Fields and Galois theory

Spanning. For $x \in L$ write $x = \sum_j b_j l_j$ with $b_j \in M$, and each $b_j = \sum_i a_{ij} m_i$ with $a_{ij} \in K$. Then $x = \sum_{i,j} a_{ij}\, m_i l_j$. Independence. If $\sum_{i,j} a_{ij} m_i l_j = 0$ with $a_{ij} \in K$, then $\sum_j \big(\sum_i a_{ij} m_i\big) l_j = 0$ with coefficients in $M$, so each $\sum_i a_{ij} m_i = 0$, and so each $a_{ij} = 0$. Count the basis. (If either factor is infinite, the same argument produces arbitrarily large independent sets in $L$.)

The evaluation map $\varepsilon : K[x] \to L$, $f \mapsto f(\alpha)$, is a ring homomorphism with image $K[\alpha]$. Its kernel is a non-zero ideal of the PID $K[x]$, so it is $(m)$ for the monic polynomial $m$ of least degree in it, which is $m_\alpha$. This gives the divisibility in (a). The image $K[\alpha]$ is a subring of a field, hence a domain, so $(m_\alpha)$ is a prime ideal; it is non-zero, hence maximal in the PID $K[x]$, and $m_\alpha$ is irreducible. By the first isomorphism theorem $K[\alpha] \cong K[x]/(m_\alpha)$, which is a field. So $K[\alpha]$ is a field containing $K$ and $\alpha$, and therefore equals $K(\alpha)$. Part (c) is Corollary 10 of the quotients lesson. Finally, if $f$ is monic irreducible with $f(\alpha) = 0$, then $m_\alpha \mid f$ forces $f = m_\alpha$.

A cubic that factors over $\Q$ has a linear factor, hence a rational root. Rational root test: if $\frac{r}{s}$ in lowest terms is a root of $a_3x^3 + \dots + a_0 \in \Z[x]$, then $r \mid a_0$ and $s \mid a_3$. For $x^3 - 2$ the candidates $\pm 1, \pm 2$ give $-1, -3, 6, -10$, never $0$. For $8x^3 - 6x - 1$ the candidates are $\pm 1, \pm \frac12, \pm\frac14, \pm\frac18$, and evaluation gives $1, -3, -3, 1, -\frac{19}{8}, \frac{3}{8}, -\frac{111}{64}, -\frac{17}{64}$: never $0$. Finally $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ with $\theta = 20^\circ$ gives $4c^3 - 3c = \frac12$, i.e. $8c^3 - 6c - 1 = 0$ for $c = \cos 20^\circ$. Apply Theorem 2.

A construction produces points step by step, each an intersection of two lines or circles drawn through points already constructed. Let $F_k$ be the field generated over $\Q$ by the coordinates of the points after $k$ steps. A line through two points with coordinates in $F$ has an equation $ax + by = c$ with $a, b, c \in F$. A circle has an equation $x^2 + y^2 + dx + ey + f = 0$ with $d, e, f \in F$. Intersecting two lines means solving linear equations, so the new coordinates lie in $F$. Intersecting a line and a circle means substituting the line into the circle, which gives a quadratic equation over $F$. Two circles reduce to this case, since subtracting their equations gives a line. So each step has $[F_{k+1} : F_k] \in \{1, 2\}$, and by the tower law $[F_n : \Q] = 2^m$. If $c$ is a coordinate of a constructed point, then $\Q \subseteq \Q(c) \subseteq F_n$, and $[\Q(c) : \Q]$ divides $2^m$ (tower law again).

Doubling a cube of volume $1$ means constructing $\sqrt[3]2$, of degree $3$. Trisecting $60^\circ$ would construct $\cos 20^\circ$, of degree $3$. A regular heptagon would give $\cos\frac{2\pi}{7}$, a root of the irreducible cubic $8x^3 + 4x^2 - 4x - 1$. None of these degrees is a power of $2$.

Existence (Kronecker). Let $g$ be an irreducible factor of $f$. Then $K_1 = K[x]/(g)$ is a field containing $K$ and a root $\bar x$ of $f$, with $[K_1 : K] = \deg g \le n$. Over $K_1$, $f = (x - \bar x)f_1$ with $\deg f_1 = n - 1$. Induct on $n$ and use the tower law. Uniqueness (sketch). An isomorphism $\varphi : K \to K'$ extends to $K[x]/(g) \to K'[x]/(g^\varphi)$, sending the root to any chosen root of $g^\varphi$. Inducting on the degree, it extends to an isomorphism of splitting fields. Full details: Dummit and Foote, Theorem 13.27.

(1) The characteristic is not $0$, since $F$ is finite, so it is a prime $p$, and the image of $\Z$ is $\mathbb{F}_p$. As an $\mathbb{F}_p$-vector space of dimension $n$, $F$ has $p^n$ elements.

(2) $(xy)^p = x^py^p$, and $(x + y)^p = \sum_k \binom{p}{k}x^ky^{p-k} = x^p + y^p$, because $p \mid \binom pk$ for $0 \lt k \lt p$. A field homomorphism is injective, and an injective self-map of a finite set is bijective.

(3) $f = X^q - X$ has derivative $qX^{q-1} - 1 = -1$, so $\gcd(f, f') = 1$ and $f$ has $q$ distinct roots in a splitting field $E$. The roots form the fixed set of the ring homomorphism $\phi^n : x \mapsto x^{q}$, so they are closed under $+$, $\times$ and inverses: they form a subfield with $q$ elements, which contains all roots of $f$ and hence equals $E$. For uniqueness, if $|F| = q$, then every $x \in F^\times$ satisfies $x^{q-1} = 1$ by Lagrange's theorem in the group $F^\times$, so every element of $F$ is a root of $f$. As $f$ has at most $q$ roots, $F$ is a splitting field of $f$ over $\mathbb{F}_p$, and Theorem 5 applies.

(4) Let $N = q - 1$, and let $\psi(d)$ be the number of elements of order $d$ in $F^\times$. If $\psi(d) \ne 0$, take $g$ of order $d$. The $d$ powers of $g$ are roots of $X^d - 1$, which has at most $d$ roots in a field, so every element of order $d$ is a power of $g$, and exactly $\varphi(d)$ of those have order $d$. Thus $\psi(d) \le \varphi(d)$ for all $d \mid N$. But $\sum_{d \mid N}\psi(d) = N = \sum_{d \mid N}\varphi(d)$ (groups lesson, Proposition 4), so $\psi(d) = \varphi(d)$ for every $d \mid N$. In particular $\psi(N) = \varphi(N) \ge 1$: there is an element of order $N$.

(5) If $K \subseteq \mathbb{F}_{p^n}$ with $[K : \mathbb{F}_p] = d$, then $n = d\,[\mathbb{F}_{p^n} : K]$, so $d \mid n$, and $K$ is the set of roots of $X^{p^d} - X$ by (3), so it is unique. Conversely, if $d \mid n$, then $p^d - 1 \mid p^n - 1$. So $X^{p^d} - X$ divides $X^{p^n} - X$, which splits into distinct linear factors in $\mathbb{F}_{p^n}$. Its $p^d$ roots form a subfield, by the argument in (3).

(6) If $\phi^k = \mathrm{id}$ with $1 \le k \lt n$, all $p^n$ elements would be roots of $X^{p^k} - X$, which has only $p^k$ roots. So $\phi$ has order $n$, and $|\operatorname{Aut}(\mathbb{F}_{p^n}/\mathbb{F}_p)| \le [\mathbb{F}_{p^n} : \mathbb{F}_p] = n$ by Theorem 7. Hence the group is $\langle\phi\rangle \cong \Z/n\Z$.

Simple case. An automorphism $\sigma$ over $K$ is determined by $\sigma(\alpha)$, because $L = K[\alpha]$. And $m_\alpha(\sigma(\alpha)) = \sigma(m_\alpha(\alpha)) = 0$. Conversely, for each root $\beta \in L$ of $m_\alpha$, $K(\alpha) \cong K[x]/(m_\alpha) \cong K(\beta) \subseteq L$ gives an embedding $\alpha \mapsto \beta$. Since $[K(\beta) : K] = \deg m_\alpha = [L : K]$, it is onto $L$. So $|\operatorname{Aut}(L/K)|$ is the number of roots of $m_\alpha$ in $L$, which is at most $\deg m_\alpha = [L : K]$.

General case. Distinct automorphisms $\sigma_1, \dots, \sigma_r$ are linearly independent over $L$ as functions $L \to L$ (Dedekind's lemma: take a shortest dependence $\sum c_i\sigma_i = 0$; substituting $yx$ for $x$ and subtracting $\sigma_1(y)$ times the original produces a shorter one). They are $K$-linear maps of the $K$-vector space $L$ of dimension $d = [L : K]$. The space of such maps, viewed as an $L$-vector space by post-multiplication, has dimension $d$ (a $K$-linear map is determined by its values on a basis). So $r \le d$.

Let $n = |H|$ and suppose $\alpha_1, \dots, \alpha_m \in L$ with $m \gt n$. The linear system $\sum_{j=1}^m \sigma(\alpha_j)\,x_j = 0$ ($\sigma \in H$) has $n$ equations and $m$ unknowns, so it has a non-zero solution in $L^m$. Choose one with the fewest non-zero entries, scaled so that some entry equals $1$. For $\tau \in H$, applying $\tau$ to the equations permutes them (as $\sigma$ runs over $H$, so does $\tau\sigma$), so $\tau(x)$ is again a solution. Then $x - \tau(x)$ is a solution with fewer non-zero entries (the scaled entry cancels), so it is $0$. Hence every $x_j \in L^H$, and the equation for $\sigma = \mathrm{id}$ is a non-trivial linear relation $\sum \alpha_j x_j = 0$ over $L^H$. So any $m \gt n$ elements are dependent, and $[L : L^H] \le n$. On the other hand $H \subseteq \operatorname{Aut}(L/L^H)$, so by Theorem 7, $n \le |\operatorname{Aut}(L/L^H)| \le [L : L^H] \le n$. All are equal, and $H = \operatorname{Aut}(L/L^H)$.

Both maps reverse inclusions: a larger group fixes less, and a larger field is fixed by fewer automorphisms. (1) For a subgroup $H$, Artin's theorem gives $\operatorname{Gal}(L/L^H) = H$. For an intermediate field $M$: $L$ is the splitting field over $K$ of a separable $f$, hence also over $M$, so $L/M$ is Galois and $L^{\operatorname{Gal}(L/M)} = M$ by Theorem 9(d). (2) Theorem 9(c) for $L/M$ gives the first equation; the tower law and Lagrange's theorem give the second.

(3) (Sketch.) For $\sigma \in G$, $\operatorname{Gal}(L/\sigma M) = \sigma \operatorname{Gal}(L/M)\sigma^{-1}$. If $M/K$ is normal, then $\sigma M = M$ for all $\sigma$, since $\sigma$ permutes the roots of minimal polynomials of elements of $M$, which all lie in $M$. So $\operatorname{Gal}(L/M)$ is normal. Restriction $\sigma \mapsto \sigma|_M$ is then a homomorphism $G \to \operatorname{Aut}(M/K)$ with kernel $\operatorname{Gal}(L/M)$. Its image has order $[G : \operatorname{Gal}(L/M)] = [M : K]$, so $M/K$ is Galois and the first isomorphism theorem gives the isomorphism. Conversely, if $\operatorname{Gal}(L/M)$ is normal, then $\sigma M = M$ for all $\sigma$, and restriction gives $[M : K]$ automorphisms of $M/K$, so $M/K$ is Galois by Theorem 9(c).

We show that if $N \trianglelefteq H \le S_n$ with $H/N$ abelian and $H$ containing every 3-cycle, then $N$ contains every 3-cycle. Then no chain from $S_n$ down to $\{e\}$ with abelian quotients can exist. Given a 3-cycle $(a\,b\,c)$, choose $d, e$ distinct from $a, b, c$ (possible as $n \ge 5$). Following each of the five points shows that $(a\,b\,c) = x y x^{-1}y^{-1}$ with $x = (a\,b\,d)$ and $y = (a\,c\,e)$ (composing right to left). Since $H/N$ is abelian, every commutator of elements of $H$ lies in $N$ (Proposition 6 of quotients), so $(a\,b\,c) \in N$. Now in a chain $S_n = G_s \trianglerighteq G_{s-1} \trianglerighteq \dots \trianglerighteq G_0$ with abelian quotients, induction down the chain shows that every $G_i$ contains all 3-cycles, so $G_0 \ne \{e\}$.