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Fundamental theorem of Galois theory

The Fundamental Theorem of Galois Theory is a central result in abstract algebra that establishes, for a finite Galois extension L/K of fields with Galois group G = \mathrm{Gal}(L/K), a bijective, inclusion-reversing correspondence between the subfields of L containing K and the subgroups of G. Under this correspondence, each subgroup H \leq G maps to its fixed field L^H = \{ \alpha \in L \mid h(\alpha) = \alpha \ \forall h \in H \}, and each intermediate field F (with K \subseteq F \subseteq L) maps to the subgroup G_F = \{ g \in G \mid g(\beta) = \beta \ \forall \beta \in F \}, with the maps being inverses of each other; moreover, the index [G : H] = [L : L^H] and normality of subfields over K corresponds to normality of subgroups in G.^[1] This theorem provides a profound lattice isomorphism between the structure of field extensions and group theory, enabling the translation of problems about polynomials and field adjunctions into questions about group actions and subgroups.^[2] A key application is the criterion for solvability of polynomials by radicals: an irreducible polynomial over a field of characteristic zero is solvable by radicals if and only if the Galois group of its splitting field is a solvable group, meaning it possesses a composition series with abelian factors.^[2] Important corollaries include the fact that if L/K is Galois, then for any intermediate field F, the extension L/F is always Galois, and F/K is Galois if and only if G_F is normal in G, with \mathrm{Gal}(F/K) \cong G / G_F.^[1] The theorem originated in the work of Évariste Galois (1811–1832), who, building on earlier attempts by Paolo Ruffini and Niels Henrik Abel to address the unsolvability of general quintic equations, introduced the concept of the Galois group as the group of automorphisms of the splitting field fixing the base field, linking group structure directly to radical solvability.^[2] Galois' ideas, outlined in his 1831 memoir and published posthumously in 1846 by Joseph Liouville, were initially overlooked but were rigorously formalized and popularized by Camille Jordan in his 1870 treatise Traité des substitutions et des équations algébriques.^[2] Further refinements came from Richard Dedekind in the 1870s, who clarified the correspondence in the context of infinite extensions and modular reductions, solidifying the theorem's role in modern algebra.^[3]

Background Concepts

Galois Extensions

A Galois extension is a finite field extension K/F that is both normal and separable.^[4] This definition captures the extensions where the structure of intermediate fields corresponds closely to subgroups of the automorphism group, laying the groundwork for deeper results in field theory.^[5] Separability of an extension K/F means that the minimal polynomial over F of every element in K has distinct roots in an algebraic closure.^[4] Normality requires that every irreducible polynomial in F having at least one root in K splits completely into linear factors in K.^[5] Equivalently, a finite extension K/F is Galois if and only if it is the splitting field over F of some separable polynomial in F.^[4] For a finite Galois extension K/F, the degree [K:F] equals the order of the automorphism group \mathrm{Aut}(K/F).^[5] This automorphism group, often denoted \mathrm{Gal}(K/F), consists of all F-automorphisms of K.^[4] The term "Galois extension" was coined in the 1940s to honor the 19th-century mathematician Évariste Galois, whose work on the solvability of polynomial equations by radicals inspired the modern theory of field extensions.^[6]

Galois Groups

In the context of a Galois extension K/F, where K is a finite, normal, and separable extension of the base field F, the Galois group G = \mathrm{Aut}_F(K) is defined as the group of all field automorphisms of K that fix every element of F pointwise.^[7] This group operation is given by composition of automorphisms, with the identity automorphism serving as the neutral element.^[8] For such extensions, the order of the Galois group equals the degree of the extension, so |G| = [K:F].^[7] Small Galois groups often exhibit familiar structures. For a quadratic extension K = F(\sqrt{d}) with d \in F not a square in F, the Galois group is cyclic of order 2, generated by the automorphism sending \sqrt{d} to -\sqrt{d}.^[9] In contrast, for the splitting field of an irreducible cubic polynomial over F (of characteristic not 2 or 3) whose discriminant is not a square in F, the Galois group is the symmetric group S_3 of order 6.^[10] The Galois group acts faithfully on the roots of any irreducible separable polynomial f \in F whose splitting field is K, embedding G as a transitive permutation group on those roots via the action \sigma(\alpha) = \sigma(\alpha) for root \alpha.^[11] This permutation representation arises because automorphisms permute roots while preserving algebraic relations over F, providing a concrete realization of the abstract group structure.^[12]

Statement of the Theorem

The Bijection Between Subfields and Subgroups

Let K/F be a finite Galois extension with Galois group G = \Gal(K/F). The fundamental theorem of Galois theory asserts that there is a bijection between the set of intermediate fields L such that F \subseteq L \subseteq K and the set of subgroups H \leq G. Specifically, each subgroup H corresponds to the intermediate field L = K^H, and each intermediate field L corresponds to the subgroup H = \Gal(K/L).^[13] The map from subgroups to subfields sends a subgroup H \leq G to its fixed field K^H = \{ x \in K \mid \sigma(x) = x \ \forall \sigma \in H \}, which is the set of all elements of K fixed pointwise by every automorphism in H. Note that F = K^G, since the full Galois group fixes the base field. The inverse map sends an intermediate field L with F \subseteq L \subseteq K to the subgroup \Gal(K/L) = \{ \sigma \in G \mid \sigma(y) = y \ \forall y \in L \}, consisting of all automorphisms in G that fix L pointwise. This subgroup is normal in G if and only if L/F is Galois, but the bijection holds regardless.^[13] To establish the bijection, one shows that the maps are inverses. For injectivity, suppose K^H = K^{H'}; then H and H' both consist of precisely the automorphisms fixing K^H, so H = H'. This follows from Artin's lemma, which states that if a finite group of automorphisms G acts on a field E with fixed field E^G, then [E : E^G] \leq |G|, with equality if the action is faithful, ensuring distinct subgroups yield distinct fixed fields. For surjectivity, given any intermediate field L, the subgroup \Gal(K/L) has fixed field exactly L, as K is Galois over L (since it is Galois over F and L contains F), and the fixed field of \Gal(K/L) coincides with L.^[13] The correspondence is order-reversing: if H_1 \leq H_2 \leq G, then K^{H_2} \subseteq K^{H_1}, and conversely, if F \subseteq L_1 \subseteq L_2 \subseteq K, then \Gal(K/L_1) \leq \Gal(K/L_2). Moreover, for any intermediate field L, the degree [K : L] = |\Gal(K/L)|. This holds because the extension K/L is Galois.^[13]

Fixed Fields and Normal Closures

In the context of a finite Galois extension K/F with Galois group G = \Gal(K/F), for any subgroup H \leq G, the fixed field K^H is defined as the subfield of K consisting of all elements \alpha \in K such that \sigma(\alpha) = \alpha for every \sigma \in H.^[13] This fixed field satisfies F \subseteq K^H \subseteq K, and the extension K / K^H is Galois; moreover, K^H / F is Galois if and only if H is normal in G. Moreover, if H is a normal subgroup of G, then \Gal(K^H/F) \cong G/H.^[13] A fundamental property is the degree formula [K : K^H] = |H|, which follows from the fact that the natural action of H on K yields a faithful representation of dimension equal to the index.^[7] The fixed field construction provides the inverse to the map sending intermediate fields to their corresponding Galois subgroups. Specifically, if L is an intermediate field with F \subseteq L \subseteq K and H = \Gal(K/L), then L = K^H.^[13] This mutual inverse relationship ensures that the correspondence between subgroups of G and subfields of K containing F is bijective, as established by the fundamental theorem.^[7] For a finite extension L/F that is separable but not necessarily normal, the normal closure of L/F is the smallest Galois extension of F containing L, obtained as the compositum of all distinct F-conjugates of L.^[14] This closure is generated by adjoining all roots of the minimal polynomials over F of a basis for L over F, ensuring it is the splitting field of a separable polynomial over F.^[15] The normal closure plays a crucial role in extending the Galois correspondence to non-normal extensions by embedding them into a Galois setting.^[13]

Properties of the Correspondence

Lattice Structure and Inclusion Reversals

The fundamental theorem of Galois theory provides a bijection between the intermediate fields of a finite Galois extension K/F and the subgroups of its Galois group G = \mathrm{Gal}(K/F), which induces a lattice anti-isomorphism between the poset of subfields ordered by inclusion and the poset of subgroups ordered by reverse inclusion. Under this correspondence, the fixed field map H \mapsto K^H sends subgroups to subfields, while the Galois group map L \mapsto \mathrm{Gal}(K/L) sends subfields to subgroups, preserving the lattice structure but reversing the order relations. This anti-isomorphism ensures that the entire lattice of intermediate subfields is in one-to-one correspondence with the lattice of all subgroups of G, allowing the algebraic structure of the extension to be analyzed through group-theoretic terms.^[16]^[2] A key feature of this correspondence is the reversal of inclusions. Specifically, if F \subseteq L \subseteq M \subseteq K, then the associated subgroups satisfy G \supseteq \mathrm{Gal}(K/M) \supseteq \mathrm{Gal}(K/L), meaning that larger subfields correspond to smaller subgroups. This inclusion-reversing property holds because the Galois group of a larger extension is naturally a subgroup of the Galois group of a smaller one, reflecting how automorphisms fix more elements in bigger fields. For instance, the full group G corresponds to the base field F, while the trivial subgroup corresponds to the top field K. This reversal is fundamental to understanding the hierarchical structure of the extension.^[2]^[17] The correspondence also relates field degrees to group indices quantitatively. For any intermediate field L with F \subseteq L \subseteq K, the degree of the extension [L : F] equals the index of the corresponding subgroup in G, that is, [L : F] = |G : \mathrm{Gal}(K/L)|. This relation follows from the fact that the cosets of \mathrm{Gal}(K/L) in G parametrize the distinct embeddings of L into K fixing F, aligning the multiplicative structure of field extensions with the combinatorial structure of group actions. Moreover, since the overall degree [K : F] = |G|, the degrees of subextensions multiply accordingly along chains, mirroring the indices of subgroup chains.^[17]^[16] Finally, the lattices are fully mirrored under this anti-isomorphism, with joins and meets preserved in reverse. The join (smallest containing subfield) of two subfields L_1 and L_2 corresponds to the intersection of their Galois groups \mathrm{Gal}(K/L_1) \cap \mathrm{Gal}(K/L_2), whose fixed field is L_1 L_2. Conversely, the meet (largest contained subfield) of L_1 and L_2 corresponds to the subgroup generated by \mathrm{Gal}(K/L_1) and \mathrm{Gal}(K/L_2), whose fixed field is L_1 \cap L_2. This duality allows the subgroup lattice of G to encode the complete poset of subfields, providing a powerful tool for classifying extensions via group theory.^[16]^[2]

Normal Subgroups and Galois Subextensions

In the context of a finite Galois extension K/F with Galois group G = \Gal(K/F), there is a one-to-one correspondence between the normal subgroups of G and the Galois subextensions of K/F. Specifically, for a subgroup H \leq G, the fixed field K^H forms a Galois extension over F if and only if H is normal in G.^[18] In this case, the Galois group of the subextension is isomorphic to the quotient group: \Gal(K^H/F) \cong G/H.^[13] This correspondence preserves the group structure, allowing the quotient G/H to act faithfully on K^H.^[18] A key result establishing this normality condition is Dedekind's theorem, which characterizes intermediate extensions within a Galois extension. For an intermediate field L with F \subseteq L \subseteq K, the extension L/F is Galois if and only if the subgroup \Gal(K/L) is normal in G.^[13] When this holds, the Galois group \Gal(L/F) is isomorphic to G / \Gal(K/L).^[18] This theorem, formalized by Dedekind in his 1894 supplement to Dirichlet's Vorlesungen über Zahlentheorie, underscores the structural interplay between field normality and group normality.^[19] Artin's theorem complements this by focusing on the fixed fields directly. It states that if H is a normal subgroup of G, then the fixed field K^H is a Galois extension over the base field F, with the degree [K^H : F] = |G/H|.^[13] This result, attributed to Emil Artin in his modern reformulation of Galois theory during the 1920s and 1930s, relies on the separability and normality of the overall extension K/F.^[13] The theorem ensures that the correspondence restricts to a bijection between normal subgroups of G and Galois subextensions of K/F. The quotient group G/H inherits the action of G on K, but restricted to the fixed field K^H, enabling a recursive decomposition of the extension. Elements of G/H act as automorphisms of K^H over F, preserving the Galois structure and allowing the fundamental theorem to apply iteratively to the quotient extension.^[13] This recursive property facilitates the analysis of composite extensions and subgroup lattices in finite Galois theory.^[18]

Finite Examples

Quadratic Extension Example

A quintessential illustration of the fundamental theorem of Galois theory arises in quadratic extensions of the rational numbers. Consider the field extension K = \mathbb{Q}(\sqrt{d}) over the base field F = \mathbb{Q}, where d is a square-free integer not equal to 0 or 1. This extension has degree [K : \mathbb{Q}] = 2 and is Galois, as it is the splitting field of the separable irreducible polynomial x^2 - d \in \mathbb{Q}.^[13] The Galois group G = \mathrm{Gal}(K / \mathbb{Q}) thus has order 2, making it isomorphic to \mathbb{Z}/2\mathbb{Z}.^[13] It is generated by the automorphism \sigma: \sqrt{d} \mapsto -\sqrt{d}, which extends to the non-trivial element of G while fixing \mathbb{Q}.^[9] The subgroups of G are precisely the trivial subgroup \{ \mathrm{id} \} and the full group G. The fixed field of G is K^G = \{ \alpha \in K \mid \sigma(\alpha) = \alpha \ \forall \sigma \in G \} = \mathbb{Q}, since any element outside \mathbb{Q} involves \sqrt{d} and is moved by \sigma.^[13] Conversely, the fixed field of \{ \mathrm{id} \} is K^{\{ \mathrm{id} \}} = K, as the identity fixes everything. These yield exactly the intermediate fields \mathbb{Q} and K, with no proper subfields in between due to the prime degree of the extension. This establishes the bijection between subgroups of G and subfields of K containing \mathbb{Q}, as predicted by the theorem.^[13] The lattice of subfields and subgroups forms a simple chain: \mathbb{[Q](/page/Q)} \subset K corresponds to G \supset \{ \mathrm{[id](/page/id)} \}, with inclusion reversed and no intermediate elements. This structure highlights the one-to-one correspondence without further complexity, serving as the minimal non-trivial case of the Galois correspondence.^[7]

Cubic Irreducible Polynomial Example

Consider the irreducible cubic polynomial f(x) = x^3 + x + 1 over \mathbb{[Q](/page/Q)}. By the rational root theorem, the possible rational roots are \pm 1, but f(1) = 3 \neq 0 and f(-1) = -1 \neq 0, so f(x) has no rational roots and is thus irreducible over \mathbb{[Q](/page/Q)}.^[20] Let K be the splitting field of f(x) over \mathbb{Q}, with roots \alpha, \beta, \gamma. The discriminant of f(x) is -4(1)^3 - 27(1)^2 = -31, which is negative and not a square in \mathbb{Q}; hence, f(x) has one real root and two complex conjugate roots, and the Galois group G = \mathrm{Gal}(K/\mathbb{Q}) \cong S_3 with [K : \mathbb{Q}] = 6.^[20]^[13] The group S_3 has six subgroups: the trivial subgroup \{e\}, the full group S_3, the normal alternating subgroup A_3 (cyclic of order 3 and index 2), and three subgroups of order 2 generated by transpositions (e.g., \langle (1\ 2) \rangle, \langle (1\ 3) \rangle, \langle (2\ 3) \rangle, labeling roots 1, 2, 3). By the fundamental theorem of Galois theory, these subgroups correspond bijectively to subfields of [K](/page/K) containing \mathbb{Q}.^[13] The fixed field of A_3 is a quadratic subfield L of K with [L : \mathbb{Q}] = 2 and [K : L] = 3; explicitly, L = \mathbb{Q}(\sqrt{-31}), the quadratic subextension arising from the square root of the discriminant. The fixed fields of the order-2 subgroups are three isomorphic cubic subfields \mathbb{Q}(\alpha), \mathbb{Q}(\beta), \mathbb{Q}(\gamma), each of degree 3 over \mathbb{Q} and index 2 in K. For instance, the subgroup generated by the transposition swapping \beta and \gamma (fixing \alpha) has fixed field \mathbb{Q}(\alpha), since elements of this subgroup permute the other roots but leave \alpha invariant.^[13]^[20] This correspondence yields a lattice of six subfields: \mathbb{Q} (fixed by S_3), the three cubics \mathbb{Q}(\alpha), \mathbb{Q}(\beta), \mathbb{Q}(\gamma) (fixed by the order-2 subgroups), the quadratic L (fixed by A_3), and K (fixed by \{e\}). The inclusions reverse the subgroup lattice: for example, \mathbb{Q} \subset L \subset K corresponds to S_3 \supset A_3 \supset \{e\}, while each cubic lies between \mathbb{Q} and K but is not contained in L, reflecting the non-normal order-2 subgroups.^[13]

Cyclotomic Field Example

The cyclotomic extensions serve as a canonical family of abelian Galois extensions that exemplify the fundamental theorem of Galois theory. Consider the nth cyclotomic field K = \mathbb{Q}(\zeta_n), where \zeta_n is a primitive nth root of unity. This field is the splitting field over \mathbb{Q} of the irreducible nth cyclotomic polynomial \Phi_n(X), so the extension K/\mathbb{Q} is Galois with degree [K:\mathbb{Q}] = \phi(n), where \phi denotes Euler's totient function. The Galois group G = \mathrm{Gal}(K/\mathbb{Q}) is isomorphic to the multiplicative group (\mathbb{Z}/n\mathbb{Z})^\times, which is abelian and has order \phi(n). The elements of G act by sending \zeta_n to \zeta_n^k for k \in (\mathbb{Z}/n\mathbb{Z})^\times.^[21] By the fundamental theorem, the subgroups of G are in bijective correspondence with the intermediate fields \mathbb{Q} \subseteq F \subseteq K, where each subgroup H \leq G fixes a unique field K^H of degree |G:H| over \mathbb{Q}, and the inclusion of fields reverses the inclusion of subgroups. Since G is abelian, every subgroup is normal, implying that every intermediate extension F/\mathbb{Q} is Galois.^[21] A specific illustration arises with n=7, the prime case where K = \mathbb{Q}(\zeta_7) has degree $6over\mathbb{Q}, and G \cong \mathbb{Z}/6\mathbb{Z}is cyclic, generated by the [automorphism](/page/Automorphism)\sigmawith\sigma(\zeta_7) = \zeta_7^3. The [subgroups](/page/Subgroup) of G

are unique for each [divisor](/page/Divisor) of &#36;6

: the trivial subgroup \{e\} (order $1), the full group G

([order](/page/Order) &#36;6

), a subgroup H_2 of order $2

(index &#36;3

), and a subgroup H_3 of order $3

(index &#36;2

). These correspond to fixed fields of degrees $6

, &#36;1

, $3

, and &#36;2

over \mathbb{Q}, respectively. The fixed field of H_3 (index $2) is the quadratic subfield \mathbb{Q}(\sqrt{-7}), while the fixed field of H_2

(index &#36;3

) is the maximal real subfield \mathbb{Q}(\zeta_7 + \zeta_7^{-1}), a degree-$3extension generated by the trace of\zeta_7$ to the reals.^[21]^[21] In the general case, certain fixed fields are themselves cyclotomic subfields \mathbb{Q}(\zeta_d) for divisors d \mid n. Specifically, \mathbb{Q}(\zeta_d) is the fixed field of the kernel of the natural surjection (\mathbb{Z}/n\mathbb{Z})^\times \twoheadrightarrow (\mathbb{Z}/d\mathbb{Z})^\times, which has index \phi(d); this subgroup corresponds to the conductor d in the cyclotomic setting. For n=7, the only such cyclotomic subfields are \mathbb{Q} = \mathbb{Q}(\zeta_1) and the full K = \mathbb{Q}(\zeta_7), but the additional subgroups yield the non-cyclotomic quadratic and cubic subfields noted above. This structure highlights how the lattice of subgroups of the abelian group G governs the tower of subextensions, with all inclusions reversing under the Galois correspondence.^[22]^[21]

Applications

Solvability by Radicals

The fundamental theorem of Galois theory provides a criterion for determining when a polynomial equation over the rationals \mathbb{Q} can be solved using radicals, linking the structure of field extensions to group-theoretic properties of their Galois groups. Specifically, an irreducible polynomial f(x) \in \mathbb{Q} of degree n is solvable by radicals if and only if its Galois group \mathrm{Gal}(K/\mathbb{Q}), where K is the splitting field of f over \mathbb{Q}, is a solvable group.^[23]^[24]^[25]^[26] This equivalence arises because the theorem establishes a bijection between subfields of K and subgroups of the Galois group, allowing the analysis of radical towers through the group's subgroup lattice. A radical extension is obtained by adjoining an nth root of an element from the base field, such as K(\sqrt{a}) for a \in K, and assuming the characteristic does not divide n and the base field contains the nth roots of unity, the Galois group of such an extension is cyclic (hence abelian).^[23]^[26] Solvability by radicals corresponds to embedding the splitting field in a tower of such simple radical extensions, where each step yields an abelian Galois group; the composite Galois group over \mathbb{Q} is then solvable, possessing a subnormal series with abelian factor groups.^[24]^[25] Conversely, if the Galois group is solvable, the correspondence theorem guarantees a chain of subfields corresponding to the subnormal series, each with abelian Galois groups that can be realized as radical extensions.^[23]^[26] Évariste Galois's key insight in the early 19th century was to connect the solvability of polynomial equations by radicals to the solvability of their associated permutation groups, resolving a problem dating back to efforts by mathematicians like Ruffini and Abel.^[24]^[25] For instance, the general quintic equation x^5 + a x^4 + b x^3 + c x^2 + d x + e = 0 over \mathbb{Q} has Galois group isomorphic to the symmetric group S_5 (or the alternating group A_5 in some cases), both of which are nonsolvable because A_5 is simple and nonabelian, lacking a nontrivial normal subgroup with abelian quotients.^[23]^[24] This demonstrates that general quintics are not solvable by radicals, a result building on Abel's earlier proof for degree 5 but generalized through Galois's group-theoretic framework.^[26] The Galois correspondence facilitates practical checks for solvability: one computes the Galois group of the polynomial (often via its resolvent or cycle type analysis) and examines its composition series for abelian factors.^[25]^[26] For polynomials with solvable Galois groups, such as quadratics (S_2 \cong \mathbb{Z}/2\mathbb{Z}) or cubics with three real roots (A_3 \cong \mathbb{Z}/3\mathbb{Z}), explicit radical solutions exist, underscoring the theorem's role in classifying solvable cases.^[24]

Constructibility in Field Extensions

A real number \alpha is constructible with straightedge and compass starting from the rationals if and only if \alpha lies in a field extension of \mathbb{Q} obtained by a tower of quadratic extensions, which is equivalent to the Galois group of the splitting field of the minimal polynomial of \alpha over \mathbb{Q} being a 2-group (i.e., of order $2^n for some n \geq 0).^[27] This characterization follows from the fundamental theorem of Galois theory, which establishes a bijection between subfields of the splitting field and subgroups of the Galois group, with degrees corresponding to subgroup indices.^[13] Since straightedge and compass constructions correspond to adjoining square roots—yielding quadratic extensions—the overall extension degree must be a power of 2, implying the Galois group has order $2^n.^[28] Conversely, if the Galois group G of the splitting field K/\mathbb{Q} is a finite 2-group, then G admits a chain of subgroups G = G_0 \triangleright G_1 \triangleright \cdots \triangleright G_n = \{e\} where each [G_i : G_{i+1}] = 2, corresponding via the fundamental theorem to a tower of quadratic extensions \mathbb{Q} = K_0 \subset K_1 \subset \cdots \subset K_n = K containing \alpha, making \alpha constructible.^[27] This structure ensures that every element in such an extension can be reached through successive quadratic adjunctions, aligning with the operations of ruler-and-compass geometry.^[13] A classic application is the impossibility of trisecting a general angle, such as 60°, which would require constructing \cos(20^\circ), a root of the irreducible cubic $8x^3 - 6x - 1 over \mathbb{Q}. The splitting field of this polynomial has Galois group S_3, of order 6 (not a power of 2), so \cos(20^\circ) is not constructible.^[13] Similarly, doubling the cube—constructing a side length of

\sqrt{{grok:render&&&type=render_inline_citation&&&citation_id=3&&&citation_type=wikipedia}}{2}

, a root of the irreducible cubic x^3 - 2 over \mathbb{Q}—fails because its splitting field has Galois group S_3, again of order 6, precluding a tower of quadratic extensions.^[13] These impossibilities highlight how non-2-group Galois structures block constructibility, even for low-degree extensions.^[29]

Infinite Extensions

The Infinite Galois Correspondence

In the case of infinite Galois extensions, the fundamental theorem of Galois theory generalizes to establish a bijection between intermediate fields and certain subgroups of the Galois group, requiring the introduction of a topology on the group to handle the infinite structure. Specifically, consider an infinite algebraic extension K/F that is Galois, meaning it is normal and separable, with the property that every finite subextension of K/F is also Galois. The Galois group G = \Gal(K/F) consists of all F-automorphisms of K, and it can be endowed with the Krull topology, making G a profinite group—that is, a compact, Hausdorff, totally disconnected topological group isomorphic to the inverse limit \varprojlim \Gal(L/F), where the limit is taken over all finite Galois subextensions L/F of K/F.^[30]^[31] The infinite Galois correspondence asserts that there is a bijection between the intermediate fields F \subseteq L \subseteq K and the closed subgroups of G. The maps defining this correspondence are L \mapsto \Gal(K/L) (the subgroup consisting of automorphisms fixing L pointwise, a closed subgroup) and H \mapsto K^H (which sends a closed subgroup to its fixed field). This bijection is inclusion-reversing: if H_1 \subseteq H_2 are closed subgroups, then K^{H_2} \subseteq K^{H_1}, and conversely for intermediate fields. Moreover, L/F is Galois if and only if \Gal(K/L) is a normal closed subgroup of G, in which case \Gal(L/F) \cong G / \Gal(K/L). This extends the finite case by restricting to closed subgroups, as the topology ensures the correspondence is well-behaved.^[30]^[31] Krull's theorem formalizes this bijection, stating that for an infinite Galois extension K/F, the map H \mapsto K^H induces an anti-isomorphism between the lattice of closed subgroups of G and the lattice of intermediate fields F \subseteq L \subseteq K. A key distinction from the finite case is that not all subgroups of G are closed; Krull proved that every infinite Galois extension admits non-closed subgroups, which do not correspond to intermediate fields under the naive map without topology, as their fixed fields would not yield the full inverse. The profinite structure and closure condition thus ensure continuity and bijectivity, preserving the core insights of the theorem while accommodating infinite degree.^[30]^[31]

Profinite Groups and Topology

In the context of infinite Galois extensions, the Galois group G = \Gal(K^\sep / K), where K^\sep denotes the separable closure of the base field K, is equipped with the profinite topology, also known as the Krull topology. This topology is defined by taking as a basis of open neighborhoods of the identity element the subgroups \Gal(K^\sep / L), where L / K ranges over all finite Galois extensions. Each such subgroup is open and normal in G, and the cosets of these subgroups form a basis for the open sets, making the topology the coarsest one that renders all the natural projection maps G \to \Gal(L / K) continuous. With this topology, G becomes a profinite group, characterized as a compact, Hausdorff, and totally disconnected topological group. The compactness arises from the inverse limit structure of G as \lim_{\leftarrow} \Gal(L / K) over finite Galois extensions L / K, while total disconnectedness follows from the basis of clopen (closed and open) normal subgroups. Closed subgroups in profinite groups are those that are closed in the topological sense and can be characterized as intersections of open subgroups containing them, ensuring that the Galois correspondence remains bijective. The group operations in G, including the action on roots of unity or algebraic elements, are continuous with respect to the profinite topology. All automorphisms in G are continuous with respect to the profinite topology on G and the corresponding topology on the separable closure. This continuity property ensures that the Galois correspondence between closed subgroups of G and intermediate fields preserves topological closure, linking the algebraic and topological aspects of infinite extensions. A prominent example is the absolute Galois group G_\mathbb{Q} = \Gal(\overline{\mathbb{Q}} / \mathbb{Q}), which is profinite and non-abelian under the Krull topology. This group encodes the symmetries of all finite extensions of the rationals and serves as a foundational object in number theory, with its profinite structure enabling the study of infinite towers of extensions.

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[10]
[PDF] galois groups of cubics and quartics (not in characteristic 2)
We will describe a procedure for figuring out the Galois groups of separable irreducible polynomials in degrees 3 and 4 over fields not of characteristic 2.
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[PDF] Galois groups as permutation groups - Keith Conrad
A Galois group is a group of field automorphisms under composition. By looking at the effect of a Galois group on field generators we can interpret the Galois ...
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[PDF] The computation of Galois groups over function fields
Hence Galk(f) acts on the roots α1,...,αn of f by permutation. In this way Galk(f) is a subgroup of Sn , the symmetric group on n letters. However, this ...
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Below is a merged summary of the Fundamental Theorem of Galois Theory based on the provided segments. To retain all information in a dense and organized manner, I will use a combination of narrative text and a table to capture details efficiently. The summary integrates all key points, including statements, fixed fields, maps, proofs, properties, and references, while avoiding redundancy and ensuring completeness.
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[PDF] 9. Normal and Separable extensions - UCSD Math
Definition 9.4. Let L/K be a field extension. A normal closure for L/K is a field N/L such that N/K is nor- mal, and there are no proper intermediary fields, ...
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[PDF] Math 210B. Normal field extensions 1. A definition In Exercise 7 of ...
2. Normal closure. If k//k is a general algebraic extension, a normal closure of k//k is an algebraic extension E/k/ normal over k with the “minimality” ...
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[PDF] Galois Theory
Theorem 25.10. Let π : Y → X be a Galois cover with Galois group G. There is a lattice anti-isomorphism. H. // (Y → Y/H → X) subgroups of G. // intermediate.
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[PDF] the galois anti-isomorphism
Aug 25, 2011 · That is, a separable closure is normal, hence Galois. Definition 3.10. The absolute Galois group Gal(k) of a field k is Gal(ks|k), where ks is ...<|separator|>
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Fundamental theorem of Galois theory—The Stacks project
The normal subgroups H of G correspond exactly to those subextensions M with M/K Galois. Proof. By Lemma 9.21.4 given a subextension L/M/K the extension L/M is ...
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[PDF] Dedekind's treatment of Galois theory in the Vorlesungen
Dec 14, 2009 · At the end of this section, we indicate the precise extent to which Dedekind can be said to have formulated the fundamental theorem of. Galois ...
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[PDF] Cubic Equations - LSU Math
3 if L. 6 if L < K'. K. (2.12) Example. The polynomial f(x) x3 + 3x + 1 is irreducible over Q, and it has only one real root. To see that there is only one ...
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[PDF] Fields and Galois Theory - James Milne
THEOREM 3.16 (FUNDAMENTAL THEOREM OF GALOIS THEORY) Let E be a Galois ex- tension of F with Galois group G. The map H 7! E. H is a bijection from the set of.
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[PDF] M345P11: Cyclotomic fields. 1 Introduction.
Then L is called a cyclotomic field and there is an amazingly simple and beautiful answer to the question – in this case Gal(L/Q) = (Z/nZ)×; the Galois group is ...
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[PDF] 5.3 Solvability by Radicals - math.binghamton.edu
Apr 30, 2020 · We say that a polynomial f(x) 2 K[x] is solvable by radicals, if all its roots can be expressed by radicals over K. Definition 5.8 A Galois ...
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[PDF] Galois Theory and Solvability - Whitman People
Feb 2, 2023 · Suppose that f is solvable by radicals, and suppose that the fields Ji, elements bi and exponents mi, for i = 1,...,s, are as described by the ...
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[PDF] Chapter 7 Galois theory
It will be lynchpin of our argument showing that not all polynomials over Q are solvable by radicals over Q. Also, from here on, we will exclusively focus on ...<|control11|><|separator|>
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[PDF] Lecture 24 - Math 5111 (Algebra 1)
Dec 7, 2020 · fundamental theorem of Galois theory G = Gal(K/F) is a quotient of Gal(L/F). Thus G is a quotient of a solvable group, hence is solvable as.
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[PDF] Lecture 22 - Math 5111 (Algebra 1)
Nov 30, 2020 · choose for the coefficients, the Galois group will always be cyclic, since every extension of finite fields is Galois with cyclic. Galois group.
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[PDF] Constructibility and Galois Theories
Identify the plane with C. The set C ⊆ C of constructible numbers is the collection of numbers which can be realized, starting from 0 and 1, and applying.
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[PDF] galois theory at work - keith conrad
The Galois group of (X2 − 2)(X2 − 3) over Q is Z/2Z × Z/2Z. Its Galois group over R is trivial. Page 10. 10. KEITH CONRAD.
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[PDF] 26 The idele group, profinite groups, infinite Galois theory
Dec 3, 2018 · Theorem 26.22 (Fundamental theorem of Galois theory). Let L/K be a Galois extension and let G := Gal(L/K) be endowed with the Krull topology.
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### Summary of Infinite Galois Theory Notes by J. Ruiter, Michigan State University