The central mystery of classical algebra is the following: given a polynomial equation, what does the structure of its roots tell you about the symmetries of the equation, and vice versa? For centuries, mathematicians could solve quadratics, cubics, and quartics by radicals — by a sequence of root extractions — but the quintic resisted every attempt. Galois, at nineteen, found the reason: the answer depends not on the degree of the polynomial, but on the symmetry group of its roots. The Galois correspondence is the precise statement of this relationship. It is a bijection between the intermediate fields of a field extension and the subgroups of an associated group — a perfect duality between algebra and symmetry that is one of the most beautiful results in all of mathematics. To see what the problem really is, consider the polynomial $f(x) = x^4 - 5x^2 + 6$. This factors over $\mathbb{Q}$ as $(x^2 - 2)(x^2 - 3)$, so its roots are $\pm\sqrt{2}$ and $\pm\sqrt{3}$. To adjoin all four roots to $\mathbb{Q}$, we form the field $\mathbb{Q}(\sqrt{2}, \sqrt{3})$. This field sits between $\mathbb{Q}$ and $\mathbb{R}$, and it is not alone: the field $\mathbb{Q}(\sqrt{6})$ also sits between them, as does $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$. There is an entire lattice of intermediate fields. What organizes this lattice? The answer is the group of field automorphisms of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ that fix $\mathbb{Q}$ pointwise. The Galois correspondence says that subgroups of this group and intermediate fields correspond to each other in a precise, order-reversing way. [example: The Biquadratic Extension] Let $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$. Every element of $K$ can be written as $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$ with $a, b, c, d \in \mathbb{Q}$. We have $[K : \mathbb{Q}] = 4$, since $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}(\sqrt{2})] = 2$ (because $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$, as $\sqrt{3} = a + b\sqrt{2}$ would give $3 = a^2 + 2b^2 + 2ab\sqrt{2}$, forcing $ab = 0$ and then $3 = a^2$ or $3 = 2b^2$, neither of which has rational solutions) and $[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] = 2$. A field automorphism $\sigma: K \to K$ fixing $\mathbb{Q}$ is determined by where it sends $\sqrt{2}$ and $\sqrt{3}$. Since $\sigma(\sqrt{2})^2 = \sigma(2) = 2$, we need $\sigma(\sqrt{2}) = \pm\sqrt{2}$. Similarly $\sigma(\sqrt{3}) = \pm\sqrt{3}$. This gives four automorphisms: \begin{align*} e &: \sqrt{2} \mapsto \sqrt{2},\quad \sqrt{3} \mapsto \sqrt{3} \\ \sigma &: \sqrt{2} \mapsto -\sqrt{2},\quad \sqrt{3} \mapsto \sqrt{3} \\ \tau &: \sqrt{2} \mapsto \sqrt{2},\quad \sqrt{3} \mapsto -\sqrt{3} \\ \sigma\tau &: \sqrt{2} \mapsto -\sqrt{2},\quad \sqrt{3} \mapsto -\sqrt{3} \end{align*} These form the group $G = \operatorname{Gal}(K/\mathbb{Q}) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, the Klein four-group. The subgroups of $G$ are $\{e\}$, $\langle\sigma\rangle$, $\langle\tau\rangle$, $\langle\sigma\tau\rangle$, and $G$ itself. The Galois correspondence associates each to an intermediate field: \begin{align*} G &\longleftrightarrow \mathbb{Q} \\ \langle\sigma\tau\rangle &\longleftrightarrow \mathbb{Q}(\sqrt{6}) \\ \langle\tau\rangle &\longleftrightarrow \mathbb{Q}(\sqrt{2}) \\ \langle\sigma\rangle &\longleftrightarrow \mathbb{Q}(\sqrt{3}) \\ \{e\} &\longleftrightarrow K \end{align*} For instance, $\mathbb{Q}(\sqrt{6})$ is exactly the fixed field of $\sigma\tau$: an element $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$ is fixed by $\sigma\tau$ iff $\sigma\tau(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a - b\sqrt{2} - c\sqrt{3} + d\sqrt{6}$ equals the original, which forces $b = c = 0$. The surviving elements $a + d\sqrt{6}$ are precisely $\mathbb{Q}(\sqrt{6})$. [/example] This example is not a coincidence. The Galois correspondence holds in precisely the setting where the group of automorphisms is "large enough" — this is the content of the theory of Galois extensions. ## Definition The right setting for the Galois correspondence is a Galois extension. Before defining this, we need the group whose structure we will study. An automorphism of a field extension $K/k$ is a field isomorphism $\sigma: K \to K$ that fixes every element of the base field $k$ pointwise. Any composition of such automorphisms is again one, and the identity is trivially one, so these form a group. [definition: Galois Group] Let $K/k$ be a field extension. The **Galois group** of $K$ over $k$ is \begin{align*} \operatorname{Gal}(K/k) := \{\sigma: K \xrightarrow{\sim} K \mid \sigma \text{ is a field automorphism and } \sigma(a) = a \text{ for all } a \in k\}. \end{align*} The group operation is composition of maps. [/definition] The Galois group always exists, but it may be very small. For the extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$, the only automorphism fixing $\mathbb{Q}$ is the identity, because any automorphism must send $\sqrt[3]{2}$ to another cube root of $2$ in the field, and $\sqrt[3]{2}$ is the only real cube root of $2$ — the other two are complex and not in $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}$. This Galois group has order $1$, which is much smaller than the degree $[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 3$. The Galois correspondence will not apply to this extension because it is not Galois. The condition for the correspondence to be a perfect bijection is that the Galois group be as large as possible — that its order equal the degree of the extension. This is precisely the definition of a Galois extension. [definition: Galois Extension] A finite field extension $K/k$ is **Galois** if \begin{align*} |\operatorname{Gal}(K/k)| = [K : k]. \end{align*} [/definition] Equivalently — and this equivalence is a theorem, not part of the definition — a finite extension $K/k$ is Galois if and only if it is both normal and separable. Normality means that $K$ is the splitting field of some polynomial $f \in k[x]$ over $k$; separability means that every element of $K$ has a minimal polynomial over $k$ with no repeated roots. Over a field of characteristic zero (such as $\mathbb{Q}$), every algebraic extension is automatically separable, so Galoisness reduces to normality. This is why splitting fields over $\mathbb{Q}$ are always Galois. [quotetheorem:3310] Condition 4 is especially important: it says that the automorphism group "sees" $k$ completely, that no element outside $k$ is accidentally fixed by every automorphism. This is what breaks down for $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$: the trivial Galois group fixes everything, but $\sqrt[3]{2} \notin \mathbb{Q}$ is fixed by all (zero non-identity) automorphisms. Now we can state the fixed field construction, which is the other half of the correspondence. For any subgroup $H \le \operatorname{Gal}(K/k)$, we can ask which elements of $K$ are left unchanged by every element of $H$. These elements form a subfield of $K$ containing $k$. [definition: Fixed Field] Let $K/k$ be a Galois extension and $H \le \operatorname{Gal}(K/k)$ a subgroup. The **fixed field** of $H$ is \begin{align*} K^H := \{\alpha \in K \mid \sigma(\alpha) = \alpha \text{ for all } \sigma \in H\}. \end{align*} [/definition] The fixed field $K^H$ is indeed a field: if $\sigma(\alpha) = \alpha$ and $\sigma(\beta) = \beta$ for all $\sigma \in H$, then $\sigma(\alpha + \beta) = \sigma(\alpha) + \sigma(\beta) = \alpha + \beta$ and $\sigma(\alpha\beta) = \alpha\beta$, so $K^H$ is closed under addition and multiplication, and similar reasoning shows inverses are preserved. We have now assembled the two directions of the Galois correspondence. Given a Galois extension $K/k$: - A subgroup $H \le \operatorname{Gal}(K/k)$ maps to the fixed field $K^H$, an intermediate field between $k$ and $K$. - An intermediate field $k \subset F \subset K$ maps to the relative Galois group $\operatorname{Gal}(K/F)$, a subgroup of $\operatorname{Gal}(K/k)$. The fundamental theorem says these two maps are mutually inverse bijections. ## The Fundamental Theorem Before stating the theorem, it is worth sketching why it is true — why should a bijection between subgroups and intermediate fields exist at all? The key is a result of Artin: if $H$ is a finite group of automorphisms of a field $K$, then $[K : K^H] = |H|$. The proof uses the linear independence of distinct field automorphisms (Dedekind's theorem on independence of characters): if $\sigma_1, \ldots, \sigma_n \in H$ are distinct, there is no nontrivial $k$-linear relation $\sum c_i \sigma_i = 0$. This independence forces the fixed field $K^H$ to be small — the larger $H$ is, the harder it is for an element to survive all the automorphisms. Combining this with a dimension count, one obtains $[K : K^H] = |H|$ exactly. Once this degree formula is in hand, the bijectivity of the correspondence follows from a straightforward order argument: $H \mapsto K^H \mapsto \operatorname{Gal}(K/K^H)$ starts from $H$ and ends at a group whose size $|\operatorname{Gal}(K/K^H)| = [K : K^H] = |H|$ forces it to equal $H$. The Galois correspondence is not merely a bijection — it is an order-reversing bijection, meaning it flips inclusions. Larger subgroups correspond to smaller intermediate fields, and vice versa. This reversal is the key structural feature. [quotetheorem:1274] Let us unpack what this theorem is saying, one piece at a time. The bijection $\Phi$ assigns to each subgroup its fixed field. The content of the theorem is that $\Phi$ is truly a bijection — it hits every intermediate field exactly once. The map $\Psi$ going the other direction takes an intermediate field $F$ and returns the group of automorphisms of $K$ that fix $F$. The extension $K/F$ is itself Galois (normality and separability are inherited from $K/k$ up the tower), so $\operatorname{Gal}(K/F)$ is the Galois group of this sub-extension. The order-reversal says: the bigger the subgroup, the more elements it fixes, but the more elements it fixes, the fewer elements are in the fixed field. Adding constraints to a group (making it larger by adding more automorphisms) reduces the fixed field. Conversely, a small subgroup has few automorphisms, so many elements survive — giving a large fixed field. The degree formula is precise quantitative content: the degree $[K : K^H]$ equals the order of $H$, not just in some qualitative sense but exactly. This means the "size" of the subgroup and the "size" of the sub-extension above its fixed field are identical. The complementary formula $[K^H : k] = [G : H]$ follows from the tower law $[K : k] = [K : K^H][K^H : k] = |H| \cdot [K^H : k]$, combined with $|G| = [K : k]$. Part 3 — the correspondence between normal subgroups and Galois sub-extensions — is one of the deepest aspects of the theorem. The extension $K^H/k$ need not be Galois: it is Galois if and only if $H \trianglelefteq G$. When it is, the Galois group $\operatorname{Gal}(K^H/k)$ is the quotient $G/H$. This is because automorphisms of $K$ fixing $k$ can be restricted to automorphisms of $K^H$ fixing $k$, and the kernel of this restriction map is $H$ (the automorphisms of $K$ fixing $K^H$). The image is all of $\operatorname{Gal}(K^H/k)$ precisely when $H$ is normal. [illustration:galois-correspondence-lattice] ## Normal Subgroups and Galois Sub-extensions The relationship between normal subgroups and Galois sub-extensions is so important that it deserves its own treatment. The question is: when does an intermediate field $F$ with $k \subset F \subset K$ itself form a Galois extension of the base field $k$? The answer involves the action of $G = \operatorname{Gal}(K/k)$ on the set of intermediate fields. Given $\sigma \in G$ and an intermediate field $F$, we can form the image $\sigma(F) = \{\sigma(\alpha) : \alpha \in F\}$. This is again an intermediate field (automorphisms send subfields to subfields), but it need not equal $F$. Not every intermediate field behaves symmetrically under the Galois group. An automorphism $\sigma \in G$ can map an intermediate field $F$ to a different intermediate field $\sigma(F)$ — and when it does, $F$ cannot be Galois over the base field $k$, because Galoisness of $F/k$ requires that all embeddings of $F$ into the algebraic closure that fix $k$ send $F$ into itself. The fields that resist this displacement — those carried into themselves by every element of $G$ — are precisely the ones that can form Galois extensions of $k$. This motivates singling them out. [definition: Galois-Stable Intermediate Field] Let $K/k$ be a Galois extension with group $G$. An intermediate field $F$ with $k \subset F \subset K$ is **$G$-stable** (or **Galois-stable**) if $\sigma(F) = F$ for all $\sigma \in G$. [/definition] A $G$-stable intermediate field is one that the Galois group of the big extension permutes into itself. The theorem tells us that $F$ is $G$-stable if and only if $H = \operatorname{Gal}(K/F)$ is a normal subgroup of $G$, and in that case $F/k$ is Galois with group $G/H$. To see why stability and normality are linked, note that for $\sigma \in G$, the subgroup corresponding to $\sigma(F)$ under the Galois correspondence is $\sigma H \sigma^{-1}$ (property 4 of the theorem). For $\sigma(F) = F$, we need $\sigma H \sigma^{-1} = H$, i.e., $H \trianglelefteq G$. [example: Normal and Non-Normal Subextensions] Let $K = \mathbb{Q}(\zeta_8)$ where $\zeta_8 = e^{2\pi i/8}$ is a primitive eighth root of unity. Then $K$ is the splitting field of $x^8 - 1$ over $\mathbb{Q}$, and $[K : \mathbb{Q}] = 4$. The Galois group is $G = \operatorname{Gal}(\mathbb{Q}(\zeta_8)/\mathbb{Q}) \cong (\mathbb{Z}/8\mathbb{Z})^\times \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, where the automorphism $\sigma_j$ (for $\gcd(j, 8) = 1$) sends $\zeta_8 \mapsto \zeta_8^j$. The three non-trivial elements of $G$ are $\sigma_3, \sigma_5, \sigma_7$ (corresponding to $j = 3, 5, 7$). Since $G$ is abelian, every subgroup is normal, so every intermediate field is Galois over $\mathbb{Q}$. Now compare with $L = \mathbb{Q}(\sqrt[4]{2})$ and its Galois closure $M = \mathbb{Q}(\sqrt[4]{2}, i)$, which has degree $8$ over $\mathbb{Q}$ and $\operatorname{Gal}(M/\mathbb{Q}) \cong D_8$ (the dihedral group of order $8$). The subfield $L = \mathbb{Q}(\sqrt[4]{2}) \subset M$ corresponds to a subgroup $H \le D_8$ of order $[M : L] = 2$. However, this subgroup is not normal in $D_8$ (since $D_8$ is non-abelian and not every order-$2$ subgroup is normal). As a consequence, the extension $L/\mathbb{Q}$ is not Galois — indeed $L = \mathbb{Q}(\sqrt[4]{2})$ is not normal over $\mathbb{Q}$ because the minimal polynomial $x^4 - 2$ has roots $\sqrt[4]{2}, i\sqrt[4]{2}, -\sqrt[4]{2}, -i\sqrt[4]{2}$, but $L \subset \mathbb{R}$ contains only $\pm\sqrt[4]{2}$, missing the complex roots. This shows concretely why the normal subgroup condition is necessary: a non-normal subgroup corresponds to an intermediate field that is not fixed under conjugation by the Galois group, and hence cannot be Galois over the base. [/example] [example: A Non-Galois Extension and What Goes Wrong] Let $k = \mathbb{Q}$ and consider $f(x) = x^3 - 2$, with roots $\sqrt[3]{2}$, $\omega\sqrt[3]{2}$, $\omega^2\sqrt[3]{2}$ where $\omega = e^{2\pi i/3}$. The splitting field is $K = \mathbb{Q}(\sqrt[3]{2}, \omega)$, which has degree $[K:\mathbb{Q}] = 6$ and Galois group $\operatorname{Gal}(K/\mathbb{Q}) \cong S_3$. Consider the intermediate field $F = \mathbb{Q}(\sqrt[3]{2})$. This is not a Galois extension of $\mathbb{Q}$: the minimal polynomial $x^3 - 2$ of $\sqrt[3]{2}$ over $\mathbb{Q}$ has roots $\omega\sqrt[3]{2}$ and $\omega^2\sqrt[3]{2}$ that do not lie in $F \subset \mathbb{R}$, so $F$ fails to be a normal extension. The corresponding subgroup $H = \operatorname{Gal}(K/F)$ is generated by complex conjugation $\sigma$ (fixing $\sqrt[3]{2}$ and sending $\omega \mapsto \omega^2$), a subgroup of order $2$. There are three such order-$2$ subgroups in $S_3$, each conjugate to the others, so none is normal. The practical failure: if you try to restrict automorphisms in $\operatorname{Gal}(K/\mathbb{Q})$ to the subfield $\mathbb{Q}(\sqrt[3]{2})$, the result need not be an automorphism of $\mathbb{Q}(\sqrt[3]{2})$ — the automorphism $\tau: \sqrt[3]{2} \mapsto \omega\sqrt[3]{2}$, $\omega \mapsto \omega$ moves $\sqrt[3]{2}$ outside $F$. This is precisely what it means for the extension to be non-normal and for the subgroup to be non-normal: the Galois group does not act on $F$ as a group of automorphisms of $F$. We can trace through the maps $\Phi$ and $\Psi$ to see concretely where the naive bijection fails. Suppose we try to run the Galois correspondence for the non-Galois extension $F/\mathbb{Q}$ where $F = \mathbb{Q}(\sqrt[3]{2})$. The Galois group of $F$ over $\mathbb{Q}$ is $\operatorname{Gal}(F/\mathbb{Q}) = \{e\}$: only the identity automorphism fixes $\mathbb{Q}$ and stays inside $F$ (any automorphism of $F$ fixing $\mathbb{Q}$ must send $\sqrt[3]{2}$ to another root of $x^3 - 2$, but those roots are not in $F$). With only the trivial group, the map $\Psi$ can only output $\{e\}$, so $\Phi(\Psi(F)) = K^{\{e\}} = F$ for the single intermediate field $F$ — seemingly correct. But now consider $\Phi$: it should output all intermediate fields, yet the only subgroup of $\{e\}$ is $\{e\}$ itself, giving only the single intermediate field $F$ and the base $\mathbb{Q}$. The extension $F/\mathbb{Q}$ has degree $3$, and a genuine Galois correspondence would predict $[F:\mathbb{Q}] = |\operatorname{Gal}(F/\mathbb{Q})| = 3$ — but $|\{e\}| = 1 \ne 3$. The Galois group is too small by a factor of $3$, so $\Psi(\Phi)$ cannot possibly be the identity on the full lattice: there are no non-trivial subgroups for $\Phi$ to output, even though the degree-$3$ extension should have (if it were Galois) a cyclic Galois group with a nontrivial element of order $3$. The bijection $\Psi \circ \Phi$ is not the identity because $\Psi$ maps the single intermediate field $\mathbb{Q}(\sqrt[3]{2})$ to $\{e\}$, and $\Phi(\{e\}) = \mathbb{Q}(\sqrt[3]{2})$, which recovers $F$ — but the intermediate fields predicted by a group of size $1$ are only $\{\mathbb{Q}, F\}$, missing the two intermediate fields a degree-$3$ Galois extension would have. In short: the Galois group is not large enough to generate the full correspondence, so $\Psi \circ \Phi$ is not surjective onto subgroups of the expected size, and the bijection fails entirely. [/example] ## Degree and Index Calculations The degree formula in the Galois correspondence is a powerful computational tool. It allows us to determine degrees of intermediate extensions purely from the group theory, without having to find explicit generators or minimal polynomials. The key principle: if $H \le G$ corresponds to $F = K^H$, then \begin{align*} [F : k] = [G : H] = \frac{|G|}{|H|}. \end{align*} Every subfield computation becomes a coset counting problem in the group. [example: Subfields of a Cyclotomic Extension] The splitting field of $x^7 - 1$ over $\mathbb{Q}$ is $K = \mathbb{Q}(\zeta_7)$, where $\zeta_7 = e^{2\pi i / 7}$. The degree is $[K : \mathbb{Q}] = \varphi(7) = 6$, and the Galois group is \begin{align*} G = \operatorname{Gal}(\mathbb{Q}(\zeta_7)/\mathbb{Q}) \cong (\mathbb{Z}/7\mathbb{Z})^\times \cong \mathbb{Z}/6\mathbb{Z}, \end{align*} where the automorphism $\sigma_j$ sends $\zeta_7 \mapsto \zeta_7^j$ for $j \in \{1, 2, 3, 4, 5, 6\}$. Since $G \cong \mathbb{Z}/6\mathbb{Z}$, its subgroups are cyclic groups of orders $1, 2, 3, 6$ (one subgroup of each order, since $G$ is cyclic). By the Galois correspondence, the intermediate fields $k \subset F \subset K$ have degrees $[F : k] = [G:H] = 6/|H|$, giving degrees $6, 3, 2, 1$. - The subgroup $H = \{1\}$ of order $1$ corresponds to $K^{\{1\}} = K$, with $[K:\mathbb{Q}] = 6$. - The subgroup $H \cong \mathbb{Z}/2\mathbb{Z}$ of order $2$ corresponds to a field $F_3$ of degree $3$ over $\mathbb{Q}$. The generator is $\sigma_6$, sending $\zeta_7 \mapsto \zeta_7^6 = \bar\zeta_7$. The fixed field consists of elements symmetric under complex conjugation, which is $\mathbb{Q}(\zeta_7 + \zeta_7^6) = \mathbb{Q}(2\cos(2\pi/7))$. - The subgroup $H \cong \mathbb{Z}/3\mathbb{Z}$ of order $3$ (generated by $\sigma_2$, since $2$ has order $3$ in $(\mathbb{Z}/7\mathbb{Z})^\times$: $2^1 = 2, 2^2 = 4, 2^3 = 8 \equiv 1$) corresponds to the quadratic subfield $F_2$ with $[F_2:\mathbb{Q}] = 2$. This is $\mathbb{Q}(\sqrt{-7})$ — the quadratic subfield of the $7$th cyclotomic field, which exists because $-7 \equiv 1 \pmod 4$ and the discriminant of $\mathbb{Q}(\zeta_7)/\mathbb{Q}$ involves $7$. - The full group $G$ corresponds to $\mathbb{Q}$ itself. The tower $\mathbb{Q} \subset \mathbb{Q}(\sqrt{-7}) \subset \mathbb{Q}(\zeta_7)$ has degrees $2$ and $3$, consistent with $[K:\mathbb{Q}] = 6 = 2 \times 3$. [/example] The example illustrates how the group-theoretic structure directly constrains what intermediate fields can look like. A degree-$6$ cyclic extension has exactly one intermediate field of each degree dividing $6$, and we can identify each field by analyzing the fixed-field of the corresponding cyclic subgroup. [remark: Galois Correspondence is Order-Reversing, Not Covariant] A common source of confusion: the correspondence reverses the order of inclusion. The subgroup $\{e\}$ corresponds to the largest intermediate field $K$, and the full group $G$ corresponds to the smallest intermediate field $k$. When you add more elements to a subgroup (making it larger), the fixed field shrinks. This reversal is not a quirk — it is intrinsic to the construction and deeply analogous to the reversal in Pontryagin duality and other duality theorems in mathematics. [/remark] ## Applications to Solvability by Radicals The Galois correspondence was originally developed to answer the question of when a polynomial equation can be solved by radicals — that is, when its roots can be expressed using the field operations ($+$, $-$, $\times$, $\div$) starting from the coefficients, together with the extraction of $n$th roots. What does "solvable by radicals" mean in terms of field extensions? If a root $\alpha$ of $f \in k[x]$ can be expressed by a radical formula, then there is a tower of extensions \begin{align*} k = F_0 \subset F_1 \subset F_2 \subset \cdots \subset F_r \end{align*} where each step $F_{i+1} = F_i(\beta_i)$ adjoins an $n_i$th root: $\beta_i^{n_i} \in F_i$. Such a tower is called a radical tower, and a polynomial is solvable by radicals iff its splitting field is contained in the top of some radical tower over $k$. [definition: Solvable Extension] A polynomial $f \in k[x]$ is **solvable by radicals** if its splitting field is contained in the top of a **radical tower**: a tower of extensions \begin{align*} k = F_0 \subset F_1 \subset \cdots \subset F_r \end{align*} where each step $F_{i+1} = F_i(\beta_i)$ is obtained by adjoining an element $\beta_i$ satisfying $\beta_i^{n_i} \in F_i$ for some positive integer $n_i$. A Galois extension $K/k$ is called **solvable** if $K$ is contained in the top of some radical tower over $k$. [/definition] The bridge between solvability of extensions and solvability of groups is the following: [quotetheorem:3327] The connection works in both directions via the Galois correspondence. If $f$ is solvable by radicals, the radical tower gives — after adjoining appropriate roots of unity — a tower of cyclic extensions. By the correspondence, these cyclic extensions correspond to a subnormal series in the Galois group with cyclic (hence abelian) quotients. The converse uses the fact that cyclic groups correspond to cyclic extensions. Now we can explain the insolubility of the quintic. The general degree-$5$ polynomial (with coefficients that are transcendental indeterminates over $\mathbb{Q}$) has Galois group $S_5$. The alternating group $A_5$ is simple and non-abelian — it has no proper normal subgroups. Any subnormal series for $S_5$ must pass through $A_5$, and since $A_5$ is simple, the only subnormal series is $\{e\} \trianglelefteq A_5 \trianglelefteq S_5$, with quotients $A_5$ (non-abelian, order $60$) and $\mathbb{Z}/2\mathbb{Z}$. Since $A_5$ is not abelian, $S_5$ is not solvable, and hence the general quintic is not solvable by radicals. [example: A Specific Insolvable Quintic] Consider $f(x) = x^5 - 6x + 3 \in \mathbb{Q}[x]$. One can verify that $f$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion with $p = 3$ (since $3 \nmid 1$, $3 \mid (-6)$, $3 \mid 3$, $3^2 = 9 \nmid 3$). By Eisenstein, $f$ is irreducible over $\mathbb{Q}$, so $[\mathbb{Q}(\alpha) : \mathbb{Q}] = 5$ for any root $\alpha$. Now we determine the number of real roots. Differentiating, $f'(x) = 5x^4 - 6$, which vanishes at $x = \pm(6/5)^{1/4} \approx \pm 1.046$. Since $f(-1.046) \approx (-1.046)^5 - 6(-1.046) + 3 \approx -1.3 + 6.28 + 3 = 7.98 > 0$ and $f(1.046) \approx 1.3 - 6.28 + 3 = -1.98 < 0$, the polynomial has exactly three real roots (it starts at $-\infty$, rises to a local max above zero, dips to a local min below zero, then rises to $+\infty$). Since $f$ has degree $5$, three real roots, and two complex conjugate roots, the Galois group $G = \operatorname{Gal}(K/\mathbb{Q})$ must contain: - A $5$-cycle (since $f$ is irreducible of degree $5$, $G$ acts transitively on the roots, and by Cauchy's theorem applied to a Sylow $5$-subgroup, $G$ contains an element of order $5$), - A transposition (complex conjugation, restricted to the splitting field, permutes the two complex roots and fixes the three real roots, so it is a $2$-cycle in $S_5$). A subgroup of $S_5$ containing a $5$-cycle and a transposition must be all of $S_5$ — this is a standard result in permutation group theory. Therefore $G \cong S_5$, which is not solvable, so $f$ is not solvable by radicals. [/example] [remark: Solvable Polynomials Can Have Any Degree] The insolvability of the general quintic does not mean every quintic is insolvable — it means there exist specific quintics (like $x^5 - 6x + 3$) whose Galois group is $S_5$. A quintic with Galois group, say, $\mathbb{Z}/5\mathbb{Z}$ or $D_{10}$ (both solvable) would be solvable by radicals. What fails is the existence of a general formula that works for all quintics with arbitrary rational coefficients. [/remark] ## The Lattice of Subfields The Galois correspondence does more than establish a bijection — it converts the lattice of subgroups of a group into the lattice of intermediate fields, with all the order-theoretic structure preserved (but reversed). This has rich consequences for understanding the combinatorics of field extensions. Two subgroups $H_1, H_2 \le G$ have a meet (greatest lower bound) $H_1 \cap H_2$ and a join (least upper bound) $\langle H_1, H_2 \rangle$ in the lattice of subgroups. Under the Galois correspondence: \begin{align*} K^{H_1 \cap H_2} &= K^{H_1} \cdot K^{H_2} \quad \text{(compositum of fields)} \\ K^{\langle H_1, H_2 \rangle} &= K^{H_1} \cap K^{H_2}. \end{align*} The meet of subgroups corresponds to the join of fields (compositum), and the join of subgroups corresponds to the meet of fields (intersection). This is the algebraic content of the order-reversal. When two subgroups $H_1$ and $H_2$ meet in the lattice, their intersection $H_1 \cap H_2$ corresponds — under the order-reversing correspondence — to a field that is larger than either $K^{H_1}$ or $K^{H_2}$. What is this larger field? Given two intermediate fields $F_1$ and $F_2$, what is the smallest field containing both? Neither $F_1$ alone nor $F_2$ alone will do — we need a field that contains every element of $F_1$ and every element of $F_2$, closed under all field operations. Such a field must exist (the ambient field $K$ itself contains both), so among all intermediate fields containing both $F_1$ and $F_2$, there is a smallest one. This is their compositum — the field-theoretic join — and it plays the role that the group-theoretic meet plays on the subgroup side. We now define it precisely. [definition: Compositum] Let $K/k$ be a Galois extension and $F_1, F_2$ two intermediate fields. The **compositum** of $F_1$ and $F_2$ is the smallest subfield of $K$ containing both $F_1$ and $F_2$: \begin{align*} F_1 \cdot F_2 := k(F_1 \cup F_2). \end{align*} [/definition] The compositum is the smallest field containing all elements of both $F_1$ and $F_2$. It corresponds, under the Galois correspondence, to the intersection $H_1 \cap H_2$ of the corresponding subgroups — the elements fixed by both $H_1$ and $H_2$ are fixed by their intersection. [quotetheorem:3336] The theorem transforms questions about intermediate fields into questions about subgroups. Finding the smallest field containing $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ inside $\mathbb{Q}(\sqrt{2},\sqrt{3})$ becomes the question: what is the intersection of the subgroups $\langle\sigma\rangle$ and $\langle\tau\rangle$ in $G \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$? Since $\sigma$ and $\tau$ are distinct elements of order $2$, $\langle\sigma\rangle \cap \langle\tau\rangle = \{e\}$, and $K^{\{e\}} = K$ — so the compositum is all of $K = \mathbb{Q}(\sqrt{2},\sqrt{3})$. This can be verified directly: $K = \mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2}) \cdot \mathbb{Q}(\sqrt{3})$. [example: Computing Composita via Group Theory] Inside the cyclotomic field $\mathbb{Q}(\zeta_{12})$ of degree $\varphi(12) = 4$, with \begin{align*} G = \operatorname{Gal}(\mathbb{Q}(\zeta_{12})/\mathbb{Q}) \cong (\mathbb{Z}/12\mathbb{Z})^\times = \{1, 5, 7, 11\} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}, \end{align*} consider the three quadratic subfields: - $F_1 = \mathbb{Q}(\sqrt{-3})$ corresponding to $H_1 = \langle\sigma_{11}\rangle$ (where $\sigma_{11}(\zeta_{12}) = \zeta_{12}^{11}$), - $F_2 = \mathbb{Q}(\sqrt{3})$ corresponding to $H_2 = \langle\sigma_7\rangle$ (where $\sigma_7(\zeta_{12}) = \zeta_{12}^7$), - $F_3 = \mathbb{Q}(i)$ corresponding to $H_3 = \langle\sigma_5\rangle$ (where $\sigma_5(\zeta_{12}) = \zeta_{12}^5$). First we verify the identification of fixed fields. Since $\zeta_{12} = e^{2\pi i/12}$, we have $\zeta_{12}^3 = i$ and $\zeta_{12} + \zeta_{12}^{-1} = 2\cos(\pi/6) = \sqrt{3}$, so $\mathbb{Q}(\zeta_{12})$ contains both $i$ and $\sqrt{3}$. The three quadratic subfields of this degree-$4$ extension are $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{3})$, and $\mathbb{Q}(\sqrt{-3}) = \mathbb{Q}(i\sqrt{3})$. Now compute the compositum $F_1 \cdot F_2$. By the lattice anti-isomorphism, this corresponds to $H_1 \cap H_2$. In the Klein four-group $G = \{e, \sigma_5, \sigma_7, \sigma_{11}\}$, the elements $\sigma_{11}$ and $\sigma_7$ are distinct non-identity elements, so $\langle\sigma_{11}\rangle \cap \langle\sigma_7\rangle = \{e\}$. Therefore $F_1 \cdot F_2 = K^{\{e\}} = \mathbb{Q}(\zeta_{12})$. This is consistent with a direct check: $\mathbb{Q}(\sqrt{-3}, \sqrt{3})$ contains $\sqrt{-3}$ and $\sqrt{3}$, hence their quotient \begin{align*} \frac{\sqrt{-3}}{\sqrt{3}} = \sqrt{\frac{-3}{3}} = \sqrt{-1} = i. \end{align*} So $\mathbb{Q}(\sqrt{-3}, \sqrt{3})$ contains $i$ and $\sqrt{3}$, giving $\mathbb{Q}(i, \sqrt{3}) = \mathbb{Q}(\zeta_{12})$ (since $\zeta_{12} = \frac{\sqrt{3}}{2} + \frac{1}{2}i \in \mathbb{Q}(i, \sqrt{3})$). Thus $\mathbb{Q}(\sqrt{-3}) \cdot \mathbb{Q}(\sqrt{3}) = \mathbb{Q}(\zeta_{12})$, as predicted by the group-theoretic computation. [/example] ## Infinite Galois Theory The Galois correspondence we have developed applies to finite Galois extensions. For infinite extensions — such as the extension $\mathbb{Q}^{ab}/\mathbb{Q}$ of $\mathbb{Q}$ by all abelian extensions (the Kronecker-Weber theorem says this equals $\mathbb{Q}(\{\zeta_n : n \ge 1\})$) — a more refined version of the correspondence is needed. The issue is that for an infinite Galois extension $K/k$, there can be subgroups of $G = \operatorname{Gal}(K/k)$ that do not arise as fixed-field stabilizers of any subfield. The failure is topological in character: without some notion of proximity or limit, we cannot distinguish subgroups that are genuinely the stabilizer of a subfield from those that are merely dense approximations to one. The correct fix is to put a topology on $G$ that reflects the structure of the extension — one where convergence of automorphisms means convergence on every finite piece of the extension. An automorphism $\sigma$ should be "close to" the identity if it fixes a large finite Galois sub-extension pointwise. This is exactly what the Krull topology captures: continuity in this topology is the algebraic content of compatibility with finite approximations, and compactness encodes the fact that automorphisms are determined by their values on algebraic elements. Only closed subgroups — those that are limits of their own approximating sequences — arise as stabilizers of actual subfields. [definition: Krull Topology] Let $K/k$ be a Galois extension (possibly infinite) with group $G = \operatorname{Gal}(K/k)$. The **Krull topology** on $G$ is defined by taking as a neighborhood basis of the identity the collection of subgroups $\operatorname{Gal}(K/F)$ for $F/k$ ranging over all finite Galois subextensions of $K/k$. This makes $G$ a topological group, and $G$ is compact, Hausdorff, and totally disconnected — that is, $G$ is a **profinite group**. [/definition] The Krull topology is constructed so that the open subgroups are precisely the subgroups of finite index that arise as Galois groups of finite sub-extensions. A subgroup $H$ is closed (in the Krull topology) if and only if $H = \operatorname{Gal}(K/K^H)$, i.e., $H$ is the full stabilizer of its own fixed field. [quotetheorem:2387] The failure of non-closed subgroups to appear in the bijection can be illustrated with the extension $\overline{\mathbb{Q}}/\mathbb{Q}$ (the algebraic closure): the absolute Galois group $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ is an enormous profinite group with uncountably many subgroups, but most of them are not closed. For instance, one can construct an algebraically independent (over $\mathbb{F}_p$) subset of $\overline{\mathbb{F}_p}$, but the corresponding subgroup of $\operatorname{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p) \cong \hat{\mathbb{Z}}$ would be dense and non-closed. The Galois correspondence only sees the closed subgroups, which correspond to the actual intermediate fields $\mathbb{F}_{p^n}$ for $n \ge 1$. [explanation: Why Profinite Groups Appear] A profinite group is a projective limit (inverse limit) of finite groups. The Galois group $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ is the limit of the system $\{\operatorname{Gal}(K/\mathbb{Q})\}$ ranging over all finite Galois extensions $K/\mathbb{Q}$, with transition maps given by restriction. An element $\sigma \in \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ is a coherent system of automorphisms: for each finite Galois extension $K/\mathbb{Q}$, it restricts to some $\sigma|_K \in \operatorname{Gal}(K/\mathbb{Q})$, and these restrictions are compatible: if $K_1 \subset K_2$, then $\sigma|_{K_1} = (\sigma|_{K_2})|_{K_1}$. This inverse limit structure explains why the Krull topology is natural: a basic open neighborhood of the identity is the set of automorphisms that agree with the identity on some finite Galois sub-extension, which is exactly the kernel of the restriction map to that sub-extension. The Krull topology makes the restriction maps continuous, and it is the coarsest topology that does so. The compactness of $G$ follows from Tychonoff's theorem: each finite Galois group $\operatorname{Gal}(K/\mathbb{Q})$ is finite (hence compact with the discrete topology), and $G$ embeds as a closed subset of the product $\prod_{K} \operatorname{Gal}(K/\mathbb{Q})$, which is compact. [/explanation] ## The Covering Space Analogy One of the most illuminating perspectives on the Galois correspondence comes from algebraic topology. The formal resemblance is not accidental — both theories describe how a structured object with a symmetry group relates to its quotients. Let $X$ be a path-connected, locally path-connected, and semi-locally simply connected topological space, and fix a basepoint $x_0 \in X$. There is a bijection between connected covering spaces $p: \tilde{X} \to X$ (up to isomorphism) and subgroups of the fundamental group $\pi_1(X, x_0)$. This is the classification theorem for covering spaces. The analogy with Galois theory is precise: | Galois theory | Covering spaces | |---|---| | Galois extension $K/k$ | Universal cover $\tilde{X} \to X$ | | Galois group $G = \operatorname{Gal}(K/k)$ | Fundamental group $\pi_1(X, x_0)$ | | Subgroup $H \le G$ | Subgroup $H \le \pi_1(X, x_0)$ | | Fixed field $K^H$ | Intermediate covering $\tilde{X}/H \to X$ | | $[K : K^H] = |H|$ | Number of sheets of $\tilde{X}/H \to X$ equals $|H|$ | | Normal subgroup $H \trianglelefteq G$ | Regular (normal) covering | | Quotient $G/H = \operatorname{Gal}(K^H/k)$ | Deck transformation group of $\tilde{X}/H$ | Both correspondences are order-reversing: a larger subgroup (more symmetry imposed) produces a smaller intermediate object (fewer points or a smaller field). Both have a distinguished "universal" object at the top ($K$ or $\tilde{X}$) that covers everything. Both require a maximality condition: the Galois condition ($|\operatorname{Gal}(K/k)| = [K:k]$) corresponds to $\tilde{X}$ being simply connected (the universal cover), and passing to a quotient by a non-normal subgroup gives a non-Galois/non-regular cover. The analogy runs deeper than a table of correspondences. The étale fundamental group in algebraic geometry — defined via finite étale covers of a scheme — literally encompasses the Galois group of the field as a special case. For the spectrum of a field $k$, the étale covers are exactly the finite separable field extensions, and the étale fundamental group is the absolute Galois group $\operatorname{Gal}(k^{\mathrm{sep}}/k)$. In this sense, Galois theory and covering space theory are not merely analogous — they are the same theorem stated in different geometric languages. [illustration:covering-space-analogy] ## References - Emil Artin, *Galois Theory* (1944). The foundational modern account, emphasizing the role of automorphisms. - Serge Lang, *Algebra* (3rd ed., 2002). Chapters V and VI for a comprehensive treatment of field theory and Galois theory. - Michael Artin, *Algebra* (2nd ed., 2010). Chapters 15 and 16 for a geometrically motivated presentation. - David S. Dummit and Richard M. Foote, *Abstract Algebra* (3rd ed., 2004). Chapters 13 and 14 for an encyclopedic treatment with many examples. - Jean-Pierre Serre, *Local Fields* (1979). For the infinite Galois theory and connections to number theory. - Patrick Morandi, *Field and Galois Theory* (1996). A clean modern presentation with detailed proofs.