4Field Extensions and Galois Theory

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 $K$). ∎

Proof. Evaluation $K[X] \to L$, $P \mapsto P(\alpha)$, has image $K[\alpha]$ and kernel $(\pi_\alpha)$; since $\pi_\alpha$ is irreducible, $K[X]/(\pi_\alpha)$ is a field (Proposition 2.4: in the PID $K[X]$, irreducible generates a maximal ideal), 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. 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. ∎

Proof. $K(\alpha, \beta) = (K(\alpha))(\beta)$ is finite over $K(\alpha)$ ($\beta$ algebraic over $K \subseteq K(\alpha)$) and $K(\alpha)/K$ is finite: by the tower law $[K(\alpha,\beta):K] < \infty$, and every element of $K(\alpha, \beta)$ — including the four listed — is algebraic (Theorem 4.4). Transitivity: if $\beta$ is algebraic over $L$ and $L/K$ is 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), and $F(\beta)/F$ is finite: $[F(\beta):K] < \infty$, so $\beta$ is algebraic over $K$. ∎

Proof. Existence, by induction on $\deg P$: pick an irreducible 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 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 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 factor $Q$ of $P$ with $\deg Q \geq 2$; $Q^\sigma$ is an irreducible factor of $P^\sigma$, with a root $\beta \in L'$; then

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

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 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 $E$ 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; Zorn) and $K_1 = R/\mathfrak m$: a field extension of $K$ in which every nonconstant $f \in K[X]$ has a root, namely $\bar X_f$, and which is algebraic over $K$ (it is generated by the $\bar X_f$, each 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 factor of $g$ over $K_n$ has a root in $K_{n+1} \subseteq \Omega$: $\Omega$ is algebraically closed, and algebraic over $K$ (each $K_n$ is, by transitivity, Corollary 4.5): $\Omega$ is an algebraic closure.

Uniqueness. Let $\Omega, \Omega'$ be two algebraic closures. 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 $\Omega'$, 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, and $\Omega'$ is 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. ∎

Proof. (1) A finite field $E$ 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 of it over $\mathbb F_p$ — determining $E$ up to isomorphism (Theorem 4.7). Conversely, in a splitting field $L$ 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: 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 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 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)). ∎

Proof. Let $G \leq K^\times$ be finite. By the structure theorem (Corollary 3.13), $G \cong \Z/d_1 \times \dots \times \Z/d_s$ 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. ∎

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, every element is a $p$-th power (Frobenius is onto), so $Q(X^p) = \sum b_k^p X^{pk} = (\sum b_kX^k)^p$ is not irreducible: $P' = 0$ cannot happen for irreducible $P$ there either. ∎

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 (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). In the separable case both counts are equalities: minimal polynomials over the bigger field $K(\alpha_1)$ divide those over $K$, hence remain separable. ∎

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

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 $L = K(\gamma)$. ∎

Proof. Let $L$ split the separable $P$ over $K$, and fix an algebraic closure $\Omega \supseteq L$. $L/K$ is 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 applied to the generators (roots of the separable $P$, whose minimal polynomials divide $P$) yields exactly $[L:K]$ extensions of $K \hookrightarrow \Omega$ (in the inductive step, the minimal polynomial of a root of $P$ over an intermediate field still divides $P$, hence is 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 of $P$ over $F$, and $P$ remains separable: $L/F$ is Galois; the same argument gives normality over $F$. ∎

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 $L$,

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: $[L:K] \leq n$. ∎

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 $\pi_\alpha$ over $K$ has degree $\geq 2$ and is separable ($L/K$ separable, Proposition 4.19), so it has another root $\beta \neq \alpha$ in an algebraic 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); by normality (Proposition 4.19) $\tau(L) = L$, so $\tau \in G$, $\beta = \tau(\alpha) \in L$, and $\tau(\alpha) \neq \alpha$.

(2) For a subgroup $H$: $L/L^H$ is Galois (Proposition 4.19), 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 gives $L^{\operatorname{Gal}(L/F)} = F$ by (1) applied to $L/F$. The two maps are mutually inverse; they reverse inclusions evidently. Degrees: $[L:L^H] = \abs H$ 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): $\abs{\operatorname{Aut}_K(F)} = [F:K]$ and $\rho$ is onto. It remains to see $F/K$ is Galois: $F$ is separable over $K$ (inside the separable $L/K$), and $F = K(\gamma)$ (Theorem 4.17); 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 of $K[X]$ split by $F$, and its roots generate $F$: $F/K$ is Galois. Conversely, if $F = L^H$ with $F/K$ Galois, normality of $F$ (Proposition 4.19, applied to embeddings $F \to \Omega$ restricted from elements of $G$) gives $\sigma(F) = F$ for all $\sigma \in G$, i.e. $H \trianglelefteq G$. ∎

Proof. Let $f = \pi_{\zeta_n}$, so $\Phi_n = fg$ with $f, g \in \Z[X]$ monic (Gauss’s lemma Lemma 2.23: 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, then Gauss again). Reduce mod $p$: $\bar g(X^p) = \bar g(X)^p$ (Frobenius 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 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. Consequently $[\Q(\zeta_n):\Q] = \varphi(n)$, and $\Q(\zeta_n)$ is the splitting field of the separable $X^n - 1$ (all roots are powers of $\zeta_n$): 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. ∎

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 consequence: $[\Q(z):\Q]$ divides $[F_r : \Q] = 2^r$ (tower law).

($\Leftarrow$) The constructible numbers form a field: sums and differences by parallelograms (parallels are 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 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 $0$). ∎