--- title: The Galois Correspondence module: Galois Theory moduleNumber: 11 lessonNumber: 1 order: 1101 summary: > Galois theory attaches to a field extension its group of symmetries and shows that, for the right extensions, the subgroups of that group are in exact order-reversing correspondence with the intermediate fields. The automorphism group, Artin's theorem, the characterization of Galois extensions, and the Fundamental Theorem together turn questions about fields into questions about finite groups. topics: [Galois Theory] sources: - book: Dummit & Foote ref: "Ch. 14 — Galois Theory; §14.1 Basic Definitions" - book: Dummit & Foote ref: "§14.2 The Fundamental Theorem of Galois Theory" - book: Judson ref: "Ch. 23 — Galois Theory; §23.1 Field Automorphisms, §23.2 The Fundamental Theorem" draft: false --- A [field extension](/abstract-algebra/field-theory/field-extensions-and-algebraic-elements) $K/F$ records that $K$ is a larger field containing $F$. Galois theory studies its _symmetries_: the ways $K$ can be rearranged that leave $F$ untouched. These symmetries form a group, and for a well-behaved class of extensions the internal structure of that group reproduces, exactly and in reverse, the lattice of fields lying between $F$ and $K$. A question about fields (which subfields exist, which are themselves symmetric over $F$) becomes a question about a finite group (which subgroups exist, which are normal). The [insolvability of the quintic](/abstract-algebra/galois-theory/solvability-by-radicals-and-the-quintic) and the three [classical construction problems](/abstract-algebra/field-theory/straightedge-and-compass-constructions) are settled by reading group structure off the extension. ## Automorphisms and the group they form > **Definition (Automorphism).** An isomorphism $\sigma : K \to K$ of a field with > itself is an **automorphism** of $K$. The set of all automorphisms is written > $\operatorname{Aut}(K)$. For $\alpha \in K$ we abbreviate $\sigma(\alpha)$ as > $\sigma\alpha$. An automorphism $\sigma$ **fixes** $\alpha$ if $\sigma\alpha = \alpha$, > and fixes a subset $F \subseteq K$ if it fixes every element of $F$. Composition of automorphisms is again an automorphism, and the inverse of an isomorphism is an isomorphism, so $\operatorname{Aut}(K)$ is a group under composition. The identity map $1$ is always present. Every automorphism sends $1 \mapsto 1$ and $0 \mapsto 0$, hence fixes the entire prime subfield generated by $1$. So $\mathbb{Q}$ and $\mathbb{F}_p$ admit only the identity: $\operatorname{Aut}(\mathbb{Q}) = \{1\}$ and $\operatorname{Aut}(\mathbb{F}_p) = \{1\}$. > **Definition (Automorphisms fixing $F$).** For an extension $K/F$, write > $\operatorname{Aut}(K/F)$ for the automorphisms of $K$ that fix $F$ pointwise. It is > a subgroup of $\operatorname{Aut}(K)$. The single fact that makes these groups computable is that an automorphism cannot move a root of a polynomial anywhere except to another root of the same polynomial. > **Proposition (Automorphisms permute roots).** Let $K/F$ be an extension and let > $\alpha \in K$ be algebraic over $F$ with minimal polynomial $m_\alpha(x) \in F[x]$. > For any $\sigma \in \operatorname{Aut}(K/F)$, the image $\sigma\alpha$ is again a root > of $m_\alpha$. More generally, $\sigma$ permutes the roots of every polynomial with > coefficients in $F$. The proof is a one-line consequence of $\sigma$ being a ring homomorphism that fixes $F$. If $\sum_i a_i \alpha^i = 0$ with $a_i \in F$, then applying $\sigma$ and using $\sigma(a_i) = a_i$ gives $\sum_i a_i (\sigma\alpha)^i = 0$, so $\sigma\alpha$ satisfies the same equation.[^df-perm] Two corollaries follow immediately: - **Automorphisms are determined by their action on generators.** If $K = F(\alpha_1, \dots, \alpha_k)$, then $\sigma \in \operatorname{Aut}(K/F)$ is determined by the values $\sigma\alpha_1, \dots, \sigma\alpha_k$, each of which is a root of the corresponding minimal polynomial. - **$\operatorname{Aut}(K/F)$ is finite for finite extensions.** There are finitely many generators, each with finitely many conjugate roots, so only finitely many candidate maps exist. Not every assignment of generators to conjugate roots yields an automorphism — the generators may satisfy algebraic relations that any automorphism must respect. This is the recurring subtlety in computing a Galois group by hand. ### First examples > **Worked example.** For $K = \mathbb{Q}(\sqrt{2})$, an automorphism must send > $\sqrt{2}$ to a root of $x^2 - 2$, so $\sqrt{2} \mapsto \pm\sqrt{2}$. Both choices > give automorphisms: the identity, and $\sigma : a + b\sqrt{2} \mapsto a - b\sqrt{2}$. > So $\operatorname{Aut}(\mathbb{Q}(\sqrt{2})/\mathbb{Q}) = \{1, \sigma\} \cong \mathbb{Z}/2\mathbb{Z}$, > and the automorphism count equals the degree $2$. > > For $K = \mathbb{Q}(\sqrt[3]{2})$, an automorphism must send $\sqrt[3]{2}$ to a root > of $x^3 - 2$. The other two roots are complex and $\mathbb{Q}(\sqrt[3]{2}) \subset > \mathbb{R}$ contains neither, so the only option is $\sqrt[3]{2} \mapsto \sqrt[3]{2}$. > Here $\operatorname{Aut}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}) = \{1\}$ is trivial even > though $[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 3$: the automorphism group is smaller > than the degree, which the definition of a Galois extension rules out. ## Fixed fields and the Galois connection The construction runs both ways. From a subfield we produced a group of automorphisms; from a group of automorphisms we recover a subfield. > **Definition (Fixed field).** For a subgroup $H \leq \operatorname{Aut}(K)$, the set > $K^H = \{\alpha \in K : h\alpha = \alpha \text{ for all } h \in H\}$ of elements fixed > by every automorphism in $H$ is a subfield of $K$, called the **fixed field** of $H$. That $K^H$ is closed under the field operations follows because each $h$ is a homomorphism: if $h$ fixes $\alpha$ and $\beta$ then it fixes $\alpha \pm \beta$, $\alpha\beta$, and $\alpha^{-1}$. The two constructions — subfield $\rightsquigarrow$ fixing group and subgroup $\rightsquigarrow$ fixed field — are **inclusion-reversing**. > **Proposition (Order reversal).** For subfields $F_1 \subseteq F_2 \subseteq K$, > $\operatorname{Aut}(K/F_2) \leq \operatorname{Aut}(K/F_1)$. For subgroups > $H_1 \leq H_2 \leq \operatorname{Aut}(K)$ with fixed fields $F_1, F_2$, we have > $F_2 \subseteq F_1$. A larger field imposes more conditions to fix, so fewer automorphisms qualify; a larger group of automorphisms fixes fewer elements. This pair of arrows between the poset of subfields and the poset of subgroups is a **Galois connection**. The content of the theory is that, restricted to the right extensions, the connection is a bijection. $$ % caption: The two inclusion-reversing maps between intermediate fields and subgroups: % a field goes to the automorphisms fixing it, a subgroup goes to its fixed field. \begin{tikzpicture}[scale=1.0, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} % fields column \node (K) at (0,3) {$K$}; \node (E) at (0,1.5) {$E$}; \node (F) at (0,0) {$F$}; \draw[black] (F) -- (E) -- (K); \node[anchor=south, font=\footnotesize] at (0,3.2) {extensions}; % groups column \node (one) at (5,3) {$1$}; \node (H) at (5,1.5) {$H$}; \node (G) at (5,0) {$G$}; \draw[black] (G) -- (H) -- (one); \node[anchor=south, font=\footnotesize] at (5,3.2) {subgroups}; % correspondence arrows \draw[->, acc, thick] (E.east) to[bend left=18] node[above, font=\footnotesize, black]{$\mathrm{Aut}$} (H.west); \draw[->, acc, thick] (H.south west) to[bend left=18] node[below, font=\footnotesize, black]{$K^H$} (E.south east); \end{tikzpicture} $$ ## Bounding the automorphism count The example $\mathbb{Q}(\sqrt[3]{2})$ showed the automorphism group can be smaller than the degree. It is never larger. The bound comes from counting extensions of an isomorphism to a splitting field. > **Proposition (Counting bound).** Let $E$ be the splitting field over $F$ of > $f(x) \in F[x]$. Then $|\operatorname{Aut}(E/F)| \leq [E : F]$, with equality when > $f$ is separable over $F$. The proof is an induction on $[E:F]$ using the extension theorem for isomorphisms of base fields: an isomorphism $\varphi : F \to F'$ extends to at most $[E:F]$ isomorphisms of the splitting fields, and to exactly $[E:F]$ when the irreducible factors have distinct roots.[^df-count] Taking $F = F'$ and $\varphi = 1$ specializes the count to automorphisms. The same argument, refined in the next section, gives $|\operatorname{Aut}(K/F)| \leq [K:F]$ for _every_ finite extension. > **Definition (Galois extension).** A finite extension $K/F$ is **Galois** if > $|\operatorname{Aut}(K/F)| = [K : F]$. When $K/F$ is Galois, the automorphism group is > the **Galois group**, written $\operatorname{Gal}(K/F)$. > **Corollary.** The splitting field of a separable polynomial over $F$ is Galois over > $F$. In particular, every splitting field over $\mathbb{Q}$ is Galois, since one may > discard repeated factors without changing the splitting field, leaving a separable > polynomial.[^df-galois] ## Artin's theorem The counting bound goes one direction. The reverse — that a finite group of automorphisms is the full automorphism group of its fixed field, and the degree equals the group order — is a theorem of Artin, and it rests on a lemma about the independence of characters. > **Definition (Character).** A **character** of a group $G$ with values in a field $L$ > is a homomorphism $\chi : G \to L^\times$ into the multiplicative group of $L$. Distinct > characters $\chi_1, \dots, \chi_n$ are **linearly independent** over $L$ if no nontrivial > $L$-linear combination $\sum_i a_i \chi_i$ vanishes as a function on $G$. An embedding $\sigma : K \hookrightarrow L$ restricts to a character $K^\times \to L^\times$, and this character carries all the information about $\sigma$ (only $0$ is omitted, and $\sigma(0) = 0$). So the following theorem applies directly to field automorphisms. > **Theorem (Linear independence of characters).** Distinct characters $\chi_1, \dots, > \chi_n$ of a group $G$ with values in a field $L$ are linearly independent over $L$. The proof takes a minimal-length dependence relation $\sum_{i=1}^m a_i \chi_i = 0$, picks $g_0$ with $\chi_1(g_0) \neq \chi_m(g_0)$, evaluates the relation at $g_0 g$, and subtracts $\chi_m(g_0)$ times the original. The $\chi_m$ term cancels while the $\chi_1$ term remains with nonzero coefficient, producing a shorter relation, contradicting minimality.[^df-char] The application to fields is Artin's theorem. > **Theorem (Artin).** Let $G = \{1 = \sigma_1, \sigma_2, \dots, \sigma_n\}$ be a finite > subgroup of $\operatorname{Aut}(K)$ with fixed field $F = K^G$. Then $[K : F] = n = |G|$. Both inequalities are proved by manufacturing a linear dependence among the $\sigma_i$ where none can exist. If $n > [K:F]$, a homogeneous system with more unknowns than equations has a nonzero solution, which assembles into a dependence of the $\sigma_i$ over $K$ — impossible by independence of characters. If $n < [K:F]$, one takes $n+1$ elements independent over $F$ and, using that $G$ is closed under composition, again extracts a shorter-than-minimal dependence.[^df-artin] Three corollaries follow. - **Sharp bound.** $|\operatorname{Aut}(K/F)| \leq [K:F]$ for any finite extension, with equality if and only if $F$ is the fixed field of $\operatorname{Aut}(K/F)$. So $K/F$ is Galois exactly when $F = K^{\operatorname{Aut}(K/F)}$. - **No hidden automorphisms.** If $G$ is a finite group of automorphisms with fixed field $F$, then $\operatorname{Aut}(K/F) = G$ exactly; $K/F$ is Galois with Galois group $G$. - **Distinct groups, distinct fixed fields.** Different finite subgroups of $\operatorname{Aut}(K)$ have different fixed fields. The last corollary is the injectivity that the Fundamental Theorem upgrades to a bijection: taking fixed fields is a one-to-one map from subgroups to intermediate fields. ## Characterizations of a Galois extension Artin's theorem also delivers the converse of the counting bound and, with it, a structural characterization. > **Theorem (Characterization of Galois extensions).** $K/F$ is Galois if and only if > $K$ is the splitting field of some separable polynomial over $F$. When this holds, > every irreducible polynomial in $F[x]$ with a root in $K$ is separable and splits > completely in $K$. The forward direction shows that for any $\alpha \in K$, the distinct images $\{\sigma\alpha : \sigma \in G\}$ are the roots of a polynomial $f(x) = \prod_i (x - \alpha_i)$ whose coefficients, being symmetric under $G$, lie in the fixed field $F$. Since the minimal polynomial of $\alpha$ divides $f$ and $f$ divides it, they are equal — so the minimal polynomial is separable and splits in $K$.[^df-normal] This yields four interchangeable descriptions. | # | Characterization | Emphasis | | --- | --- | --- | | 1 | Splitting field of a separable polynomial over $F$ | existence of enough roots | | 2 | $F$ equals the fixed field of $\operatorname{Aut}(K/F)$ | nothing outside $F$ is fixed | | 3 | $[K:F] = \lvert\operatorname{Aut}(K/F)\rvert$ | maximal automorphism count | | 4 | Finite, **normal**, and **separable** | roots and repeated-root conditions | A **normal** extension is one that, containing a single root of an irreducible polynomial, contains all of them; a **separable** extension has minimal polynomials with [distinct roots](/abstract-algebra/field-theory/separable-and-cyclotomic-extensions). An extension fails to be Galois the moment one irreducible polynomial has a root in $K$ but not all of them; $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ fails on this very point. ## The Fundamental Theorem Let $K/F$ be Galois with group $G = \operatorname{Gal}(K/F)$. > **Theorem (Fundamental Theorem of Galois Theory).** There is an inclusion-reversing > bijection between the intermediate fields $F \subseteq E \subseteq K$ and the subgroups > $H \leq G$, given by $E \mapsto \operatorname{Aut}(K/E)$ and $H \mapsto K^H$. Under this > correspondence: > - **(degrees)** $[K : E] = |H|$ and $[E : F] = [G : H]$; > - **(lower extension)** $K/E$ is always Galois, with $\operatorname{Gal}(K/E) = H$; > - **(normality)** $E/F$ is Galois if and only if $H \trianglelefteq G$, and then > $\operatorname{Gal}(E/F) \cong G/H$; > - **(lattice duality)** $E_1 \cap E_2$ corresponds to $\langle H_1, H_2 \rangle$ and > the composite $E_1 E_2$ corresponds to $H_1 \cap H_2$. The bijection is the two Artin corollaries: distinct subgroups have distinct fixed fields (injective), and every intermediate field $E$ is the fixed field of $\operatorname{Aut}(K/E)$ because $K$ is the splitting field of the same separable polynomial viewed over $E$ (surjective). The degree statements are Artin applied to $K/E$ and the tower law. $$ % caption: The Galois correspondence for a Galois extension: intermediate fields on the % left, subgroups on the right. Degrees on the left edges equal $|H|$ and $[G:H]$; the % dashed links pair each field with its fixing subgroup. \begin{tikzpicture}[scale=1.0, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} % field lattice \node (K) at (0,3.4) {$K$}; \node (E) at (0,1.7) {$E$}; \node (F) at (0,0) {$F$}; \draw[black] (F) -- (E); \draw[black] (E) -- (K); \node[acc, font=\footnotesize, anchor=east] at (-0.2,2.55) {$[K:E]$}; \node[acc, font=\footnotesize, anchor=east] at (-0.2,0.85) {$[E:F]$}; % group lattice (inverted) \node (one) at (6,3.4) {$1$}; \node (H) at (6,1.7) {$H$}; \node (G) at (6,0) {$G$}; \draw[black] (G) -- (H); \draw[black] (H) -- (one); \node[acc, font=\footnotesize, anchor=west] at (6.2,2.55) {order $H$}; \node[acc, font=\footnotesize, anchor=west] at (6.2,0.85) {index $[G:H]$}; % correspondence \draw[<->, dashed, black] (K) -- (one); \draw[<->, dashed, black] (E) -- (H); \draw[<->, dashed, black] (F) -- (G); \end{tikzpicture} $$ The normality clause has an operational reading: conjugating $E$ by $\sigma \in G$ yields $\sigma(E) = K^{\sigma H \sigma^{-1}}$, so $E$ is stable under all of $G$ exactly when $\sigma H \sigma^{-1} = H$ for all $\sigma$, i.e. $H \trianglelefteq G$. When $E/F$ is not Galois, the elements of $G$ still act, but they carry $E$ to _conjugate fields_ isomorphic to it rather than back to itself. $$ % caption: When $H$ is not normal (left), the Galois group carries the field $E = K^H$ % among distinct conjugate fields $\sigma(E) = K^{\sigma H \sigma^{-1}}$; when $H$ is % normal (right), every automorphism maps $E$ back to itself and $E/F$ is Galois. \begin{tikzpicture}[scale=1.0, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} % left panel: non-normal, three conjugate fields permuted \node[draw, minimum size=8mm] (E1) at (0,1.4) {$E_1$}; \node[draw, minimum size=8mm] (E2) at (-1.2,-0.7) {$E_2$}; \node[draw, minimum size=8mm] (E3) at (1.2,-0.7) {$E_3$}; \draw[->, acc, thick] (E1) to[bend left=20] (E3); \draw[->, acc, thick] (E3) to[bend left=20] (E2); \draw[->, acc, thick] (E2) to[bend left=20] (E1); \node[font=\footnotesize, align=center] at (0,-1.9) {$H$ not normal:\\conjugates vary}; % right panel: normal, E fixed \begin{scope}[xshift=6.5cm] \node[draw, minimum size=8mm, draw=acc] (E) at (0,0.3) {$E$}; \draw[->, acc, thick] (E) to[out=40,in=110,looseness=6] (E); \node[acc, font=\footnotesize, anchor=south] at (0,1.35) {$G$}; \node[font=\footnotesize, align=center] at (0,-1.9) {$H$ normal:\\$E$ is stable}; \end{scope} \end{tikzpicture} $$ ## A worked correspondence: $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ The field $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is the splitting field of $(x^2 - 2)(x^2 - 3)$, hence Galois of degree $4$ over $\mathbb{Q}$. An automorphism sends $\sqrt{2} \mapsto \pm\sqrt{2}$ and $\sqrt{3} \mapsto \pm\sqrt{3}$ independently, giving four maps, all of which are automorphisms. Writing $\sigma$ for the sign flip on $\sqrt{2}$ and $\tau$ for the sign flip on $\sqrt{3}$, $$ \operatorname{Gal}(K/\mathbb{Q}) = \{1, \sigma, \tau, \sigma\tau\} \cong \mathbb{Z}/2 \times \mathbb{Z}/2, $$ the Klein four-group. Every subgroup of an abelian group is normal, so every intermediate field is Galois over $\mathbb{Q}$. The three subgroups of order $2$ correspond to the three quadratic subfields. | Subgroup $H$ | Fixed field $K^H$ | $[K^H : \mathbb{Q}]$ | | --- | --- | --- | | $\{1\}$ | $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ | $4$ | | $\{1, \sigma\}$ | $\mathbb{Q}(\sqrt{3})$ | $2$ | | $\{1, \tau\}$ | $\mathbb{Q}(\sqrt{2})$ | $2$ | | $\{1, \sigma\tau\}$ | $\mathbb{Q}(\sqrt{6})$ | $2$ | | $\{1, \sigma, \tau, \sigma\tau\}$ | $\mathbb{Q}$ | $1$ | The composite $\sigma\tau$ fixes $\sqrt{6} = \sqrt{2}\sqrt{3}$ because it flips both factors. Displaying the two lattices side by side, inverted so that inclusions match, shows the bijection concretely. $$ % caption: For the Klein four-group, the lattice of subfields of $K = \mathbb{Q}(a,b)$ % (left) mirrors the inverted lattice of subgroups (right); each quadratic field pairs % with an order-two subgroup. Here $a = \sqrt{2}$, $b = \sqrt{3}$, $c = \sqrt{6}$, and % $H_1, H_2, H_3$ are the subgroups fixing $\mathbb{Q}(a), \mathbb{Q}(c), \mathbb{Q}(b)$. \begin{tikzpicture}[scale=1.0, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} % field lattice: top K, three quadratics, bottom Q \node (K) at (0,3) {$K$}; \node (Qa) at (-2.2,1.5) {$\mathbb{Q}(a)$}; \node (Qc) at (0,1.5) {$\mathbb{Q}(c)$}; \node (Qb) at (2.2,1.5) {$\mathbb{Q}(b)$}; \node (Q) at (0,0) {$\mathbb{Q}$}; \draw[black] (Q) -- (Qa) -- (K); \draw[black] (Q) -- (Qc) -- (K); \draw[black] (Q) -- (Qb) -- (K); % group lattice inverted: top {1}, three order-2, bottom G \begin{scope}[xshift=8cm] \node (one) at (0,3) {$1$}; \node (Ht) at (-2.2,1.5) {$H_1$}; \node (Hst) at (0,1.5) {$H_2$}; \node (Hs) at (2.2,1.5) {$H_3$}; \node (G) at (0,0) {$G$}; \draw[black] (G) -- (Ht) -- (one); \draw[black] (G) -- (Hst) -- (one); \draw[black] (G) -- (Hs) -- (one); \end{scope} % linking arrow \draw[<->, acc, thick] (3.3,1.5) -- (4.7,1.5); \node[font=\footnotesize, anchor=south] at (4,1.55) {duality}; \end{tikzpicture} $$ ## A non-abelian correspondence: the splitting field of $x^3 - 2$ The splitting field of $x^3 - 2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[3]{2}, \rho)$, where $\rho = e^{2\pi i / 3}$ is a primitive cube root of unity. It has degree $6$, and the Galois group permutes the three roots $\sqrt[3]{2}, \rho\sqrt[3]{2}, \rho^2\sqrt[3]{2}$ faithfully, so $\operatorname{Gal}(\mathbb{Q}(\sqrt[3]{2}, \rho)/\mathbb{Q}) \cong S_3$. Generators are $\sigma$ (a $3$-cycle of the roots, sending $\sqrt[3]{2} \mapsto \rho\sqrt[3]{2}$ and fixing $\rho$) and $\tau$ (complex conjugation, fixing $\sqrt[3]{2}$ and sending $\rho \mapsto \rho^2$), with $\sigma^3 = \tau^2 = 1$ and $\tau\sigma = \sigma^2\tau$. > **Worked example.** Write $\alpha = \sqrt[3]{2}$, so $\sigma : \alpha \mapsto > \rho\alpha,\ \rho \mapsto \rho$ and $\tau : \alpha \mapsto \alpha,\ \rho \mapsto \rho^2$. > Track both sides of $\tau\sigma = \sigma^2\tau$ on the generators $\alpha, \rho$ of $K$: > $$ > \tau\sigma(\alpha) = \tau(\rho\alpha) = \rho^2\alpha, \qquad > \sigma^2\tau(\alpha) = \sigma^2(\alpha) = \rho^2\alpha, > $$ > $$ > \tau\sigma(\rho) = \tau(\rho) = \rho^2, \qquad > \sigma^2\tau(\rho) = \sigma^2(\rho^2) = \rho^2. > $$ > The two maps agree on $\alpha$ and $\rho$, which generate $K$, so $\tau\sigma = > \sigma^2\tau$: the group is the nonabelian $S_3$, not $\mathbb{Z}/6$. The subgroups of $S_3$ dictate the intermediate fields. Complex conjugation $\tau = (\alpha_2\,\alpha_3)$ fixes the real root $\alpha_1 = \sqrt[3]{2}$, and the $3$-cycle $\sigma = (\alpha_1\,\alpha_2\,\alpha_3)$ generates the unique normal subgroup $\langle\sigma\rangle = A_3$. The full correspondence has six matched pairs. | Subgroup $H \leq S_3$ | Order | Fixed field $K^H$ | $[K^H : \mathbb{Q}]$ | Normal | | --- | --- | --- | --- | --- | | $\{1\}$ | $1$ | $\mathbb{Q}(\sqrt[3]{2}, \rho)$ | $6$ | yes | | $\langle\sigma\rangle = A_3$ | $3$ | $\mathbb{Q}(\rho)$ | $2$ | yes | | $\langle\tau\rangle$ | $2$ | $\mathbb{Q}(\sqrt[3]{2})$ | $3$ | no | | $\langle\sigma^2\tau\rangle$ | $2$ | $\mathbb{Q}(\rho\sqrt[3]{2})$ | $3$ | no | | $\langle\sigma\tau\rangle$ | $2$ | $\mathbb{Q}(\rho^2\sqrt[3]{2})$ | $3$ | no | | $S_3$ | $6$ | $\mathbb{Q}$ | $1$ | yes | The three normal subgroups $1, A_3, S_3$ match the three fields Galois over $\mathbb{Q}$: $K$ itself, the quadratic $\mathbb{Q}(\rho) = \mathbb{Q}(\sqrt{-3})$ with $\operatorname{Gal}(\mathbb{Q}(\rho)/\mathbb{Q}) \cong S_3/A_3 \cong \mathbb{Z}/2$, and the base $\mathbb{Q}$. The three conjugate order-$2$ subgroups match the three cubic fields $\mathbb{Q}(\sqrt[3]{2}), \mathbb{Q}(\rho\sqrt[3]{2}), \mathbb{Q}(\rho^2\sqrt[3]{2})$ — isomorphic but distinct, permuted by $G$ as it conjugates the subgroups. None is Galois over $\mathbb{Q}$, since each holds one root of $x^3 - 2$ and misses the other two. $$ % caption: The Galois correspondence for $x^3 - 2$ over $\mathbb{Q}$: the six intermediate % fields (left) pair with the six subgroups of $S_3$ (right, drawn upside down) in reverse % order. The normal (Galois) pairs are $K$ with $1$, $\mathbb{Q}(r)$ with $A_3$, and % $\mathbb{Q}$ with $S_3$; the three cubic fields pair with the three conjugate order-2 % subgroups. Here $c = \sqrt[3]{2}$, $r = e^{2\pi i/3}$, and $H_1, H_2, H_3$ are generated % by $\tau$, $\sigma^2\tau$, $\sigma\tau$. \begin{tikzpicture}[scale=0.95, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} % field lattice \node (K) at (0,3.2) {$K$}; \node (Qr) at (-2.7,1.6) {$\mathbb{Q}(r)$}; \node (Qc) at (-0.9,1.6) {$\mathbb{Q}(c)$}; \node (Qrc) at (0.9,1.6) {$\mathbb{Q}(rc)$}; \node (Qr2c) at (2.7,1.6) {$\mathbb{Q}(r^2c)$}; \node (Q) at (0,0) {$\mathbb{Q}$}; \draw[black] (Q) -- (Qr); \draw[black] (Q) -- (Qc); \draw[black] (Q) -- (Qrc); \draw[black] (Q) -- (Qr2c); \draw[black] (Qr) -- (K); \draw[black] (Qc) -- (K); \draw[black] (Qrc) -- (K); \draw[black] (Qr2c) -- (K); % group lattice, inverted \begin{scope}[xshift=8.6cm] \node (one) at (0,3.2) {$1$}; \node (A3) at (-2.7,1.6) {$A_3$}; \node (H1) at (-0.9,1.6) {$H_1$}; \node (H2) at (0.9,1.6) {$H_2$}; \node (H3) at (2.7,1.6) {$H_3$}; \node (S3) at (0,0) {$S_3$}; \draw[black] (S3) -- (A3); \draw[black] (S3) -- (H1); \draw[black] (S3) -- (H2); \draw[black] (S3) -- (H3); \draw[black] (A3) -- (one); \draw[black] (H1) -- (one); \draw[black] (H2) -- (one); \draw[black] (H3) -- (one); \end{scope} % duality link \draw[<->, acc, thick] (3.3,1.6) -- (5.3,1.6); \node[acc, font=\footnotesize, anchor=south] at (4.3,1.65) {duality}; \end{tikzpicture} $$ [^df-perm]: **Dummit & Foote**, _Abstract Algebra_, §14.1, Proposition 2 — automorphisms fixing $F$ permute the roots of any polynomial over $F$; an automorphism of $K/F$ is determined by its action on a generating set. [^df-count]: **Dummit & Foote**, _Abstract Algebra_, §14.1, Proposition 5 — the number of automorphisms of a splitting field $E/F$ is at most $[E:F]$, with equality for separable polynomials, proved by induction via the isomorphism-extension theorem (Theorem 13.27). [^df-galois]: **Dummit & Foote**, _Abstract Algebra_, §14.1, Corollary 6 and the Galois-extension definition — $|\operatorname{Aut}(K/F)| = [K:F]$ defines Galois; splitting fields of separable polynomials attain the bound. [^df-char]: **Dummit & Foote**, _Abstract Algebra_, §14.2, Theorem 7 and Corollary 8 (linear independence of characters) — distinct characters, and hence distinct embeddings and automorphisms, are linearly independent as functions. [^df-artin]: **Dummit & Foote**, _Abstract Algebra_, §14.2, Theorem 9 (Artin) and Corollaries 10–12 — for a finite group $G$ of automorphisms with fixed field $F$, $[K:F] = |G|$; distinct finite subgroups have distinct fixed fields. [^df-normal]: **Dummit & Foote**, _Abstract Algebra_, §14.2, Theorem 13 and Theorem 14 (Fundamental Theorem) — the four characterizations of Galois extensions and the inclusion-reversing bijection between subfields and subgroups, with the normality clause $\operatorname{Gal}(E/F) \cong G/H$.