In mathematics, the fundamental theorem of Galois theory stands as a pivotal result, elegantly articulating the intricate structural relationships within particular categories of field extensions when examined through the lens of groups. This profound insight was meticulously developed and rigorously proven by Évariste Galois himself, forming a cornerstone of what we now know as Galois theory. One might even say it's the kind of foundational truth that makes the rest of the universe seem a little less chaotic, at least for a moment.

At its most distilled and fundamental level, the theorem posits that for a given field extension, denoted as $E/F$, which possesses the properties of being both finite and Galois, there exists a precise one-to-one correspondence between the collection of its intermediate fields and the subgroups of its associated Galois group. For those who prefer their definitions explicit, intermediate fields are simply fields $K$ that comfortably reside between the base field and the extension field, satisfying the inclusion $F \subseteq K \subseteq E$. These are often referred to, with a certain lack of imagination, as subextensions of $E/F$. This correspondence isn't merely a convenient mapping; it's a deep structural isomorphism that allows one to translate complex questions about field structures into more manageable problems within group theory. It's almost as if Galois, in his brief and tumultuous life, decided to hand us a universal decoder ring for algebraic extensions.

Explicit description of the correspondence

For finite extensions, a context where one might expect things to be somewhat less complicated, the correspondence can be delineated with a rather straightforward, if not entirely intuitive, precision.

• Mapping Subgroups to Intermediate Fields: For any given subgroup $H$ of the Galois group $\text{Gal}(E/F)$, the theorem defines a corresponding fixed field. This field, often denoted as $E^H$, is comprised solely of those elements within the larger field $E$ that remain utterly invariant under the action of every single automorphism belonging to the subgroup $H$. In essence, $E^H$ collects all the elements that $H$ simply cannot touch, a testament to their inherent stability under specific transformations. It's a field of quiet resistance, if you will.

• Mapping Intermediate Fields to Subgroups: Conversely, for any intermediate field $K$ nestled between $F$ and $E$, the theorem associates a specific subgroup of the Galois group. This subgroup is precisely $\text{Aut}(E/K)$, which is defined as the collection of all automorphisms within $\text{Gal}(E/F)$ that leave every element of $K$ fixed. These are the transformations that respect the structure of $K$, acting as its silent guardians within the larger field $E$.

The true brilliance, or perhaps the ultimate convenience, of the fundamental theorem of Galois theory lies in its assertion that this two-way mapping constitutes a perfect one-to-one correspondence. This holds true if and only if the original extension $E/F$ is indeed a Galois extension. This condition isn't merely a technicality; it's the bedrock upon which the entire elegant structure rests.

Consider the extremities of this correspondence, which often serve as useful anchors. The topmost field, $E$ itself, corresponds to the trivial subgroup of $\text{Gal}(E/F)$ – the subgroup containing only the identity automorphism. This makes perfect sense; only the identity transformation leaves all elements of $E$ fixed. Conversely, the base field $F$ corresponds to the entire group $\text{Gal}(E/F)$. This implies that all automorphisms in the Galois group must fix the elements of the base field, which is, by definition, a fundamental property of Galois extensions.

It's worth noting, for the sake of completeness and to avoid any naive assumptions, that the notation $\text{Gal}(E/F)$ is strictly reserved for Galois extensions. If $E/F$ happens to be a Galois extension, then, rather conveniently, $\text{Gal}(E/F) = \text{Aut}(E/F)$. However, if one dares to venture into extensions that are not Galois, this beautiful symmetry predictably crumbles. In such cases, the "correspondence" becomes a rather one-sided affair, offering only an injective (but decidedly not surjective) map from the set of subgroups of $\text{Aut}(E/F)$ to the set of intermediate subfields of $E/F$. The reverse mapping, from subfields to subgroups, is then merely surjective (but not injective). This asymmetry means that the neat, perfect pairing is lost. Specifically, if $E/F$ is not Galois, the base field $F$ itself cannot be recovered as the fixed field of any subgroup of $\text{Aut}(E/F)$, underscoring the critical importance of the Galois condition for the theorem's full power. It's a stark reminder that not all extensions are created equal, and some simply refuse to play by the rules.

Properties of the correspondence

The fundamental theorem of Galois theory isn't just a statement of bijection; it provides a framework with several profoundly useful properties that translate structural information between fields and groups with remarkable fidelity.

• Inclusion-Reversing Nature: This correspondence exhibits an elegant, almost counter-intuitive, inclusion-reversing property. Specifically, the inclusion of subgroups $H_1 \subseteq H_2$ holds if and only if the inclusion of their corresponding fixed fields is reversed: $E^{H_1} \supseteq E^{H_2}$. This means that if you have a larger subgroup of automorphisms, it will fix fewer elements in the field $E$, resulting in a smaller fixed field. Conversely, a smaller subgroup, being less restrictive, will leave more elements untouched, thus defining a larger fixed field. It's like a mathematical seesaw, where more group structure means less field structure, and vice-versa.

• Relationship Between Degrees and Orders: The magnitudes of these structures are also intimately connected, in a manner perfectly consistent with the inclusion-reversing property mentioned above. If $H$ is a subgroup of $\text{Gal}(E/F)$, then the order of the subgroup, $|H|$, is precisely equal to the degree of the field extension $[E:E^H]$. This tells us exactly how much larger $E$ is than the subfield fixed by $H$. Furthermore, the ratio of the order of the entire Galois group to the order of the subgroup, $|\text{Gal}(E/F)| / |H|$, gives us the degree of the extension of the fixed field over the base field: $[E^H:F]$. These formulas provide a direct quantitative link, allowing one to calculate dimensions of fields from the sizes of groups, and vice versa. It’s a beautifully efficient algebraic accounting system.

• Normality and Quotient Structures: Perhaps one of the most profound aspects of this correspondence relates to the concept of normality. The intermediate field $E^H$ is a normal extension of $F$ (which, for a separable extension, is equivalent to being a Galois extension) if and only if its corresponding subgroup $H$ is a normal subgroup of the entire Galois group $\text{Gal}(E/F)$. This is not a mere coincidence; it reflects a deep structural harmony. When this condition holds, the restriction of the automorphisms from $\text{Gal}(E/F)$ to the subfield $E^H$ naturally induces an isomorphism between the Galois group of the normal subextension, $\text{Gal}(E^H/F)$, and the quotient group $\text{Gal}(E/F)/H$. This means that the "factor group" captures the Galois structure of the sub-extension over the base field, providing a powerful tool for understanding nested field structures. It's a testament to how profoundly group theory mirrors the internal symmetries of fields.

Example 1

Let's consider an illustrative case, often presented as a first foray into the practical application of this theorem. We examine the field $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$, which can be systematically constructed by first adjoining $\sqrt{2}$ to the field of rational numbers $\mathbb{Q}$, and then subsequently adjoining $\sqrt{3}$ to the resulting field $\mathbb{Q}(\sqrt{2})$. Each element within this extension field $K$ can be uniquely expressed in the form $(a+b\sqrt{2}) + (c+d\sqrt{2})\sqrt{3}$, where $a, b, c, d$ are all elements of $\mathbb{Q}$. This construction implies a clear structure for the elements we're working with.

The Galois group $G = \text{Gal}(K/\mathbb{Q})$ consists of all automorphisms of $K$ that leave elements of $\mathbb{Q}$ fixed. Such automorphisms are entirely determined by where they send $\sqrt{2}$ and $\sqrt{3}$. Since these are roots of the irreducible polynomials $x^2-2$ and $x^2-3$ respectively, any automorphism must permute their roots. Thus, $\sqrt{2}$ must map to either $\sqrt{2}$ or $-\sqrt{2}$, and similarly, $\sqrt{3}$ must map to either $\sqrt{3}$ or $-\sqrt{3}$. This gives us four distinct possibilities for how these automorphisms can act:

1. Identity ($e$): $\sqrt{2} \mapsto \sqrt{2}$, $\sqrt{3} \mapsto \sqrt{3}$. This automorphism fixes every element in $K$.
2. Automorphism ($f$): $\sqrt{2} \mapsto -\sqrt{2}$, $\sqrt{3} \mapsto \sqrt{3}$. Its action on a generic element is $f\left((a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3}\right) = (a-b\sqrt{2})+(c-d\sqrt{2})\sqrt{3} = a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$. This transformation leaves $\sqrt{3}$ fixed, but flips the sign of anything involving $\sqrt{2}$.
3. Automorphism ($g$): $\sqrt{2} \mapsto \sqrt{2}$, $\sqrt{3} \mapsto -\sqrt{3}$. Its action is $g\left((a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3}\right) = (a+b\sqrt{2})-(c+d\sqrt{2})\sqrt{3} = a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6}$. Here, $\sqrt{2}$ is fixed, while $\sqrt{3}$ changes sign.
4. Composition ($fg$): $\sqrt{2} \mapsto -\sqrt{2}$, $\sqrt{3} \mapsto -\sqrt{3}$. This is the composition of $f$ and $g$, resulting in $fg\left((a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3}\right) = (a-b\sqrt{2})-(c-d\sqrt{2})\sqrt{3} = a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}$. Both radicals have their signs inverted.

Since the degree of the field extension $[K:\mathbb{Q}]$ is 4 (as $[K:\mathbb{Q}(\sqrt{2})]=2$ and $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$, so $2 \times 2 = 4$), the order of the Galois group must also be 4. This confirms that we have identified all possible automorphisms: $G = \{1, f, g, fg\}$. This group is, rather tellingly, isomorphic to the Klein four-group, an abelian group where every non-identity element has order 2.

The fundamental theorem of Galois theory then dictates that the five subgroups of $G$ correspond directly to the five intermediate fields situated between the base field $\mathbb{Q}$ and the extension field $K$.

• The trivial subgroup $\{1\}$: As expected, this corresponds to the entire extension field $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$, as only the identity automorphism fixes all elements of $K$.
• The entire group $G$: This corresponds to the base field $\mathbb{Q}$, since all elements of $G$ fix every rational number by definition.
• The subgroup $\{1, f\}$: This subgroup contains the identity and the automorphism $f$ that flips $\sqrt{2}$ to $-\sqrt{2}$ but leaves $\sqrt{3}$ fixed. Therefore, its fixed field consists of all elements in $K$ that are invariant under $f$. This is precisely the subfield $\mathbb{Q}(\sqrt{3})$, as elements of the form $a+b\sqrt{3}$ (where $a, b \in \mathbb{Q}$) are fixed by $f$.
• The subgroup $\{1, g\}$: Similarly, this subgroup contains the identity and the automorphism $g$ that flips $\sqrt{3}$ to $-\sqrt{3}$ but leaves $\sqrt{2}$ fixed. Its fixed field is thus $\mathbb{Q}(\sqrt{2})$, comprising elements of the form $a+b\sqrt{2}$ (where $a, b \in \mathbb{Q}$).
• The subgroup $\{1, fg\}$: This subgroup contains the identity and the automorphism $fg$ which flips both $\sqrt{2}$ and $\sqrt{3}$. To find its fixed field, we look for elements that are invariant under $fg$. Observe that $\sqrt{6} = \sqrt{2}\sqrt{3}$ is transformed by $fg$ to $(-\sqrt{2})(-\sqrt{3}) = \sqrt{6}$. Thus, elements of the form $a+b\sqrt{6}$ are fixed. This corresponds to the subfield $\mathbb{Q}(\sqrt{6})$.

This example clearly demonstrates the inverse relationship between the lattice of subgroups and the lattice of subfields, a visual representation of the theorem's core assertion.

Example 2

Moving beyond the comfortable confines of abelian groups, let us consider a more intricate scenario where the Galois group is not abelian. This typically signifies a richer, more complex interplay of symmetries within the field extension.

We examine the splitting field $K$ of the irreducible polynomial $x^3-2$ over $\mathbb{Q}$. The roots of this polynomial are $\theta, \omega\theta, \omega^2\theta$, where $\theta = \sqrt[3]{2}$ (the real cube root of 2) and $\omega = -{\tfrac {1}{2}}+i{\tfrac {\sqrt {3}}{2}}$ is a primitive cube root of 1 (specifically, $\omega \neq 1$). The field $K$ is therefore given by $K = \mathbb{Q}(\theta, \omega)$. When viewed within the complex numbers, this construction becomes particularly clear.

The minimal polynomial for $\omega$ over $\mathbb{Q}$ is $x^2+x+1$. Consequently, the degree of the extension $[\mathbb{Q}(\omega):\mathbb{Q}]$ is 2. The minimal polynomial for $\theta$ over $\mathbb{Q}$ is $x^3-2$, giving $[\mathbb{Q}(\theta):\mathbb{Q}]$ a degree of 3. Since $\omega$ is not in $\mathbb{Q}(\theta)$ (because $\mathbb{Q}(\theta)$ is a subfield of the real numbers), the degree of the full extension $[K:\mathbb{Q}]$ is the product of these degrees: $[K:\mathbb{Q}] = [K:\mathbb{Q}(\theta)] \cdot [\mathbb{Q}(\theta):\mathbb{Q}] = 2 \cdot 3 = 6$. This implies that the Galois group $G = \text{Gal}(K/\mathbb{Q})$ must have six elements.

These six elements are entirely determined by how they permute the three roots of $x^3-2$: $\alpha_1=\theta, \alpha_2=\omega\theta, \alpha_3=\omega^2\theta$. Since there are precisely $3! = 6$ possible permutations of three distinct objects, $G$ must be isomorphic to the symmetric group $S_3$. This means its structure is decidedly non-abelian.

The group $G$ can be generated by two key automorphisms:

• $f$: This automorphism cyclically permutes the roots of $x^3-2$. It is defined by $f(\theta) = \omega\theta$ and $f(\omega) = \omega$. In cycle notation, $f=(123)$, meaning $\alpha_1 \mapsto \alpha_2 \mapsto \alpha_3 \mapsto \alpha_1$. It fixes $\omega$.
• $g$: This automorphism swaps two of the complex roots while fixing the real root. It is defined by $g(\theta) = \theta$ and $g(\omega) = \omega^2$. In cycle notation, $g=(23)$, meaning $\alpha_2 \mapsto \alpha_3 \mapsto \alpha_2$. It fixes $\theta$. Notably, $g$ acts as the complex conjugation mapping within $K$, as $\omega^2$ is the complex conjugate of $\omega$.

These generators satisfy the relations $f^3 = g^2 = (gf)^2 = 1$, which are the defining relations for $S_3$. The full group is then $G = \{1, f, f^2, g, gf, gf^2\}$.

The subgroups of $G$ and their corresponding subfields, as dictated by the fundamental theorem of Galois theory, are as follows:

• Trivial group $\{1\}$: Corresponds to the entire field $K$.

• Entire group $G$: Corresponds to the base field $\mathbb{Q}$.

• The unique subgroup of order 3: $H = \{1, f, f^2\}$. This subgroup is generated by the cyclic permutation $f$. Since $f$ fixes $\omega$, this subgroup corresponds to the subfield $\mathbb{Q}(\omega)$. The degree of this subfield over $\mathbb{Q}$ is $[\mathbb{Q}(\omega):\mathbb{Q}] = |G|/|H| = 6/3 = 2$, which is consistent with the minimal polynomial of $\omega$ being $x^2+x+1$. Furthermore, $H$ is a normal subgroup of $G$ (as it's the only subgroup of order 3 in $S_3$, and any subgroup of prime index is normal). This normality implies that $\mathbb{Q}(\omega)$ is a normal extension (and thus Galois) over $\mathbb{Q}$. Indeed, $\mathbb{Q}(\omega)$ is the splitting field of $x^2+x+1$. Its Galois group over $\mathbb{Q}$ is isomorphic to the quotient group $G/H = \{[1], [g]\}$, where $[g]$ represents the coset of $g$ modulo $H$. This group has order 2, and its only non-trivial automorphism is the complex conjugation $g$.

• Three subgroups of order 2: These are $\{1, g\}$, $\{1, gf\}$, and $\{1, gf^2\}$.

  • The subgroup $\{1, g\}$ corresponds to the subfield $\mathbb{Q}(\theta)$. This is because $g$ fixes $\theta$. The degree of this subfield over $\mathbb{Q}$ is $6/2 = 3$, consistent with $x^3-2$ being the minimal polynomial of $\theta$.
  • The subgroup $\{1, gf\}$ corresponds to the subfield $\mathbb{Q}(\omega\theta)$.
  • The subgroup $\{1, gf^2\}$ corresponds to the subfield $\mathbb{Q}(\omega^2\theta)$. These three subgroups are not normal subgroups in $G$. Consequently, their corresponding subfields $\mathbb{Q}(\theta)$, $\mathbb{Q}(\omega\theta)$, and $\mathbb{Q}(\omega^2\theta)$ are not Galois extensions (and thus not normal extensions) over $\mathbb{Q}$. This is evident from the fact that each of these fields contains only one of the roots of $x^3-2$. For an extension to be normal, it must contain all conjugates of any of its elements. Since $\mathbb{Q}(\theta)$ contains $\theta$ but not $\omega\theta$ or $\omega^2\theta$, it cannot be normal. Indeed, each of these subfields has no non-trivial automorphisms that fix $\mathbb{Q}$, reflecting their non-normal group counterparts. The lack of internal symmetry is directly mirrored in the group structure.

Example 3

Let us now delve into a more abstract, yet equally illuminating, example involving fields of rational functions. Consider $E = \mathbb{Q}(\lambda)$, which is the field of rational functions in the indeterminate $\lambda$ with coefficients from $\mathbb{Q}$. Within this field, we can define a fascinating group of automorphisms:

$G = \left\{\lambda, {\frac{1}{1-\lambda}}, {\frac{\lambda-1}{\lambda}}, {\frac{1}{\lambda}}, {\frac{\lambda}{\lambda-1}}, 1-\lambda \right\} \subset \mathrm{Aut}(E);$

Here, an automorphism $\phi: E \to E$ is conveniently represented by its action on the indeterminate $\lambda$, specifically $\phi(\lambda)$. So, if you apply an automorphism $\phi$, any function $f(\lambda)$ transforms into $f(\phi(\lambda))$. This group $G$ is, remarkably, isomorphic to the symmetric group $S_3$, the same group encountered in Example 2. These six specific transformations are, in fact, the six cross-ratios of four distinct points.

Let $F$ be the fixed field of this entire group $G$. By the fundamental theorem of Galois theory, this implies that $\mathrm{Gal}(E/F)=G$. The elements of $F$ are precisely those rational functions of $\lambda$ that remain invariant under all six transformations in $G$.

If $H$ is any subgroup of $G$, then the fixed field corresponding to $H$ can be constructed in a rather elegant way. Consider the polynomial $P(T) := \prod_{h \in H}(T-h) \in E[T]$. The coefficients of this polynomial, which are symmetric polynomials in the elements of $H$, generate the fixed field of $H$. The Galois correspondence guarantees that every subfield of $E/F$ can be constructed via this method.

Let's illustrate with a couple of specific subgroups:

• For the subgroup $H = \{\lambda, 1-\lambda\}$, the polynomial would be $(T-\lambda)(T-(1-\lambda)) = T^2 - (\lambda + 1-\lambda)T + \lambda(1-\lambda) = T^2 - T + \lambda(1-\lambda)$. The coefficients are $1$ and $\lambda(1-\lambda)$. Thus, the fixed field of $H$ is $\mathbb{Q}(\lambda(1-\lambda))$. Any rational function of $\lambda(1-\lambda)$ will be fixed by both $\phi(\lambda)=\lambda$ and $\phi(\lambda)=1-\lambda$.

• For the subgroup $H = \{\lambda, {\tfrac{1}{\lambda}}\}$, the polynomial is $(T-\lambda)(T-{\tfrac{1}{\lambda}}) = T^2 - (\lambda + {\tfrac{1}{\lambda}})T + 1$. The coefficients are $1$, $1$, and $\lambda + {\tfrac{1}{\lambda}}$. Consequently, the fixed field is $\mathbb{Q}(\lambda + {\tfrac{1}{\lambda}})$. This is because the expression $\lambda + {\tfrac{1}{\lambda}}$ is invariant under the transformation $\lambda \mapsto {\tfrac{1}{\lambda}}$.

The fixed field of the entire group $G$ is the base field $F = \mathbb{Q}(j)$, where $j$ is the famous j-invariant, expressed here in terms of the modular lambda function:

$j = {\frac {256(1-\lambda(1-\lambda))^{3}}{(\lambda(1-\lambda))^{2}}} = {\frac {256(1-\lambda+\lambda^{2})^{3}}{\lambda^{2}(1-\lambda)^{2}}} .$

This connection highlights the deep interplay between Galois theory, function theory, and modular forms. The j-invariant is notoriously fixed by a much larger group of transformations, the modular group, but these six transformations are a finite subgroup of that.

Similar examples can be elegantly constructed for each of the symmetry groups of the platonic solids. These groups also possess faithful actions on the projective line $\mathbb{P}^1(\mathbb{C})$, and consequently on $\mathbb{C}(x)$, leading to analogous fixed fields and Galois correspondences. It's almost as if the universe is designed to have these symmetries, and Galois theory is just pointing them out.

Example 4

It is often equally instructive to examine cases where a theorem fails to hold, as this clarifies the critical conditions for its applicability. Here, we present an example of a finite extension $E/F$ that is not Galois, and with it, we demonstrate unequivocally that the fundamental theorem of Galois theory simply does not apply in its stated form.

Let $E = \mathbb{Q}(\sqrt[3]{2})$ and $F = \mathbb{Q}$. This is undoubtedly a finite extension, as its degree is $[E:F] = 3$. However, $E/F$ is emphatically not a splitting field over $F$. The minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$ is $x^3-2$. Its roots are $\sqrt[3]{2}$, $\omega\sqrt[3]{2}$, and $\omega^2\sqrt[3]{2}$, where $\omega$ is a primitive cube root of unity. While $\sqrt[3]{2}$ is in $E$, the other two roots, $\omega\sqrt[3]{2}$ and $\omega^2\sqrt[3]{2}$, are complex numbers and do not lie within $E$ (since $E$ is a subfield of the real numbers). For an extension to be a splitting field, it must contain all the roots of the polynomial. Since $E$ fails this condition, it cannot be a Galois extension.

Now, let's consider the Galois group $G = \mathrm{Gal}(E/F)$. Any automorphism $f \in G$ is entirely determined by where it maps $\sqrt[3]{2}$. Since $f$ must preserve the field operations and fix $\mathbb{Q}$, we must have $f(\sqrt[3]{2})^3 = f((\sqrt[3]{2})^3) = f(2) = 2$. This means $f(\sqrt[3]{2})$ must be another root of $x^3-2$. However, since $f$ maps elements of $E$ to elements of $E$, and $E$ is a real field, $f(\sqrt[3]{2})$ must be real. The only real root of $x^3-2$ is $\sqrt[3]{2}$ itself. Therefore, $f(\sqrt[3]{2}) = \sqrt[3]{2}$. This implies that the only possible automorphism in $G$ is the identity map. Thus, $G = \{1\}$, the trivial group.

The order of the Galois group is $|G|=1$. This is strictly less than the degree of the extension, $[E:F]=3$. This inequality, $|G| < [E:F]$, is a definitive indicator that $E/F$ is not a Galois extension.

Now, let's observe the consequences for the Galois correspondence. The group $G=\{1\}$ has only one subgroup: itself. According to a functioning correspondence, there should be only one intermediate field. However, the intermediate fields between $F=\mathbb{Q}$ and $E=\mathbb{Q}(\sqrt[3]{2})$ are $\mathbb{Q}$ itself and $\mathbb{Q}(\sqrt[3]{2})$ itself (as there are no fields strictly between them, given the prime degree of the extension). We have two intermediate fields, but only one subgroup. This clear mismatch demonstrates that the Galois correspondence, in its finite form, utterly fails when the extension is not Galois. It's almost as if the universe decided to pull the rug out from under the theorem, simply because we neglected a crucial condition.

Applications

The profound utility of the fundamental theorem of Galois theory extends far beyond merely categorizing field structures. Its power lies in its ability to translate complex questions about fields into the more tractable domain of group theory. This algebraic Rosetta Stone is absolutely central to some of the most celebrated results in the history of mathematics.

Perhaps the most famous application is its role in proving the insolvability of the general quintic equation by radicals, a monumental achievement detailed in the Abel–Ruffini theorem. Before Galois, mathematicians struggled for centuries to find a general formula for the roots of quintic (and higher-degree) polynomials, analogous to the quadratic formula. Galois's insight provided the definitive answer: no such general formula exists using only arithmetic operations and root extractions. The proof hinges on first determining the structure of the Galois groups associated with radical extensions – extensions formed by adjoining $n$-th roots of elements. These groups are shown to possess a specific algebraic property: they are solvable groups. The fundamental theorem of Galois theory then establishes a direct link: a polynomial equation is solvable by radicals if and only if its Galois group is a solvable group. Since the symmetric group $S_5$, which is the Galois group for the general quintic, is not solvable, the theorem immediately implies the insolvability of the general quintic. It's a rather elegant way to definitively close a chapter that had frustrated generations of mathematicians.

Beyond this foundational result, the theorem underpins entire branches of modern number theory and algebra. Theories such as Kummer theory and class field theory are deeply predicated on the principles established by the fundamental theorem of Galois theory. Kummer theory, for instance, provides a detailed description of abelian extensions in terms of roots of unity, heavily relying on the correspondence between fields and abelian Galois groups. Class field theory, a vast and intricate area of number theory, generalizes the concepts of Galois theory to describe abelian extensions of number fields, using sophisticated notions of ideal class groups and other arithmetic invariants, all built upon the foundational insights of Galois's work. These advanced theories demonstrate the theorem's enduring power as a conceptual bedrock for exploring the deepest symmetries in mathematics.

Infinite case

While the finite case of the fundamental theorem of Galois theory is remarkably elegant and powerful, the universe of algebraic extension fields is, regrettably, not always finite. When dealing with infinite algebraic extensions, we can still define them to be Galois if they satisfy the conditions of being normal and separable. However, a significant complication arises: the beautiful bijection observed in the finite theorem generally breaks down. The problem is that, if one simply considers every subgroup of the infinite Galois group, one can often find distinct subgroups that, rather inconveniently, fix precisely the same intermediate field. This "too many subgroups" issue necessitates a more refined approach.

To salvage the correspondence in this infinite realm, we introduce a topology on the Galois group. This additional structure, known as the Krull topology, provides the necessary "filter" to identify the relevant subgroups.

Let $E/F$ be a Galois extension, which may be infinite in degree. Let $G = \text{Gal}(E/F)$ be its Galois group. We then consider the set $\text{Int}_{\text{F}}(E/F) = \{G_i = \text{Gal}(L_i/F) \mid L_i/F \text{ is a finite Galois extension and } L_i \subseteq E\}$, which comprises the Galois groups of all finite intermediate Galois extensions contained within $E$. For each index $i \in I$, we define the restriction maps $\varphi_i: G \rightarrow G_i$ by $\sigma \mapsto \sigma_{|L_i}$. This map essentially "zooms in" on the action of an automorphism from $G$ on a specific finite sub-extension $L_i$.

The Krull topology on $G$ is then defined as the weakest topology such that all these restriction maps $\varphi_i: G \rightarrow G_i$ are continuous, assuming each $G_i$ is endowed with the discrete topology. More formally, $G$ is isomorphic to the inverse limit $\varprojlim G_i$ of these finite Galois groups, where each $G_i$ again carries the discrete topology. This construction makes $G$ a profinite group. In fact, it's a rather satisfying truth that every profinite group can be realized as the Galois group of some Galois extension (see, for instance, Ribes–Zalesskii [1]). It is important to note that when $E/F$ is a finite extension, the Krull topology simply reduces to the discrete topology, meaning the finite case is a special instance of the infinite one.

With the Krull topology firmly established on the Galois group, we can now elegantly restate the fundamental theorem of Galois theory for infinite Galois extensions.

Let $\mathcal{F}(E/F)$ denote the set of all intermediate field extensions of $E/F$, and let $\mathcal{C}(G)$ denote the set of all closed subgroups of $G = \text{Gal}(E/F)$ when $G$ is endowed with the Krull topology. Then, a perfect bijection exists between $\mathcal{F}(E/F)$ and $\mathcal{C}(G)$. This bijection is established by two maps:

• The map $\Phi: \mathcal{F}(E/F) \rightarrow \mathcal{C}(G)$: This map takes an intermediate field $L$ and associates it with its corresponding Galois group $\text{Gal}(E/L)$.
• The map $\Gamma: \mathcal{C}(G) \rightarrow \mathcal{F}(E/F)$: This map takes a closed subgroup $N$ of $G$ and associates it with its fixed field, defined as $\text{Fix}_{E}(N) := \{a \in E \mid \sigma(a)=a \text{ for all } \sigma \in N\}$.

A crucial step in validating this extended theorem is to ensure that $\Phi$ is indeed a well-defined map. Specifically, one must verify that $\Phi(L) = \text{Gal}(E/L)$ always results in a closed subgroup of $G$ for any intermediate field $L$. This essential property is rigorously proven in sources such as Ribes–Zalesskii, Theorem 2.11.3 [1]. The introduction of the Krull topology thus provides the necessary analytical framework to extend the profound insights of Galois to the vast and intricate landscape of infinite field extensions, ensuring that the correspondence, much like time, remains unbroken, albeit with an added layer of complexity.

See also

• Galois connection