$ , $\sigma$ is uniquely determined by $\sigma(\zeta_n) = \zeta_n^{i_\sigma}$ for $i_\sigma \in \mathbb Z_n^*$ . Define the desired monomorphism by $\sigma \mapsto i_\sigma$ .

Definition 2. Let $\{\alpha_1,\dots,\alpha_{\varphi(n)}\}$ denote the primitive n-th roots of unity. Define the n-th cyclotomic polynomial $\Phi_n$ by

$\displaystyle \Phi_n(x) = \prod_{i=1}^{\varphi(n)} (x-\alpha_i).$

We remark that $x^n - 1 = \prod_{d \mid n} \Phi_d(x)$ .

Problem 3. Show that $\Phi_n \in \mathbb Z[x]$ is irreducible in $\mathbb Q[x]$ . Furthermore, $\mathrm{Gal}(\mathbb Q(\zeta_n)/\mathbb Q) \cong \mathbb Q$ .

(Click for Solution)

Solution. Firstly, $\Phi_n \in \mathbb Q[x]$ . Fix a root $u$ of $\phi_n$ . Then for any prime $p \nmid n$ , $f_{\mathbb Q}^u = f_{\mathbb Q}^{u^p}$ . For any $1 \leq k\leq n-1$ , write

$\displaystyle k = \prod_{i=1}^r p_i^{\alpha_i},$

In particular,

$\begin{aligned} f_{\mathbb Q}^u(u) = 0 \quad &\Rightarrow \quad f_{\mathbb Q}^u(u^{p_i}) = f_{\mathbb Q}^{u^p}(u^{p_i}) = 0 \\ &\Rightarrow \quad f_{\mathbb Q}^{u}(u^k) = f_{\mathbb Q}^{u}(u^{p_i}) = 0.\end{aligned}$

Therefore, there are at least $\varphi(n)$ roots of $f$ , so that $\deg(f) = \varphi(n)$ . Therefore, $\Phi_n = f$ is irreducible, and $\mathbb Q(\zeta_n) = \mathcal Z(\Phi_n)$ . Using Problem 2, the monomorphism $\eta : \mathrm{Gal}(\mathbb Q(\zeta_n)/\mathbb Q) \to \mathbb Z_n^*$ defined by $\eta(\sigma) = i_\sigma$ has kernel $\ker(\eta) = \{\mathrm{id}_{\mathbb Q}\}$ , so that $\mathrm{Gal}(\mathbb Q(\zeta_n)/\mathbb Q) \cong \mathbb Q$ .

Problem 4. Let $\Omega$ be an algebraically closed field with $\mathrm{char}(\Omega) = p \neq 0$ . Suppose $n$ is a positive integer with $p \nmid n$ . Show that $\mathcal Z(\Phi_n)$ contains exactly the primitive $n$ -th roots of unity in $\Omega$ .

(Click for Solution)

Solution. Let $\mathcal Z_n \subseteq \mathcal Z(f_n)$ denote the subset of the primitive $n$ -th roots of unity in $\Omega$ . We aim to show that $\mathcal Z(\Phi_n) = \mathcal Z_n$ . Fix $u \in \mathcal Z(\Phi_d)$ for any $d \mid n$ . Then $\Phi_d(x) \mid (x^d - 1)$ implies that $u^d - 1 = 0$ , so that $u \in \mathcal Z_d$ . Setting $d = n$ , $\mathcal Z(\Phi_n) \subseteq \mathcal Z_n$ . Furthermore,

$\varphi(n) \leq |\mathcal Z(\Phi_n)| \leq \deg(\Phi_n) = \varphi(n) = |\mathcal Z_n|.$

Therefore, $\mathcal Z(\Phi_n) = \mathcal Z_n$ .

Problem 5. Prove Dirichlet’s theorem: for every positive integer $n$ , there exists infinitely many prime numbers $p$ such that $p \equiv_n 1$ .

(Click for Solution)

Solution. Fix $k \in \mathbb N$ . Since $\Phi_n(x) \to \infty$ as $x \to \infty$ , there exists $m_k \in \mathbb N$ such that the integer $\Phi_n(m_k k n) > 1$ . Hence, there exists a prime

$p_k \mid \Phi_n(m_k k m) \mid (m_k k n)^n - 1,$

so that $[m_k k n]_{p_k}^n = [1]_{p_k}$ . In particular, $\Phi_n(m_k k n) = 0$ so that $\mathbb Z_{p_k}^*$ is cyclic with order $p_k - 1$ . Therefore, $n \mid p_k - 1$ , so that $[p_k]_n = [1]_n$ . Define the infinite set $\{p_k : k \in \mathbb N \}$ as desired.

—Joel Kindiak, 4 May 26, 1226H

July 8, 2026
Euclidean Constructibility
Definition 1. A real number $u$ is said to be constructible if we can construct a line segment of length $u$ using just a straightedge and a compass. By assumption, $1$ is constructible. We allow constructible numbers to be negative.

Problem 1. Show that the set $\frak C$ of constructible numbers forms a subfield of $\mathbb R$ that contains $\mathbb Q$ .

(Click for Solution)

Solution. We first show that $\mathbb Q \subseteq \frak C$ . Fix $p / q \in \mathbb Q^+$ with $\gcd(p, q) = 1$ without loss of generality. Given a point $(0, 0)$ , it is not hard to construct the point $(q, p)$ . The line that intersects both points has equation $y = px/q$ . In particular, it intersects the line $x = 1$ and the point $(1, p/q)$ . Therefore, the line segment connecting $(0,p/q)$ and $(1, p/q)$ have length $p/q$ , so that $p/q \in \frak C$ , as required.

To check that $\frak C$ forms a field, suppose $u, v \in \frak C$ . It suffices to check that $u+v \in \frak C$ , $uv \in \frak C$ , and $u/v \in \frak C$ .

For addition, let $s_1$ denote the line segment connecting $(0, 0)$ and $(u,0)$ , and $s_2$ denote the line segment connecting $(0,0)$ and $(v,0)$ . Use the parallel postulate to translate the line segment by $(0, 1)^{\mathrm T}$ , so that $s_2$ now passes through $(0, 1)$ and $(v, 1)$ . Use the parallel postulate to translate the line segment by $(u,-1)^{\mathrm T}$ . Then the line segment obtained by combining $s_1$ and $s_2$ has a total length of $u + v$ , as required.

For multiplication, let $s_1$ denote the line segment connecting $(0, 0)$ and $(u,0)$ , and $s_2$ denote the line segment connecting $(0,0)$ and $(0,v)$ . Let $s_3$ denote the line segment connecting $(0,0)$ and $(0,1)$ , and $s_4$ connect $(0,1)$ to $(u,0)$ . Using the parallel postulate, draw a line parallel to $s_4$ passing through $(0, v)$ . Extend $s_4$ to a line segment $s_5$ until it intersects said line. Let $\ell$ denote the length of $s_5$ . Using similar triangles, $\ell/u = v/1$ , so that $\ell = uv$ .

For division, let $s_1$ denote the line segment connecting $(0, 0)$ and $(u,0)$ , and $s_2$ denote the line segment connecting $(0,0)$ and $(0,v)$ . Let $s_3$ denote the line segment connecting $(0,0)$ and $(0,1)$ , and $s_4$ connect $(0,v)$ to $(u,0)$ . Using the parallel postulate, draw a line parallel to $s_4$ passing through $(0, 1)$ and intersecting $s_1$ . Denote the shorterned line segment by $s_5$ . Let $\ell$ denote the length of $s_5$ . Using similar triangles, $1/v = \ell/u$ , so that $\ell = u/v$ , as required.

Problem 2. Show that $u \in \frak C$ if and only if there exist real numbers $\alpha_1,\dots,\alpha_n$ such that $\alpha_i \in \mathbb F_i$ for each $i$ , and $u \in \mathbb F_{n+1},$ where

$\mathbb F_i := \mathbb Q(\sqrt{\alpha_1},\dots,\sqrt{\alpha_{i-1}}).$

In particular, we say that the angle $\theta$ is constructible if and only if $\cos(\theta)$ is constructible.
(Click for Solution)

Solution. Denote $\mathbb F_0 := \mathbb Q$ . Given $\mathbb F_i$ , iew numbers can be constructed via one of three methods:
- Intersecting lines
- Intersecting a line and a circle
- Intersecting circles
Carrying out the first and third operations drawn using points in $\mathbb F_i^2$ produces that points belong to $\mathbb F_i^2$ , and thus constructs numbers that remain in $\mathbb F_i$ . Carrying out the second operation produces points that belong to $\mathbb F_{i+1} := \mathbb F_i(\sqrt{\alpha_i})$ for some $\alpha_i \in \mathbb F_i$ . Now, $u \in \frak C$ if and only if it can be constructed using a finite number of one of these three steps. This is equivalent to the existence of a chain

$\mathbb F_0 \subseteq \mathbb F_1 \subseteq \cdots \subseteq \mathbb F_n \subseteq \mathbb F_{n+1}$

such that $u \in \mathbb F_{n+1}$ , as required.
Problem 3. If $u \in \frak C$ , show that there exists a positive integer $r$ such that $[\mathbb Q(u) : \mathbb Q] = 2^r$ .

(Click for Solution)

Solution. Using Problem 2, write $\mathbb Q(u) = \mathbb F_{n+1}$ . As subfields, $[\mathbb F_{n+1} : \mathbb F_n] = 2$ . Therefore, by the tower rule,

$\begin{aligned} [\mathbb Q(u): \mathbb Q] &= [\mathbb F_{n+1} : \mathbb F_0] \\ &= [\mathbb F_{n+1} : \mathbb F_n] \cdot [\mathbb F_n : \mathbb F_{n-1}] \cdot \cdots \cdot [\mathbb F_1 : \mathbb F_0] \\ &= \underbrace{ 2 \cdot 2 \cdot \cdots \cdot 2 }_{n+1} \\ &= 2^{n+1}. \end{aligned}$

Set $r:=n+1$ and we are done.

Problem 4. Show that the angle $\pi/9$ is not constructible. Therefore, we cannot, in general, trisect the angle.

(Click for Solution)

Solution. Firstly, we recall that $\cos(\pi/3) = 1/2$ , so that the angle $\pi/3$ is constructible. Denote $u := \cos(\pi/9)$ . Using the triple angle identity,

$\cos(\pi/3) = 4 \cos^3(\pi/9) - 3 \cos(\pi/9).$

Substituting, $8u^3 - 6u - 1 = 0$ . Using the rational root test, the polynomial $f(x) = 8x^3 - 6x - 1$ is irreducible in $\mathbb Q[x]$ . By definition,

$[\mathbb Q(u) : \mathbb Q] = \deg(f_{\mathbb Q}^u) = 3 \neq 2^r$

for any positive integer $r$ . By the contrapositive of Problem 3, $u \notin \frak C$ .

Problem 5. Show that $\sqrt[3]{2}$ is not constructible. Therefore, we cannot double the cube.

(Click for Solution)

Solution. Use the rational root test to show that the polynomial $x^3 - 2 \in \mathbb Q[x]$ is irreducible. Conclude using Problem 3.

—Joel Kindiak, 2 May 26, 2110H
July 6, 2026
Abelian Groups
Let $(G, +)$ denote an Abelian group with additive identity $0$ .

Definition 1. We say that $G$ is finitely generated if there exists a set $\{x_1,\dots, x_n\} \subseteq G$ , called a generating set that spans $G$ , such that

$G = \langle x_1 \rangle + \cdots + \langle x_n \rangle.$

For any $x \in G$ and $n \in \mathbb Z$ , define the map $\cdot : \mathbb Z \times G \to G$ by

$n \cdot x = \begin{cases} 0, & n = 0, \\ \underbrace{x+\cdots+x}_n, & n > 0, \\ (-n, x), & n < 0. \end{cases}$

Denote $nx = n \cdot x$ for brevity. It is not hard to check that $(-n) \cdot x = -(nx)$ for any integer $n$ . Furthermore,

$\langle x \rangle = \{nx : n \in \mathbb Z\} \equiv \mathbb Z\{x\}.$

More generally, the set $\{x_1,\dots, x_n\} \subseteq G$ is linearly independent if for integers $c_1,\dots, c_n$ ,

$\displaystyle \sum_{i=1}^n v_i x_i = 0 \quad \Rightarrow \quad v_i = 0.$

A generating set for $G$ that is linearly independent is called a basis for $G$ .

Definition 2. We say that the Abelian group $G$ is free if there exists a positive integer $n$ such that $G \cong \mathbb Z^n$ . Here, the right-hand side is a direct product in the following sense: letting $\odot_i$ denote the binary operation in the set $K_i$ , define $\odot$ on $K_1\times \cdots \times K_n$ by

$\begin{bmatrix} u_1 \\ \vdots \\ u_n \end{bmatrix} \odot \begin{bmatrix} v_1 \\ \vdots \\ v_n \end{bmatrix} := \begin{bmatrix} u_1 \odot_1 v_1 \\ \vdots \\ u_n \odot_n v_n \end{bmatrix}.$

More precisely then, the isomorphism means $(G, + ) \cong (\mathbb Z^n, +)$ . In this case, we call $n$ the rank of $G$ .

Problem 1. Suppose $G$ is free. Show that $G$ is finitely generated by a basis $\{x_1,\dots, x_n\}$ , and

$\displaystyle G \cong \langle x_1 \rangle \oplus \cdots \oplus \langle x_n \rangle.$

(Click for Solution)

Solution. Since $G$ is free, let $\phi : \mathbb Z^n \to G$ denote the group isomorphism. Define $x_i := \phi(\mathbf e_i)$ . We claim that $\{x_1,\dots, x_n\}$ forms the desired basis.

For any $x \in G$ , there exists $\mathbf v \in \mathbb Z^n$ such that $x = \phi(\mathbf v)$ . By definition, there exists integers $v_i$ such that $\mathbf v = \sum_{i=1}^n v_i \mathbf e_i$ . Since $\phi$ is a homomorphism,

$\displaystyle x = \phi(\mathbf v) = \sum_{i=1}^n v_i \phi(\mathbf e_i) = \sum_{i=1}^n v_i x_i.$

Linear independence follows suit, since the isomorphism implies that

$\displaystyle \sum_{i=1}^n v_i x_i = 0 \quad \Rightarrow \quad \sum_{i=1}^n v_i \mathbf e_i = \mathbf 0,$

so that $v_i = 0$ for each $i$ . For the direct product claim, we check that the map

$\displaystyle (v_1 x_1, \dots, v_n x_n) \mapsto \sum_{i=1}^n v_i x_i$

is an isomorphism similar as before.

Remark 1. The spanning property holds as long as $\phi$ is surjective, while the linear independence property holds as long as $\phi$ is injective.

Problem 2. Suppose $G$ is free and $H \leq G$ . Show the following:
- $H$ is free of rank $m \leq n$ .
- There exists a basis $\{u_1,\dots, u_n\}$ for $G$ and positive integers $c_1,\dots ,c_m$ such that $\{c_1u_1,\dots, c_mu_m\}$ forms a basis for $H$ .
- $[G : H]$ is finite if and only if $m = n$ .
(Click for Solution)

Solution. Let $G$ be free with rank $n$ . Let $\phi : \mathbb Z^n \to G$ denote the given isomorphism. Since $H \leq G$ , $\phi^{-1}(H) \leq \mathbb Z^n$ .

It suffices to assume $G = \mathbb Z^n$ . We prove by induction on $n$ . For $n = 1$ , $H \leq \mathbb Z$ implies that

$H = \alpha \mathbb Z = \langle \alpha \rangle \cong \mathbb Z$

for some $\alpha \in \mathbb Z^+$ . Then $\{\alpha\}$ serves as the desired basis for $H$ .

Suppose the result holds for free subgroups with rank $< n$ . Let $\pi : \mathbb Z^n \to \mathbb Z$ denote projection onto the last coordinate, i.e. $(v_1,\dots,v_n)^{\mathrm T} \mapsto v_n$ . It is clear that $\pi$ is a surjective linear transformation, even when restricted to $\mathbb Z$ as the underlying set of scalars. Then $\ker(\pi) \cong \mathbb Z^{n-1}$ via the identification

$(v_1,\dots, v_{n-1},0) \cong (v_1,\dots, v_{n-1}).$

Since $\ker(\pi) \leq G$ and $H \leq G$ ,

$H \cap \ker(\pi) \leq \ker(\pi) \cong \mathbb Z^{n-1}.$

By induction, $H \cap \ker(\phi) \cong \mathbb Z^k$ with $k \leq n-1$ . Similarly, $\pi(H) \cong \mathbb Z^\ell$ is free with $\ell \leq 1$ . Check that

$H \cong (H \cap \ker(\pi)) \times \pi(H),$

so that

$\begin{aligned} H \cong (H \cap \ker(\pi)) \times \pi(H) &\cong \mathbb Z^k \times \mathbb Z^\ell \\ &\cong \mathbb Z^{k+\ell} \\ &\leq \mathbb Z^n, \end{aligned}$

as required.

Now we prove the second two propositions. Let $\mathbf x := (x_1,\dots, x_n)^{\mathrm T}$ be an (ordered) basis for $G$ and $\mathbf y := (y_1, \dots, y_m)^{\mathrm T}$ be an (ordered) basis for $H$ . By writing each $y_i$ in terms of the $x_j$ , we can obtain a matrix $\mathbf A \in \mathcal M_{m \times n}(\mathbb Z)$ such that

$\mathbf y = \mathbf A \mathbf x.$

Let $\mathbf M$ denote the invertible matrix capturing the process of reducing an augmented matrix into reduced row-echelon form (RREF), so that $\mathbf R = \mathbf M \mathbf A$ denotes the RREF of $\mathbf A$ . Then

$\mathbf M \mathbf y = \mathbf R \mathbf x.$

Then the elements $\{y_1,\dots, y_m\}$ that correspond to the $m$ pivot-columns in $\mathbb R$ yields the desired basis for $H$ . If $m < n$ , then $\mathbb Z \leq G/H$ , and $[G : H] = \infty$ .

Problem 3. Now suppose instead that $G$ is finitely generated. Show that there exists an integer $r \geq 0$ , primes $p_s$ , and integers $\alpha_s > 0$ such that

$\displaystyle G \cong \mathbb Z^r \oplus \bigoplus_{s=1}^k \mathbb Z/p_s^{\alpha_s} \mathbb Z,$

where $\mathbb Z/n\mathbb Z = \mathbb Z_n$ by definition. This result is one version of the fundamental theorem of finitely generated Abelian groups.

(Click for Solution)

Solution. Let $\{x_1,\dots, x_n\}$ denote a basis for $G$ and $\mathcal I$ denote the collection of indices such that $|\langle x_i \rangle| = \infty$ . For $i \in \mathcal I$ , $\langle x_i \rangle \cong \mathbb Z$ . For $j \in \{1,\dots, n\} \backslash \mathcal I =: \mathcal J$ , there exists $k_j \in \mathbb Z^+$ such that $\langle x_j \rangle \cong \mathbb Z/k_j \mathbb Z$ . Since

$\displaystyle G \cong \bigoplus_{i \in \mathcal I} \langle x_i \rangle \oplus \bigoplus_{j \in \mathcal J} \langle x_j \rangle,$

intuitively, we would like the first piece to be isomorphic to $\mathbb Z^{|\mathcal I|}$ , and the second to be isomorphic to a product of $\mathbb Z/k_j\mathbb Z$ . If $\mathcal J = \emptyset$ , then $G \cong \mathbb Z^n$ trivially. Otherwise, define

$\displaystyle H := \bigoplus_{j \in \mathcal J} \langle x_j \rangle \leq G.$

By Problem 2, $H$ is free of rank $m := |\mathcal J| \leq n$ , and finite. Hence, there exists a basis $\{u_1,\dots, u_m\}$ for $G$ and positive integers $c_1,\dots, c_m$ such that $\{c_1u_1,\dots, c_mu_m\}$ forms a basis for $H$ . In particular,

$\begin{aligned} G \cong \bigoplus_{r=1}^m \langle u_r \rangle \oplus \bigoplus_{r=m+1}^n \langle u_r \rangle,\quad H \cong \bigoplus_{r=1}^m c_r \langle u_r \rangle. \end{aligned}$

Then, the required isomorphism

$\displaystyle \phi : G \to \bigoplus_{r=1}^m \mathbb Z/c_r \mathbb Z \times \mathbb Z^{n-m}.$

is defined by

$\displaystyle \sum_{r=1}^n a_r u_r \mapsto ([a_1]_{c_1},\dots, [a_m]_{c_m}, a_{m+1} , \dots, a_n).$

where $[a]_c := a\ \text{mod}\ c$ . Since $\{c_1u_1,\dots, c_mu_m\}$ forms a basis, $\gcd(c_i,c_j) = 1$ for $i \neq j$ . By the fundamental theorem of arithmetic, there exists unique primes $p_i$ and integers $\alpha_i > 0$ such that

$\displaystyle N := c_1 c_2\cdot \cdots \cdot c_m = p_1^{\alpha_1} p_2^{\alpha_2} \cdot \cdots \cdot p_k^{\alpha_k}.$

By the Chinese remainder theorem,

$\displaystyle \bigoplus_{r=1}^m \mathbb Z/c_r \mathbb Z \cong \mathbb Z/N\mathbb Z \cong \bigoplus_{s=1}^k \mathbb Z/p_s^{\alpha_s} \mathbb Z.$

Therefore, after permuting,

$\displaystyle G \cong \mathbb Z^{n-m} \oplus \bigoplus_{s=1}^k \mathbb Z/p_s^{\alpha_s} \mathbb Z.$

Therefore, set $r := n-m$ , as required.

—Joel Kindiak, 2 May 26, 1432H
July 3, 2026
Sylow Theorems

Let $G$ be a finite group. Recall the class formula:

$\displaystyle |G| = |Z(G)| + \sum_{|\mathcal O_x| > 1} [G : G_x].$

Problem 1. Let $p$ be a prime number such that $p \mid |G|$ . Use induction on $|G|$ to show that there exists a subgroup of $G$ with order $p^\alpha$ , where $p^\alpha \mid |G|$ and $p^{\alpha+1} \nmid |G|$ . We call such a subgroup a $p$ -Sylow subgroup of $G$ .

(Click for Solution)

Solution. The result is vacuously true for $|G| = 1$ . Suppose the result holds for any group $H$ with $|H| < |G|$ . Write

$|G| = p^\alpha \cdot m,\quad \gcd(p, m) = 1.$

In particular, $p^{\alpha+1} \nmid |G|$ . There are two cases: $p \mid |Z(G)|$ or $p \nmid |Z(G)|$ .

If $p \mid |Z(G)|$ , then $Z(G)$ contains an Abelian (and thus, normal) subgroup $H$ with order $p$ by Cauchy’s theorem, since $Z(G)$ is Abelian. Denote $\tilde G := G/H$ so that

$\displaystyle |\tilde G| = \frac{ |G| }{ |H| } = p^{\alpha - 1} \cdot m < |G|.$

By induction, $\tilde G$ contains a subgroup $K$ with order $p^{\alpha-1}$ . Now, the projection map $\pi : G \to \tilde G$ defined by $g \mapsto gH$ is a homomorphism, so that $\pi^{-1}(K) \triangleleft G$ . By definition,

$\pi^{-1}(K)/H = K.$

Therefore,

$\displaystyle \frac{ |\pi^{-1}(K)| }{ |H| }= |\pi^{-1}(K) / H| = |K|.$

Consequently, $|\pi^{-1}(K)| = |H| \cdot |K| = p \cdot p^{\alpha-1} = p^{\alpha}$ , as required.

Suppose $p \nmid |Z(G)|$ . By the class formula, write

$\displaystyle |G| = |Z(G)| + \sum_{i=1}^n [G : G_{x_i}]$

without loss of generality, where $|\mathcal O_{x_i}| > 1$ . If $p \mid [G : G_{x_i}]$ for each $i$ , then $p \mid |Z(G)|$ , a contradiction. Therefore, there exists some $i$ such that $p \nmid [G : G_{x_i}]$ . Note that $G_{x_i} \leq G$ and $[G : G_{x_i}] > 1$ , implies that

$\displaystyle |G_{x_i}| = \frac{ |G| }{ [G : G_{x_i}] } = p^\alpha \cdot \frac{ m }{ [G : G_{x_i}] } < |G|.$

By induction, $H$ contains a subgroup $K$ with order $p^{\alpha}$ . In particular, $K \leq H \leq G$ , as required.

Definition 1. Call a subgroup $H \leq G$ a $p$ -subgroup of $G$ if there exists an integer $n$ such that $|H| = p^n \mid |G|$ .

Problem 2. For any $p$ -Sylow subgroup $P$ , define

$\mathcal S := \{gPg^{-1} : g \in G\} = \{g_1 Pg_1^{-1}, \dots, g_n P g_n^{-1}\}.$

Show that $n \equiv_p 1$ .

(Click for Solution)

Solution. Define the collection

$\mathcal S := \{gPg^{-1} : g \in G\} = \{g_1 Pg_1^{-1}, \dots, g_n P g_n^{-1}\},$

since $G$ is finite. For any subgroup $Q$ , define the conjugation map $Q \times \mathcal S \to \mathcal S$ by

$(v, g_i P g_i^{-1}) \mapsto v (g_i P g_i^{-1}) v^{-1} = (vg_i)P(vg_i)^{-1}.$

Write $\mathcal S$ as a disjoint union of orbits under this action:

$\displaystyle \mathcal S := \bigsqcup_{j=1}^m \mathcal O_j, \quad \sum_{j=1}^m |\mathcal O_j| = n.$

Suppose for simplicity and without loss of generality that $g_j P g_j^{-1} \in \mathcal O_j$ for $1 \leq j \leq m$ . Then $|\mathcal O_j| = [Q : Q_{g_j P g_j^{-1}}]$ , where

$Q_H := \{ v \in Q : vHv^{-1} = H \}.$

By construction,

$Q_{g_j P g_j^{-1}} = G_{g_j P g_j^{-1}} \cap Q = g_j P g_j^{-1} \cap Q.$

Therefore, $|\mathcal O_j| = [Q : g_j P g_j^{-1} \cap Q]$ .

Now, we can show that $o(g_j P g_j^{-1}) = o(P)$ , so that each $g_j P g_j^{-1}$ is also a $p$ -Sylow subgroup. In particular, $o(g_1 P g_1^{-1}) = o(P)$ , so that $g_1 P g_1^{-1}$ is a $p$ -Sylow subgroup, and hence, a $p$ -subgroup. Replacing $Q$ with $g_1 P g_1^{-1}$ ,

$\displaystyle |\mathcal O_j| = [g_1 P g_1^{-1} : g_1 P g_1^{-1} \cap g_j P g_j^{-1}].$

In particular, $|\mathcal O_i| = 1$ , and

$g_1 P g_1^{-1} \neq g_j P g_j^{-1},\quad j \neq 1,$

so that $|\mathcal O_j| > 1$ . Since $P$ is a $p$ -Sylow subgroup, $g_j P g_j^{-1}$ is a $p$ -subgroup, so that there exists $\alpha_i$ such that $|g_j P g_j^{-1}| = p^{\alpha_i}$ . In particular, $p \mid p^{\alpha_j} \mid |\mathcal O_j|$ . Therefore,

$\displaystyle n = |\mathcal O_1| + \sum_{j=2}^m |\mathcal O_j| = 1 + kp.$

Therefore, $n \equiv_p 1$ .

Problem 3. Show that for any $p$ -subgroup $Q$ of $G$ , there exists a $p$ -Sylow subgroup $P$ such that $Q \leq P$ .

(Click for Solution)

Solution. Let $Q$ be any $p$ -subgroup of $G$ . By Theorem 1, there exists a $p$ -Sylow subgroup $P$ . Define $\mathcal S$ as per Problem 2. Suppose $Q \not\subseteq g_i P g_i^{-1}$ for any $i$ . Then $Q \cap g_i P g_i^{-1} \leq Q$ and $Q \cap g_i P g_i^{-1} \neq Q$ , so defining the orbits as in Problem 2, for any $1 \leq j \leq m$ ,

$|\mathcal O_j| = [Q : Q \cap g_j P g_j^{-1}] > 1.$

In particular, $p \mid |\mathcal O_j|$ for all $j$ , so that $n \equiv_p 0$ , a contradiction. Therefore, $Q \leq g_j P g_j^{-1}$ , and since the latter is a $p$ -Sylow subgroup, we are done.

Problem 4. Show that for any two $p$ -Sylow subgroups $H_1, H_2$ of $G$ , there exists $g \in G$ such that $H_2 = gH_1 g^{-1}$ . In this case, we say that $H_1,H_2$ are conjugate.

(Click for Solution)

Solution. Fix $p$ -Sylow subgroups $H_1, H_2$ . Write $P := H_1$ and define

$\mathcal S = \{g_1 P g_1^{-1}, \dots, u_n P u_n^{-1}\}$

as in Problem 2. Since $H_2$ is a $p$ -subgroup, there exists $j$ such that

$H_2 \leq g_j P g_j^{-1}.$

In particular, $p^\alpha = |H_2| \mid |g_j P g_j^{-1}|$ . If $|g_j Pg_j^{-1}| = p^{\alpha+1}$ , then $p^{\alpha + 1} \mid |G|$ , a contradiction. Therefore,

$|g_j Pg_j^{-1}| = p^{\alpha} = |H_2|$

implies that $H_2 = g_j P g_j^{-1} = g_j H_1 g_j^{-1}$ .

—Joel Kindiak, 1 May 26, 1142H

July 2, 2026
Group Actions
Let $K$ be any set and $G$ be a group with binary operation $\cdot$ and identity element $1$ .

Definition 1. A group action on $K$ by $G$ is an induced map $* : G \times K \to K$ such that
- $(g \cdot h) * x = g * (h * x)$ for $g,h \in G$ and $x \in K$ ,
- $1 * x = x$ for $x \in K$ .
In this case, we say that $G$ acts on $K$ via the map $*$ . If $K = G$ , we say that $G$ acts on itself via the map $*$ .

Denote $\mathrm{Sub}(G) := \{H \in \mathcal P(G) : H \leq G\}$ . For any $H \in \mathrm{Sub}(G)$ , recall that $G/H$ consists of all sets of the form $g H$ . If $K \triangleleft G$ , then $G/H$ forms a subgroup.

Problem 1. Show that $G$ acts on $G/H$ via the map $u * vH = (u \cdot v)H$ .

(Click for Solution)

Solution. We verify Definition 1. Fix $u,v,w \in G$ so that $wH \in G/H$ . Then

$\begin{aligned} (u \cdot v) * wH &= ((u \cdot v) \cdot w)H \\ &= (u \cdot (v \cdot w))H \\ &= u * (v \cdot w)H \\ &= u *(v * wH). \end{aligned}$

Similarly, $1 * wH = (1 \cdot w)H = wH$ .

Remark 1. Setting $H = \{1\}$ , $G$ acts on itself via the map $g * h = g \cdot h$ .

Problem 2. Show that $G$ acts on $\mathrm{Sub}(G)$ via the map $g * H = g H g^{-1}$ .

(Click for Solution)

Solution. We verify Definition 1. Fix $u, v \in G$ and $H \in K$ . Then

$\begin{aligned} (u \cdot v) * H &= (u \cdot v) H (u \cdot v)^{-1} \\ &= (u \cdot v) H (v^{-1} \cdot u^{-1}) \\ &= u \cdot (v H v^{-1}) \cdot u^{-1} \\ &= u \cdot (v * H) \cdot u^{-1} \\ &= u * (v * H). \end{aligned}$

Similarly, $1 * H = 1 H 1^{-1} = H$ .

Remark 2. Setting $H = \{h\}$ and applying similar reasoning, $G$ acts on itself via the map $g * h = g \cdot h \cdot g^{-1}$ .

Problem 3. Given $x , y \in K$ , denote $x \sim y$ to mean that there exists $g \in G$ such that $y = g * x$ . Show that $\sim$ forms an equivalence relation on $K$ .

(Click for Solution)

Solution. We check the three properties of an equivalence relation.

For reflexivity, for any $x \in K$ , $x = 1 * x$ by Definition 1, so that $x \sim x$ , as required.

For symmetry, suppose $x \sim y$ . Then there exists $g \in G$ such that $y = g*x$ . By Definition 1,

$\begin{aligned} x = 1 * x = (g^{-1} \cdot g) * x = g^{-1} * (g * x) = g^{-1} * y \end{aligned}$

implies that $y \sim x$ , as required.

For transitivity, suppose $x \sim y$ and $y \sim z$ . Then there exist $g, h \in G$ such that

$y = g * x,\quad z = h * y.$

By composition and Definition 1,

$z = h * y = h * (g * x) = (h \cdot g) * x$

implies that $x \sim z$ , as required.

Using Problem 3, for each $x \in K$ , define the orbit $\mathcal O_x := [x] \in K/{\sim}$ . It is not hard to check that

$\mathcal O_x = \{g * x : g \in G\} .$

Then $\sim$ partitions $K$ into

$\displaystyle K = \bigsqcup_{\mathcal O_x \in K/{\sim}} \mathcal O_x.$

Problem 4. For any $x \in K$ , show that $G_x \leq G$ , where

$G_x := \{g \in G : g * x = x\}.$

Furthermore, show that for any $g \in G$ ,

$g \cdot G_x \cdot g^{-1} = G_{g * x}.$

Deduce that $|\mathcal O_x| = |G/G_x|$ , i.e., the orbit-stabilizer formula.

(Click for Solution)

Solution. We verify the group axioms for $G_x$ .

For closure, given $g,h \in G_x$ ,

$(g \cdot h)*x = g * (h*x) = g*x = x$

implies that $g \cdot h \in G_x$ . The other three properties follow similarly.

For the conjugate-like claim, fix $g \in G$ . Expand both sets:

$\begin{aligned} g \cdot G_x \cdot g^{-1} &= \{g \cdot h \cdot g^{-1} : h \in G_x\}, \\ G_{g*x} &= \{k \in G : k * (g * x) = g * x\}. \end{aligned}$

We observe that for any $h \in G_x$ ,

$\begin{aligned} (g\cdot h\cdot g^{-1}) * (g*x) &= g *(h * (g^{-1} * (g*x))) \\ &= g *(h * ((g^{-1} \cdot g) *x )) \\ &= g *(h * (1 *x )) \\ &= g *(h * x) \\ &= g * x. \end{aligned}$

We leave it as an exercise to therefore check that equality holds.

Finally, define the map $f : G \to K$ by $f(g) = g * x$ . Then $f(g) = f(h)$ if and only if $h \in g G_x$ , inducing a bijection $\phi : G/G_x \to f(G)$ . By definition,

$f(G) = \{ f(g) : g \in G \} = \{ g * x : g \in G \} = \mathcal O_x.$

Therefore, $|\mathcal O_x| = |G/G_x|$ .

Remark 3. In particular, if $K$ is finite, then

$\displaystyle |K| = \sum_{\mathcal O_x \in K/{\sim}} |G/G_x|.$

Now suppose that $G$ is finite and Abelian. Denote its identity element by $1_G$ .

Problem 5. Let $p$ be a prime number such that $p \mid |G|$ . Show that $G$ contains an element of order $p$ . The subgroup it generates, clearly, also has order $p$ . This result is known as Cauchy’s theorem.

(Click for Solution)

Solution. For each $g \in G$ , denote its order by $o(g)$ . If there exists an element $g$ such that $p \mid o(g)$ , then defining $n := o(g)/p$ , the element $g^n$ is the desired element. Suppose no such element exists. Define

$n_G := \mathrm{lcm}\{o(g) : g \in G\}.$

In particular, $p \nmid n_G$ . We prove by induction on $|G|$ that $|G| \mid n_G^k$ for some integer $k$ . The base case $|G|$ is obvious. For the inductive step, fix $g \in G \backslash \{1_G\}$ . By definition, $\langle g \rangle$ implies that $|\langle g \rangle| \mid n_G$ . By assumption, $p \nmid |\langle g \rangle|$ and $p \mid |G|$ , which implies that $|\langle g \rangle| \neq |G|$ . In particular, for any $x \in G$ , $(x \langle g \rangle)^{n_G} = 1_{G/\langle g \rangle}$ , so that $n_{G/\langle g \rangle} \mid n_G$ .

Since $|G/\langle g\rangle | < |G|$ , by the inductive hypothesis,

$|G| / |\langle g\rangle | = |G/\langle g\rangle | \mid n_{G/\langle g \rangle}^\ell$

for some integer $\ell$ . Therefore,

$|G| \mid |\langle g \rangle| \cdot n_{G/\langle g \rangle}^\ell \mid n_G \cdot n_G^\ell = n_G^{\ell+1}.$

Set $k := \ell + 1$ , as required. In particular, $p \mid |G| \mid n_G^k$ . Since $p$ is prime, $p \mid n_G$ , a contradiction.

Define the centre of $G$ by $Z(G) := \{x \in G : g \cdot x = x \cdot g, g \in G\}$ .

Problem 6. Let $*$ denote conjugation as per Remark 2, so that

$g * x = g \cdot x \cdot g^{-1},\quad g,x \in G.$

Denote $[G : G_x] = |G/G_x|$ . Prove the class formula:

$\displaystyle |G| = |Z(G)| + \sum_{|\mathcal O_x| > 1} [G : G_x].$

(Click for Solution)

Solution. By Remark 3, it suffices to check that

$\displaystyle |Z(G)| = \sum_{ |\mathcal O_x| = 1} [G : G_x].$

Now $| \mathcal O_x | = 1$ implies that $\mathcal O_x = \{x\}$ , so that $G_x = G$ , and $[G : G_x] = 1$ . For any $g \in G$ ,

$g \cdot x \cdot g^{-1} = g * x = x.$

Equivalently, $g \cdot x = x \cdot g$ . Therefore, $|\mathcal O_x| = 1 \iff x \in Z(G)$ . Therefore,

$\displaystyle \sum_{ |\mathcal O_x| = 1} [G : G_x] = \sum_{ x \in Z(G)} 1 = |Z(G)|,$

as required.

—Joel Kindiak, 30 Apr 26, 1627H
July 1, 2026
Quintic Insolvability

Remark 1. I am aware that “insolvability” isn’t technically an English word. We’re just gonna use it—it has a ring to it.

Previously, we introduced the idea of the solvability of a group. It goes a bit like this: call a group solvable if it contains a sequence $H_0, H_1,\dots, H_n$ of subgroups such that

$1 = H_0 < H_1 < \cdots < H_n = G,$

and for each $i$ , $H_i \triangleleft H_{i+1}$ , $H_i \neq H_{i+1}$ , and $H_{i+1}/H_i$ is Abelian. This latter condition is responsible for classifying how “symmetric” a polynomial is, and therefore, its likelihood (or lack thereof) to have a formula involving radicals. But speaking of radicals, let’s properly formulate that idea.

Definition 1. Call a finite extension $\mathbb F(\alpha_1,\dots,\alpha_n) \supseteq \mathbb F$ radical if for each $i$ , there exists a positive integer $\beta_i$ such that $\alpha_i^{\beta_i} \in \mathbb F$ .

Intuitively, each $\alpha_i$ is the $\beta_i$ -th root of some element in $\mathbb F$ : given $\alpha_i^{\beta_i} = \gamma_i$ , we have $\alpha_i = \gamma_i^{1/\beta_i} \equiv \sqrt[\beta_i]{\gamma_i}$ . Of course, this expression requires some work in various fields to be well-defined, but it at least informs why solving degree- $n$ polynomials requires the use of the square root function $\sqrt[n]{\cdot}$ , at least for $n = 2,3,4$ .

Now let $G$ be any group. We shall first establish some basic ideas of solvable groups before connecting them to radical extensions.

Lemma 1. For any $g, h \in G$ , define $[g,h] := ghg^{-1}h^{-1}$ . Then $gh = hg$ if and only if $[g,h] = 1$ .

Proof. The right-hand side simplifies to $[g,h] = (gh)(hg)^{-1}$ .

Definition 2. Given subgroups $H, K \leq G$ , define

$[H,K] := \{[h,k]: h \in H, k \in K\}$

and the commutator subgroup of $G$ by $G' := [G, G]$ .

Example 1. $G' = \{1\}$ if and only if $G$ is Abelian.

Example 2. $\mathcal S_4' = \mathcal A_4$ .

Lemma 2. $G' \triangleleft G$ and $G/G'$ is Abelian. For any normal subgroup $N \triangleleft G$ , $G/N$ is Abelian if and only if $N$ contains $G'$ .

Proof. The subgroup $G' = [ G, G ]$ is generated by elements of the form $[g, h]$ . Furthermore, the subset $uGu^{-1}$ is generated by elements of the form $u[g,h]u^{-1}$ . A tricky computation yields

$\begin{aligned} u[g,h]u^{-1} &= [ugu^{-1}, uhu^{-1}] \in G'. \end{aligned}$

In particular, $uG'u^{-1} \subseteq G'$ , so that $G' \triangleleft G$ . Furthermore, it is not hard to check that $(gh)G' = (hg)G'$ , so that $G/G'$ is Abelian.

Now fix $N \triangleleft G$ . If $G/N$ is Abelian, then $(gh)N =(hg)N$ implies that $[g,h] \in N$ , so that $G' \leq N$ . On the other hand, if $G' \leq N$ , then by the third isomorphism theorem,

$G/N \cong (G/G')/(N/G')$

must be Abelian.

Lemma 3. Inductively define $G^{(0)} = G$ and $G^{(i+1)} := (G^{(i)})'$ . Then $G$ is solvable if and only if there exists some integer $n$ such that $G^{(n)} = 1$ .

Proof. $(\Rightarrow)$ Suppose there exists some integer $n$ such that $G^{(n)} = 1$ . By Lemma 2,

$1 = G^{(n)} \triangleleft G^{(n-1)} < \cdots < G^{(1)} = G,$

forms a subnormal series, so that $G$ is solvable.

$(\Leftarrow)$ Suppose $G$ is solvable. Then there exists a subnormal series

$1 = H_0 < H_1 < \cdots < H_n = G.$

We show the existence of $m \in \mathbb N$ such that $G^{(m)} = 1$ by induction on $n$ . If $n = 1$ , then set $m = 1$ . Suppose the result holds for general $k$ , and that $n = k+1$ :

$1 = H_0 < H_1 < \cdots < H_k < H_{k+1} = G.$

Since $G/H_k = H_{k+1}/H_k$ is Abelian, by Lemma 2, $G^{(1)} = G' \leq H_k.$ Define $H_i' := H_i \cap G^{(1)}$ . Then using the second isomorphism theorem, we can check that

$1 = H_0' < H_1' < \cdots < H_k' = G^{(1)}$

forms a subnormal series for $G^{(1)}$ , so that there exists an integer $m$ such that $G^{(m+1)} = (G^{(1)})^{(m)} = 1$ .

Lemma 4. Let $H_0,H_1,\dots, H_n$ denote a subnormal series of $G$ . If each $H_{i+1}/H_i$ is solvable, then so is $G$ . In particular, if $H \triangleleft G$ is solvable and $G/H$ is solvable, so is $G$ .

Proof Sketch. Chain the individual subnormal series together.

Theorem 1. Consider the field extensions $\mathbb Q \subseteq \mathbb L \subseteq \mathbb K \subseteq \bar{\mathbb Q}$ . If $\mathbb K \supseteq \mathbb Q$ is Galois, $\mathbb L \supseteq \mathbb Q$ is normal, and $\mathbb K \supseteq \mathbb Q$ is radical, then $\mathrm{Gal}(\mathbb L/\mathbb Q)$ is solvable.

Proof. Under the first two hypotheses, $\mathbb L \supseteq \mathbb Q$ is Galois, so that

$\displaystyle \mathrm{Gal}(\mathbb L / \mathbb Q) \cong \frac{\mathrm{Gal}(\mathbb K/\mathbb Q)}{\mathrm{Gal}(\mathbb K/\mathbb L)}.$

By Lemma 4, it suffices to show that $\mathrm{Gal}(\mathbb K/\mathbb Q)$ is solvable. Since $\mathbb K \supseteq \mathbb Q$ is radical, there exist $\alpha_i \in \mathbb K$ and $\beta_i \in \mathbb N^+$ such that $\alpha_i^{\beta_i} \in \mathbb Q(\alpha_1,\dots,\alpha_{i-1})$ and

$\mathbb K = \mathbb Q(\alpha_1,\dots,\alpha_n).$

Define $\beta := \mathrm{lcm}(\beta_1,\dots,\beta_n)$ and let $\zeta_\beta$ denote a primitive $\beta$ -th root of unity. Then

$\displaystyle \mathrm{Gal}(\mathbb K / \mathbb Q) \cong \frac{\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb Q)}{\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb K)}.$

Likewise, it suffices to check that $\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb Q)$ is solvable. Finally,

$\displaystyle \frac{\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb Q)}{\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb Q(\zeta_\beta))} \cong \mathrm{Gal}(\mathbb Q(\zeta_\beta)/ \mathbb Q) \hookrightarrow \mathbb Z_\beta^*$

is cyclic, and thus Abelian, and thus, solvable.

It remains to check that $\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb F(\zeta_\beta))$ is solvable. Define

$\mathbb L_i := \mathbb Q(\alpha_1,\dots, \alpha_i, \zeta_\beta),$

so that $\mathbb L_0 = \mathbb Q(\zeta_\beta)$ and $\mathbb L_n = \mathbb K(\zeta_\beta)$ . Consider the series

$\mathbb Q(\zeta_\beta) = \mathbb L_0 < \mathbb L_1 < \cdots < \mathbb L_n = \mathbb K(\zeta_\beta).$

For each $i$ , $\mathbb L_i$ is a splitting field of the polynomial $x^{\beta_i} - \alpha_i^{\beta_i} \in \mathbb L_{i-1}[x]$ , and thus $\mathbb L_i \supseteq \mathbb L_{i-1}$ is Galois with $\mathrm{Gal}(\mathbb L_n/\mathbb L_i) \triangleleft \mathrm{Gal}(\mathbb L_n/\mathbb L_{i-1})$ . Therefore, the series

$1 = \mathrm{Gal}(\mathbb L_n/\mathbb L_n) < \cdots < \mathrm{Gal}(\mathbb L_n/ \mathbb L_1) < \mathrm{Gal}(\mathbb L_n/\mathbb L_0)$

is subnormal. Since

$\displaystyle \frac{\mathrm{Gal}(\mathbb L_j/\mathbb L_{i-1})}{\mathrm{Gal}(\mathbb L_j/\mathbb L_{i})} \cong \mathrm{Gal}(\mathbb L_i/\mathbb L_{i-1}) \cong \mathrm{Gal}(\mathbb L_{i-1}(\zeta_\beta)/\mathbb L_{i-1})$

is Abelian, $\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb Q(\zeta_\beta)) = \mathrm{Gal}(\mathbb L_n / \mathbb L_0)$ is solvable, as desired.

Now we have properly returned to radicals, and will define solvability by radicals in a relatively intuitive manner (relative to the content of Galois theory, that is).

Definition 3. We say that a polynomial $f \in \mathbb F[x]$ is solvable by radicals over $\mathbb F$ if there exists a radical extension $\mathbb K \supseteq \mathbb F$ such that $\mathbb F(\mathcal Z_f) \subseteq \mathbb K$ .

Lemma 5. If $\mathbb K \supseteq \mathbb Q$ is radical and $\mathbb L \supseteq \mathbb Q$ is the smallest normal extension contained in $\mathbb K$ , then $\mathbb L \supseteq \mathbb Q$ is radical.

Proof. Suppose $\bar{\mathbb F} \supseteq \mathbb K$ , so that

$\mathbb Q \subseteq \mathbb L \subseteq \mathbb K \subseteq \bar{\mathbb Q}.$

Since $\mathbb K \supseteq \mathbb Q$ is radical, write

$\mathbb K = \mathbb Q(\alpha_1,\dots,\alpha_n),\quad \alpha_i^{\beta_i} \in \mathbb Q(\alpha_1,\dots,\alpha_{i-1}).$

Let $\{\sigma_1,\dots,\sigma_m\}$ denote the set of $\mathbb Q$ -homomorphisms from $\mathbb K$ to $\bar{\mathbb Q}$ . We leave it as an exercise to check that

$\mathbb L = \sigma_1(\mathbb K) \cdot \cdots \cdot \sigma_m(\mathbb K),$

and that for each $j$ ,

$\sigma_j(\mathbb K) = \mathbb Q(\sigma_j(\alpha_1),\dots,\sigma_j(\alpha_n)),$

so that $\sigma_i(\mathbb K) \supseteq \mathbb Q$ is radical, so that $\mathbb L \supseteq \mathbb Q$ is radical too.

Finally, we can establish the insolvability of the quintic!

Theorem 2 (Abel-Ruffini Theorem). For any $n \geq 5$ , there exists a polynomial $f \in \mathbb Q[x]$ whose solution does not belong to any radical extension of $\mathbb Q$ .

Proof. Consider the irreducible (and thus, separable) polynomial

$f(x) = x^n - x - 1 \in \mathbb Q[x].$

Suppose for a contradiction that $f$ is solvable. By Definition 3, there exists a radical extension $\mathbb K \supseteq \mathbb Q$ such that

$\mathbb K \supseteq \mathbb Q (\mathcal Z_f) \supseteq \mathbb Q.$

By Lemma 5, $\mathbb Q (\mathcal Z_f) \supseteq \mathbb Q$ is radical. Since the extension $\mathbb Q (\mathcal Z_f) \supseteq \mathbb Q$ is normal and separable, it is Galois. By Theorem 1, $\mathrm{Gal}(\mathbb Q (\mathcal Z_f)/\mathbb Q)$ must be solvable. However, since $n \geq 5$ ,

$\mathrm{Gal}(\mathbb Q (\mathcal Z_f)/\mathbb Q) \cong \mathcal S_n,$

which is not solvable, a contradiction. Therefore, $f$ is not solvable by radicals.

In particular, there is no quintic formula; otherwise, $f$ would be solvable by radicals, a contradiction! Does this result mean that all quintics are not solvable by radicals? That would be too extreme of a claim. But more can be discussed about these queries and more.

But for now, we are done with (very introductory) Galois theory!

—Joel Kindiak, 25 Apr 26, 1822H

June 30, 2026
Galois Group Theory

Previously, we have proven the fundamental theorem of Galois theory, which establishes a direct correspondence between field extensions and subgroups of automorphisms. We say that a finite extension $\mathbb K \supseteq \mathbb F$ is Galois if there exists some (unique) subgroup $G \leq \mathrm{Aut}(\mathbb K)$ such that $\mathbb K^G = \mathbb F$ . In this case, we write

$G = \mathrm{Gal}(\mathbb K/\mathbb F) \leq \mathrm{Aut}(\mathbb K).$

Subsequently, for any intermediate field $\mathbb F \subseteq \mathbb L \subseteq \mathbb K$ , $\mathbb K \supseteq \mathbb L$ is Galois, and the intermediate field $\mathbb L$ is in direct correspondence with the subgroup

$\mathrm{Gal}(\mathbb K/\mathbb L) \leq \mathrm{Gal}(\mathbb K/\mathbb F).$

Conversely, the subgroup $H \leq \mathrm{Gal}(\mathbb K/\mathbb F)$ corresponds to the intermediate field $\mathbb K^H$ . Finally, the extension $\mathbb L \supseteq \mathbb F$ is Galois if and only if

$\mathrm{Gal}(\mathbb L/\mathbb F) \triangleleft \mathrm{Gal}(\mathbb K/\mathbb F).$

In this case,

$\mathrm{Gal}(\mathbb K/\mathbb F)/\mathrm{Gal}(\mathbb K/\mathbb L) \cong \mathrm{Gal}(\mathbb L/\mathbb F).$

Now let $\mathbb K \supseteq \mathbb F$ denote a finite extension.

Lemma 1. Given $f \in \mathbb F[x]$ , suppose $f$ can be written as a product of linear factors in $\mathbb F(\mathcal Z_f)[x]$ , then $\mathbb F(\mathcal Z_f) \supseteq \mathbb F$ is Galois with Galois group $\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))$ .

Proof. We remark that $\mathbb F(\mathcal Z_f) \supseteq \mathbb F$ is always normal. The linear factor criterion implies that $\mathbb F(\mathcal Z_f) \supseteq \mathbb F$ is separable. Since the extension $\mathbb F(\mathcal Z_f) \supseteq \mathbb F$ is finite, we have that $\mathbb F(\mathcal Z_f) \supseteq \mathbb F$ is Galois where, by definition, $\mathrm{Gal}(\mathbb F(\mathcal Z_f)/\mathbb F) = \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))$ .

Remark 1. Recall that

$\mathcal Z_f := \{ \alpha \in \bar{\mathbb F} : f(\alpha) = 0\},$

where $\bar{\mathbb F}$ denotes the algebraic closure. We say that a polynomial $f \in \mathbb F[x]$ is separable if it can be written as a product of linear factors in $\mathbb F(\mathcal Z_f)[x]$ . In particular, $f$ is separable if and only if $\mathbb F(\mathcal Z_f) \supseteq \mathbb F$ is separable.

Theorem 1. For any $f \in \mathbb F[x]$ , define the Galois group of $f$ over $\mathbb F$ by $\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))$ . Then there exists some $n \leq \deg(f)$ such that $\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))$ is isomorphic to some subgroup of $\mathcal S_n$ (i.e. the group of bijections on $\{1,\dots, n\}$ ).

Proof. Suppose $\mathcal Z_f = \{\alpha_1,\dots,\alpha_n\}$ where $n \leq \deg(f)$ . By definition, for any $\sigma \in \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))$ , $\sigma(\mathcal Z_f) = \mathcal Z_f$ so that $\sigma|_{\mathcal Z_f} \in \mathrm{Sym}(\mathcal Z_f)$ . In particular, for any $i$ , there exists a unique $j_i$ such that $\sigma(\alpha_i) = \alpha_{j_i}$ . For each $\sigma$ , define $\phi_\sigma \in \mathcal S_n$ by $\phi_{\sigma}(i) = j_i$ . Then the map $\phi : \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f)) \to \mathcal S_n$ defined by $\phi(\sigma) = \phi_\sigma$ forms the desired group monomorphism, so that $\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f)) \cong \phi(\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))) \leq \mathcal S_n$ .

Corollary 1. If $f$ is separable and irreducible with degree $n$ , then $n \mid |\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))|$ .

Proof. If $f$ is separable, then $\mathbb F(\mathcal Z_f) \supseteq \mathbb F$ is separable too, and thus Galois with

$\mathbb F(\mathcal Z_f)^{\mathrm{Gal}(\mathbb F(\mathcal Z_f) / \mathbb F)} = \mathbb F,\quad \mathrm{Gal}(\mathbb F(\mathcal Z_f) / \mathbb F) = \mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f)).$

If $f$ is irreducible, $[\mathbb F(\alpha_1) : \mathbb F] = n$ . By the tower rule,

$|\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))| = [\mathbb F(\mathcal Z_f) : \mathbb F(\alpha_1)] \cdot [\mathbb F(\alpha_1) : \mathbb F] = [\mathbb F(\mathcal Z_f) : \mathbb F(\alpha_1)] \cdot n,$

so that $n \mid |\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))|$ .

Remark 1. By Theorem 1, if $f$ is irreducible and separable, then $\mathrm{Gal}(\mathbb F(\mathcal Z_f) / \mathbb F)$ is isomorphic to some subgroiup of $\mathcal S_n$ , where $n = \deg(f)$ .

Example 1. If $f \in \mathbb F[x]$ is irreducible with degree $2$ , then $\mathrm{Aut}_{\mathbb F}(\mathbb F(\mathcal Z_f))$ is isomorphic to either $\mathbb Z_2$ or the trivial group.

Lemma 1. If $G \leq \mathcal S_n$ contains an $n$ -cycle and a transposition, then $G = \mathcal S_n$ .

Proof. Suppose $G$ contains $(1\ 2)$ and $(1\ 2\ \dots\ n)$ without loss of generality. Recall that

$\mathcal S_n = \langle (1\ 2), (1\ 2\ \dots\ n) \rangle.$

Hence,

$\mathcal S_n \leq \langle (1\ 2), (1\ 2\ \dots\ n) \rangle \leq G \leq \mathcal S_n.$

Recall that in a finite group $G$ and a cycle $\sigma$ , we define $\sigma^n$ to be the $n$ -fold composition of $\sigma$ , and $|\sigma| := \min\{n \in \mathbb N : \sigma^n = 1\}$ .

Theorem 2. Let $p$ be prime and $f \in \mathbb Q[x]$ be irreducible with degree $p$ . If $f$ has exactly two non-real roots in $\mathbb C$ , then $\mathrm{Aut}_{\mathbb Q}(\mathbb Q(\mathcal Z_f)) \cong \mathcal S_p$ .

Proof. Denote $\mathbb L := \mathbb Q(\mathcal Z_f)$ and $G := \mathrm{Aut}_{\mathbb Q}(\mathbb L)$ for brevity. We remark that $\mathbb L \supseteq \mathbb F$ is Galois if and only if $\mathbb L^G = \mathbb Q$ . By the fundamental theorem of algebra, $\mathcal Z_f \subseteq \mathbb C$ . By hypothesis,

$\mathcal Z_f \backslash \mathbb R = \{\alpha_1,\alpha_2\},\quad \mathcal Z_f \cap \mathbb R = \{\alpha_3,\dots,\alpha_p\}.$

By Theorem 1, we can regard $G \leq \mathcal S_p$ by passing to an isomorphism. Since $f$ is irreducible, by Corollary 1, $p \mid |G|$ . Therefore, $G$ contains a permutation $\sigma$ with $|\sigma| = p$ . Write $\sigma = \sigma_r \circ \cdots \circ \sigma_1$ for disjoint cycles $\sigma_i$ . For $r \geq 2$ , $|\sigma_i| < p$ implies that

$|\sigma| = \mathrm{lcm}( |\sigma_1|,\dots, |\sigma_r|) \nmid p,$

a contradiction. Therefore, we must have $\sigma = \sigma_1$ , that is, $\sigma$ is a $p$ -cycle.

On the other hand, since $\alpha_1,\alpha_2$ are non-real, by the conjugate root theorem, $\alpha_2 = \alpha_1^*$ , where $(\cdot)^*$ denotes complex conjugation. Therefore, $G$ contains the transposition $\tau := \phi_{12}$ . By Lemma 1, $G \cong \mathcal S_p$ , as required.

In particular, consider a general quintic polynomial $f$ with degree $n \geq 5$ . If $f$ is reducible, then we can write it in the form $f = f_k \cdot f_m$ , where $\deg(f_k) = k \geq 1$ and $k+m = n$ . Then $f$ has solutions in radicals if and only if $f_k$ and $f_m$ do. This is certainly the case for $n = 5$ .

However, the more interesting case arises if $f$ is irreducible. Then Theorem 2 tells us that $\mathrm{Aut}_{\mathbb Q}(\mathbb Q(\mathcal Z_f)) \cong \mathcal S_5$ . Surprisingly, the closely related alternating group $\mathcal A_5$ that consists of the even permutations of $\mathcal S_5$ is a simple group. This simplicity turns out to be the key to the non-solvability of the quintic.

Theorem 3. Call a Galois extension $\mathbb K \supseteq \mathbb F$ Abelian over $\mathbb F$ if $\mathrm{Gal}(\mathbb K /\mathbb F)$ is Abelian. Given $\mathbb K \supseteq \mathbb L \supseteq \mathbb F$ , if $\mathbb K \supseteq \mathbb F$ is Abelian, then $\mathbb K \supseteq \mathbb L$ and $\mathbb L \supseteq \mathbb F$ are Abelian Galois extensions.

Proof. By hypothesis, $\mathrm{Gal}(\mathbb K / \mathbb F)$ is Abelian. Therefore, $\mathrm{Aut}_{\mathbb F}(\mathbb L) \leq \mathrm{Gal}(\mathbb K / \mathbb F)$ is Abelian, and thus normal. Therefore, $\mathbb L \supseteq \mathbb F$ is Galois with

$\mathrm{Gal}(\mathbb L/\mathbb F) = \mathrm{Aut}_{\mathbb F}(\mathbb L)$

being Abelian. Furthermore, since $\mathbb K \supseteq \mathbb L$ is Galois,

$\displaystyle \mathrm{Gal}(\mathbb L/\mathbb F) \cong \frac{\mathrm{Gal}(\mathbb K/\mathbb F) }{\mathrm{Gal}(\mathbb K/\mathbb L) }$

implies that $\mathrm{Gal}(\mathbb K/\mathbb L)$ must be Abelian, so that $\mathbb K \supseteq \mathbb L$ is Abelian too.

Remark 1. Replacing ‘Abelian’ with ‘cyclic’ in Theorem 3 mutatis mutandis still renders the statement true.

It seems like there is a close connection between Galois extensions and sequences of normal subgroups. We formalise this intuition by defining the subnormal series.

Definition 1. A subnormal series of a group $G$ is a finite sequence $H_0, H_1,\dots, H_n$ of normal subgroups of $G$ such that

$\{1\} = H_0 < H_1 < \cdots < H_n = G.$

Here, we denote the group identity by $1$ and denote $H_s < H_t$ to mean that $H_s \triangleleft H_t$ and $H_s \neq H_t$ . We call this series normal if $H_i < G$ for all $i$ .

Example 1. The series $\{1\} < \mathcal A_3 < \mathcal S_3$ is normal.

Example 2. The series

$\{1\} < \{(1), (1\ 2)\} < \{ (1), (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3) \} < \mathcal A_4 < \mathcal S_4$

is subnormal but not normal.

Definition 2. We say that a group $G$ is solvable if it has a subnormal series $\{H_i : i \in I\}$ such that all the factors $H_{i+1}/H_i$ are Abelian.

Example 3. Any Abelian group is solvable.

Theorem 4. For $n \geq 5$ , $\mathcal A_n$ and $\mathcal S_n$ are not solvable.

Proof. Recalling that $\mathcal A_n$ is simple, the only subnormal series of $ltaex \mathcal S_n$ is

$\{1\} < \mathcal A_n < \mathcal S_n.$

Furthermore, $\mathcal A_n/\{1\} \cong \mathcal A_n$ is not Abelian. Even more so, the factor $\mathcal S_n/\mathcal A_n$ is congruent to the subgroup of transpositions, which is clearly not Abelian. Therefore, $\mathcal S_n$ cannot be solvable.

The naming of “solvable” should already strongly hint at its connection to the non-existence of the quintic formula. But we will only properly discuss this idea the next time, through connecting it to radical extensions.

—Joel Kindiak, 25 Apr 26, 1340H

June 29, 2026
Galois Par-Excellence

The goal of this post is to prove the fundamental theorem of Galois theory.

Let $\mathbb K \supseteq \mathbb F$ denote a finite extension. Given any finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ , define the following sets:

$\begin{aligned} \mathbb F_{\text{ext}} &:= \{ \mathbb L \subseteq \mathbb K : \mathbb L \supseteq \mathbb F\ \text{is a field extension}\}, \\ G_{\text{sub}} &:= \{H \subseteq G : H \leq G \}. \end{aligned}$

Elements of $\mathbb F_{\mathrm{ext}}$ are called intermediate extensions.

Theorem 1. Define the maps $\Phi : \mathbb F_{\text{ext}} \to G_{\text{sub}}$ and $\Psi : G_{\text{sub}} \to \mathbb F_{\text{ext}}$ by

$\Phi(\mathbb L) := \mathrm{Aut}_{\mathbb L}(\mathbb K),\quad \Psi(H) := \mathbb K^H.$

Then $\Phi, \Psi$ are inverses of each other, so that

$\mathbb K^{\mathrm{Aut}_{\mathbb L}(\mathbb K)} = \mathbb L, \quad \mathrm{Aut}_{\mathbb K^H}(\mathbb K) = H.$

Before proving this theorem, we are going to need some intermediate results.

Lemma 1. Any finite subgroup $\{\sigma_1,\dots, \sigma_n\} \leq \mathrm{Aut}(\mathbb K)$ is linearly independent over $\mathbb F$ .

Proof. We proceed by induction. The set $\{\sigma_1\}$ is trivially linearly independent over $\mathbb F$ . Now consider the subgroup

$\{\sigma_1,\dots,\sigma_k,\sigma_{k+1}\}$

and the equation $\sum_{i=1}^{k+1} c_i \sigma_i = 0$ . Since the subgroup is not a singleton, $\sigma_1 \neq \sigma_2$ trivially. Hence, there exists $u \in \mathbb K$ such that $\sigma_1(u) \neq \sigma_2(u)$ . Defining $\tau_i := \sigma_i(u) \sigma_i$ and $\omega_i := \sigma_2(u) \sigma_i$ , we have

$\displaystyle \sum_{i=1}^{k+1} c_i \tau_i = 0, \quad \sum_{i=1}^{k+1} c_i \omega_i = 0.$

Since $\tau_2 = \omega_2$ , by denoting $K_j := \{1,\dots,k+1\}\backslash \{j\}$ so that $|K_j| = k < k+1$ ,

$\begin{aligned} \sum_{i \in K_2} c_i (\sigma_i(u)-\sigma_2(u))\sigma_i &= \sum_{i \in K_2} c_i (\tau_i-\omega_i) \\ &= \sum_{i=1}^{k+1} c_i (\tau_i-\omega_i) \\ &= \sum_{i=1}^{k+1} c_i \tau_i - \sum_{i=1}^{k+1} c_i \omega_i = 0. \end{aligned}$

By the induction hypothesis, $c_i(\sigma_i(u)-\sigma_2(u)) = 0$ for each $i$ . In particular, for $i = 1$ , since $\sigma_1(u) \neq \sigma_2(u)$ ,

$\displaystyle c_1(\sigma_1(u)-\sigma_2(u)) = 0 \quad \Rightarrow \quad c_1 = 0.$

Therefore, $\sum_{i=2}^{k+1} c_i \sigma_i = 0$ . By the induction hypothesis, $c_i = 0$ for $i = 1,\dots,k+1$ , as required.

Lemma 2. For any finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ ,

$[\mathbb K : \mathbb K^G] = |G|,\quad \mathrm{Aut}_{\mathbb K^G}(\mathbb K) = G.$

Proof. Define the collection

$\mathcal G := \{ \sigma : \mathbb K \to \overline{\mathbb K^G} : \sigma\text{ is a }\mathbb K^G\text{-homomorphism} \}.$

It is not hard to check that

$G \subseteq \mathrm{Aut}_{\mathbb K^G}(\mathbb K) \subseteq \mathcal G \cap \mathrm{Aut}(\mathbb K) \subseteq \mathcal G.$

Therefore,

$|G| \leq |\mathcal G| = \{ \mathbb K : \mathbb K^G \} \leq [\mathbb K : \mathbb K^G].$

On the other hand, we claim that $|G| = [\mathbb K : \mathbb K^G]$ . Suppose for a contradiction that $|G| < [\mathbb K : \mathbb K^G]$ . Write $G = \{\sigma_1,\dots, \sigma_n\}$ , which is linearly independent over $\mathbb F$ by Lemma 1. Therefore, $\sum_{\sigma \in G} \sigma \neq 0$ , which means that there exists $u \in \mathbb K$ such that $\sum_{\sigma \in G} \sigma(u) \neq 0$ .

Furthermore, since $[\mathbb K : \mathbb K^G] \geq n + 1$ , there exists a subset $\{v_1,\dots, v_{n+1} \} \subseteq \mathbb K$ that is linearly independent over $\mathbb K^G$ . For each $j=1,\dots,n+1$ , define

$\mathbf w_j := \begin{bmatrix} \sigma_1^{-1}(v_j) \\ \vdots \\ \sigma_n^{-1}(v_j) \end{bmatrix} \in \mathbb K^n.$

The set $\{\mathbf w_1,\dots, \mathbf w_n, \mathbf w_{n+1}\} \subseteq \mathbb K^n$ must be linearly dependent over $\mathbb K$ , since it contains $n+ 1 > n$ elements. Thus, there exist scalars $c_j \in \mathbb K$ with $c_1 \neq 0$ without loss of generality such that

$\displaystyle \sum_{j=1}^{n+1} c_j\begin{bmatrix} \sigma_1^{-1}(v_j) \\ \vdots \\ \sigma_n^{-1}(v_j) \end{bmatrix} = \mathbf 0.$

By multiplying by $u/c_1 \in \mathbb K$ , we can assume $c_1 = u \neq 0$ without loss of generality. For each $i$ ,

$\displaystyle 0 = \sum_{j=1}^{n+1} c_j \sigma_i^{-1}(v_j) = \sigma_i^{-1} \left( \sum_{j=1}^{n+1} \sigma_i (c_j) v_j \right),$

so that $\sum_{j=1}^{n+1} \sigma_i(c_j) v_j = 0$ . In particular,

$\displaystyle \begin{aligned} \sum_{j=1}^{n+1} \left( \sum_{\sigma \in G} \sigma(c_j)\right) v_j &= \sum_{j=1}^{n+1} \left( \sum_{i=1}^n \sigma_i(c_j)\right) v_j \\ &= \sum_{i=1}^n \left( \sum_{j=1}^{n+1} \sigma_i(c_j) v_j \right) = 0. \end{aligned}$

Now observe that for any $t$ , $\{ \sigma_t \circ \sigma_1,\dots, \sigma_t \circ \sigma_n \} = G$ . Therefore,

$\begin{aligned} \sigma_t \left( \sum_{\sigma \in G} \sigma(c_j) \right) &= \sigma_t \left( \sum_{i=1}^n \sigma_i(c_j) \right) \\ &= \sum_{i=1}^n (\sigma_t \circ \sigma_i)(c_j) \\ &= \sum_{\sigma \in G} \sigma(c_j) = \sum_{i = 1}^n \sigma_i(c_j). \end{aligned}$

Therefore, $\sum_{\sigma \in G} \sigma(c_j) \in \mathbb K^G$ for each $j$ . By linear independence, $\sum_{\sigma \in G} \sigma(c_j) = 0$ for each $j$ . In particular, for $j = 1$ ,

$\displaystyle 0 \neq \sum_{\sigma \in G} \sigma(u)= \sum_{\sigma \in G} \sigma(c_1) = 0,$

a contradiction.

Therefore, $|G| = [\mathbb K : \mathbb K^G]$ , which implies that

$\mathcal G = G \subseteq \mathrm{Aut}_{\mathbb K^G}(\mathbb K) \subseteq \mathcal G.$

Therefore, $\displaystyle \mathrm{Aut}_{\mathbb K^G}(\mathbb K) = G$ , as required.

Remark 1. Upon proving Lemma 2, we can conclude that a field extension is Galois if and only if it is normal and separable.

Lemma 3. We have $|\mathrm{Aut}_{\mathbb F}(\mathbb K)| \leq [\mathbb K : \mathbb F]$ with equality if and only if $\mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} = \mathbb F$ .

Proof. By the usual tower rule,

$\begin{aligned} [\mathbb K : \mathbb F] &= [\mathbb K : \mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)}] \cdot [\mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} : \mathbb F] \\ &= |\mathrm{Aut}_{\mathbb F}(\mathbb K)| \cdot [\mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} : \mathbb F] \\ &\geq |\mathrm{Aut}_{\mathbb F}(\mathbb K)|. \end{aligned}$

In particular,

$\begin{aligned} |\mathrm{Aut}_{\mathbb F}(\mathbb K)| = [\mathbb K : \mathbb F] \quad &\iff \quad [\mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} : \mathbb F] = 1 \\ &\iff \quad \mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} = \mathbb F. \end{aligned}$

We are now ready to prove Theorem 1.

Theorem 1 (Restated). Define the maps $\Phi : \mathbb F_{\text{ext}} \to G_{\text{sub}}$ and $\Psi : G_{\text{sub}} \to \mathbb F_{\text{ext}}$ by

$\Phi(\mathbb L) := \mathrm{Aut}_{\mathbb L}(\mathbb K),\quad \Psi(H) := \mathbb K^H.$

Then $\Phi, \Psi$ are inverses of each other, so that

$\mathbb K^{\mathrm{Aut}_{\mathbb L}(\mathbb K)} = \mathbb L, \quad \mathrm{Aut}_{\mathbb K^H}(\mathbb K) = H.$

Elements of $G_{\mathrm{sub}}$ are called intermediate fields.

Proof of Theorem 1. By Lemma 2, for any $H \in G_{\mathrm{sub}}$ ,

$\begin{aligned} (\Phi \circ \Psi)(H) &= \Phi(\Psi(H)) \\ &= \Phi(\mathbb K^H) \\ &= \mathrm{Aut}_{\mathbb K^H}(\mathbb K) \\ &= H. \end{aligned}$

For the second claim, it suffices to check that $|\mathrm{Aut}_{\mathbb L}(\mathbb K)| = [\mathbb K : \mathbb L]$ by Lemma 3, so that

$\begin{aligned} (\Psi \circ \Phi)(\mathbb L) &= \Psi(\Phi(\mathbb L)) \\ &= \Psi(\mathrm{Aut}_{\mathbb L}(\mathbb K)) \\ &= \mathbb K^{\mathrm{Aut}_{\mathbb L}(\mathbb K)} \\ &= \mathbb L. \end{aligned}$

To that end, define the collection

$\mathcal S := \{\sigma|_{\mathbb L} : \sigma \in \mathrm{Aut}_{\mathbb F}(\mathbb K)\} \subseteq \mathrm{Aut}_{\mathbb F}(\mathbb L).$

For any $\phi \in \mathrm{Aut}_{\mathbb F}(\mathbb K)$ ,

$\begin{aligned} \phi|_{\mathbb L} = \sigma|_{\mathbb L}\quad &\iff \quad (\phi^{-1} \circ \sigma)|_{\mathbb L} = \mathrm{id}_{\mathbb L}\\ &\iff \quad \phi^{-1} \circ \sigma \in \mathrm{Aut}_{\mathbb L}(\mathbb K) \\ &\iff \quad \phi \mathrm{Aut}_{\mathbb L}(\mathbb K) = \sigma \mathrm{Aut}_{\mathbb L}(\mathbb K) .\end{aligned}$

Clearly, $|\mathcal S| \leq |\mathrm{Aut}_{\mathbb F}(\mathbb L)| = \{\mathbb L : \mathbb F\}$ , so that

$\displaystyle \{\mathbb L : \mathbb F\} \geq |\mathcal S| = |\mathrm{Aut}_{\mathbb F}(\mathbb K) / \mathrm{Aut}_{\mathbb L}(\mathbb K)| = \frac{|\mathrm{Aut}_{\mathbb F}(\mathbb K)|}{|\mathrm{Aut}_{\mathbb L}(\mathbb K)| }.$

By the tower rule, and Lemma 3 again,

$\begin{aligned} [ \mathbb K : \mathbb F ] &= [\mathbb K : \mathbb L] \cdot [\mathbb L : \mathbb F] \\ &\geq [\mathbb K : \mathbb L] \cdot \{\mathbb L : \mathbb F\} \\ &\geq |\mathrm{Aut}_{\mathbb L}(\mathbb K)| \cdot \{\mathbb L : \mathbb F\} \\ &\geq |\mathrm{Aut}_{\mathbb L}(\mathbb K)| \cdot \frac{|\mathrm{Aut}_{\mathbb F}(\mathbb K)|}{|\mathrm{Aut}_{\mathbb L}(\mathbb K)| } \\ &= |\mathrm{Aut}_{\mathbb F}(\mathbb K)| \\ &= [\mathbb K: \mathbb F].\end{aligned}$

Therefore, we must have $|\mathrm{Aut}_{\mathbb L}(\mathbb K)| = [\mathbb K : \mathbb L]$ , as required.

Definition 1. We say that a finite extension $\mathbb K \supseteq \mathbb F$ is Galois if there exists a (unique) subgroup $G \leq \mathrm{Aut}(\mathbb K)$ such that $\mathbb K^G = \mathbb L$ . In this case, we call $\mathrm{Gal}(\mathbb K / \mathbb F) := G$ the Galois group of $\mathbb K$ over $\mathbb F$ . In particular,

$\mathrm{Gal}(\mathbb K / \mathbb F) = \mathrm{Aut}_{\mathbb F}(\mathbb K),$

where the latter is defined even if $\mathbb K \supseteq \mathbb F$ is not Galois.

Suppose $\mathbb K \supseteq \mathbb F$ is Galois.

Corollary 1. Given intermediate fields $\mathbb L_1 \subseteq \mathbb L_2$ ,

$\displaystyle [\mathbb L_2 : \mathbb L_1] = \frac{|\mathrm{Aut}_{\mathbb L_2}(\mathbb K)|}{|\mathrm{Aut}_{\mathbb L_1}(\mathbb K)|}.$

Proof. By Theorem 1, $|\mathrm{Aut}_{\mathbb L_1}(\mathbb K)| = [\mathbb K : \mathbb L_1]$ . Therefore,

$\displaystyle \frac{|\mathrm{Aut}_{\mathbb L_2}(\mathbb K)|}{|\mathrm{Aut}_{\mathbb L_1}(\mathbb K)|} = \frac{[\mathbb K : \mathbb L_2]}{[\mathbb K : \mathbb L_1]} = [\mathbb L_2 : \mathbb L_1].$

Corollary 2. Suppose $\mathbb K \supseteq \mathbb F$ is Galois. For any intermediate field $\mathbb L$ , $\mathbb K \supseteq \mathbb L$ is Galois by Theorem 1. Then $\mathbb L \supseteq \mathbb F$ is Galois if and only if $\mathrm{Gal}(\mathbb L/\mathbb F) \triangleleft \mathrm{Gal}(\mathbb K / \mathbb F)$ .

Proof. Recall that $\mathcal S \subseteq \mathrm{Aut}_{\mathbb F}(\mathbb L)$ and $|\mathcal S| \geq [\mathbb L : \mathbb F]$ . It is not hard to check that

$\begin{aligned} \mathrm{Gal}(\mathbb L/\mathbb F) \triangleleft \mathrm{Gal}(\mathbb K / \mathbb F) \quad &\iff \quad \mathcal S \subseteq \mathrm{Aut}_{\mathbb L}(\mathbb K). \end{aligned}$

Furthermore,

$\mathcal S \subseteq \mathrm{Aut}_{\mathbb L}(\mathbb K) \quad \iff \quad [\mathbb L:\mathbb F] = |\mathrm{Aut}_{\mathbb F}(\mathbb L)|,$

so that $\mathbb L \supseteq \mathbb F$ is Galois.

Corollary 3. Under the assumption of Corollary 2,

$\mathrm{Gal}(\mathbb K/\mathbb F)/\mathrm{Gal}(\mathbb K/\mathbb L) \cong \mathrm{Gal}(\mathbb L/\mathbb F).$

That is, we obtain some kind of third isomorphism theorem for Galois groups.

Proof. Check that the map $\sigma \mapsto \sigma|_{\mathbb L}$ is a surjective group homomorphism with kernel $\mathrm{Gal}(\mathbb K/\mathbb L)$ , then invoke the first isomorphism theorem for groups.

The fundamental theorem of Galois theory is lovely, because it tells us that the intermediate fields in the finite extension $\mathbb K \supseteq \mathbb F$ are in direct correspondence with the subgroups of automorphisms $\mathrm{Aut}(\mathbb K)$ . Therefore, we can translate our rather complicated ideas in field theory into relatively simple ones in group theory. Paradoxically, this connection was the historical motivation to even study group theory to begin with.

Next time, we use these mathematical mountaineering tools to begin our ascent in proving that there is no quintic formula, beginning with the group-theoretic connections with Galois extensions.

—Joel KIndiak, 24 Apr 26, 1435H

June 26, 2026
Galois Extensions
We have almost reached our most important definition in abstract algebra. But to get there, we still need to discuss some polynomial-based observations.

Let $\mathbb F$ denote a field with algebraic closure $\bar{\mathbb F}$ .

Definition 1. Given $f \in \mathbb F[x] \subseteq \bar{\mathbb F}[x]$ , define the set of zeroes of $f$ by

$\mathcal Z_f := \{ \alpha \in \bar{\mathbb F} : f(\alpha) = 0 \}.$

Note that $\mathcal Z_f = \{\alpha_1,\dots,\alpha_n\}$ is finite. We call

$\mathbb F(\mathcal Z_f) \equiv \mathbb F(\alpha_1,\dots,\alpha_n)$

the splitting field of $f$ over $\mathbb F$ .

Of course, this definition generalises to families of functions rather naturally.

Definition 2. Given $\mathcal F \subseteq \mathbb F[x] \subseteq \bar{\mathbb F}[x]$ , define the set of zeroes of $\mathcal F$ by

$\displaystyle \mathcal Z_{\mathcal F} := \bigcup_{f \in \mathcal F} \mathcal Z_f.$

Recall that $\mathbb F(\mathcal Z_{\mathcal F})$ denotes the smallest field that contains $\mathbb F$ and $\mathcal Z_{\mathcal F}$ . We call $\mathbb F(\mathcal Z_{\mathcal F})$ the splitting field of $\mathcal F$ over $\mathbb F$ . Observe that $\mathbb F(\mathcal Z_f) = \mathbb F(\mathcal Z_{\{f\}})$ so that Definition 1 can be thought of as subsumed under Definition 2.

We say that a field $\mathbb K \supseteq \mathbb F$ is a splitting field over $\mathbb F$ if there exists some $\mathcal F \subseteq \mathbb F[x]$ such that $\mathbb K = \mathbb F(\mathcal Z_{\mathcal F})$ .

Example 1. The splitting field of $x^4 - 1$ over $\mathbb Q$ is $\mathbb Q(i)$ , since

$x^4 - 1 = (x-1)(x+1)(x^2+1)$

has the roots $1,-1,i,-i$ . Writing as a product of linear factors,

$x^4 - 1 = (x-1)(x+1)(x-i)(x+i).$

Hence, we can conceive the splitting field of $\mathcal F$ as the “smallest” field extension that includes all roots of the functions in $\mathcal F$ .

Now let $\mathbb K \supseteq \mathbb F$ denote any finite extension.

Definition 2. We say that the extension $\mathbb K \supseteq \mathbb F$ is normal if for any $u \in \mathbb K$ ,

$\mathbb F(\mathcal Z_{f_{\mathbb F}^u}) \subseteq \mathbb K.$

Lemma 1. The following are equivalent.
- $\mathbb K \supseteq \mathbb F$ is normal.
- There exists some nonzero $f \in \mathbb F[x]$ such that $\mathbb K = \mathbb F(\mathcal Z_f)$ .
- For any $\mathbb F$ -homomorphism $\sigma : \mathbb K \to \bar{\mathbb F}$ , $\sigma(\mathbb K) = \mathbb K$ .
Proof. We will prove in a cycle.

Suppose $\mathbb K \supseteq \mathbb F$ is normal. Since the extension is finite, write $\mathbb K = \mathbb F(\alpha_1,\dots,\alpha_n)$ for some $\alpha_i \in \mathbb K$ . By Definition 2,

$\mathbb F(\mathcal Z_{f_{\mathbb F}^{\alpha_i}}) \subseteq \mathbb K$

for each $i$ . Define the polynomial $f := \prod_{i=1}^n f_{\mathbb F}^{\alpha_i} \in \mathbb F[x] \backslash \{0\}$ , whose zeroes are characterised by

$\displaystyle \mathcal Z_f = \bigcup_{i=1}^n \mathcal Z_{f_{\mathbb F}^{\alpha_i}} \subseteq \bigcup_{i=1}^n \mathbb F(\mathcal Z_{f_{\mathbb F}^{\alpha_i}}) \subseteq \bigcup_{i=1}^n \mathbb K = \mathbb K.$

On the other hand, since each $\alpha_i \in \mathcal Z_f$ , we have

$\mathbb K = \mathbb F(\alpha_1,\dots,\alpha_n) \subseteq \mathbb F (\mathcal Z_f) \subseteq \mathbb K.$

Therefore, $\mathbb K = \mathbb F(\mathcal Z_f)$ , as required. That is, $\mathbb K$ is a splitting field over $\mathbb F$ .

Now suppose there exists some $f \in \mathbb F[x] \backslash \{0\}$ such that $\mathbb K = \mathbb F(\mathcal Z_f)$ . Then for any $\alpha \in \mathcal Z_f$ ,

$f(\sigma(\alpha)) = \sigma(f(\alpha)) = \sigma(0) = 0,$

so that $\sigma(\mathcal Z_f) \subseteq \mathcal Z_f$ and thus $|\sigma(\mathcal Z_f)| \leq |\mathcal Z_f|$ . Since $\sigma$ is injective, $|\mathcal Z_f| \leq |\sigma(\mathcal Z_f)|$ . Therefore, $\sigma(\mathcal Z_f) = \mathcal Z_f$ since the latter is finite. In particular,

$\displaystyle \sigma(\mathbb K) = \sigma(\mathbb F(\mathcal Z_f)) = \mathbb F(\sigma(\mathcal Z_f)) = \mathbb F(\mathcal Z_f) = \mathbb K.$

Finally, suppose for any $\mathbb F$ -homomorphism $\sigma : \mathbb K \to \bar{\mathbb F}$ , $\sigma(\mathbb K) = \mathbb K$ . Fix $u \in \mathbb K$ . We need to show that $\mathbb F(\mathcal Z_{f_{\mathbb F}^u}) \subseteq \mathbb K$ . It suffices to check that for any conjugate $v \in \bar{\mathbb F}$ of $u$ , $v \in \mathbb K$ . Since $u,v$ are conjugates, there exists an $\mathbb F$ -homomorphism

$\sigma : \mathbb F(u) \to \mathbb F(v), \quad \sigma(u) = v.$

Thus, there exists an extension $\tilde{\sigma} : \mathbb K \to \bar{\mathbb F}$ of $\sigma$ which is itself an $\mathbb F$ -homomorphism. By hypothesis,

$v = \sigma(u) \in \sigma(\mathbb K) = \mathbb K,$

as required.

Lemma 2. For any finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ , $[\mathbb K : \mathbb K^G] = |G|$ .

Proof. Delayed.

Theorem 1. There exists a finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ such that $\mathbb K^G = \mathbb F$ if and only if $\mathbb K \supseteq \mathbb F$ is normal and separable.

Remark 1. Recall that $\mathbb K^G$ denotes the fixed field

$\mathbb K^G := \{u \in \mathbb K : \sigma(u) = u, \sigma \in G\} \subseteq \mathbb K.$

Proof. As a finite extension, write $\mathbb K = \mathbb F(\alpha_1,\dots,\alpha_n)$ for distinct $\alpha_i \in \mathbb K$ .

$(\Rightarrow)$ Suppose $\mathbb K^G = \mathbb F$ for some finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ . Define the finite subset $\mathcal Z \subseteq \mathbb K$ and the polynomial $f \in \mathbb K[x]$ by

$\displaystyle \mathcal Z := \bigcup_{i=1}^n \bigcup_{\sigma \in G} \{ \sigma(\alpha_i) \},\quad f(x) = \prod_{ \alpha \in \mathcal Z} (x-\alpha).$

Clearly, $\mathbb K = \mathbb F(\mathcal Z_f)$ , where $\mathcal Z_f = \mathcal Z$ . We claim that $f \in \mathbb F[x]$ , so that $\mathbb K \supseteq \mathbb F$ is separable, and normal by Lemma 1. To that end, we claim that for any $\phi \in G$ , $\phi(\mathcal Z) = \mathcal Z$ . To see this result, observe that $\phi \circ G = G$ . Therefore,

$\begin{aligned} \phi(\mathcal Z) &= \bigcup_{i=1}^n \bigcup_{\sigma \in G} \{ (\phi \circ \sigma)(\alpha_i) \} \\ &= \bigcup_{i=1}^n \bigcup_{\psi \in \phi \circ G} \{ \psi (\alpha_i) \} \\ &= \bigcup_{i=1}^n \bigcup_{\psi \in G} \{ \psi (\alpha_i) \} = \mathcal Z. \end{aligned}$

In particular,

$\begin{aligned} \phi(f) &= \phi \left( \prod_{\alpha \in \mathcal Z} (\cdot - \alpha) \right) \\ &= \prod_{\alpha \in \mathcal Z} (\cdot - \phi(\alpha)) \\ &= \prod_{\beta \in \phi(\mathcal Z)} (\cdot - \beta) \\ &= \prod_{\beta \in \mathcal Z} (\cdot - \beta) = f. \end{aligned}$

Writing $f(x) = \sum_{i=0}^n c_i x^i$ , we have $\phi(c_i) = c_i$ . Since $\phi \in G$ is chosen arbitrarily, $c_i \in \mathbb K^G = \mathbb F$ , so that $f \in \mathbb F[x]$ , as required.

$(\Leftarrow)$ Suppose $\mathbb K \supseteq \mathbb F$ is normal and separable. Suppose without loss of generality that $f_{\mathbb F}^{\alpha_1},\dots, f_{\mathbb F}^{\alpha_n}$ are distinct products of degree-one terms. Define $f := \prod_{i=1}^n f_{\mathbb F}^{\alpha_i} \in \mathbb F[x]$ , clearly a product of degree-one terms. Since $\mathbb K \supseteq \mathbb F$ is normal, $\mathbb F( \mathcal Z_{f_{\mathbb F}^{\alpha_i}}) \subseteq \mathbb K$ for each $i$ . Since each $\alpha_i \in \mathcal Z$ ,

$\displaystyle \mathbb K \subseteq \mathbb F(\mathcal Z) \subseteq \bigcup_{i=1}^n \mathbb F( \mathcal Z_{f_{\mathbb F}^{\alpha_i}}) \subseteq \mathbb K.$

Since, $\mathcal Z = \mathcal Z_f$ , we have $\mathbb K = \mathbb F(\mathcal Z) = \mathbb F(\mathcal Z_f)$ . By Lemma 1, for any $\mathbb F$ -homomorphism $\sigma : \mathbb K \to \bar{\mathbb F}$ , $\sigma(\mathbb K) = \mathbb K$ , so that $\mathrm{Aut}_{\mathbb F}(\mathbb K) = \mathrm{Aut}(\mathbb K)$ . Since $\mathbb K \supseteq \mathbb F$ is separable,

$|\mathrm{Aut}_{\mathbb F}(\mathbb K)| = \{ \mathbb K : \mathbb F \} = [ \mathbb K : \mathbb F].$

By definition, $\mathbb K \supseteq \mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)}$ is Galois. By the tower rule and Lemma 2,

$\begin{aligned} |\mathrm{Aut}_{\mathbb F}(\mathbb K)| &= [ \mathbb K : \mathbb F ] \\ &= [ \mathbb K : \mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} ] \cdot [ \mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} : \mathbb F ] \\ &= |\mathrm{Aut}_{\mathbb F}(\mathbb K)| \cdot [ \mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} : \mathbb F ]. \end{aligned}$

Therefore, $[ \mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} : \mathbb F ] = 1$ implies that $\mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} = \mathbb F$ . Define $G := \mathrm{Aut}_{\mathbb F}(\mathbb K)$ , as required.

Definition 3. For any finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ , we say that the extension $\mathbb K \supseteq \mathbb K^G$ is Galois with Galois group $G =: \mathrm{Gal}(\mathbb K / \mathbb K^G)$ .

We say that an extension $\mathbb K \supseteq \mathbb F$ is Galois if there exists a finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ such that $\mathbb K^G = \mathbb F$ . In this case, $\mathbb K^{\mathrm{Gal}(\mathbb K/\mathbb F)} = \mathbb F$ . By Theorem 1, if $\mathbb K \supseteq \mathbb F$ is finite, then the extension $\mathbb K \supseteq \mathbb F$ is Galois if and only if the extension $\mathbb K \supseteq \mathbb F$ is normal and separable.

However, we haven’t proven Lemma 2 in order to arrive at this equivalence. Even worse; while Definition 3 is well defined, the Galois extension comment, strictly speaking, is not. How do we know that given two finite subgroups $G,H \leq \mathrm{Aut}(\mathbb K)$ , $\mathbb K^G = \mathbb K^H$ implies that $G = H$ ? Only after establishing this result can we properly define the Galois group $\mathrm{Gal}(\mathbb K/ \mathbb F)$ .

This is the content of the famous fundamental theorem of Galois theory, tying together the various ideas that we have explored regarding splitting fields. Once again, suppose $\mathbb K \supseteq \mathbb F$ is finite. Given any finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ , define the following sets:

$\begin{aligned} \mathbb F_{\text{ext}} &:= \{ \mathbb L \subseteq \mathbb K : \mathbb L \supseteq \mathbb F\ \text{is a field extension}\}, \\ G_{\text{sub}} &:= \{H \subseteq G : H \leq G \}. \end{aligned}$

Elements of $\mathbb F_{\mathrm{ext}}$ are called intermediate extensions.

Theorem 2. Define the maps $\Phi : \mathbb F_{\text{ext}} \to G_{\text{sub}}$ and $\Psi : G_{\text{sub}} \to \mathbb F_{\text{ext}}$ by

$\Phi(\mathbb L) := \mathrm{Aut}_{\mathbb L}(\mathbb K),\quad \Psi(H) := \mathbb K^H.$

Then $\Phi, \Psi$ are inverses of each other.

Corollary 1. Given intermediate fields $\mathbb L_1, \mathbb L_2$ ,

$\mathbb L_1 \subseteq \mathbb L_2 \quad \iff \quad \Phi(\mathbb L_1) \supseteq \Phi(\mathbb L_2).$

In this case, $[\mathbb L_2 : \mathbb L_1] = [\Phi(\mathbb L_1) : \Phi(\mathbb L_2)]$ .

Proof. Under Theorem 2,

$\begin{aligned} \mathbb L_1 \subseteq \mathbb L_2 \quad &\Rightarrow \quad \mathrm{Aut}_{\mathbb L_1}(\mathbb K) \supseteq \mathrm{Aut}_{\mathbb L_2}(\mathbb K) \\ &\Rightarrow \quad \Phi(\mathbb L_1) \supseteq \Phi(\mathbb L_2) \\ &\Rightarrow \quad \mathbb K^{\Phi(\mathbb L_1)} \subseteq \mathbb K^{\Phi(\mathbb L_2)} \\ &\Rightarrow \quad \Psi(\Phi(\mathbb L_1)) \subseteq \Psi(\Phi(\mathbb L_2)) \\ &\Rightarrow \quad (\Psi \circ \Phi)(\mathbb L_1) \subseteq (\Psi \circ \Phi)(\mathbb L_2) \\ &\Rightarrow \quad \mathbb L_1 \subseteq \mathbb L_2. \end{aligned}$

Furthermore,

Before proving this theorem, we should take some time to appreciate what this theorem is telling us.

Example 2. For any prime number $p$ , the extension $\mathbb Q(\sqrt p) \supseteq \mathbb Q$ is Galois with Galois group

$\mathrm{Gal}(\mathbb Q(\sqrt p)/\mathbb Q) \cong \mathbb Z_2.$

Proof. Since $\mathbb Q(\sqrt p)$ is the splitting field of $x^2-p$ over $\mathbb Q$ , the extension $\mathbb Q(\sqrt p) \supseteq \mathbb Q$ is normal by Lemma 1 and separable by the factorisation

$x^2-p = (x-\sqrt p)(x+\sqrt p).$

By Theorem 1, $\mathbb Q(\sqrt p) \supseteq \mathbb Q$ is Galois. For $i \in \{0, 1\}$ , define the $\mathbb Q$ -automorphisms $\sigma_i : \sqrt 2 \mapsto (-1)^i \sqrt 2$ . It is clear that

$\{ \sigma_0 , \sigma_1 \} \cong \mathbb Z_2.$

To show that $\mathrm{Aut}_{\mathbb Q}(\mathbb Q(\sqrt p)) = \{ \sigma_0, \sigma_1 \}$ , it suffices to prove the direction $(\subseteq)$ , since the direction $(\supseteq)$ is obvious. To that end, fix any $\mathbb Q$ -automorphism

$\sigma \in \mathrm{Aut}_{\mathbb Q}(\mathbb Q(\sqrt p)).$

Thus, for any $a , b \in \mathbb Q$ ,

$\begin{aligned} \sigma(a + b\sqrt 2) &= \sigma(a) + \sigma(b\sqrt 2) \\ &= \sigma(a) + \sigma(b) \sigma(\sqrt 2) \\ &= a + b \sigma(\sqrt 2). \end{aligned}$

Hence, $\sigma$ is determined by $\sigma(\sqrt p)$ . Suppose $\sigma(\sqrt p) = c + d\sqrt p$ for some $c, d \in \mathbb Q$ . Then

$\begin{aligned} \sigma(\sqrt p)^2 &= \sigma((\sqrt p)^2) \\ (c+d\sqrt p)^2 &= \sigma(2) \\ (c^2 + pd^2) + 2cd\sqrt p &= p. \end{aligned}$

If $d = 0$ , then $\sqrt p = \sigma^{-1}(c) = c \in \mathbb Q$ , a contradiction. Therefore, $2cd = 0$ implies that $c = 0$ . Therefore, $pd^2 = p$ implies that $d = \pm 1 = (-1)^i$ for $i \in \{0,1\}$ . Therefore,

$\sigma(\sqrt 2) = (-1)^i \sqrt p = \sigma_i(\sqrt p),$

and hence $\sigma \in \{\sigma_0, \sigma_1\}$ .

Example 3. For distinct primes $p, q$ , the extension $\mathbb Q(\sqrt p, \sqrt q) \supseteq \mathbb Q$ is Galois with Galois group

$\mathrm{Gal}(\mathbb Q(\sqrt p, \sqrt q)/\mathbb Q) \cong \mathbb Z_2 \times \mathbb Z_2.$

Proof. Since $\mathbb Q(\sqrt p, \sqrt q)$ is the splitting field of $(x^2-p)(x^2-q)$ over $\mathbb Q$ , the extension $\mathbb Q(\sqrt p, \sqrt q) \supseteq \mathbb Q$ is normal by Lemma 1 and separable by the factorisation

$(x^2-p)(x^2-q) = (x-\sqrt p)(x+\sqrt p)(x-\sqrt q)(x+\sqrt q).$

By Theorem 1, $\mathbb Q(\sqrt p, \sqrt q) \supseteq \mathbb Q$ is Galois. For $i,j \in \{0,1\}$ , define the $\mathbb Q$ -automorphisms

$\begin{aligned} \sigma_{ij} : (\sqrt p, \sqrt q) &\mapsto ((-1)^i \sqrt p, (-1)^j \sqrt q). \end{aligned}$

We leave it as an exercise to check in similar fashion to Example 2 that

$\mathrm{Aut}_{\mathbb Q}(\mathbb Q(\sqrt p, \sqrt q)) = \{\sigma_{ij} : i,j \in \{0, 1\}\},$

and that the map $\sigma_{ij} \mapsto ([i], [j])$ yields the desired group isomorphism.

Therefore, with finite Galois-ness, we can characterise field extensions in terms of familiar finite groups: subfields of the field extension are in direct correspondence with subgroups of the Galois group! This connection is the key insight of Galois theory that would help us encode the roots of higher-degree polynomials into familiar groups.

The grand idea to the non-existence of a quintic formula looks like this—if a quintic formula exists, then the corresponding Galois group needs to satisfy some properties. However, it fails to do so. By contrapositive or contradiction, a quintic formula cannot exist.

We will prove the rather elegant Theorem 2 in the next post. Ironically, we would need Lemma 2 first. But we will explore both ideas in the next time.

—Joel Kindiak, 23 Apr 26, 1926H
June 25, 2026
Revisiting Automorphisms
One key idea of Galois theory is how we can relate field extensions to field automorphisms. Of course, we should have a working definition of what we mean by field automorphisms first.

Let $\mathbb F$ be any field and $\mathbb K \supseteq \mathbb F$ denote a field extension.

Theorem 1. Let $u, v \in \mathbb K$ be algebraic over $\mathbb F$ . Define the map $\Psi_{u,v} : \mathbb F(u) \to \mathbb F(v)$ by

$\displaystyle \Psi_{u,v}\left( \sum_{i=0}^n a_iu^i \right) = \sum_{i=0}^n a_iv^i.$

The following are equivalent.
- $\Psi_{u,v}$ is a homomorphism.
- $\Psi_{u,v}$ is an isomorphism.
- $f_{\mathbb K}^u = f_{\mathbb K}^v$ .
In the latter, we say that $u$ and $v$ are conjugates.

Proof. If $\Psi_{u,v}$ is an isomorphism then it is immediately a homomorphism. Define the evaluation map $\Psi_v : \mathbb K[x] \to \mathbb K(v)$ by $\Psi(x) := v$ and extended to a homomorphism. Define $\Psi_u$ similarly, so that $\Psi_v = \Psi_{u, v} \circ \Psi_u$ . In particular,

$\mathbb F[x] \neq \ker(\Psi_v) \supseteq \langle f_{\mathbb K}^u \rangle \quad \Rightarrow \quad \ker(\Psi_v) = \langle f_{\mathbb K}^u \rangle,$

inducing an isomorphism $\tilde{\Psi}_v : \mathbb F[x]/ \langle f_{\mathbb K}^u \rangle \to \mathbb F(v)$ . By definition, $\tilde{\Psi}_u = \Psi_{u,v} \circ \tilde{\Psi}_v$ , so that $\Psi_{u,v} = \tilde{\Psi}_u \circ \tilde{\Psi}_v^{-1}$ is the required isomorphism if it is a homomorphism. Finally, $\Psi_{u,v}$ is an isomorphism if and only if

$\mathbb F[x]/\langle f_{\mathbb K}^u \rangle \cong \mathbb F[x]/\langle f_{\mathbb K}^v \rangle,$

which holds if and only if $f_{\mathbb K}^u = f_{\mathbb K}^v$ .

Recall that given the field extensions $\mathbb K_1 \supseteq \mathbb F$ and $\mathbb K_2 \supseteq \mathbb F$ , a field homomorphism $\phi : \mathbb K_1 \to \mathbb K_2$ is a $\mathbb F$ –homomorphism if $\phi|_{\mathbb F} = \mathrm{id}$ .

Corollary 1. Let $u$ be algebraic over a field $\mathbb F$ . Then for any conjugate $v$ of $u$ , there exists a unique $\mathbb F$ -homomorphism $\phi : \mathbb F(u) \to \bar{\mathbb F}$ such that $\phi(u) = v$ .

When homomorphisms are bijective, we call them isomorphisms. When isomorphisms map fields to themselves, we call them automorphisms. Denote the set of automorphisms on a field $\mathbb F$ by $\mathrm{Aut}(\mathbb F)$ .

Definition 1. For any automorphism $\sigma : \mathbb K \to \mathbb K$ , we say that $u \in \mathbb K$ is fixed under $\sigma$ if $\sigma(u) = u$ . Furthermore, for any subfield $\mathbb F \subseteq \mathbb K$ , $\sigma$ is called an $\mathbb F$ -automorphism if $\sigma|_{\mathbb F} = \mathrm{id}$ . Define

$\mathrm{Aut}_{\mathbb F}(\mathbb K) := \{ \sigma \in \mathrm{Aut}(\mathbb K) : \sigma \text{ is an }\mathbb F\text{-automorphism}\}.$

Example 1. For any $\sigma \in \mathrm{Aut}(\mathbb F)$ , $\sigma$ is a $\mathbb Q$ -automorphism.

Lemma 1. Given $\Sigma \subseteq \mathrm{Aut}(\mathbb F)$ , the set

$\mathbb F^{\Sigma} := \{ u \in \mathbb K : \sigma(u) = u \text{ for all } \sigma \in \Sigma\}$

forms a sub-field of $\mathbb K$ , called the fixed field of $\Sigma$ in $\mathbb F$ .

Given any isomorphism $\sigma : \mathbb F_1 \to \mathbb F_2$ , define the set

$\Sigma_{\mathbb F_1, \mathbb F_2}^{\sigma} := \{ \phi \in \mathrm{Hom}(\bar{\mathbb F}_1, \bar{\mathbb F}_2 ) : \phi|_{\mathbb F_1} = \sigma \}.$

Lemma 2. Suppose $\mathbb K \supseteq \mathbb F$ is a finite extension. For any isomorphism $\sigma : \mathbb F \to \mathbb L$ ,

$|\Sigma_{\mathbb K, \bar{\mathbb L}}^{\sigma}| = | \Sigma_{\mathbb K, \bar{\mathbb F}}^{ \mathrm{id} } |.$

Proof. Let $\phi : \bar{\mathbb L} \to \bar{\mathbb F}$ extend $\sigma^{-1} : \mathbb L \to \mathbb F$ . For any $\Psi \in \Sigma_{\mathbb K, \bar{\mathbb L}}^{\sigma}$ , $\Psi : \bar{\mathbb K} \to \bar{\mathbb L}$ is a homomorphism that extends $\sigma$ . Then, for any $x \in \mathbb F \subseteq \mathbb K$ , since $\sigma(x) \in \mathbb L$ ,

$(\phi \circ \Psi)(x) = \phi^{-1}(\Psi(x)) = \phi(\sigma(x)) = \sigma^{-1}(\sigma(x)) = x.$

Therefore, $(\phi \circ \Psi)|_{\mathbb F} = \mathrm{id}$ , i.e. $\phi \circ \Psi : \mathbb K \to \bar{\mathbb F}$ is a uniquely-defined $\mathbb F$ -homomorphism. Similarly, for any $\Psi_0 \in \Sigma_{\mathbb K, \bar{\mathbb F}}^{\mathrm id}$ , there exists a uniquely-defined extension of $\sigma$ defined by $\phi^{-1} \circ \Psi_0$ .

Definition 2. If $\mathbb K \supseteq \mathbb F$ is a finite extension, we define $\{ \mathbb K : \mathbb F\} := | \Sigma_{\mathbb K, \bar{\mathbb F}}^{ \mathrm{id} } |$ . By Lemma 2, for any isomorphism $\sigma : \mathbb F \to \mathbb L$ , $\{ \mathbb K : \mathbb F\} = |\Sigma_{\mathbb K, \bar{\mathbb L}}^{\sigma}|$ .

Theorem 2. For any $u$ algebraic over $\mathbb F$ , $\{\mathbb F(u) : \mathbb F\}$ equals the number of distinct roots of $f_{\mathbb F}^u$ .

Proof. By Definition 2, $\{\mathbb F(u) : \mathbb F\} = |\Sigma_{\mathbb F(u), \bar{\mathbb F}}^{\mathrm{id}}|$ . The right-hand side counts the number of $\mathbb F$ -homomorphisms from $\mathbb F(u)$ to $\bar{\mathbb F}$ . Define

$\mathcal R := \{v \in \bar{\mathbb F} : f_{\mathbb F}^u(v) = 0\},$

where $f_{\mathbb F}^u(x) = \sum_{j=0}^s b_j x^j$ .

We claim that $|\Sigma_{\mathbb F(u), \bar{\mathbb F}}^{\mathrm{id}}| = |\mathcal R|$ . Fix $\Psi \in \Sigma_{\mathbb F(u), \bar{\mathbb F}}^{\mathrm{id}}$ . Then

$\displaystyle f_{\mathbb F}^u(\Psi(u)) = \sum_{j=0}^s b_j \Psi(u)^j = \Psi\left( \sum_{j=0}^s b_j u^j \right) = \Psi(f_{\mathbb F}^u(u)) = \Psi(0) = 0,$

so that each $\Psi \in \Sigma_{\mathbb F(u), \bar{\mathbb F}}$ yields a unique root $\Psi(u) \in \mathcal R$ .

Conversely, given any root $v \in \mathcal R$ , define the evaluation map $\phi_v : \bar{\mathbb F}[x] \to \bar{\mathbb F}$ by

$\phi_v|_{\mathbb F} = \mathrm{id},\quad \phi_v(x) = v,$

as well as the inclusion map $\psi : \mathbb F[x] \to \bar{\mathbb F}[x]$ . Then $\phi_v \circ \psi : \mathbb F[x] \to \bar{\mathbb F}$ is a ring homomorphism such that $\langle f_{\mathbb F}^u \rangle \subseteq \ker(\phi \circ \psi)$ . Since $\langle f_{\mathbb F}^u \rangle$ is maximal in $\mathbb F[x]$ ,

$\begin{aligned} ((\phi_v \circ \psi)(1) = 1 \quad &\Rightarrow \quad \ker(\phi_v \circ \psi) \neq \mathbb F[x]) \\ \quad &\Rightarrow \quad \ker(\phi_v \circ \psi) = \langle f_{\mathbb F}^u\rangle. \end{aligned}$

By the first isomorphism theorem, $\phi_v$ induces the ring isomorphism

$\mathbb F(u) \cong \mathbb F[x]/\langle f_{\mathbb F}^u \rangle = \mathbb F[x]/{\ker(\phi_v \circ \psi)} \cong \bar{\mathbb F}.$

In particular, $\phi_v$ induces a ring isomorphism $\Phi_v : \mathbb F(u) \to \bar{\mathbb F}$ . Furthermore, $\Phi_v$ is an $\mathbb F$ -homomorphism. Therefore, each root $v \in \mathcal R$ yields a unique $\mathbb F$ -homomorphism $\Phi_v \in \Sigma_{\mathbb F(u), \bar{\mathbb F}}^{\mathrm{id}}$ .

Corollary 2. For the finite extensions $\mathbb F \subseteq \mathbb K \subseteq \mathbb L$ ,

$\{ \mathbb L : \mathbb F \} = \{ \mathbb L : \mathbb K \} \cdot \{ \mathbb K : \mathbb F \},$

i.e. the tower rule for the separable degree of finite extensions.

Proof. Denote $p:= \{ \mathbb L : \mathbb F \}$ , $q:= \{ \mathbb L : \mathbb K \}$ , and $r := \{ \mathbb K : \mathbb F \}$ . Hence,
- for $1 \leq i \leq p$ , there exists an $\mathbb F$ -homomorphism $\sigma_i : \mathbb L \to \bar{\mathbb F}$ ,
- for $1 \leq j \leq q$ , there exists an $\mathbb K$ -homomorphism $\rho_j : \mathbb L \to \bar{\mathbb K}$ ,
- for $1 \leq k \leq r$ , there exists an $\mathbb F$ -homomorphism $\mu_k : \mathbb K \to \bar{\mathbb F}$ .
For each $k$ , extend $\mu_k$ to an $\mathbb F$ -homomorphism $\tilde{\mu}_k : \bar{\mathbb K} \to \bar{\mathbb F}$ . Then $\rho_j \circ \tilde{\mu}_k : \mathbb L \to \bar{\mathbb F}$ forms an $\mathbb F$ -homomorphism. Therefore, $q \cdot r \leq p$ .

Likewise, for each $i$ , consider the $\mathbb F$ -homomorphism $\sigma_i : \mathbb L \to \bar{\mathbb F}$ . Now $\sigma_i|_{\mathbb K} : \mathbb K \to \bar{\mathbb F}$ must be an $\mathbb F$ -homomorphism, so that $\sigma_i|_{\mathbb K} = \mu_k$ for some $k$ . In particular, $\sigma_i$ extends $\mu_k$ , so that $\sigma_i = \rho_j \circ \tilde{\mu}_k$ for some $j$ . Therefore, $p \leq q \cdot r$ .

Lemma 3. For any monic and irreducible polynomial $f \in \mathbb F[x]$ , there exist $u_1,\dots,u_r$ and some positive integer $\alpha$ such that, regarded as a polynomial in $\bar{\mathbb F}$ ,

$\displaystyle f(x) = \prod_{i=1}^r (x-u_i)^\alpha.$

Proof. Write $f(x) = \sum_{i=1}^r a_i x^i$ . Factorise $f$ into its roots $u_1,\dots, u_n$ when regarded as a polynomial in $\bar{\mathbb F}$ :

$\displaystyle f(x) = \prod_{i=1}^n (x-u_i)^{\alpha_i}.$

The crucial claim is that the $\alpha_i$ terms all equal each other. To establish that claim, We first regard $f \in \mathbb F(u_1,\dots, u_n)[x]$ . For any $k \neq \ell$ , the map $\Psi_{u_k,u_\ell} : \mathbb F(u_k) \to \mathbb F(u_\ell)$ as defined by Theorem 1 is an $\mathbb F$ -isomorphism. Fix any homomorphism

$\tilde{\Psi}_{u_k,u_\ell} : \mathbb F(u_1,\dots, u_n) \to \bar{\mathbb F}$

that extends $\Psi_{u_k,u_\ell}$ . Then evaluating $\tilde{\Psi}_{u_k,u_\ell}(f)$ and simplifying in two different ways,

$\begin{aligned} \prod_{j=1}^n (x-u_j)^{\alpha_j} = \tilde{\Psi}_{u_k,u_\ell}(f) &= \prod_{j=1}^n (x- \tilde{\Psi}_{u_k,u_\ell}(u_j) )^{\alpha_j} \\ &= (x-u_\ell)^{\alpha_k } \prod_{j \neq k}^n (x- \tilde{\Psi}_{u_k,u_\ell}(u_j) )^{\alpha_j}. \end{aligned}$

Since there are $\alpha_\ell$ copies of $(x-u_\ell)$ on the left-hand side, we have $\alpha_k \leq \alpha_\ell$ . Since this inequality holds for any $k, \ell$ , we also must have $\alpha_\ell \leq \alpha_k$ . Therefore, $\alpha_k = \alpha_\ell =: \alpha$ , so that

$\displaystyle f(x) = \prod_{i=1}^n (x-u_i)^{\alpha_i} = \prod_{i=1}^n (x-u_i)^{\alpha}.$

Corollary 3. For the finite extension $\mathbb F \subseteq \mathbb K$ , $\{ \mathbb K : \mathbb F \} \mid [\mathbb K : \mathbb F]$ .

Proof. Suppose $[\mathbb K : \mathbb F] = n$ . Define $\mathbb F_0 := \mathbb F$ , $\mathbb F_r = \mathbb F(u_1,\dots, u_r)$ , so that $\mathbb F_n = \mathbb K$ . Observe that

$\mathbb F \subseteq \mathbb F(u_1) \subseteq \cdots \subseteq \mathbb F(u_1,\dots, u_{n-1}) \subseteq \mathbb K.$

By the tower rule for separable degrees,

$\displaystyle \{\mathbb K : \mathbb F\} = \prod_{i=1}^n \{ \mathbb F_i : \mathbb F_{i-1} \}.$

It suffices to show that $m := \{ \mathbb F_i : \mathbb F_{i-1} \} \mid [ \mathbb F_i : \mathbb F_{i-1} ]$ . By Theorem 2, $\mathbb F_i = \mathbb F_{i-1}(u_i)$ implies that $m = \{ \mathbb F_{i-1}(u_i): \mathbb F_{i-1} \}$ is given by the number of distinct roots of $f_{\mathbb F_{i-1}}^{u_i}$ . By Lemma 3, factorise this polynomial into

$\displaystyle f_{\mathbb F_{i-1}}^{u_i}(x) = \prod_{j=1}^m (x - v_j)^{\alpha}.$

Therefore, $\alpha \cdot m = \deg(f_{\mathbb F_{i-1}}^{u_i}) = [ \mathbb F_i : \mathbb F_{i-1} ]$ , as required.

Definition 2 helps us define the notion of a separable extension—a central notion in describing the “Galois-ness” of a field.

Definition 3. A field extension $\mathbb K \supseteq \mathbb F$ is separable if $\{\mathbb K : \mathbb F\} = [\mathbb K : \mathbb F]$ . An element $u \in \mathbb K$ is separable over $\mathbb F$ if $\mathbb F(u) \supseteq \mathbb F$ is separable.

Theorem 3. Suppose $\mathbb F \subseteq \mathbb K \subseteq \mathbb L$ is finite. Then $\mathbb L \supseteq \mathbb F$ is separable if and only if $\mathbb K \supseteq \mathbb F$ and $\mathbb L \supseteq \mathbb K$ are separable.

Proof. Recall the tower rule for usual and separable degrees:

$\begin{aligned} [ \mathbb L : \mathbb F ] &= [ \mathbb L : \mathbb K ] \cdot [ \mathbb K : \mathbb F ], \\ \{ \mathbb L : \mathbb F \} &= \{ \mathbb L : \mathbb K \} \cdot \{ \mathbb K : \mathbb F\}. \end{aligned}$

Taking ratios, each ratio is an integer by Corollary 3:

$\begin{aligned} \frac{ [ \mathbb L : \mathbb F ] }{ \{ \mathbb L : \mathbb F \} }&= \frac{ [ \mathbb L : \mathbb K ] }{ \{ \mathbb L : \mathbb K \} } \cdot \frac{ [ \mathbb K : \mathbb F ] }{ \{ \mathbb K : \mathbb F \} }. \end{aligned}$

The direction $(\Leftarrow)$ is obvious by the usual tower rule. For the direction $(\Rightarrow)$ , the left-hand side would be $1$ , so that must both ratios on the right side are also equal to $1$ , as required.

Corollary 4. Suppose $\mathbb K \supseteq \mathbb F$ is finite. Then the following statements are equivalent:
- Every element in $\mathbb K$ is separable over $\mathbb F$ .
- There exist $u_1,\dots, u_r \in \mathbb K$ separable over $\mathbb F$ such that $\mathbb K = \mathbb F(u_1,\dots, u_r)$ .
- $\mathbb K$ is separable over $\mathbb F$ .
Proof. Write $\mathbb K = \mathbb F(u_1,\dots, u_r)$ and apply Theorem 3 repeatedly.

Corollary 5. Suppose $\mathbb K \supseteq \mathbb F$ is algebraic. Then

$\mathbb K_{\mathrm s} := \{u \in \mathbb K : u\text{ is separable over }\mathbb F\}$

forms a subfield of $\mathbb K$ , called the separable closure of $\mathbb F$ in $\mathbb K$ .

Proof. Given $u,v \in \mathbb K_s$ , $\mathbb F(u,v) \supseteq \mathbb F$ is separable by Corollary 4.

Eventually, we would describe a special subset of separable extensions as Galois, but to do that we need to explore the notion of splitting fields and normal extensions.

—Joel Kindiak, 14 Apr 26, 1858H
June 22, 2026