---
title: "Field Extensions and Galois Theory"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 4
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory
---

# Chapter 4 — Field Extensions and Galois Theory

Can every equation be solved by [radicals](#def-b3-galois-radical), as the quadratic formula and Cardano’s cubic formulas suggest? Can one trisect an angle with ruler and compass? Both questions, open for centuries, are answered — negatively — by a single idea of Évariste [Galois](#def-b3-galois-galois): attach to every polynomial a finite *group* of symmetries of its roots, and read the answer off the group. This chapter builds the dictionary: [field extensions](#def-b3-galois-extension) and [degrees](#def-b3-galois-extension), [splitting fields](#thm-b3-galois-splitting) and [algebraic closures](#def-b3-galois-closure), [finite fields](#thm-b3-galois-finitefields) (a complete theory — and the promised cyclicity of $\mathbb F_q^\times$), separability, then the [Galois correspondence](#thm-b3-galois-fundamental) itself, with full proofs. We harvest: the impossibility of the classical constructions, the structure of cyclotomic fields, and the unsolvability of the quintic by [radicals](#def-b3-galois-radical) — [Chapter 1](https://one-course.com/books/math/5/en/chapter/1-group-theory#ch-b3-groups)’s simplicity of $A_5$ striking its target.

## 4.1 Extensions, degree, algebraicity

**Definition 4.1.**

A *field extension* $L/K$ is a field $L$ containing $K$ as a subfield; $L$ is then a $K$-vector space, and the *degree* $[L:K]$ is its dimension. The extension is *finite* if $[L:K] <
\infty$. The *characteristic* of a field is the generator $\geq 0$ of the kernel of $\Z \to K$, $n \mapsto n\cdot 1$: it is $0$ or a prime $p$; correspondingly $K$ contains a smallest subfield (*prime field*) isomorphic to $\Q$ or to $\mathbb
F_p = \Z/p\Z$.

**Theorem 4.2 (Tower law).**

If $K \subseteq L \subseteq M$, then $[M:K] = [M:L]\,[L:K]$: if $(e_i)$ is a basis of $L$ over $K$ and $(f_j)$ a basis of $M$ over $L$, then $(e_if_j)$ is a basis of $M$ over $K$.

**Proof.** Generating: $x \in M$ writes $x = \sum_j \lambda_jf_j$ ($\lambda_j \in L$), each $\lambda_j = \sum_i \mu_{ij}e_i$ ($\mu_{ij} \in K$): $x = \sum_{i,j}\mu_{ij}e_if_j$. Independent: $\sum_{i,j}\mu_{ij}e_if_j = 0$ rewrites $\sum_j (\sum_i
\mu_{ij}e_i)f_j = 0$; the inner sums are in $L$, so vanish ($f_j$ independent over $L$); then all $\mu_{ij} = 0$ ($e_i$ independent over $K$). ∎

**Definition 4.3.**

Let $L/K$ and $\alpha \in L$. If some nonzero $P \in K[X]$ has $P(\alpha) = 0$, $\alpha$ is *algebraic* over $K$; the monic generator $\pi_\alpha$ of the [ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal) $\{P : P(\alpha) = 0\}$ of $K[X]$ is its *minimal polynomial*, an [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) polynomial ($\pi = QR$ with $Q(\alpha) = 0$ forces $R$ constant by minimality of the [degree](#def-b3-galois-extension)). Otherwise $\alpha$ is *transcendental*. We write $K(\alpha)$ for the smallest subfield of $L$ containing $K$ and $\alpha$, and $K[\alpha]$ for the smallest subring.

**Theorem 4.4.**

If $\alpha$ is [algebraic](#def-b3-galois-algebraic) over $K$ with $d = \deg\pi_\alpha$, then

$$
K(\alpha) = K[\alpha] \cong K[X]/(\pi_\alpha),
\qquad [K(\alpha) : K] = d,
$$

with basis $1, \alpha, \dots, \alpha^{d-1}$. Conversely, if $[L:K] < \infty$, every $\alpha \in L$ is [algebraic](#def-b3-galois-algebraic) of [degree](#def-b3-galois-extension) $\deg \pi_\alpha$ dividing $[L:K]$.

**Proof.** Evaluation $K[X] \to L$, $P \mapsto P(\alpha)$, has image $K[\alpha]$ and kernel $(\pi_\alpha)$; since $\pi_\alpha$ is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility), $K[X]/(\pi_\alpha)$ is a *field* ([Proposition 2.4](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#prop-b3-rings-primemaximal): in the [PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) $K[X]$, [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) generates a [maximal ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-primemaximal)), so $K[\alpha]$ is a field containing $K$ and $\alpha$: it equals $K(\alpha)$. The classes of $1, X, \dots, X^{d-1}$ form a basis of the quotient (Euclidean division), whence the basis and the [degree](#def-b3-galois-extension). Conversely if $[L:K]
= n < \infty$: $1, \alpha, \dots, \alpha^n$ are dependent, giving an annihilating polynomial; then $[K(\alpha):K] =
\deg\pi_\alpha$ divides $n$ by the [tower law](#thm-b3-galois-tower). ∎

**Corollary 4.5.**

If $\alpha, \beta$ are [algebraic](#def-b3-galois-algebraic) over $K$, so are $\alpha \pm
\beta$, $\alpha\beta$, $\alpha/\beta$ ($\beta \ne 0$): the elements of $L$ [algebraic](#def-b3-galois-algebraic) over $K$ form a subfield of $L$. Moreover algebraicity is transitive: [algebraic](#def-b3-galois-algebraic) over [algebraic](#def-b3-galois-algebraic) is [algebraic](#def-b3-galois-algebraic).

**Proof.** $K(\alpha, \beta) = (K(\alpha))(\beta)$ is finite over $K(\alpha)$ ($\beta$ [algebraic](#def-b3-galois-algebraic) over $K \subseteq K(\alpha)$) and $K(\alpha)/K$ is finite: by the [tower law](#thm-b3-galois-tower) $[K(\alpha,\beta):K] <
\infty$, and every element of $K(\alpha, \beta)$ — including the four listed — is [algebraic](#def-b3-galois-algebraic) ([Theorem 4.4](#thm-b3-galois-simple)). Transitivity: if $\beta$ is [algebraic](#def-b3-galois-algebraic) over $L$ and $L/K$ is [algebraic](#def-b3-galois-algebraic), the coefficients $c_0, \dots, c_{m-1}$ of $\pi_{\beta/L}$ generate a finite extension $F = K(c_0, \dots,
c_{m-1})$ of $K$ (repeated [tower law](#thm-b3-galois-tower)), and $F(\beta)/F$ is finite: $[F(\beta):K] < \infty$, so $\beta$ is [algebraic](#def-b3-galois-algebraic) over $K$. ∎

**Example 4.6.**

$[\Q(\sqrt2):\Q] = 2$, $[\Q(\sqrt[3]2):\Q] = 3$ ($X^3 - 2$ is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility): [Eisenstein](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#thm-b3-rings-criteria)), $[\Q(\zeta_p):\Q] = p - 1$ for $\zeta_p = \eu^{2\iu\pi/p}$ ($\Phi_p$ is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility), [Example 2.26](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#ex-b3-rings-eisenstein)). The [tower law](#thm-b3-galois-tower) is already a weapon: $\sqrt[3]2 \notin \Q(\sqrt2)$, since $3 \nmid 2$.

## 4.2 Splitting fields; algebraic closure

**Theorem 4.7 (Splitting fields).**

Let $P \in K[X]$ be nonconstant. There exists a *splitting field* of $P$ over $K$: an extension $L =
K(\alpha_1, \dots, \alpha_n)$ generated by roots of $P$ in which $P$ splits into linear factors. It is unique up to $K$-isomorphism, and $[L:K] \leq (\deg P)!$.

**Proof.** *Existence*, by induction on $\deg P$: pick an [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) factor $Q$ of $P$; the field $K_1 = K[X]/(Q)$ contains the root $\alpha_1 = \bar X$ of $Q$, hence of $P$; write $P = (X -
\alpha_1)P_1$ over $K_1$ and apply induction to $P_1$ over $K_1$; [degrees](#def-b3-galois-extension) multiply to at most $n(n-1)\cdots = n!$.

*Uniqueness* follows from the stronger *isomorphism extension lemma*: let $\sigma \colon K \to K'$ be an isomorphism, $P \in K[X]$, $P^\sigma$ the polynomial with mapped coefficients, $L, L'$ [splitting fields](#thm-b3-galois-splitting) of $P, P^\sigma$; then $\sigma$ extends to an isomorphism $L \to L'$. Induction on $[L:K]$: if $P$ splits in $K$, then $L = K$, and $L' = K'$ ($P^\sigma$ splits in $K'$, and $L'$ is generated by its roots). Otherwise choose a root $\alpha \in L \setminus K$ of an [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) factor $Q$ of $P$ with $\deg Q \geq 2$; $Q^\sigma$ is an [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) factor of $P^\sigma$, with a root $\beta \in L'$; then

$$
K(\alpha) \cong K[X]/(Q) \xrightarrow{\ \sigma\ }
K'[X]/(Q^\sigma) \cong K'(\beta)
$$

extends $\sigma$ with $\alpha \mapsto \beta$. Now $L$ is a [splitting field](#thm-b3-galois-splitting) of $P$ over $K(\alpha)$, and $L'$ of $P^\sigma$ over $K'(\beta)$, with $[L : K(\alpha)] < [L:K]$: induction extends further to $L \to L'$. ∎

**Definition 4.8.**

A field $\Omega$ is *algebraically closed* if every nonconstant polynomial of $\Omega[X]$ has a root in $\Omega$ (hence splits). An *algebraic closure* of $K$ is an [algebraic](#def-b3-galois-algebraic) extension $\bar K/K$ with $\bar K$ [algebraically](#def-b3-galois-algebraic) closed.

**Theorem 4.9 (Steinitz).**

Every field $K$ has an [algebraic closure](#def-b3-galois-closure), unique up to $K$-isomorphism.

**Proof.** *Existence (Artin’s construction).* Let $R = K[(X_f)_f]$ be the polynomial ring with one variable $X_f$ per nonconstant monic $f \in K[X]$, and $I$ the [ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal) generated by all $f(X_f)$. $I$ is proper: a relation $1 = \sum_{i=1}^r g_i\,
f_i(X_{f_i})$ involves finitely many polynomials; in a common [splitting field](#thm-b3-galois-splitting) $E$ of $f_1\cdots f_r$ pick roots $\alpha_i$ of $f_i$ and evaluate $X_{f_i} \mapsto \alpha_i$ (other variables $\mapsto 0$): $1 = 0$, absurd. Let $\mathfrak m \supseteq I$ be maximal ([Theorem 2.8](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#thm-b3-rings-krull); Zorn) and $K_1 = R/\mathfrak
m$: a [field extension](#def-b3-galois-extension) of $K$ in which every nonconstant $f \in
K[X]$ has a root, namely $\bar X_f$, and which is [algebraic](#def-b3-galois-algebraic) over $K$ (it is generated by the $\bar X_f$, each [algebraic](#def-b3-galois-algebraic)). Iterate: $K \subseteq K_1 \subseteq K_2 \subseteq \cdots$, where $K_{n+1}$ does to $K_n$ what $K_1$ did to $K$, and let $\Omega =
\bigcup_n K_n$, a field. Any nonconstant $g \in \Omega[X]$ has its finitely many coefficients in some $K_n$; an [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) factor of $g$ over $K_n$ has a root in $K_{n+1} \subseteq
\Omega$: $\Omega$ is [algebraically closed](#def-b3-galois-closure), and [algebraic](#def-b3-galois-algebraic) over $K$ (each $K_n$ is, by transitivity, [Corollary 4.5](#cor-b3-galois-algclosed)): $\Omega$ is an [algebraic closure](#def-b3-galois-closure).

*Uniqueness.* Let $\Omega, \Omega'$ be two [algebraic closures](#def-b3-galois-closure). Consider the set of pairs $(E, \tau)$ where $K
\subseteq E \subseteq \Omega$ and $\tau\colon E \to \Omega'$ is a $K$-embedding, ordered by extension; it is nonempty ($(K,
\mathrm{id})$) and inductive (union of a chain), so Zorn gives a maximal $(E_0, \tau_0)$. If $E_0 \neq \Omega$, pick $\alpha \in
\Omega\setminus E_0$: $\pi_{\alpha/E_0}$ maps to a polynomial over $\tau_0(E_0)$ having a root $\beta$ in the [algebraically closed](#def-b3-galois-closure) $\Omega'$, and $\tau_0$ extends to $E_0(\alpha) \to
\Omega'$ ($\alpha \mapsto \beta$), contradicting maximality. So there is a $K$-embedding $\tau \colon \Omega \to \Omega'$; its image, isomorphic to $\Omega$, is [algebraically closed](#def-b3-galois-closure), and $\Omega'$ is [algebraic](#def-b3-galois-algebraic) over it: for $x \in \Omega'$, $\pi_{x/\tau(\Omega)}$ splits over $\tau(\Omega)$, so $x \in
\tau(\Omega)$. Thus $\tau$ is onto: an isomorphism. ∎

**Remark 4.10.**

For $K = \Q$ one may avoid the transfinite machinery: the [algebraic](#def-b3-galois-algebraic) numbers $\bar\Q = \{z \in \C : z \text{ algebraic over
} \Q\}$ form an [algebraic closure](#def-b3-galois-closure) — a subfield of $\C$ by [Corollary 4.5](#cor-b3-galois-algclosed), [algebraically closed](#def-b3-galois-closure) because $\C$ is (d’Alembert–Gauss, proved by complex analysis in [Chapter 16](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#ch-b3-holomorphic)) and roots of polynomials over $\bar\Q$ are [algebraic](#def-b3-galois-algebraic) over $\Q$ by transitivity.

## 4.3 Finite fields

**Theorem 4.11.**

Let $p$ be prime, $n \geq 1$, $q = p^n$.

1. A [finite field](#thm-b3-galois-finitefields) has cardinality a prime power, and for each $q$ there is exactly one field $\mathbb F_q$ with $q$ elements up to isomorphism: the [splitting field](#thm-b3-galois-splitting) of $X^q - X$ over $\mathbb F_p$ .
2. The *Frobenius* $F  \colon x \mapsto x^p$ is an automorphism of $\mathbb  F_q$ , and the automorphism group of $\mathbb F_q$ is cyclic of order $n$ , generated by $F$ .
3. $\mathbb F_{p^m}$ embeds in $\mathbb F_{p^n}$ iff $m \mid  n$ .

**Proof.** (1) A [finite field](#thm-b3-galois-finitefields) $E$ has characteristic $p > 0$ and is a finite-dimensional $\mathbb F_p$-vector space: $\abs E = p^n$. Its multiplicative group has order $q - 1$, so every $x \in E$ satisfies $x^q = x$: $E$ consists of $q$ roots of $X^q - X$, hence is a [splitting field](#thm-b3-galois-splitting) of it over $\mathbb F_p$ — determining $E$ up to isomorphism ([Theorem 4.7](#thm-b3-galois-splitting)). Conversely, in a [splitting field](#thm-b3-galois-splitting) $L$ of $X^q - X$, the set $E$ of its roots is a *subfield*: $(x + y)^q = x^q + y^q$ by iterating the freshman’s dream $(a+b)^p = a^p + b^p$ ($p \mid \binom pk$), and $(xy)^q = x^qy^q$, $(x^{-1})^q = (x^q)^{-1}$; it has exactly $q$ elements since $X^q - X$ is [separable](#def-b3-galois-separable): its derivative is $qX^{q-1} - 1 = -1$ (as $p \mid q$), coprime to it, so no repeated roots. Thus $L = E$ has $q$ elements.

(2) $F$ is a field morphism (freshman’s dream), injective (fields), hence bijective on the finite $\mathbb F_q$. $F^n =
\mathrm{id}$ ($x^q = x$), and no smaller power is the identity: $F^m = \mathrm{id}$ means all $q$ elements are roots of $X^{p^m} - X$, forcing $p^m \geq q$. So $\langle F\rangle$ is cyclic of order $n$; and there are no other automorphisms, by the bound $\abs{\operatorname{Aut}} \leq [\,\mathbb F_q :
\mathbb F_p\,] = n$ proved below ([Proposition 4.16](#prop-b3-galois-embeddings) with $L = \mathbb F_q$, $K =
\mathbb F_p$: automorphisms fix the prime field).

(3) If $\mathbb F_{p^m} \subseteq \mathbb F_{p^n}$, the [tower law](#thm-b3-galois-tower) gives $p^n = (p^m)^d$: $m \mid n$. Conversely if $m \mid n$, then $p^m - 1 \mid p^n - 1$ (geometric sum), so $X^{p^m} - X$ divides $X^{p^n} - X$ (same argument on exponents: $X^{a} - 1 \mid X^{b}
- 1$ when $a \mid b$), and the roots of the former inside $\mathbb F_{p^n}$ form the required subfield, of cardinality $p^m$ (separability as in (1)). ∎

**Theorem 4.12 (Cyclicity).**

Every finite subgroup of the multiplicative group of a field is cyclic. In particular $\mathbb F_q^\times \cong
\Z/(q-1)\Z$.

**Proof.** Let $G \leq K^\times$ be finite. By the structure theorem ([Corollary 3.13](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#cor-b3-modules-abelian)), $G \cong \Z/d_1 \times \dots
\times \Z/d_s$ with $d_1 \mid \dots \mid d_s$. Every $x \in G$ then satisfies $x^{d_s} = 1$; but $X^{d_s} - 1$ has at most $d_s$ roots in the field $K$: $\abs G = d_1\cdots d_s \leq d_s$, forcing $s = 1$: $G$ is cyclic. ∎

**Example 4.13.**

$\mathbb F_8 = \mathbb F_2[X]/(X^3 + X + 1)$: the cubic has no root in $\mathbb F_2$, hence is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility). Writing $\omega =
\bar X$: $\mathbb F_8^\times$ is cyclic of order $7$, so *every* element $\neq 0, 1$ generates. The subfields of $\mathbb F_{p^{12}}$ form the divisor lattice of $12$: $\mathbb F_p, \mathbb F_{p^2}, \mathbb F_{p^3}, \mathbb F_{p^4},
\mathbb F_{p^6}, \mathbb F_{p^{12}}$ — a first, complete instance of the [Galois correspondence](#thm-b3-galois-fundamental).

## 4.4 Separability and embeddings

**Definition 4.14.**

A polynomial $P \in K[X]$ is *separable* if it has no repeated root in a [splitting field](#thm-b3-galois-splitting) — equivalently $\gcd(P, P') = 1$ (a repeated root is a common root; conversely, over the [splitting field](#thm-b3-galois-splitting), a common root is repeated; and the gcd does not change under [field extension](#def-b3-galois-extension), [Corollary 3.17](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#cor-b3-modules-descent)’s argument). An [algebraic element](#def-b3-galois-algebraic) is separable if its [minimal polynomial](#def-b3-galois-algebraic) is; an extension $L/K$ is separable if all its elements are.

**Proposition 4.15.**

An *[irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility)* $P \in K[X]$ is [separable](#def-b3-galois-separable) unless $P' = 0$, which forces $\operatorname{char} K = p > 0$ and $P \in K[X^p]$. Consequently every [algebraic](#def-b3-galois-algebraic) extension of a field of characteristic $0$, and of a [finite field](#thm-b3-galois-finitefields), is [separable](#def-b3-galois-separable) (such fields are called *perfect*).

**Proof.** $\gcd(P, P')$ divides $P$; if it is not $1$, irreducibility forces $\gcd = P$ (up to a constant), so $P \mid P'$ with $\deg
P' < \deg P$: $P' = 0$. Writing $P = \sum a_kX^k$: $ka_k = 0$ for all $k$, so in characteristic $0$, $P$ is constant (excluded); in characteristic $p$, $a_k = 0$ unless $p \mid k$: $P =
Q(X^p)$. Over a [finite field](#thm-b3-galois-finitefields), every element is a $p$-th power ([Frobenius](#thm-b3-galois-finitefields) is onto), so $Q(X^p) = \sum b_k^p X^{pk} = (\sum
b_kX^k)^p$ is not [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility): $P' = 0$ cannot happen for [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) $P$ there either. ∎

**Proposition 4.16 (Counting embeddings).**

Let $L = K(\alpha_1, \dots, \alpha_r)$ be finite over $K$, and $\sigma \colon K \to \Omega$ an embedding into an [algebraically closed field](#def-b3-galois-closure). Then the number of extensions of $\sigma$ to $L$ is at most $[L:K]$, with equality if $L/K$ is [separable](#def-b3-galois-separable). In particular $\abs{\operatorname{Aut}_K(L)} \leq [L:K]$.

**Proof.** Induction on $[L:K]$ via simple steps. For $L = K(\alpha)$: an extension $\tau$ is determined by $\tau(\alpha)$, which must be a root in $\Omega$ of $\pi_\alpha^\sigma$; conversely each such root gives one extension ($K(\alpha) \cong K[X]/(\pi_\alpha)$). The number of extensions is the number of *distinct* roots of $\pi_\alpha^\sigma$ in $\Omega$: at most $\deg\pi_\alpha =
[K(\alpha):K]$, with equality iff $\pi_\alpha$ is [separable](#def-b3-galois-separable) (separability of $\pi^\sigma$ and $\pi$ agree: gcd with the derivative is preserved by $\sigma$). In general, factor $L =
K(\alpha_1)(\alpha_2, \dots)$: extensions of $\sigma$ to $K(\alpha_1)$ number $\leq [K(\alpha_1):K]$, and each extends in $\leq [L : K(\alpha_1)]$ ways by induction; multiply ([tower law](#thm-b3-galois-tower)). In the [separable](#def-b3-galois-separable) case both counts are equalities: [minimal polynomials](#def-b3-galois-algebraic) over the bigger field $K(\alpha_1)$ divide those over $K$, hence remain [separable](#def-b3-galois-separable). ∎

**Theorem 4.17 (Primitive element).**

Every finite *[separable](#def-b3-galois-separable)* extension is simple: $L =
K(\gamma)$ for some $\gamma$.

**Proof.** If $K$ is finite, so is $L$, and a generator $\gamma$ of the cyclic group $L^\times$ ([Theorem 4.12](#thm-b3-galois-cyclic)) does it. Let $K$ be infinite; by induction it suffices to treat $L =
K(\alpha, \beta)$. Let $n = [L:K]$; by [Proposition 4.16](#prop-b3-galois-embeddings) there are $n$ distinct $K$-embeddings $\sigma_1, \dots, \sigma_n \colon L \to \Omega$ ($\Omega$ an [algebraic closure](#def-b3-galois-closure)). The polynomial

$$
D(T)=\prod_{i<j}\bigl[\bigl(\sigma_i(\alpha)-\sigma_j(\alpha)
\bigr) + T\bigl(\sigma_i(\beta) - \sigma_j(\beta)\bigr)\bigr]
$$

is not identically zero: a factor vanishes identically only if $\sigma_i, \sigma_j$ agree on both $\alpha$ and $\beta$, hence on $L$ — excluded for $i \neq j$. As $K$ is infinite, pick $c \in
K$ with $D(c) \ne 0$: then the $n$ elements $\sigma_i(\alpha +
c\beta)$ are pairwise distinct, so $\gamma = \alpha + c\beta$ has at least $n$ distinct conjugates in $\Omega$, i.e. $\deg \pi_\gamma \geq n$: $[K(\gamma):K] \geq n = [L:K]$ forces $L = K(\gamma)$. ∎

## 4.5 The Galois correspondence

**Definition 4.18.**

A finite extension $L/K$ is *Galois* if it is the [splitting field](#thm-b3-galois-splitting) of a *[separable](#def-b3-galois-separable)* polynomial over $K$. Its *Galois group* is $\operatorname{Gal}(L/K) = \operatorname{Aut}_K(L)$, the group of field automorphisms of $L$ fixing $K$ pointwise.

**Proposition 4.19.**

If $L/K$ is [Galois](#def-b3-galois-galois), then $\abs{\operatorname{Gal}(L/K)} =
[L:K]$; moreover $L/F$ is [Galois](#def-b3-galois-galois) for every intermediate field $K
\subseteq F \subseteq L$, and every $F$-embedding $L \to \Omega
\supseteq L$ has image $L$ (*normality*).

**Proof.** Let $L$ split the [separable](#def-b3-galois-separable) $P$ over $K$, and fix an [algebraic closure](#def-b3-galois-closure) $\Omega \supseteq L$. $L/K$ is [separable](#def-b3-galois-separable): it is generated by roots of $P$; separability of every element follows from the equality case below, but let us argue directly — [Proposition 4.16](#prop-b3-galois-embeddings) applied to the generators (roots of the [separable](#def-b3-galois-separable) $P$, whose [minimal polynomials](#def-b3-galois-algebraic) divide $P$) yields exactly $[L:K]$ extensions of $K \hookrightarrow \Omega$ (in the inductive step, the [minimal polynomial](#def-b3-galois-algebraic) of a root of $P$ over an intermediate field still divides $P$, hence is [separable](#def-b3-galois-separable)). Each such embedding $\tau \colon L \to \Omega$ permutes the roots of $P$ ($\tau$ fixes the coefficients), and $L$ is generated by them: $\tau(L) = L$. Hence embeddings $=$ automorphisms: $\abs{\operatorname{Gal}(L/K)} = [L:K]$. For intermediate $F$: $L$ is also the [splitting field](#thm-b3-galois-splitting) of $P$ over $F$, and $P$ remains [separable](#def-b3-galois-separable): $L/F$ is [Galois](#def-b3-galois-galois); the same argument gives normality over $F$. ∎

**Lemma 4.20 (Artin).**

Let $G$ be a finite group of automorphisms of a field $L$ and $K = L^G = \{x : \sigma(x) = x\ \forall\sigma \in G\}$ its fixed field. Then $[L : L^G] \leq \abs G$.

**Proof.** Let $n = \abs G$, $G = \{\sigma_1, \dots, \sigma_n\}$, and suppose $x_1, \dots, x_{n+1} \in L$ are linearly independent over $K$. The homogeneous linear system of $n$ equations in $n+1$ unknowns $(c_j)$ over $L$,

$$
\sum_{j=1}^{n+1} c_j\,\sigma_i(x_j) = 0
\qquad (i = 1, \dots, n),
$$

has a nonzero solution; choose one with the *fewest* nonzero entries, say $c_1, \dots, c_r \neq 0$ (renumbering), $r \geq 2$ (a single $c_j\sigma_i(x_j) = 0$ is impossible), normalized $c_r = 1$. Not all $c_j$ lie in $K$: the equation for $\sigma_i =
\mathrm{id}$ would contradict independence; say $c_1 \notin K$, so $\tau(c_1) \ne c_1$ for some $\tau \in G$. Apply $\tau$ to all equations: since $\tau\sigma_i$ runs over $G$, the vector $(\tau(c_j))_j$ is another solution; subtracting, $(c_j -
\tau(c_j))_j$ is a solution with fewer nonzero entries (the $r$-th entry $1 - 1 = 0$ vanishes, the first does not) and not zero: contradiction. So any $n+1$ elements are dependent: $[L:K]
\leq n$. ∎

**Theorem 4.21 (Fundamental theorem of Galois theory).**

Let $L/K$ be a [Galois extension](#def-b3-galois-galois) with group $G =
\operatorname{Gal}(L/K)$.

1. $L^G = K$ .
2. The maps $H \mapsto L^H$ and $F \mapsto  \operatorname{Gal}(L/F)$ are mutually inverse, inclusion-reversing bijections between subgroups of $G$ and intermediate fields $K \subseteq F \subseteq L$ ; moreover $[L : L^H] = \abs H$ and $[L^H : K] = [G : H]$ .
3. $H \trianglelefteq G$ iff $L^H/K$ is [Galois](#def-b3-galois-galois) , and then restriction induces $\operatorname{Gal}(L^H/K) \cong  G/H$ .

**Proof.** (1) Clearly $K \subseteq L^G$. Conversely let $\alpha \in L
\setminus K$; we exhibit $\sigma \in G$ with $\sigma(\alpha) \ne
\alpha$. The [minimal polynomial](#def-b3-galois-algebraic) $\pi_\alpha$ over $K$ has [degree](#def-b3-galois-extension) $\geq 2$ and is [separable](#def-b3-galois-separable) ($L/K$ [separable](#def-b3-galois-separable), [Proposition 4.19](#prop-b3-galois-galoisorder)), so it has another root $\beta \neq \alpha$ in an [algebraic closure](#def-b3-galois-closure) $\Omega \supseteq L$. Extend the $K$-embedding $K(\alpha) \to \Omega$, $\alpha \mapsto
\beta$, to an embedding $\tau\colon L \to \Omega$ ([Proposition 4.16](#prop-b3-galois-embeddings)); by normality ([Proposition 4.19](#prop-b3-galois-galoisorder)) $\tau(L) = L$, so $\tau \in
G$, $\beta = \tau(\alpha) \in L$, and $\tau(\alpha) \neq
\alpha$.

(2) For a subgroup $H$: $L/L^H$ is [Galois](#def-b3-galois-galois) ([Proposition 4.19](#prop-b3-galois-galoisorder)), and $\operatorname{Gal}(L/L^H) \supseteq H$ trivially, so $[L:L^H] =
\abs{\operatorname{Gal}(L/L^H)} \geq \abs H$; Artin’s lemma gives $[L:L^H] \leq \abs H$: equality, and $\operatorname{Gal}(L/L^H) = H$. For an intermediate field $F$: $L/F$ [Galois](#def-b3-galois-galois) gives $L^{\operatorname{Gal}(L/F)} = F$ by (1) applied to $L/F$. The two maps are mutually inverse; they reverse inclusions evidently. [Degrees](#def-b3-galois-extension): $[L:L^H] = \abs H$ just proved, and $[L^H:K] = [L:K]/[L:L^H] = \abs G/\abs H$.

(3) For $\sigma \in G$ and $H \leq G$: $\sigma(L^H) =
L^{\sigma H\sigma^{-1}}$ (direct check). By the bijection, $\sigma(L^H) = L^H$ for all $\sigma$ iff $H \trianglelefteq G$. Now if $H \trianglelefteq G$, set $F = L^H$: every $\sigma \in
G$ restricts to an automorphism of $F$, giving a morphism $\rho
\colon G \to \operatorname{Aut}_K(F)$ with kernel $\{\sigma :
\sigma\restriction_F = \mathrm{id}\} = \operatorname{Gal}(L/F) =
H$. So $G/H$ embeds in $\operatorname{Aut}_K(F)$, whence $\abs{\operatorname{Aut}_K(F)} \geq [G:H] = [F:K]$; the reverse inequality always holds ([Proposition 4.16](#prop-b3-galois-embeddings)): $\abs{\operatorname{Aut}_K(F)} = [F:K]$ and $\rho$ is onto. It remains to see $F/K$ is [Galois](#def-b3-galois-galois): $F$ is [separable](#def-b3-galois-separable) over $K$ (inside the [separable](#def-b3-galois-separable) $L/K$), and $F = K(\gamma)$ ([Theorem 4.17](#thm-b3-galois-primitive)); the polynomial $\prod_{\sigma \in G/H}\bigl(X - \sigma(\gamma)\bigr)$ (product over the distinct images, which lie in $F$: $\sigma(F) =
L^{\sigma H\sigma^{-1}} = L^H = F$ by normality of $H$) has coefficients fixed by $G$, hence in $K$ by (1): it is a [separable polynomial](#def-b3-galois-separable) of $K[X]$ split by $F$, and its roots generate $F$: $F/K$ is [Galois](#def-b3-galois-galois). Conversely, if $F = L^H$ with $F/K$ [Galois](#def-b3-galois-galois), normality of $F$ ([Proposition 4.19](#prop-b3-galois-galoisorder), applied to embeddings $F \to \Omega$ restricted from elements of $G$) gives $\sigma(F) = F$ for all $\sigma \in G$, i.e. $H
\trianglelefteq G$. ∎

![The Galois correspondence for the splitting field L of X3 - 2 over ℚ (j = 2 π/3): subgroups of Gal(L/ℚ) S_3 (left, order reversed) match intermediate fields (right). The unique normal proper subgroup (1\,2\,3) corresponds to the unique subextension ℚ( √3)/ℚ that is Galois; the three conjugate subgroups (i\,j) correspond to the three conjugate cubic fields ℚ(jk√(3)2), none of them normal over ℚ.](https://one-course.com/images/onecourse/chapters/math-5/b3-galois/fig-94b147c908a8.svg)

*The [Galois correspondence](#thm-b3-galois-fundamental) for the [splitting field](#thm-b3-galois-splitting) $L$ of $X^3 - 2$ over $\Q$ ($j = \eu^{2\iu\pi/3}$): subgroups of $\operatorname{Gal}(L/\Q) \cong S_3$ (left, order reversed) match intermediate fields (right). The unique normal proper subgroup $\langle(1\,2\,3)\rangle$ corresponds to the unique subextension $\Q(\iu\sqrt3)/\Q$ that is [Galois](#def-b3-galois-galois); the three conjugate subgroups $\langle(i\,j)\rangle$ correspond to the three conjugate cubic fields $\Q(j^k\sqrt[3]2)$, none of them normal over $\Q$.*

## 4.6 Cyclotomic extensions

**Definition 4.22.**

Let $n \geq 1$ and $\zeta_n = \eu^{2\iu\pi/n}$. The $n$-th *cyclotomic polynomial* is $\Phi_n = \prod_{\gcd(k,n)=1,\ 1 \le k \le n} \bigl(X -
\zeta_n^k\bigr)$, of [degree](#def-b3-galois-extension) $\varphi(n)$; grouping the roots of $X^n - 1$ by exact order, $X^n - 1 = \prod_{d \mid n}\Phi_d$, which shows inductively that $\Phi_n \in \Z[X]$ (Euclidean division of monic integer polynomials).

**Theorem 4.23.**

$\Phi_n$ is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) over $\Q$; hence $[\Q(\zeta_n) : \Q] =
\varphi(n)$ and

$$
\operatorname{Gal}\bigl(\Q(\zeta_n)/\Q\bigr) \;\cong\;
(\Z/n\Z)^\times,
\qquad \sigma_a(\zeta_n) = \zeta_n^a .
$$

The extension $\Q(\zeta_n)/\Q$ is thus [Galois](#def-b3-galois-galois) with *abelian* group.

**Proof.** Let $f = \pi_{\zeta_n}$, so $\Phi_n = fg$ with $f, g \in \Z[X]$ monic (Gauss’s lemma [Lemma 2.23](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#lem-b3-rings-gauss): contents multiply, all polynomials monic). *Claim: if $\zeta$ is a root of $f$ and $p \nmid n$ is prime, then $\zeta^p$ is a root of $f$.* Otherwise $\zeta^p$ is a root of $g$ (it is a primitive $n$-th root of unity), so $\zeta$ is a root of $g(X^p)$, and $f
\mid g(X^p)$ in $\Z[X]$ ([minimal polynomial](#def-b3-galois-algebraic), then Gauss again). Reduce mod $p$: $\bar g(X^p) = \bar g(X)^p$ ([Frobenius](#thm-b3-galois-finitefields) on $\mathbb F_p[X]$: coefficientwise $a^p = a$, and freshman’s dream), so $\bar f \mid \bar g^{\,p}$: $\bar f$ and $\bar g$ share an [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) factor, and $\bar\Phi_n = \bar f\bar g$ has a repeated factor. Then so does $X^n - \bar 1$; but its derivative $\bar nX^{n-1}$ is coprime to it ($p \nmid n$, and $0$ is not a root): contradiction.

Every primitive root $\zeta_n^k$ ($\gcd(k, n) = 1$) is obtained from $\zeta_n$ by successive prime powers not dividing $n$ (factor $k$): the claim propagates, so every primitive root is a root of $f$: $f = \Phi_n$, [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility). Consequently $[\Q(\zeta_n):\Q] = \varphi(n)$, and $\Q(\zeta_n)$ is the [splitting field](#thm-b3-galois-splitting) of the [separable](#def-b3-galois-separable) $X^n - 1$ (all roots are powers of $\zeta_n$): [Galois](#def-b3-galois-galois). An automorphism $\sigma$ sends $\zeta_n$ to another primitive root $\zeta_n^{a(\sigma)}$, and $\sigma
\mapsto a(\sigma)$ is an injective morphism into $(\Z/n\Z)^\times$; both groups have order $\varphi(n)$: isomorphism. ∎

## 4.7 Ruler and compass

**Definition 4.24.**

Identify the plane with $\C$; start from $\{0, 1\}$. A point is *constructible* if it is obtainable by finitely many intersections of lines through two already-constructed points and circles centered at a constructed point with radius a distance of two constructed points.

**Theorem 4.25 (Wantzel).**

$z \in \C$ is [constructible](#def-b3-galois-constructible) iff there is a tower $\Q = F_0
\subseteq F_1 \subseteq \dots \subseteq F_r$ with $[F_{i+1} :
F_i] = 2$ and $z \in F_r$. In particular, a [constructible number](#def-b3-galois-constructible) is [algebraic](#def-b3-galois-algebraic) of [degree](#def-b3-galois-extension) a power of $2$ over $\Q$.

**Proof.** ($\Rightarrow$) The coordinates of the intersection of two lines through points with coordinates in a subfield $F \subseteq \R$ solve a linear system over $F$: they stay in $F$. Line–circle and circle–circle intersections lead, after eliminating the linear part (subtracting the two circle equations gives a line), to a quadratic equation over $F$: the new coordinates lie in $F$ or in $F(\sqrt d)$ for some $d \in F$, $d > 0$. By induction, every constructed point has coordinates in a tower of quadratic extensions of $\Q$; and $z = x + \iu y$ is in a quadratic tower too (adjoin $\iu$: one more quadratic step). The [degree](#def-b3-galois-extension) consequence: $[\Q(z):\Q]$ divides $[F_r : \Q] = 2^r$ ([tower law](#thm-b3-galois-tower)).

($\Leftarrow$) The [constructible numbers](#def-b3-galois-constructible) form a field: sums and differences by parallelograms (parallels are [constructible](#def-b3-galois-constructible): drop and raise perpendiculars twice — the classical perpendicular through a point uses one circle and two arcs); products and quotients by Thales’ intercept configurations (given lengths $a, b$ construct $ab$ and $a/b$ with similar triangles on two rays). And the field is closed under square roots: for $a > 0$, the circle of diameter $1 + a$ and the perpendicular at the junction point meet at height $\sqrt a$ (altitude-geometric-mean relation in a right triangle); for a complex $w = \rho\eu^{\iu\theta}$, construct $\sqrt\rho$ and bisect $\theta$ (angle bisection is a compass construction). Real and imaginary parts of members of a quadratic tower are therefore [constructible](#def-b3-galois-constructible) by induction on the tower: each step adjoins roots of a quadratic, expressible by field operations and one square root of an already-constructed number (the quadratic formula; in characteristic $0$). ∎

**Corollary 4.26.**

The three classical problems are unsolvable by ruler and compass:

1. *Duplication of the cube* : $\sqrt[3]2$ has [degree](#def-b3-galois-extension) $3$ , not a power of $2$ .
2. *Trisection of the angle* : trisecting $60^\circ$ requires $\cos 20^\circ$ , a root of the [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) $8X^3  - 6X - 1$ : [degree](#def-b3-galois-extension) $3$ .
3. *Squaring the circle* : $\sqrt\pi$ is transcendental ( $\pi$ is — Lindemann’s theorem, admitted here: its proof belongs to a course in transcendence theory).

Also, the regular $n$-gon is [constructible](#def-b3-galois-constructible) iff $\varphi(n)$ is a power of $2$ (Gauss–Wantzel; the “if” uses the weekend problem’s method, the “only if” is [Theorem 4.25](#thm-b3-galois-wantzel) applied to $\zeta_n$, of [degree](#def-b3-galois-extension) $\varphi(n)$). For $n = 7$: $\varphi(7) = 6$: the regular heptagon is impossible; for $n = 17$: $\varphi(17) = 16 = 2^4$: [constructible](#def-b3-galois-constructible) — the weekend problem constructs it.

**Proof.** (1) $X^3 - 2$ is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) ([Eisenstein](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#thm-b3-rings-criteria)). (2) From $\cos
3\theta = 4\cos^3\theta - 3\cos\theta$ with $3\theta =
60^\circ$: $8c^3 - 6c = 1$ for $c = \cos 20^\circ$; the cubic $8X^3 - 6X - 1$ has no rational root (candidates $\pm1, \pm\frac
12, \pm\frac14, \pm\frac18$ fail), hence is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility): [degree](#def-b3-galois-extension) $3$. A general $60^\circ$ angle is [constructible](#def-b3-galois-constructible), so a trisector would construct $c$. (3) If $\sqrt\pi$ were [constructible](#def-b3-galois-constructible) it would be [algebraic](#def-b3-galois-algebraic), hence also $\pi$. The $n$-gon statement: the [degree](#def-b3-galois-extension) of $\zeta_n$ is $\varphi(n)$ ([Theorem 4.23](#thm-b3-galois-cyclotomicirred)); necessity follows from Wantzel; for sufficiency, the [Galois group](#def-b3-galois-galois), abelian of order $2^m$, admits a chain of index-$2$ subgroups (a finite $2$-group does: [Exercise 1.10](https://one-course.com/books/math/5/en/chapter/1-group-theory#exo-b3-groups-10)), whose fixed fields form a quadratic tower ending at $\Q(\zeta_n)$ ([Theorem 4.21](#thm-b3-galois-fundamental)); conclude by [Theorem 4.25](#thm-b3-galois-wantzel). ∎

## 4.8 Solvability by radicals

**Definition 4.27.**

An extension $L/K$ (characteristic $0$ throughout this section) is *radical* if there is a tower $K = F_0 \subseteq \dots
\subseteq F_r = L$ with $F_{i+1} = F_i(\alpha_i)$, $\alpha_i^{n_i} \in F_i$: each step adjoins an $n_i$-th root. A polynomial $P \in K[X]$ is *solvable by radicals* if its [splitting field](#thm-b3-galois-splitting) is contained in some radical extension of $K$.

**Lemma 4.28.**

Let $K$ contain a primitive $n$-th root of unity $\zeta$, i.e. $\zeta$ of order $n$ in $K^\times$, and $a \in K^\times$. Then $K(\sqrt[n]a)/K$ is [Galois](#def-b3-galois-galois) with *cyclic* group. Conversely — not needed below — every cyclic extension of [degree](#def-b3-galois-extension) $n$ is of this form. Moreover $K(\zeta_n)/K$ is [Galois](#def-b3-galois-galois) with *abelian* group, for any $K$ of characteristic $0$.

**Proof.** $X^n - a$ is [separable](#def-b3-galois-separable) ($\gcd$ with $nX^{n-1}$: $a \neq 0$) and splits in $K(\alpha)$, $\alpha^n = a$: its roots are the $\zeta^k\alpha \in K(\alpha)$. So $K(\alpha)/K$ is [Galois](#def-b3-galois-galois); the map $\sigma \mapsto \sigma(\alpha)/\alpha \in \mu_n = \langle
\zeta\rangle$ is an injective morphism ($\sigma\tau(\alpha) =
\sigma(\tau(\alpha)/\alpha \cdot \alpha) =
\tau(\alpha)/\alpha\cdot\sigma(\alpha)$, the quotient being in $K$), into a cyclic group: $\operatorname{Gal}$ is cyclic. The converse is Kummer theory, which we shall not need (see the remark below). For $K(\zeta_n)$: it splits the [separable](#def-b3-galois-separable) $X^n -
1$, and $\sigma \mapsto a(\sigma)$ with $\sigma(\zeta_n) =
\zeta_n^{a(\sigma)}$ embeds the group in the abelian $(\Z/n\Z)^\times$ as in [Theorem 4.23](#thm-b3-galois-cyclotomicirred) (injectivity only needs $\zeta_n$ to generate the roots of unity involved). ∎

**Theorem 4.29 (Galois).**

Let $K$ be of characteristic $0$ and $P \in K[X]$ with [splitting field](#thm-b3-galois-splitting) $L$. If $P$ is [solvable by radicals](#def-b3-galois-radical), then $\operatorname{Gal}(L/K)$ is a [solvable group](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived). (The converse is also true; we shall not need it.)

**Proof.** Step 1: *enlarge the [radical](#def-b3-galois-radical) tower to a [Galois](#def-b3-galois-galois) one.* Let $L
\subseteq M$ with $M/K$ [radical](#def-b3-galois-radical), with root exponents $n_1, \dots,
n_r$ and $n = n_1\cdots n_r$. First adjoin $\zeta_n$: the tower $K \subseteq K(\zeta_n) \subseteq M(\zeta_n)$ is still [radical](#def-b3-galois-radical) ($\zeta_n$ is a root of unity: a [radical](#def-b3-galois-radical) step, $\zeta_n^n = 1$), and its steps beyond the first happen over fields containing the needed roots of unity. Next, replace $M(\zeta_n)$ by the composite $N$ of all $\sigma(M(\zeta_n))$, $\sigma$ ranging over the (finitely many) $K$-embeddings of $M(\zeta_n)$ into a fixed [algebraic closure](#def-b3-galois-closure): $N$ is the [splitting field](#thm-b3-galois-splitting) of the product of [minimal polynomials](#def-b3-galois-algebraic) of a generating set (characteristic $0$: finite and [separable](#def-b3-galois-separable)), hence $N/K$ is [Galois](#def-b3-galois-galois); and $N$ is [radical](#def-b3-galois-radical) over $K$: each $\sigma(M(\zeta_n))$ is [radical](#def-b3-galois-radical) over $K$ (apply $\sigma$ to a [radical](#def-b3-galois-radical) tower), and a composite of [radical](#def-b3-galois-radical) extensions is [radical](#def-b3-galois-radical) (concatenate the towers: if $F'/K$ is [radical](#def-b3-galois-radical) with tower adjoining $\beta_j$, then $F''(\beta_j)$-type steps remain [radical](#def-b3-galois-radical) over any bigger base).

Step 2: *read [solvability](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) off the [Galois](#def-b3-galois-galois) tower.* So assume $L \subseteq N$, $N/K$ [Galois](#def-b3-galois-galois) and [radical](#def-b3-galois-radical) with tower $K
\subseteq K(\zeta_n) = E_0 \subseteq E_1 \subseteq \dots
\subseteq E_s = N$, each $E_{i+1} = E_i(\sqrt[n_i]{a_i})$ with $\zeta_{n_i} \in E_0 \subseteq E_i$. Let $G =
\operatorname{Gal}(N/K)$ and $G_i = \operatorname{Gal}(N/E_i)$: a decreasing chain $G \supseteq G_0 \supseteq G_1 \supseteq
\dots \supseteq G_s = \{e\}$. Each $E_{i+1}/E_i$ is [Galois](#def-b3-galois-galois) with cyclic group ([Lemma 4.28](#lem-b3-galois-cyclicsteps)), so by the fundamental theorem applied to the [Galois extension](#def-b3-galois-galois) $N/E_i$ ([Theorem 4.21](#thm-b3-galois-fundamental)(3), with ambient group $G_i$): $G_{i+1} \trianglelefteq G_i$ with $G_i/G_{i+1} \cong
\operatorname{Gal}(E_{i+1}/E_i)$ cyclic. Similarly $E_0/K$ is [Galois](#def-b3-galois-galois) with abelian group $G/G_0$ ([Lemma 4.28](#lem-b3-galois-cyclicsteps)). The chain exhibits $G$ as [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) ([Proposition 1.29](https://one-course.com/books/math/5/en/chapter/1-group-theory#prop-b3-groups-derived)). Finally $\operatorname{Gal}(L/K)$ is a *quotient* of $G$: $L/K$ is [Galois](#def-b3-galois-galois) ($P$ [separable](#def-b3-galois-separable) in characteristic $0$) and restriction $G
\to \operatorname{Gal}(L/K)$ is onto ([Theorem 4.21](#thm-b3-galois-fundamental)(3) with $H =
\operatorname{Gal}(N/L)$); quotients of [solvable groups](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) are [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived). ∎

**Corollary 4.30 (Unsolvability of the quintic).**

There are polynomials of [degree](#def-b3-galois-extension) $5$ over $\Q$ that are not [solvable by radicals](#def-b3-galois-radical): for instance $X^5 - 4X + 2$, whose [Galois group](#def-b3-galois-galois) is $S_5$ ([Exercise 4.11](#exo-b3-galois-11)), a [non-solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) group ([Corollary 1.34](https://one-course.com/books/math/5/en/chapter/1-group-theory#cor-b3-groups-snnotsolvable)). No general formula in [radicals](#def-b3-galois-radical) can exist for [degree](#def-b3-galois-extension) $\geq 5$.

**Remark 4.31.**

The converse of [Theorem 4.29](#thm-b3-galois-solvable) — a [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) [Galois group](#def-b3-galois-galois) implies [solvability](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) by [radicals](#def-b3-galois-radical) — is proved by descending the [derived series](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) and showing every cyclic extension (with enough roots of unity) is [radical](#def-b3-galois-radical), via *Lagrange resolvents*; it explains why [degrees](#def-b3-galois-extension) $2, 3, 4$ have formulas: $S_2, S_3, S_4$ are [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) ([Example 1.30](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-solvableexamples)). We leave it admitted at this level; a full treatment belongs to a master’s course, but [Exercise 4.8](#exo-b3-galois-8) makes it concrete for the cubic.

## 4.9 Exercises

**Exercise 4.1 ★.**

Show $[\Q(\sqrt2, \sqrt3):\Q] = 4$, that $\Q(\sqrt2 + \sqrt3) =
\Q(\sqrt2, \sqrt3)$, and compute the [minimal polynomial](#def-b3-galois-algebraic) of $\sqrt2 + \sqrt3$ over $\Q$.

**Solution of Exercise 4.1.**

$\sqrt3 \notin \Q(\sqrt2)$: from $\sqrt3 = a + b\sqrt2$ ($a, b
\in \Q$), squaring gives $3 = a^2 + 2b^2 + 2ab\sqrt2$, so $ab =
0$; $b = 0$ makes $\sqrt3$ rational, $a = 0$ gives $\sqrt6 = 2b
\in \Q$ — both false (standard prime-factorization arguments). Hence $[\Q(\sqrt2,\sqrt3) : \Q(\sqrt2)] = 2$ and the [tower law](#thm-b3-galois-tower) gives [degree](#def-b3-galois-extension) $4$.

Let $\gamma = \sqrt2 + \sqrt3$. Then $\gamma^2 = 5 + 2\sqrt6$ and $(\gamma^2 - 5)^2 = 24$: $\gamma$ annihilates $X^4 - 10X^2 + 1$. Moreover $\gamma^3 = 11\sqrt2 + 9\sqrt3$, so $\gamma^3 - 9\gamma
= 2\sqrt2$: $\sqrt2 \in \Q(\gamma)$, then $\sqrt3 = \gamma -
\sqrt2 \in \Q(\gamma)$: $\Q(\gamma) = \Q(\sqrt2,\sqrt3)$, of [degree](#def-b3-galois-extension) $4$. The annihilating quartic, having the [degree](#def-b3-galois-extension) of the [minimal polynomial](#def-b3-galois-algebraic), *is* the [minimal polynomial](#def-b3-galois-algebraic): $X^4 -
10X^2 + 1$ (in particular it is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) over $\Q$).

**Exercise 4.2 ★.**

Let $\alpha = \sqrt[3]2$. Show that $\Q(\alpha)/\Q$ is not normal (exhibit an embedding $\Q(\alpha) \to \C$ whose image is not $\Q(\alpha)$), determine the [splitting field](#thm-b3-galois-splitting) $L$ of $X^3 -
2$ and $[L:\Q]$, and check $\operatorname{Aut}_\Q(\Q(\alpha)) =
\{\mathrm{id}\}$: for [non-Galois](#def-b3-galois-galois) extensions, the automorphism group can be far smaller than the [degree](#def-b3-galois-extension).

**Solution of Exercise 4.2.**

The three roots of $X^3 - 2$ in $\C$ are $\alpha, j\alpha,
j^2\alpha$ with $j = \eu^{2\iu\pi/3}$. The map $\alpha \mapsto
j\alpha$ defines a $\Q$-embedding $\Q(\alpha) \to \C$ ([Theorem 4.4](#thm-b3-galois-simple): both generate degree-$3$ extensions with the same [minimal polynomial](#def-b3-galois-algebraic)), whose image $\Q(j\alpha)
\not\subseteq \R$ differs from $\Q(\alpha) \subseteq \R$: $\Q(\alpha)/\Q$ is not normal. The [splitting field](#thm-b3-galois-splitting) is $L =
\Q(\alpha, j)$, with $[L:\Q] = [L:\Q(\alpha)]\,[\Q(\alpha):\Q] =
2 \cdot 3 = 6$ ($j$ satisfies $X^2 + X + 1$, [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) over the real field $\Q(\alpha)$). An automorphism of $\Q(\alpha)$ must send $\alpha$ to a root of $X^3 - 2$ *inside* $\Q(\alpha) \subseteq \R$: only $\alpha$ qualifies, so $\operatorname{Aut}_\Q(\Q(\alpha)) = \{\mathrm{id}\}$, of order $1 < 3$.

**Exercise 4.3 ★.**

Construct $\mathbb F_9$ as $\mathbb F_3[X]/(X^2+1)$ and find a generator of $\mathbb F_9^\times$. List the monic [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) polynomials of [degrees](#def-b3-galois-extension) $1, 2, 3$ over $\mathbb F_2$, and verify $X^8 - X = X(X+1)(X^3+X+1)(X^3+X^2+1)$ over $\mathbb F_2$.

**Solution of Exercise 4.3.**

$X^2 + 1$ has no root in $\mathbb F_3$ ($0, 1, 2 \mapsto 1, 2,
2$), so $\mathbb F_9 = \mathbb F_3[X]/(X^2+1)$ is a field with $9$ elements; write $\omega = \bar X$, $\omega^2 = -1$. The group $\mathbb F_9^\times$ is cyclic of order $8$; $\omega$ has order $4$, but $1 + \omega$ works: $(1+\omega)^2 = 1 + 2\omega +
\omega^2 = 2\omega$, $(1+\omega)^4 = 4\omega^2 = \omega^2 = -1
\ne 1$: order $8$.

Over $\mathbb F_2$ — [degree](#def-b3-galois-extension) $1$: $X$, $X + 1$; [degree](#def-b3-galois-extension) $2$: $X^2 + X + 1$ (the other three quadratics have roots); [degree](#def-b3-galois-extension) $3$: $X^3 + X + 1$ and $X^3 + X^2 + 1$ (no roots in $\mathbb
F_2$; the other six cubics have roots). Verification:

$$
(X^3{+}X{+}1)(X^3{+}X^2{+}1) = X^6 + X^5 + X^4 + X^3 + X^2 + X
+ 1,
$$

and $X(X{+}1)(X^6 + \dots + 1) = X(X^7 + 1) = X^8 + X = X^8 -
X$ over $\mathbb F_2$ — exactly the [irreducibles](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) of [degree](#def-b3-galois-extension) dividing $3$, as [Exercise 4.6](#exo-b3-galois-6) predicts ([degree](#def-b3-galois-extension) $2$ is absent: $2 \nmid 3$).

**Exercise 4.4 ★★.**

(a) Find all primitive roots modulo $7$ and modulo $11$ (i.e. generators of $\mathbb F_7^\times$, $\mathbb F_{11}^\times$). (b) Show that for $p$ odd, $x \in \mathbb F_p^\times$ is a square iff $x^{(p-1)/2} = 1$ (Euler’s criterion), and recover the criterion for $-1$ of [Problem 2.1](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#pb-b3-rings-1).

**Solution of Exercise 4.4.**

(a) Mod $7$: the powers of $3$ are $3, 2, 6, 4, 5, 1$: order $6$, a generator; the primitive roots are the $3^k$ with $\gcd(k, 6) = 1$: $3$ and $3^5 = 5$. Mod $11$: powers of $2$: $2, 4, 8, 5, 10, 9, 7, 3, 6, 1$: a generator; primitive roots $2^k$, $\gcd(k, 10) = 1$: $2, 2^3 = 8, 2^7 = 7, 2^9 = 6$.

(b) Write $x = g^k$ with $g$ a generator ([Theorem 4.12](#thm-b3-galois-cyclic)). Then $x$ is a square iff $k$ is even (squares are the $g^{2l}$, and $g^{2l} = g^{k}$ iff $k
\equiv 2l \bmod p-1$, [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) iff $k$ even, $p - 1$ being even). And $x^{(p-1)/2} = g^{k(p-1)/2} = 1$ iff $(p-1) \mid
k\frac{p-1}2$ iff $k$ even: the two conditions agree. For $x =
-1 = g^{(p-1)/2}$: it is a square iff $\frac{p-1}2$ is even, iff $p \equiv 1 \pmod 4$ — [Problem 2.1](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#pb-b3-rings-1) again.

**Exercise 4.5 ★★.**

Show that $\mathbb F_{p^m} \cap \mathbb F_{p^n} = \mathbb
F_{p^{\gcd(m,n)}}$ and $\mathbb F_{p^m}\mathbb F_{p^n} = \mathbb
F_{p^{\operatorname{lcm}(m,n)}}$ inside a fixed [algebraic closure](#def-b3-galois-closure) $\bar{\mathbb F}_p$, and describe $\operatorname{Gal}(\mathbb F_{p^n}/\mathbb F_{p^m})$ for $m
\mid n$.

**Solution of Exercise 4.5.**

Inside $\bar{\mathbb F}_p$, $\mathbb F_{p^k} = \{x : x^{p^k} =
x\}$ is the fixed set of $F^k$. The intersection $\mathbb F_{p^m}
\cap \mathbb F_{p^n}$ is fixed by $F^m$ and $F^n$, hence by $F^{\gcd(m,n)}$ ($\gcd = am + bn$: on a fixed element, $F^{am +
bn} = (F^m)^a(F^n)^b$ acts trivially — exponents may be taken positive by periodicity); so it lies in $\mathbb
F_{p^{\gcd(m,n)}}$, which is conversely contained in both ([Theorem 4.11](#thm-b3-galois-finitefields)(3)). The composite $\mathbb
F_{p^m}\mathbb F_{p^n}$: any field containing both has [degree](#def-b3-galois-extension) divisible by $m$ and $n$, hence by $\operatorname{lcm}(m,n)$; and $\mathbb F_{p^{\operatorname{lcm}}}$ contains both: it is the composite. For $m \mid n$: $\operatorname{Gal}(\mathbb
F_{p^n}/\mathbb F_{p^m})$ consists of the powers of $F$ fixing $\mathbb F_{p^m}$, i.e. of $F^m$: cyclic of order $n/m$, generated by $F^m \colon x \mapsto x^{p^m}$ (order as in [Theorem 4.11](#thm-b3-galois-finitefields)(2)).

**Exercise 4.6 ★★.**

Let $I_d(q)$ be the number of monic [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) polynomials of [degree](#def-b3-galois-extension) $d$ over $\mathbb F_q$. Prove

$$
X^{q^n} - X \;=\; \prod_{d \mid n}\ \prod_{P \text{ irred.
monic, } \deg P = d} P ,
\qquad\text{hence}\qquad
q^n = \sum_{d \mid n} d\, I_d(q).
$$

Deduce $I_1, I_2, I_3, I_4$ explicitly, and $I_d(q) \geq 1$ for every $d$ (so extensions $\mathbb F_{q^d}/\mathbb F_q$ exist as quotients $\mathbb F_q[X]/(P)$ for all $d$).

**Solution of Exercise 4.6.**

$X^{q^n} - X$ is [separable](#def-b3-galois-separable) (derivative $-1$) with root set $\mathbb F_{q^n}$. Let $P$ be monic [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) of [degree](#def-b3-galois-extension) $d$. If $d \mid n$: $\mathbb F_q[X]/(P) \cong \mathbb F_{q^d}
\subseteq \mathbb F_{q^n}$, so $P$ has a root $\alpha \in
\mathbb F_{q^n}$; $\alpha^{q^n} = \alpha$, and $P =
\pi_\alpha$ divides $X^{q^n} - X$. If $P \mid X^{q^n} - X$: a root $\alpha \in \mathbb F_{q^n}$ generates $\mathbb F_{q^d}
\subseteq \mathbb F_{q^n}$, so $d \mid n$ ([Theorem 4.11](#thm-b3-galois-finitefields)(3)). Distinct [irreducibles](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) are coprime and the product is [separable](#def-b3-galois-separable): each $P$ appears with exponent exactly $1$, and every root of $X^{q^n}-X$ is a root of its [minimal polynomial](#def-b3-galois-algebraic): the factorization holds. Comparing [degrees](#def-b3-galois-extension): $q^n = \sum_{d\mid n} d\,I_d(q)$.

Consequently $I_1 = q$; $q^2 = I_1 + 2I_2$ gives $I_2 =
\frac{q^2 - q}2$; $q^3 = I_1 + 3I_3$ gives $I_3 = \frac{q^3 -
q}3$; $q^4 = I_1 + 2I_2 + 4I_4$ gives $I_4 = \frac{q^4 -
q^2}4$. Existence: $nI_n = q^n - \sum_{d \mid n,\, d < n}
dI_d \geq q^n - \sum_{d \leq n/2} q^d > q^n - q^{n/2 + 1} \geq
0$ for $n \geq 2$ (and $I_1 = q \geq 1$): $I_n \geq 1$ always.

**Exercise 4.7 ★★.**

Determine $\operatorname{Gal}(\Q(\sqrt2,\sqrt3)/\Q)$ and the complete lattice of intermediate fields. Same question for the [splitting field](#thm-b3-galois-splitting) of $(X^2-2)(X^2-3)(X^2-6)$ — what do you notice?

**Solution of Exercise 4.7.**

$L = \Q(\sqrt2, \sqrt3)$ is the [splitting field](#thm-b3-galois-splitting) of $(X^2 -
2)(X^2 - 3)$, [separable](#def-b3-galois-separable): [Galois](#def-b3-galois-galois) of [degree](#def-b3-galois-extension) $4$ ([Exercise 4.1](#exo-b3-galois-1)). An automorphism sends $\sqrt2 \mapsto
\pm\sqrt2$ and $\sqrt3 \mapsto \pm\sqrt3$: at most $4$ choices, and $\abs G = 4$ realizes all: $G \cong (\Z/2\Z)^2$, with elements $\mathrm{id}, \sigma (\sqrt2 \mapsto -\sqrt2), \tau
(\sqrt3\mapsto-\sqrt3), \sigma\tau$. Subgroups of order $2$: $\langle\sigma\rangle, \langle\tau\rangle,
\langle\sigma\tau\rangle$, with fixed fields $\Q(\sqrt3)$, $\Q(\sqrt2)$, $\Q(\sqrt6)$ (note $\sigma\tau$ fixes $\sqrt6 =
\sqrt2\sqrt3$). The lattice: $\Q$ below, the three quadratic fields in the middle, $L$ on top — and nothing else ([Theorem 4.21](#thm-b3-galois-fundamental)). For $(X^2-2)(X^2-3)(X^2-6)$: the [splitting field](#thm-b3-galois-splitting) is the *same* $L$ ($\sqrt6 =
\sqrt2\sqrt3$), so the answer is identical: the [Galois correspondence](#thm-b3-galois-fundamental) is an invariant of the *extension*, not of the polynomial chosen to present it.

**Exercise 4.8 ★★.**

(The cubic, solved by its group) Let $P = X^3 + pX + q \in
\Q[X]$ be [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) with roots $x_1, x_2, x_3$ and [splitting field](#thm-b3-galois-splitting) $L$. Let $\delta = (x_1 - x_2)(x_1 - x_3)(x_2 - x_3)$ and $\Delta = \delta^2 = -4p^3 - 27q^2$ (admit this classical identity or verify it by expanding symmetric functions). (a) Show $\operatorname{Gal}(L/\Q) \cong A_3$ or $S_3$, according to whether $\Delta$ is or is not a square in $\Q$. (b) With $j = \zeta_3$, define the Lagrange resolvents $u = x_1
+ jx_2 + j^2x_3$ and $v = x_1 + j^2x_2 + jx_3$. Show $u^3 + v^3
= -27q$ and $uv = -3p$, and solve for $u^3, v^3$: Cardano’s formulas drop out. Where did [solvability](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) of $S_3$ get used?

**Solution of Exercise 4.8.**

(a) $G$ acts faithfully and transitively (irreducibility) on the three roots: $G \hookrightarrow S_3$ with $3 \mid \abs G$: $G
\cong A_3$ or $S_3$. Every $\sigma \in G$ permutes the $x_i$, and $\sigma(\delta) = \varepsilon(\sigma)\,\delta$ ($\delta$ is alternating in the roots). If $\Delta$ is a square in $\Q$: $\delta \in \Q^\times$ (note $\delta \neq 0$: [separable](#def-b3-galois-separable)), so $\varepsilon(\sigma) = 1$ for all $\sigma$: $G \subseteq A_3$, hence $= A_3$. If not: $\delta \notin \Q$, so some $\sigma$ has $\varepsilon(\sigma) = -1$: $G = S_3$. (In both cases $\Q(\delta)
= L^{G \cap A_3}$.)

(b) With $x_1 + x_2 + x_3 = 0$: $u + v = 2x_1 - (x_2 + x_3) =
3x_1$. Also

$$
uv = \sum_i x_i^2 + (j + j^2)\sum_{i<k}x_ix_k
= \Bigl(\sum x_i\Bigr)^2 - 3\sum_{i<k}x_ix_k = -3p,
$$

using $j + j^2 = -1$ and $\sum_{i<k}x_ix_k = p$. Then

$$
u^3 + v^3 = (u+v)^3 - 3uv(u+v) = 27x_1^3 + 9p\cdot 3x_1
= 27\,(x_1^3 + px_1) = -27q .
$$

So $u^3, v^3$ are the roots of $Y^2 + 27qY - 27p^3 = 0$ (product $(uv)^3 = -27p^3$): $u^3 = \frac{-27q +
\sqrt{729q^2 + 108p^3}}2$, and $x_1 = \frac{u + v}3$ with $v =
-3p/u$: Cardano. [Solvability](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) of $S_3$ is the skeleton: the tower $\Q \subseteq \Q(\delta) \subseteq \Q(\delta, j, u)$ adjoins first a square root ($\delta$, fixed field of $A_3$: the step $S_3 \to S_3/A_3$), then a cube root ($u$, since $u^3
\in \Q(\delta, j)$: the step $A_3 \to \{e\}$) — the [derived series](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) $S_3 \supset A_3 \supset \{e\}$ made flesh.

**Exercise 4.9 ★★.**

In $\Q(\zeta_5)$: show that the unique quadratic subfield is $\Q(\sqrt5)$, via the Gauss sums $\eta_0 = \zeta_5 + \zeta_5^4$, $\eta_1 = \zeta_5^2 + \zeta_5^3$: compute $\eta_0 + \eta_1$ and $\eta_0\eta_1$, and deduce $\cos\frac{2\pi}5 = \frac{\sqrt5 -
1}4$. Conclude that the regular pentagon is [constructible](#def-b3-galois-constructible).

**Solution of Exercise 4.9.**

$\eta_0 + \eta_1 = \zeta_5 + \zeta_5^2 + \zeta_5^3 + \zeta_5^4 =
-1$ (sum of all $5$-th roots of unity is $0$). $\eta_0\eta_1 =
(\zeta + \zeta^4)(\zeta^2 + \zeta^3) = \zeta^3 + \zeta^4 +
\zeta^6 + \zeta^7 = \zeta^3 + \zeta^4 + \zeta + \zeta^2 = -1$ (indices mod $5$). So $\eta_0, \eta_1$ are the roots of $Y^2 + Y
- 1$: $\frac{-1 \pm \sqrt5}2$. Since $\eta_0 =
2\cos\frac{2\pi}5 > 0$: $\eta_0 = \frac{\sqrt5 - 1}2$, whence $\cos\frac{2\pi}5 = \frac{\sqrt5 - 1}4$, and $\eta_1 =
\frac{-1-\sqrt5}2$. The group $\operatorname{Gal}(\Q(\zeta_5)/\Q) \cong (\Z/5\Z)^\times$ is cyclic of order $4$: it has a *unique* subgroup of order $2$ ($\{\pm 1\}$, i.e. $\zeta \mapsto \zeta^{\pm1}$), hence $\Q(\zeta_5)$ has a unique quadratic subfield ([Theorem 4.21](#thm-b3-galois-fundamental)), which contains $\eta_0
\notin \Q$: it is $\Q(\eta_0) = \Q(\sqrt5)$. [Constructibility](#def-b3-galois-constructible): $\cos\frac{2\pi}5$ lies in the quadratic tower $\Q \subseteq
\Q(\sqrt5)$, and $\zeta_5$ one quadratic step above: [Theorem 4.25](#thm-b3-galois-wantzel) constructs the pentagon.

**Exercise 4.10 ★★★.**

Let $K = \mathbb F_p(S, T)$ (rational functions in two indeterminates) and $L = K(S^{1/p}, T^{1/p})$. (a) Show $[L:K] = p^2$ and that $\alpha^p \in K$ for every $\alpha \in L$. (b) Deduce that $L/K$ is *not* simple: no primitive element exists — inseparability is fatal to [Theorem 4.17](#thm-b3-galois-primitive).

**Solution of Exercise 4.10.**

(a) Write $s = S^{1/p}$, $t = T^{1/p}$ (elements of a chosen [algebraic closure](#def-b3-galois-closure) with $s^p = S$, $t^p = T$). $X^p - S$ is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) over $K = \mathbb F_p(S, T) = (\mathbb
F_p(T))(S)$-fractions: [Eisenstein](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#thm-b3-rings-criteria) at the [prime element](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) $S$ of the [UFD](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) $\mathbb F_p(T)[S]$ ([Theorem 2.25](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#thm-b3-rings-criteria)). So $[K(s):K] = p$; likewise $X^p - T$ is [Eisenstein](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#thm-b3-rings-criteria) at $T$ over $K(s) = \mathbb F_p(s)(T)$-fractions — $T$ remains prime in $\mathbb F_p(s)[T]$ — giving $[L : K(s)] = p$ and $[L:K] =
p^2$. For $\alpha \in L$: $L = K[s, t]$, so $\alpha = \sum
c_{ij}s^it^j$ ($c_{ij} \in K$), and by the [Frobenius](#thm-b3-galois-finitefields) morphism $\alpha^p = \sum c_{ij}^p S^iT^j \in K$.

(b) If $L = K(\alpha)$, then $[K(\alpha):K] = p^2$; but $\alpha^p = a \in K$ means $\alpha$ annihilates $X^p - a$, so $\deg\pi_\alpha \leq p < p^2$: contradiction. No primitive element: [Theorem 4.17](#thm-b3-galois-primitive) genuinely needs separability (here every $\pi_\alpha$ divides some $X^p - a =
(X - \alpha)^p$: purely inseparable).

**Exercise 4.11 ★★★.**

Let $P = X^5 - 4X + 2$ and $G$ its [Galois group](#def-b3-galois-galois) over $\Q$, acting on the $5$ roots. (a) Show $P$ is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility), and deduce $5 \mid \abs G$; conclude that $G$ contains a $5$-cycle (Cauchy, [Theorem 1.13](https://one-course.com/books/math/5/en/chapter/1-group-theory#thm-b3-groups-cauchy)). (b) Show, by studying the variations of $x \mapsto x^5 - 4x +
2$, that $P$ has exactly $3$ real roots; deduce that complex conjugation restricts to a transposition in $G$. (c) Show that a subgroup of $S_5$ containing a transposition and a $5$-cycle is $S_5$ *(conjugate the transposition by powers of the cycle)*. Conclude $G \cong S_5$ and, with [Theorem 4.29](#thm-b3-galois-solvable), that $P$ is not [solvable by radicals](#def-b3-galois-radical).

**Solution of Exercise 4.11.**

(a) [Eisenstein](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#thm-b3-rings-criteria) at $2$ ($2 \mid 4, 2$; $4 \nmid 2$): $P$ [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility). If $\alpha$ is a root, $[\Q(\alpha):\Q] = 5$ divides $[L:\Q] = \abs G$ ($L$ the [splitting field](#thm-b3-galois-splitting)): Cauchy ([Theorem 1.13](https://one-course.com/books/math/5/en/chapter/1-group-theory#thm-b3-groups-cauchy)) gives an element of order $5$ in $G \leq S_5$; in $S_5$, only $5$-cycles have order $5$ (orders are lcms of cycle lengths).

(b) $P'(x) = 5x^4 - 4$ vanishes at $\pm(4/5)^{1/4} \approx \pm
0.946$: one local maximum then one local minimum. Values: $P(-2)
= -22 < 0$, $P(0) = 2 > 0$, $P(1) = -1 < 0$, $P(2) = 26 > 0$: three sign changes, and at most three real roots (two critical points): exactly $3$ real roots, hence one pair of complex conjugate roots. Take the [splitting field](#thm-b3-galois-splitting) $L$ inside $\C$: complex conjugation maps $L$ to itself (it permutes the roots, which generate $L$) and fixes $\Q$, so it defines an element of $G$; it fixes the three real roots and swaps the other two: a transposition.

(c) Let $\tau = (a\,b)$ and $\sigma$ a $5$-cycle in $G$. Some power $\sigma^k$ sends $a$ to $b$ ($k \ne 0 \bmod 5$), and $\sigma^k$ is again a $5$-cycle: renaming, assume $\sigma =
(1\,2\,3\,4\,5)$ and $\tau = (1\,2)$. Conjugating, $\sigma^m\tau\sigma^{-m} = (\sigma^m(1)\ \sigma^m(2))$: the adjacent transpositions $(1\,2), (2\,3), (3\,4), (4\,5),
(5\,1)$ all lie in $G$; adjacent transpositions generate $S_5$ (every transposition $(i\,j)$ is a product of adjacent ones, and transpositions generate). So $G = S_5$, not [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) ([Corollary 1.34](https://one-course.com/books/math/5/en/chapter/1-group-theory#cor-b3-groups-snnotsolvable)), and [Theorem 4.29](#thm-b3-galois-solvable) concludes: $X^5 - 4X + 2$ is not [solvable by radicals](#def-b3-galois-radical).

**Exercise 4.12 ★★★.**

(The dihedral quartic) Let $\alpha = \sqrt[4]2$ and $L =
\Q(\alpha, \iu)$, the [splitting field](#thm-b3-galois-splitting) of $X^4 - 2$ over $\Q$. (a) Show $[L : \Q] = 8$ and that $G =
\operatorname{Gal}(L/\Q)$ is generated by $\sigma\colon
\alpha \mapsto \iu\alpha,\ \iu \mapsto \iu$ and complex conjugation $\tau$, with $\sigma^4 = \tau^2 = e$ and $\tau\sigma\tau = \sigma^{-1}$: $G \cong D_4$. (b) List the subgroup lattice of $D_4$ (ten subgroups) and match each to its fixed field; verify in particular that $\Q(\sqrt2)$, $\Q(\iu)$, $\Q(\iu\sqrt2)$ are the three quadratic subfields, and locate $\Q(\alpha)$, $\Q(\iu\alpha)$, $\Q(\sqrt2, \iu)$. (c) Which intermediate fields are [Galois](#def-b3-galois-galois) over $\Q$? Match your answer against the [normal subgroups](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) of $D_4$, and explain why $\Q(\alpha)/\Q$ fails while $\Q(\sqrt2)/\Q$ succeeds.

**Solution of Exercise 4.12.**

(a) $X^4 - 2$ is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) ([Eisenstein](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#thm-b3-rings-criteria) at $2$): $[\Q(\alpha):\Q] = 4$; $\iu \notin \Q(\alpha) \subseteq \R$, so $[L : \Q(\alpha)] = 2$ and $[L:\Q] = 8$. The extension is [Galois](#def-b3-galois-galois) ([splitting field](#thm-b3-galois-splitting) of a [separable polynomial](#def-b3-galois-separable): the roots are $\iu^k\alpha$), so $\abs G = 8$. An automorphism sends $\alpha$ to one of the four roots and $\iu$ to $\pm\iu$: at most $8$ maps, all realized. The stated $\sigma$ (order $4$: $\sigma^2(\alpha) = -\alpha$, $\sigma^4 = e$) and $\tau$ (order $2$) satisfy

$$
\tau\sigma\tau(\alpha) = \tau\sigma(\alpha) =
\tau(\iu\alpha) = -\iu\alpha = \sigma^{-1}(\alpha),
\qquad \tau\sigma\tau(\iu) = \iu\cdot(-1)(-1) = \iu ,
$$

more carefully: $\tau\sigma\tau(\iu) = \tau\sigma(-\iu) =
\tau(-\iu) = \iu = \sigma^{-1}(\iu)$. So $\tau\sigma\tau =
\sigma^{-1}$: the presentation of $D_4$.

(b) The ten subgroups of $D_4 = \langle\sigma,
\tau\rangle$: $\{e\}$; five of order $2$: $\langle\sigma^2\rangle$, $\langle\tau\rangle$, $\langle\sigma^2\tau\rangle$, $\langle\sigma\tau\rangle$, $\langle\sigma^3\tau\rangle$; three of order $4$: $\langle\sigma\rangle$, $\{e, \sigma^2, \tau,
\sigma^2\tau\}$, $\{e, \sigma^2, \sigma\tau,
\sigma^3\tau\}$; and $D_4$. Fixed fields ([degree](#def-b3-galois-extension) = index): $\{e\} \leftrightarrow L$; order-$2$ subgroups $\leftrightarrow$ the five quartic fields

$$
\langle\tau\rangle \leftrightarrow \Q(\alpha),\quad
\langle\sigma^2\tau\rangle \leftrightarrow \Q(\iu\alpha),
\quad
\langle\sigma^2\rangle \leftrightarrow \Q(\sqrt2, \iu),\quad
\langle\sigma\tau\rangle \leftrightarrow
\Q\bigl((1+\iu)\alpha\bigr),\quad
\langle\sigma^3\tau\rangle \leftrightarrow
\Q\bigl((1-\iu)\alpha\bigr) .
$$

Checks: $\tau$ fixes the real $\alpha$; $\sigma^2\tau$ sends $\alpha \mapsto -\alpha$ and $\iu \mapsto -\iu$, fixing $\iu\alpha$; and since $\sigma\tau(\alpha) = \iu\alpha$, $\sigma\tau(\iu) = -\iu$:

$$
\sigma\tau\bigl((1+\iu)\alpha\bigr) =
(1 - \iu)\,\iu\alpha = (1 + \iu)\alpha,
\qquad
\sigma^3\tau\bigl((1-\iu)\alpha\bigr) =
(1 + \iu)(-\iu)\alpha = (1 - \iu)\alpha :
$$

each reflection fixes its generator, and the fixed field, of [degree](#def-b3-galois-extension) $4 =$ index, is exactly the field it generates (the generator is a root of $X^4 + 8$, [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility)). Order-$4$ subgroups $\leftrightarrow$ the three quadratic fields: $\langle\sigma\rangle \leftrightarrow \Q(\iu)$ ($\sigma$ fixes $\iu$); $\{e, \sigma^2, \tau, \sigma^2\tau\}
\leftrightarrow \Q(\sqrt2)$ (all four fix $\alpha^2$ up to sign checks: $\tau(\sqrt2) = \sqrt2$, $\sigma^2(\alpha^2) =
(-\alpha)^2$); $\{e, \sigma^2, \sigma\tau, \sigma^3\tau\}
\leftrightarrow \Q(\iu\sqrt2)$ ($\sigma\tau(\iu\alpha^2) =
(-\iu)(\iu\alpha)^2 = \iu\alpha^2$).

(c) [Galois](#def-b3-galois-galois) over $\Q$ $\leftrightarrow$ [normal subgroups](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) of $D_4$: $\{e\}$, $\langle\sigma^2\rangle$ (the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions)), the three subgroups of order $4$, and $D_4$ — so the [Galois](#def-b3-galois-galois) intermediate fields are $L$, $\Q(\sqrt2, \iu)$, the three quadratic fields, and $\Q$. The five quartic fields fixed by non-normal reflections are not [Galois](#def-b3-galois-galois): $\Q(\alpha)$ contains one root of $X^4 - 2$ but not $\iu\alpha$ (it is real) — conjugation by $\sigma$ moves $\langle\tau\rangle$ to $\langle\sigma^2\tau\rangle$, exactly as it moves $\Q(\alpha)$ to $\Q(\iu\alpha)$: non-normality of the subgroup *is* the existence of a conjugate field.

## 4.10 Problem: Gauss and the regular 17-gon

**Problem 4.1.**

Weekend problem — constructibility of the 17-gon

On March 30, 1796, the nineteen-year-old Gauss showed that the regular $17$-gon is [constructible](#def-b3-galois-constructible) — the first progress on the question since antiquity. We reconstruct his computation with the tools of this chapter. Set $\zeta = \eu^{2\iu\pi/17}$, $L =
\Q(\zeta)$, $G = \operatorname{Gal}(L/\Q)$.

**Part I — The group and its filtration.**

1. Justify: $[L:\Q] = 16$ , $G \cong (\Z/17\Z)^\times$ , cyclic of order $16$ . Verify that $3$ is a generator of $(\Z/17\Z)^\times$ *(compute the powers of $3$ modulo $17$: $3, 9, 10, 13, 5, 15, 11, 16, \dots$)* .
2. Let $\sigma \in G$ with $\sigma(\zeta) = \zeta^3$ , and $H_k = \langle \sigma^{2^k}\rangle$ for $k = 0, \dots,  4$ . Show that $G = H_0 \supset H_1 \supset H_2 \supset  H_3 \supset H_4 = \{e\}$ with each index $[H_k :  H_{k+1}] = 2$ , and that the fixed fields $\Q = L_0  \subset L_1 \subset L_2 \subset L_3 \subset L_4 = L$ form a tower of quadratic extensions.
3. Conclude *a priori* , using [Theorem 4.25](#thm-b3-galois-wantzel) , that $\zeta$ — hence the $17$ -gon — is [constructible](#def-b3-galois-constructible) . The rest of the problem makes the tower explicit.

**Part II — The periods of length 8.** Define the *Gauss periods*

$$
\eta_0 = \sum_{k \text{ even}} \zeta^{3^k \bmod 17}
= \zeta^{1} + \zeta^{9} + \zeta^{13} + \zeta^{15} + \zeta^{16} +
\zeta^{8} + \zeta^{4} + \zeta^{2},
\qquad
\eta_1 = \sum_{k \text{ odd}} \zeta^{3^k \bmod 17}.
$$

4. Show that $\eta_0, \eta_1$ are fixed by $H_1$ and swapped by $\sigma$ ; deduce $\eta_0, \eta_1 \in L_1$ and that they are the two roots of a quadratic over $\Q$ .
5. Compute $\eta_0 + \eta_1 = -1$ . Show $\eta_0\eta_1 = -4$ *(each product $\zeta^a\zeta^b$ is some $\zeta^c$, $c \neq 0$; count how many times each $c$ occurs, or argue that the product is a rational integer fixed by $G$, equal to the sum over all $64$ products, and use that each nonzero residue appears equally often)* .
6. Deduce $\eta_0 = \frac{-1 + \sqrt{17}}2$ , $\eta_1 =  \frac{-1-\sqrt{17}}2$ (identify which is which numerically: $\eta_0 \approx 1.56$ ), and $L_1 =  \Q(\sqrt{17})$ .

**Part III — Periods of length 4 and 2.** Define

$$
\beta_0 = \zeta + \zeta^{13} + \zeta^{16} + \zeta^{4},\quad
\beta_1 = \zeta^3 + \zeta^5 + \zeta^{14} + \zeta^{12},\quad
\beta_2 = \zeta^9 + \zeta^{15} + \zeta^{8} + \zeta^{2},\quad
\beta_3 = \zeta^{10} + \zeta^{11} + \zeta^{7} + \zeta^{6}.
$$

7. Show $\beta_0 + \beta_2 = \eta_0$ , $\beta_1 + \beta_3 =  \eta_1$ , and that $\beta_0, \beta_2$ are fixed by $H_2$ , swapped by $\sigma^2$ .
8. Compute $\beta_0\beta_2 = -1$ and $\beta_1\beta_3 = -1$ *(expand: the sixteen exponents obtained cover $1, \dots, 16$ exactly once)* .
9. Deduce $\beta_0 = \frac{\eta_0 + \sqrt{\eta_0^2 + 4}}2$ (check the sign numerically: $\beta_0 \approx 2.05$ ) and the analogous formula for $\beta_1$ ; hence $L_2 =  \Q(\beta_0)$ , quadratic over $L_1$ .
10. Let $\gamma_0 = \zeta + \zeta^{16} =  2\cos\frac{2\pi}{17}$ and $\gamma_1 = \zeta^{13} +  \zeta^4$ . Show $\gamma_0 + \gamma_1 = \beta_0$ and $\gamma_0\gamma_1 = \beta_1$ , so that $\gamma_0 =  \frac{\beta_0 + \sqrt{\beta_0^2 - 4\beta_1}}2$ .
11. Assemble the chain of formulas expressing $\cos\frac{2\pi}{17}$ by nested square roots, and give a decimal check ( $\cos\frac{2\pi}{17} \approx 0.93247$ ).

**Part IV — Epilogue.**

12. Where exactly did the argument use that $17$ is a *Fermat prime* ( $17 = 2^{2^2} + 1$ )? Show that for a prime $p$ , the regular $p$ -gon is [constructible](#def-b3-galois-constructible) iff $p  = 2^{2^t} + 1$ for some $t$ *(if $p - 1 = 2^m$, show $m$ must itself be a power of $2$)* .
13. Deduce the complete list of [constructible](#def-b3-galois-constructible) regular $n$ -gons for $n \leq 20$ , using the Gauss–Wantzel criterion of [Corollary 4.26](#cor-b3-galois-impossible) .

**Part V — Gauss sums and quadratic reciprocity.** The periods of Part II hide a treasure. For an odd prime $p$, the *Legendre symbol* $\bigl(\frac
ap\bigr)$ is $+1$ if $a$ is a nonzero square mod $p$, $-1$ if it is not, $0$ if $p \mid a$; [Exercise 4.4](#exo-b3-galois-4)(b) (Euler’s criterion) gives $\bigl(\frac ap\bigr) \equiv
a^{(p-1)/2} \pmod p$, whence multiplicativity. Write $\zeta =
\eu^{2\iu\pi/p}$, $p^* = (-1)^{(p-1)/2}p$, and define the *Gauss sum*

$$
g \;=\; \sum_{a=1}^{p-1}\Bigl(\frac ap\Bigr)\zeta^a .
$$

14. Show $\sum_{a=1}^{p-1}\bigl(\frac ap\bigr) = 0$ (as many squares as nonsquares), and prove the alternative form $g = \sum_{a=0}^{p-1}\zeta^{a^2}$ *(each nonzero square is hit twice, and $\sum_{a}\zeta^a = 0$)* . For $p = 17$ : relate $g$ to the periods of Part II — show $g = \eta_0 - \eta_1$ *(the squares mod $17$ are exactly the even powers of the generator $3$)* .
15. Prove $g^2 = p^*$: expand $$g^2 = \sum_{a,b\neq0}\Bigl(\frac{ab}p\Bigr)  \zeta^{a+b}  = \sum_{c}\ \sum_{a \neq 0}\Bigl(\frac{a(c -  a)}p\Bigr)\zeta^{c}$$ (set $b = c - a$), substitute $c - a = at$ to evaluate the inner sum as $\bigl(\frac{-1}p\bigr)(p - 1)$ for $c = 0$ and $-\bigl(\frac{-1}p\bigr)$ otherwise, and conclude with question 14. Check numerically: for $p =  17$, $(\eta_0 - \eta_1)^2 = 17$ (Part II).
16. Deduce $\sqrt{p^*} \in \Q(\zeta_p)$ , and conclude that the *unique* quadratic subfield of $\Q(\zeta_p)$ is $\Q(\sqrt{p^*})$ — unique because $\operatorname{Gal}(\Q(\zeta_p)/\Q)$ is cyclic ( [Theorem 4.23](#thm-b3-galois-cyclotomicirred) ) and a cyclic group has exactly one subgroup of index $2$ . (Every quadratic field embeds in some cyclotomic field — this is the first case of the Kronecker–Weber theorem, whose general form lies far ahead.)
17. Now let $q \neq p$ be another odd prime. Working in the ring $\Z[\zeta]$ modulo $q$, prove $$g^q \equiv \Bigl(\frac qp\Bigr)\,g \pmod{q\Z[\zeta]}$$ *(freshman’s dream: $(x + y)^q \equiv x^q + y^q$ mod $q$ in any commutative ring; then $g^q \equiv  \sum_a\bigl(\frac ap\bigr)^q\zeta^{aq}$, reindex $b =  aq$ and pull out $\bigl(\frac{q^{-1}}p\bigr) =  \bigl(\frac qp\bigr)$)*.
18. On the other hand, $g^q = g\,(g^2)^{(q-1)/2} =  g\,(p^*)^{(q-1)/2}$; using Euler’s criterion mod $q$, deduce $g^q \equiv \bigl(\frac{p^*}q\bigr)g  \pmod{q\Z[\zeta]}$, then — multiplying the two expressions for $g^q$ by $g$ and using $g^2 = p^*$, invertible mod $q$ — conclude $$\Bigl(\frac qp\Bigr) = \Bigl(\frac{p^*}q\Bigr) .$$ *(Why does a congruence between the integers $\pm p^*$ modulo $q\Z[\zeta]$ imply their equality? Intersect with $\Z$.)*
19. Unfold $\bigl(\frac{p^*}q\bigr) =  \bigl(\frac{-1}q\bigr)^{(p-1)/2}\bigl(\frac pq\bigr)$ and $\bigl(\frac{-1}q\bigr) = (-1)^{(q-1)/2}$ to obtain the *law of quadratic reciprocity*: $$\Bigl(\frac pq\Bigr)\Bigl(\frac qp\Bigr)  = (-1)^{\frac{p-1}2\cdot\frac{q-1}2} .$$ Verify it on $(p, q) = (17, 3)$ by listing the squares mod $17$ and mod $3$, and use it to decide in three lines whether $x^2 \equiv 219 \pmod{383}$ is [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) ($383$ is prime, $219 = 3\cdot73$).

**Part VI — Counting [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) polynomials: the prime number theorem of $\mathbb F_q[X]$.** Fix a prime power $q$ and let $N_q(n)$ be the number of *monic [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility)* polynomials of [degree](#def-b3-galois-extension) $n$ over $\mathbb F_q$; recall from [Exercise 4.6](#exo-b3-galois-6) the factorization of $X^{q^n} - X$ and the identity $q^n = \sum_{d\mid n}d\,N_q(d)$, which we now invert, reinterpret, and exploit.

20. (Words) Call a word $w \in \mathbb F_q^n$ *primitive* if it is not a power $u^{n/d} =  u\cdots u$ of a strictly shorter word $u$ , and let $A(d)$ be the number of primitive words of length $d$ . Show that every word of length $n$ is uniquely a power of a primitive word of some length $d \mid n$ , so that $q^n = \sum_{d \mid n}A(d)$ ; comparing with [Exercise 4.6](#exo-b3-galois-6) , conclude $A(d) = d\,N_q(d)$ for every $d$ , and explain this coincidence by an explicit bijection: an element $\alpha \in  \mathbb F_{q^d}$ of [degree](#def-b3-galois-extension) $d$ has [Frobenius](#thm-b3-galois-finitefields) [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) $(\alpha, \alpha^q, \dots, \alpha^{q^{d-1}})$ of exactly $d$ distinct elements, and elements of [degree](#def-b3-galois-extension) $d$ correspond $d$ -to-one to [irreducibles](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) of [degree](#def-b3-galois-extension) $d$ .
21. Prove the *Möbius inversion formula*: if $f(n) =  \sum_{d\mid n}g(d)$ for all $n$, then $g(n) =  \sum_{d\mid n}\mu(d)\,f(n/d)$, where $\mu$ is the Möbius function ($\mu(m) = (-1)^{\#\text{prime  factors}}$ if $m$ is squarefree, $0$ otherwise) *(key lemma: $\sum_{d \mid m}\mu(d) = 0$ for $m >  1$ — pair the divisors with and without a fixed prime factor)*. Deduce $$N_q(n) = \frac1n\sum_{d \mid n}\mu(d)\,q^{n/d} .$$
22. Show $N_q(n) \geq \frac1n\bigl(q^n - 2q^{n/2}\bigr) > 0$ for every $n \geq 1$ : a new proof that $\mathbb F_{q^n}$ exists for all $n$ . Interpret the leading term: a random monic polynomial of [degree](#def-b3-galois-extension) $n$ is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) with probability $\sim \frac1n$ — the [perfect](#prop-b3-galois-perfect) analogue of the prime number theorem, with $\log x$ traded for $n$ ; verify numerically for $q = 2$ , $n \leq 4$ ( [Exercise 4.6](#exo-b3-galois-6) lists the counts).
23. Prove the multiplicative companion of question 21: $$\prod_{\substack{\pi \text{ monic irred.}\\ \deg\pi =  n}}\pi \;=\; \prod_{d \mid n}  \bigl(X^{q^d} - X\bigr)^{\mu(n/d)}$$ *(Möbius inversion in the abelian group of nonzero rational functions)*; verify it by hand for $q  = 2$, $n = 2$: $(X^4 - X)/(X^2 - X) = X^2 + X + 1$.

**Part VII — Two codas.**

24. (The second supplement) Part V’s method also computes $\bigl(\frac2q\bigr)$. Let $\omega = \eu^{2\iu\pi/8}$ and $g = \omega + \omega^{-1}$. Show $g^2 = 2$ *($\omega^2 = \iu$)*; then, for an odd prime $q$, prove in $\Z[\omega]$ modulo $q$ that $$g^q \equiv \omega^q + \omega^{-q} \pmod{q\Z[\omega]},$$ and that the right side equals $g$ if $q \equiv \pm1  \pmod 8$ and $-g$ if $q \equiv \pm3 \pmod 8$. Comparing with $g^q = g\,(g^2)^{(q-1)/2} \equiv  \bigl(\frac2q\bigr)g$ as in question 18, conclude $$\Bigl(\frac2q\Bigr) = (-1)^{(q^2-1)/8},$$ checking that $(q^2 - 1)/8$ is even exactly when $q  \equiv \pm1 \pmod 8$. Verify: $2$ is a square mod $7$ and mod $17$ ($3^2$ and $6^2$), not mod $3$ nor mod $5$.
25. (The zeta function of $\mathbb F_q[X]$) Prove the identity of formal power series in $t$: $$\prod_{n \geq 1}\bigl(1 - t^n\bigr)^{-N_q(n)}  = \frac1{1 - qt}$$ *(unique factorization into monic [irreducibles](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility): expand each factor as a geometric series and count monic polynomials of [degree](#def-b3-galois-extension) $n$)*. Recover the identity $q^m = \sum_{d \mid m}d\,N_q(d)$ of [Exercise 4.6](#exo-b3-galois-6) by taking logarithms. Check the coefficient of $t^2$ by hand for $q = 2$, and use question 21’s formula to compute $N_2(6) = 9$, verifying $2^6 = 1\cdot2 + 2\cdot1 + 3\cdot2 +  6\cdot9$.

**Solution of Problem 4.1.**

**1.** $\Phi_{17}$ is [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) ([Theorem 4.23](#thm-b3-galois-cyclotomicirred), or [Example 2.26](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#ex-b3-rings-eisenstein) for prime index): $[L:\Q] =
\varphi(17) = 16$ and $G \cong (\Z/17\Z)^\times$, cyclic of order $16$ ([Theorem 4.12](#thm-b3-galois-cyclic)). Powers of $3$ mod $17$:

$$
3,\ 9,\ 10,\ 13,\ 5,\ 15,\ 11,\ 16,\ 14,\ 8,\ 7,\ 4,\ 12,\ 2,\
6,\ 1
$$

— sixteen distinct values: $3$ generates.

**2.** $G = \langle\sigma\rangle$ cyclic of order $16$; $H_k = \langle\sigma^{2^k}\rangle$ has order $2^{4-k}$, and $[H_k : H_{k+1}] = 2$. By the fundamental theorem ([Theorem 4.21](#thm-b3-galois-fundamental)), $L_k = L^{H_k}$ satisfy $[L_k : \Q] = [G : H_k] = 2^k$: each $[L_{k+1}:L_k] = 2$.

**3.** $\zeta \in L = L_4$ sits atop a tower of quadratic extensions of $\Q$: by [Theorem 4.25](#thm-b3-galois-wantzel), $\zeta$ is [constructible](#def-b3-galois-constructible); the $17$-gon has vertices $\zeta^k$.

**4.** $\sigma^2$ multiplies exponents by $9$; the exponents of $\eta_0$ are the even powers of $3$,

$$
\{3^{2k} \bmod 17\} = \{1, 9, 13, 15, 16, 8, 4, 2\},
$$

a set stable under multiplication by $9 = 3^2$; so $\eta_0$ (and likewise $\eta_1$) is fixed by $H_1 =
\langle\sigma^2\rangle$: $\eta_0, \eta_1 \in L_1$, a quadratic field. $\sigma$ maps even powers to odd: it swaps $\eta_0,
\eta_1$. Hence $\eta_0 + \eta_1$ and $\eta_0\eta_1$ are fixed by all of $G$: rational; $\eta_0, \eta_1$ are the roots of a rational quadratic.

**5.** $\eta_0 + \eta_1 = \sum_{c=1}^{16}\zeta^c = -1$. The product expands into $64$ terms $\zeta^{a + b}$, $a$ in the even set, $b$ in the odd set. No term is $\zeta^0$: $b = -a$ is impossible, because $-1 = 16 = 3^8$ is an *even* power, so $-a$ stays in the even set. Thus $\eta_0\eta_1 = \sum_{c \neq 0}
n_c\zeta^c$ with $\sum n_c = 64$; applying $\sigma$ fixes $\eta_0\eta_1$ (it swaps the factors) and permutes the $\zeta^c$ transitively over all $c \neq 0$, so all $n_c$ are equal: $n_c
= 4$ and $\eta_0\eta_1 = 4\sum_{c\neq0}\zeta^c = -4$.

**6.** $\eta_{0,1}$ solve $Y^2 + Y - 4 = 0$: $\frac{-1 \pm
\sqrt{17}}2$. Numerically, pairing conjugate exponents, $\eta_0 = 2\bigl(\cos\tfrac{2\pi}{17} + \cos\tfrac{4\pi}{17} +
\cos\tfrac{8\pi}{17} + \cos\tfrac{16\pi}{17}\bigr) \approx 1.56
> 0$: $\eta_0 = \frac{-1+\sqrt{17}}2$, $\eta_1 =
\frac{-1-\sqrt{17}}2$, and $L_1 = \Q(\eta_0) = \Q(\sqrt{17})$.

**7.** The exponent sets: $\beta_0$: $\{1, 13, 16, 4\}$ = powers $3^{4k}$; $\beta_2$: $\{9, 15, 8, 2\}$ = $9 \times$ that set. Union: the even set: $\beta_0 + \beta_2 = \eta_0$; likewise $\beta_1 + \beta_3 = \eta_1$. Multiplication by $13 = 3^4$ stabilizes each $\beta_i$’s exponent set: fixed by $H_2 =
\langle\sigma^4\rangle$; and $\sigma^2$ ($\times 9$) sends $\{1,13,16,4\}$ to $\{9, 15, 8, 2\}$: swaps $\beta_0,
\beta_2$.

**8.** Expanding $\beta_0\beta_2$, the sixteen exponent sums

$$
\{1,13,16,4\} + \{9,15,8,2\} =
\{10,16,9,3,\ 5,11,4,15,\ 8,14,7,1,\ 13,2,12,6\}
$$

cover $1, \dots, 16$ exactly once: $\beta_0\beta_2 =
\sum_{c\ne0}\zeta^c = -1$. Applying $\sigma$ (which maps $\beta_0 \mapsto \beta_1$, $\beta_2 \mapsto \beta_3$: exponents $\times 3$): $\beta_1\beta_3 = \sigma(\beta_0\beta_2)
= -1$.

**9.** $\beta_0, \beta_2$ solve $Y^2 - \eta_0 Y - 1 = 0$, so $\beta_0 = \frac{\eta_0 + \sqrt{\eta_0^2 + 4}}2$ (numerically $\beta_0 = 2\cos\frac{2\pi}{17} + 2\cos\frac{8\pi}{17} \approx
2.05 > 0$, the $+$ sign). Likewise $\beta_1 = \frac{\eta_1 +
\sqrt{\eta_1^2 + 4}}2 \approx 0.344$ (numerical check fixes the sign again). $L_2 = L_1(\beta_0)$, quadratic over $L_1$.

**10.** $\gamma_0 + \gamma_1 = \zeta + \zeta^{16} +
\zeta^{13} + \zeta^4 = \beta_0$. And

$$
\gamma_0\gamma_1 = (\zeta + \zeta^{16})(\zeta^{13} + \zeta^4)
= \zeta^{14} + \zeta^{5} + \zeta^{12} + \zeta^{3} = \beta_1 .
$$

So $\gamma_0, \gamma_1$ solve $Y^2 - \beta_0Y + \beta_1 = 0$; numerically $\gamma_0 = 2\cos\frac{2\pi}{17} \approx 1.865 >
\gamma_1 \approx 0.185$: $\gamma_0 = \frac{\beta_0 +
\sqrt{\beta_0^2 - 4\beta_1}}2$.

**11.** Chaining:

$$
\eta_0 = \frac{-1 + \sqrt{17}}2,
\quad
\beta_0 = \frac{\eta_0 + \sqrt{\eta_0^2 + 4}}2,
\quad
\beta_1 = \frac{\eta_1 + \sqrt{\eta_1^2 + 4}}2,
\quad
\cos\frac{2\pi}{17} = \frac{\beta_0 + \sqrt{\beta_0^2 -
4\beta_1}}4 .
$$

Numerically: $\sqrt{17} \approx 4.1231$, $\eta_0 \approx
1.5616$, $\eta_1 \approx -2.5616$, $\beta_0 \approx 2.0494$, $\beta_1 \approx 0.3441$, $\beta_0^2 - 4\beta_1 \approx
2.8234$, and $\cos\frac{2\pi}{17} \approx \frac{2.0494 +
1.6803}4 \approx 0.93242$ — against $\cos\frac{2\pi}{17} =
0.93247\dots$: the small discrepancy is rounding in the intermediate displays; carrying more digits reproduces $0.932472$.

**12.** The construction needed $[L:\Q] = p - 1$ to be a power of $2$, so that a full chain of index-$2$ subgroups exists. If $p = 2^m + 1$ is prime and $m = ab$ with $a$ odd $>
1$: $x + 1 \mid x^a + 1$ at $x = 2^b$ shows $2^b + 1$ properly divides $p$ — impossible. So $m$ is a power of $2$: $p =
2^{2^t} + 1$, a *Fermat prime* ($3, 5, 17, 257, 65537$, …). Conversely for such $p$, $\varphi(p) = 2^{2^t}$ and the argument of questions 1–3 (or [Corollary 4.26](#cor-b3-galois-impossible)) applies: the regular $p$-gon is [constructible](#def-b3-galois-constructible) iff $p$ is a Fermat prime.

**13.** $\varphi(n)$ is a power of $2$ exactly when $n =
2^a p_1\cdots p_r$ with distinct Fermat primes $p_i$ (multiplicativity of $\varphi$; an odd prime power $p^k$, $k
\geq 2$, contributes the factor $p \nmid 2^m$). For $n \leq
20$, the [constructible](#def-b3-galois-constructible) regular $n$-gons are

$$
n = 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20
$$

with respective values

$$
\varphi(n) = 2,\ 2,\ 4,\ 2,\ 4,\ 4,\ 4,\ 8,\ 8,\ 16,\ 8 .
$$

The impossible ones are $n = 7, 9, 11, 13, 14, 18, 19$, where $\varphi(n) = 6, 6, 10, 12, 6, 6, 18$ has an odd prime factor.

**14.** The squares form the image of the squaring morphism on the cyclic $(\Z/p\Z)^\times$, of index $2$: $\frac{p-1}2$ squares, $\frac{p-1}2$ nonsquares, so the symbols sum to $0$. Then

$$
\sum_{a=0}^{p-1}\zeta^{a^2} = 1 + 2\sum_{b \text{ square}
\neq 0}\zeta^b
= 1 + \sum_{b\neq0}\Bigl(1 + \Bigl(\frac bp\Bigr)\Bigr)
\zeta^b = \sum_{b}\zeta^b + g = g,
$$

using $1 + \bigl(\frac bp\bigr) = \#\{a : a^2 = b\}$ and $\sum_{b=0}^{p-1}\zeta^b = 0$. For $p = 17$: the squares mod $17$ are the even powers of the generator $3$, i.e. the exponents appearing in $\eta_0$ (Part II), so $g =
\sum_{\text{even }k}\zeta^{3^k} - \sum_{\text{odd }k}
\zeta^{3^k} = \eta_0 - \eta_1$.

**15.** With $b = c - a$ ($a, b$ run over nonzero residues, $c = a + b$ over all residues):

$$
g^2 = \sum_c\zeta^c\sum_{a\neq0,\,a\neq c}
\Bigl(\frac{a(c-a)}p\Bigr) .
$$

For $c = 0$: $\bigl(\frac{-a^2}p\bigr) =
\bigl(\frac{-1}p\bigr)$, summed over $p - 1$ values. For $c \neq 0$: substitute $c - a = at$, i.e. $t = c/a -
1$; as $a$ runs over the nonzero residues, $t$ runs bijectively over the residues $\neq -1$ (invert: $a = c/(1 +
t)$). The summand becomes $\bigl(\frac{a^2t}p\bigr) =
\bigl(\frac tp\bigr)$, and

$$
\sum_{t \neq -1}\Bigl(\frac tp\Bigr)
= -\Bigl(\frac{-1}p\Bigr)
$$

(the full sum vanishes by question 14). Hence

$$
g^2 = \Bigl(\frac{-1}p\Bigr)\Bigl[(p-1) -
\sum_{c\neq0}\zeta^c\Bigr]
= \Bigl(\frac{-1}p\Bigr)\,p = p^* ,
$$

using $\sum_{c\neq0}\zeta^c = -1$ and Euler’s criterion $\bigl(\frac{-1}p\bigr) = (-1)^{(p-1)/2}$. For $p = 17$: $(\eta_0 - \eta_1)^2 = (\eta_0 + \eta_1)^2 - 4\eta_0\eta_1 =
1 + 16 = 17$, matching Part II.

**16.** $g^2 = p^*$ exhibits $\sqrt{p^*} = \pm g \in
\Q(\zeta_p)$, so $\Q(\sqrt{p^*})$ is a quadratic subfield. Uniqueness: subfields of [degree](#def-b3-galois-extension) $2$ correspond, by the [Galois correspondence](#thm-b3-galois-fundamental), to subgroups of index $2$ of the cyclic $\operatorname{Gal}(\Q(\zeta_p)/\Q) \cong (\Z/p\Z)^\times$, and a cyclic group of even order has exactly one such subgroup (the squares). Every quadratic field is $\Q(\sqrt{d})$ with $d$ squarefree, and combining the fields $\Q(\sqrt{p^*})$, $\Q(\iu) \subseteq \Q(\zeta_4)$ and $\Q(\sqrt2) \subseteq \Q(\zeta_8)$ inside a common $\Q(\zeta_N)$ captures every $\sqrt d$: the quadratic case of Kronecker–Weber.

**17.** In any commutative ring, $(x + y)^q = x^q + y^q
+ q(\cdots)$: the binomial coefficients $\binom qk$, $0 < k <
q$, are divisible by the prime $q$. Iterating on the $p - 1$ terms of $g$:

$$
g^q \equiv \sum_{a}\Bigl(\frac ap\Bigr)^{q}\zeta^{aq}
= \sum_a\Bigl(\frac ap\Bigr)\zeta^{aq}
\pmod{q\Z[\zeta]},
$$

($q$ odd: the symbol is unchanged). Reindex $b = aq$: $a =
q^{-1}b$ and $\bigl(\frac{q^{-1}b}p\bigr) =
\bigl(\frac{q}p\bigr)\bigl(\frac bp\bigr)$ (multiplicativity; $\bigl(\frac{q^{-1}}p\bigr) = \bigl(\frac qp\bigr)$ since the symbol of an inverse equals the symbol): $g^q \equiv
\bigl(\frac qp\bigr)g$.

**18.** $g^q = g\,(g^2)^{(q-1)/2} = g\,(p^*)^{(q-1)/2}$ exactly (question 15), and Euler’s criterion in $\Z$ gives $(p^*)^{(q-1)/2} \equiv \bigl(\frac{p^*}q\bigr) \pmod q$, hence mod $q\Z[\zeta]$: $g^q \equiv \bigl(\frac{p^*}q\bigr)g$. Comparing with question 17 and multiplying by $g$:

$$
\Bigl(\frac qp\Bigr)p^* \equiv \Bigl(\frac{p^*}q\Bigr)p^*
\pmod{q\Z[\zeta]} .
$$

Both sides are rational integers; their difference, $0$ or $\pm2p^*$, lies in $q\Z[\zeta] \cap \Z = q\Z$ (an integer $m
\in q\Z[\zeta]$ has $m/q \in \Q \cap \Z[\zeta] = \Z$, the latter because $1, \zeta, \dots, \zeta^{p-2}$ is a $\Q$-basis with rational coordinates reading off integrality). Since $q \nmid 2p^*$ ($q$ odd, $q \neq p$), the difference is $0$: $\bigl(\frac qp\bigr) =
\bigl(\frac{p^*}q\bigr)$.

**19.** By multiplicativity, $\bigl(\frac{p^*}q\bigr) =
\bigl(\frac{-1}q\bigr)^{(p-1)/2}\bigl(\frac pq\bigr) =
(-1)^{\frac{q-1}2\cdot\frac{p-1}2}\bigl(\frac pq\bigr)$, so question 18 reads $\bigl(\frac qp\bigr)\bigl(\frac
pq\bigr) = (-1)^{\frac{p-1}2\frac{q-1}2}$: reciprocity. Check $(17, 3)$: the exponent $\frac{16}2\cdot\frac22 = 8$ is even, so the two symbols must agree; squares mod $3$ are $\{1\}$ and $17 \equiv 2$: $\bigl(\frac{17}3\bigr) = -1$; squares mod $17$ are $\{1, 4,
9, 16, 8, 2, 15, 13\}$ and $3$ is absent: $\bigl(\frac3{17}\bigr) = -1$. Product $+1$, as predicted. For $x^2
\equiv 219 \pmod{383}$: $\bigl(\frac{219}{383}\bigr) =
\bigl(\frac3{383}\bigr)\bigl(\frac{73}{383}\bigr)$. First: $383 \equiv 3 \pmod4$ and $3 \equiv 3$: reciprocity gives $\bigl(\frac3{383}\bigr) = -\bigl(\frac{383}3\bigr) =
-\bigl(\frac23\bigr) = -(-1) = +1$. Second: $73 \equiv 1
\pmod 4$: $\bigl(\frac{73}{383}\bigr) =
\bigl(\frac{383}{73}\bigr) = \bigl(\frac{18}{73}\bigr) =
\bigl(\frac2{73}\bigr)$ ($18 = 2\cdot3^2$), and $73 \equiv 1
\pmod 8$ makes $2$ a square mod $73$ (supplementary law, provable by $g = \zeta_8 + \zeta_8^{-1} = \sqrt2$ in $\Q(\zeta_8)$ by the same method): $+1$. Total $+1$: the congruence is [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived).

**20.** Existence and uniqueness of the primitive root: if $w$ has period set $\{d : w = u^{n/d},\ \abs u = d\}$, the minimal such $d_0$ divides every other period $d$ (if $w$ is both a $d$-power and a $d'$-power, it is a $\gcd(d, d')$-power: compare letters at indices agreeing modulo the gcd, via Bézout), and the length-$d_0$ block is primitive. Sorting the $q^n$ words by the length of their primitive root: $q^n = \sum_{d \mid n}A(d)$. Since $A$ and $d\,N_q(d)$ satisfy the same recursion with the same values for $n = 1$ (both determine each other inductively from $q^n
= \sum_{d\mid n}(\cdot)$), they are equal: $A(d) =
d\,N_q(d)$. Bijection: an element $\alpha$ of [degree](#def-b3-galois-extension) $d$ yields the word $w_\alpha$ of the coefficients of… better, directly: elements of [degree](#def-b3-galois-extension) $d$ in $\overline{\mathbb F_q}$ are the roots of the $N_q(d)$ [irreducibles](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) of [degree](#def-b3-galois-extension) $d$, each contributing its $d$ distinct roots (separability): $d\,N_q(d)$ elements of [degree](#def-b3-galois-extension) $d$, matching the count $q^n = \sum_{d\mid n}\#\{
\text{elements of degree } d \text{ in } \mathbb F_{q^n}\}$ — the same sieve, once on words, once on field elements.

**21.** Lemma: $\sum_{d \mid m}\mu(d) = \mathbf
1_{m=1}$. For $m > 1$ fix a prime $p \mid m$: squarefree divisors of $m$ pair off as $\{d, pd\}$ with $p \nmid d$, and $\mu(pd) = -\mu(d)$: the sum cancels. Then, for $f(n) =
\sum_{e\mid n}g(e)$:

$$
\sum_{d \mid n}\mu(d)\,f\bigl(\tfrac nd\bigr)
= \sum_{d \mid n}\mu(d)\!\!\sum_{e \mid n/d}\!\!g(e)
= \sum_{e \mid n}g(e)\!\!\sum_{d \mid n/e}\!\!\mu(d)
= g(n) .
$$

With $f(n) = q^n$ and $g(n) = nN_q(n)$ ([Exercise 4.6](#exo-b3-galois-6)): $N_q(n) =
\frac1n\sum_{d\mid n}\mu(d)\,q^{n/d}$.

**22.** The $d = 1$ term is $q^n$; every other term has $\abs{\mu(d)q^{n/d}} \leq q^{n/2}$, and crudely $\sum_{d \mid n, d > 1}q^{n/d} \leq \sum_{j \leq
n/2}q^j < 2q^{n/2}$ (geometric, $q \geq 2$). So $nN_q(n) >
q^n - 2q^{n/2} \geq 0$ for $n \geq 1$: [irreducibles](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) of every [degree](#def-b3-galois-extension) exist, and $\mathbb F_q[X]/(\pi) = \mathbb F_{q^n}$ is (re)built — existence with a census. The proportion of [irreducibles](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) among the $q^n$ monic degree-$n$ polynomials is $\frac1n(1 + O(q^{-n/2}))$: the prime number theorem of $\mathbb F_q[X]$, with $n$ playing $\log x$. For $q = 2$ the counts $2, 1, 2, 3$ of [Exercise 4.6](#exo-b3-galois-6) match the formula: e.g. $N_2(4) = \frac14(2^4 - 2^2) = 3$.

**23.** In the multiplicative abelian group of nonzero rational functions over $\mathbb F_q$, set $F(n) = X^{q^n} -
X$ and $G(n) = \prod_{\deg\pi = n}\pi$; [Exercise 4.6](#exo-b3-galois-6) says $F(n) = \prod_{d\mid n}G(d)$. The Möbius argument of question 21, written multiplicatively (exponents add exactly as the sums did), gives $G(n) = \prod_{d \mid n}F(d)^{\mu(n/d)}$. For $q = 2$, $n = 2$: $G(2) = \frac{X^4 - X}{X^2 - X} = \frac{X(X^3 -
1)}{X(X - 1)} = X^2 + X + 1$, the unique [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) quadratic over $\mathbb F_2$, as it must be.

**24.** $\omega^2 = \iu$ and $\omega^{-2} = -\iu$, so $g^2 = \omega^2 + 2 + \omega^{-2} = 2$. Freshman’s dream in the commutative ring $\Z[\omega]/q\Z[\omega]$: $g^q =
(\omega + \omega^{-1})^q \equiv \omega^q + \omega^{-q}$. The value of $\omega^q + \omega^{-q}$ depends only on $q \bmod 8$: for $q \equiv \pm1$, $\omega^q + \omega^{-q} = \omega^{\pm1} +
\omega^{\mp1} = g$; for $q \equiv \pm3$, using $\omega^4 =
-1$, $\omega^{3} = -\omega^{-1}$ and $\omega^{-3} = -\omega$, so $\omega^q + \omega^{-q} = -g$. On the other hand $g^q =
g\,(g^2)^{(q-1)/2} = g\,2^{(q-1)/2} \equiv \bigl(\frac2q\bigr)
g \pmod{q\Z[\omega]}$ by Euler’s criterion mod $q$. Comparing and multiplying by $g$: $2\bigl(\frac2q\bigr) \equiv
\pm2 \pmod{q\Z[\omega]}$; if the signs disagreed, $q$ would divide $4$ in $\Z[\omega]$, hence in $\Z$ ($q\Z[\omega] \cap
\Z = q\Z$: coordinates on the basis $1, \omega, \omega^2,
\omega^3$), impossible for $q$ odd. So $\bigl(\frac2q\bigr) =
+1$ iff $q \equiv \pm1 \pmod 8$. Parity check: $q = 8k \pm 1$ gives $(q^2 - 1)/8 = 2k(4k \pm 1)$, even; $q = 8k \pm 3$ gives $(q^2 - 1)/8 = 8k^2 \pm 6k + 1$, odd: the formula $(-1)^{(q^2-1)/8}$ encodes the case split. Numerically: $3^2
= 9 \equiv 2 \pmod 7$ ($7 \equiv -1$), $6^2 = 36 \equiv 2
\pmod{17}$ ($17 \equiv 1$); the squares mod $3$ are $\{0,
1\}$ and mod $5$ are $\{0, 1, 4\}$, neither containing $2$ ($3 \equiv 3$, $5 \equiv -3 \pmod 8$).

**25.** Every monic $f \in \mathbb F_q[X]$ factors uniquely as $\prod_\pi\pi^{e_\pi}$ over the monic [irreducibles](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility): sorting by [degree](#def-b3-galois-extension),

$$
\sum_{f \text{ monic}}t^{\deg f}
= \prod_\pi\ \sum_{e \geq 0}t^{e\deg\pi}
= \prod_\pi\bigl(1 - t^{\deg\pi}\bigr)^{-1}
= \prod_{n \geq 1}\bigl(1 - t^n\bigr)^{-N_q(n)},
$$

all products t-adically legitimate (only [degrees](#def-b3-galois-extension) $\leq m$ touch the coefficient of $t^m$, and there are finitely many [irreducibles](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) of each [degree](#def-b3-galois-extension)). The left side is $\sum_m
q^mt^m = (1 - qt)^{-1}$: the identity. Logarithms: $-\log(1 -
qt) = \sum_m\frac{q^m}mt^m$, while $\sum_nN_q(n)\bigl(-\log(1
- t^n)\bigr) = \sum_nN_q(n)\sum_k\frac{t^{nk}}k$; the coefficient of $t^m$ gives $\frac{q^m}m = \sum_{dk = m}
\frac{N_q(d)}k = \frac1m\sum_{d \mid m}d\,N_q(d)$, i.e. $q^m = \sum_{d\mid m}d\,N_q(d)$. Hand check, $q = 2$, coefficient of $t^2$: $N_2(1) = 2$, $N_2(2) = 1$, and $(1 - t)^{-2}(1 - t^2)^{-1} = (1 + 2t + 3t^2 + \dots)(1 + t^2
+ \dots)$ has $t^2$-coefficient $3 + 1 = 4 = 2^2$. Finally, question 21’s formula with the divisors $1, 2, 3, 6$:

$$
N_2(6) = \tfrac16\bigl(2^6 - 2^3 - 2^2 + 2\bigr)
= \tfrac{54}6 = 9,
$$

and indeed $1\cdot2 + 2\cdot1 + 3\cdot2 + 6\cdot9 = 2 + 2 +
6 + 54 = 64 = 2^6$.