Grothendieck Galois Theory (Draft)

This note is the result of a discussion of Grothendieck Galois theory with Claude and GPT o4-mini.

Guide for reading:Since in Galois Theory we only consider the etale morphism to $\mathrm{Spec}F$$\mathrm{Spec} F$ in category of affine scheme, hence. we only need to deal with the eatle $F$$F$ algebra, whcih is isomorphic to the finite product of finite separe extension of $F$$F$. So you could jump the section 1 if you want.(That is the reason I set section 1 at the appendix.)

The main line of this notes is

$$\text{Étale Algebra}⟶\text{Fiber functor}\stackrel{\text{Aut}}{\to }\text{Étale Fudamental Group}\cong \text{Absolute Galois Group}⟶\text{Cov}(F)\cong {\pi }_1(F)\text{-FinSet}$$

 

2 Étale Morphisms over Fields2.1 Analysis of Field Extensions2.1.1 Finite Type Condition2.1.2 Flatness Condition2.1.3 Unramified Condition2.2 Detailed Computation of Separability and Kähler Differentials for Field Extensions3 Concrete Examples and the Correspondence with Polynomial Roots3.1 Detailed Analysis of Quadratic Field Extensions3.2 Detailed Analysis of Inseparable Field Extensions in Characteristic $p$3.3 Detailed Analysis of Direct Products of Finite Field Extensions3.4 Detailed Analysis of Cubic Polynomials3.5 Example of a Non-Simple Field Extension4 Connection between Étale Morphisms and Galois Theory4.1 The Absolute Galois Group4.2 The Fiber Functor: Geometric Meaning and Algebraic Correspondence4.2.1 Geometric Perspective: Fibers and Polynomial Roots4.2.2 The Fiber Functor: A Geometric Definition4.2.3 Structure of the Fiber and Correspondence with Polynomial Root4.3 The isomorphism between Étale Fundamental Group and Absolute Galois Group via Yoneda Lemma. The Fundamental Group as the Automorphism Group of a Fiber Functor: Proof and AnalysisIntroductionPrerequisitesCovering Spaces and Fiber FunctorFundamental Group ActionMain TheoremProofStep 1: Constructing the Map $\Phi$Step 2: Verifying $\Phi([\gamma])$ is a Natural TransformationStep 3: Verifying $\Phi$ is a Group HomomorphismStep 4: Proving $\Phi$ is InjectiveStep 5: Proving $\Phi$ is SurjectiveConclusionÉtale Fundamental Group4.4 Galois Theory Correspondence4.4.1 Concrete Examples of Galois Correspondence with Polynomial Roots 5.3 Fiber Functor and Category Equivalence5.3.1 Construction and Intuitive Understanding of the Fiber Functor5.3.2 Category Equivalence Theorem and Its Rigorous Proo21. Construction of the Inverse Functor $G_x$2. Proving $F_x \circ G_x \cong \text{Id}_{\mathbf{FinSet}^G}$3. Proving $G_x \circ F_x \cong \text{Id}_{\mathbf{FEt}_F}$ 4. Constructive Understanding of Category Equivalence5.4 From the duality to classical Galois CorrespondenceNormal Extensions and Normal SubgroupsClassical Galois CorrespondenceDegree and Galois Group Order Relationship6 ConclusionAppendix1 Étale Morphisms in Affine Schemes1.1 Definition of Étale Morphisms1.2 Detailed Explanation of Flatness1.3 Kähler Differential Module and the Unramified Condition$\Omega^1_{(-)/A}$ as left adjoint functor.1.4 Geometric Interpretation of Étale Morphisms over General Commutative RingsProof of the Adjunction for Kähler DifferentialsPrerequisitesTrivial ExtensionsComma Category $A\text{-Alg}/B$Kähler DifferentialsProof of the AdjunctionForward Direction: $\Phi: \mathrm{Hom}_{A\text{-Alg}/B}(C,B\oplus M) \to \mathrm{Hom}_{B\text{-Mod}}(\Omega_{C/A},M)$Reverse Direction: $\Psi: \mathrm{Hom}_{B\text{-Mod}}(\Omega_{C/A},M) \to \mathrm{Hom}_{A\text{-Alg}/B}(C,B\oplus M)$Verification of Inverse RelationshipNaturalityImplicationsConclusion1.4 Geometric Interpretation of Étale Morphisms over General Commutative Rings

2 Étale Morphisms over Fields

Now we focus on the case where $A=F$$A = F$ is a field. In this case, $X=\mathrm{Spec}(F)$$X = \operatorname{Spec}(F)$ has only one point (the closed point, corresponding to the zero ideal $(0)$$(0)$).

2.1 Analysis of Field Extensions

Let $A=F$$A = F$ be a field and $B$$B$ a finitely generated algebra over $F$$F$. We analyze the conditions for $f:\mathrm{Spec}(B)\to \mathrm{Spec}(F)$$f: \operatorname{Spec}(B) \to \operatorname{Spec}(F)$ to be étale.

2.1.1 Finite Type Condition

Since $F$$F$ is a field, any finitely generated $F$$F$-algebra automatically satisfies the finite type condition.

2.1.2 Flatness Condition

If $f$$f$ is flat, then $B$$B$ cannot have $F$$F$-torsion elements. Since $F$$F$ is a field, this means that zero divisors in $B$$B$ cannot occur in the image of invertible elements in $F$$F$. In particular, if $B$$B$ is an integral domain, then $B$$B$ must be a torsion-free $F$$F$-module, which is always satisfied in the case of fields.

For finitely generated algebras $B$$B$ over a field $F$$F$, flatness is equivalent to being torsion-free, which is further equivalent to $B$$B$ not containing $F$$F$-linear zero divisors. This implicitly means that $B$$B$ is a free module over $F$$F$.

2.1.3 Unramified Condition

The unramified condition ${\mathrm{\Omega }}_{B/F}^1=0$$\Omega^1_{B/F} = 0$ is crucial. $B{\otimes }_{F}F\cong B\cong \prod _{i=1}^r{L}_i$$B \otimes_F F \cong B \cong \prod_{i=1}^r L_i$ where each $L_i$$L_i$ is a finite separable field extension of $F$$F$.

Indeed, a $F$$F$-algebra $B$$B$ is etale iff it is finite product of finite separable field extension of $F$$F$. Hence the category of etale $F$$F$-algebra is the finite product completion of finite separable field extension of $F$$F$.

2.2 Detailed Computation of Separability and Kähler Differentials for Field Extensions

To gain a deeper understanding of the relationship between separability of field extensions and Kähler differentials, we provide detailed computations:

Theorem: Let $L/F$$L/F$ be a finite field extension. Then $L/F$$L/F$ is separable if and only if ${\mathrm{\Omega }}_{L/F}^1=0$$\Omega^1_{L/F} = 0$.

Proof: First, assume that $L=F(\alpha )$$L = F(\alpha)$ is a simple extension, with $\alpha$$\alpha$ having minimal polynomial $P(X)\in F[X]$$P(X) \in F[X]$. We have $L\cong F[X]/(P(X))$$L \cong F[X]/(P(X))$.

For such a quotient ring, it can be proven that ${\mathrm{\Omega }}_{L/F}^1\cong L\cdot dX/(P^{\prime }(\alpha )\cdot dX)\cong L/(P^{\prime }(\alpha ))$$\Omega^1_{L/F} \cong L \cdot dX / (P'(\alpha) \cdot dX) \cong L/(P'(\alpha))$ where $P^{\prime }$$P'$ is the formal derivative of $P$$P$, and $dX$$dX$ represents the formal differential of $X$$X$.

• If $L/F$$L/F$ is separable, then by the definition of separability, $P^{\prime }(\alpha )\ne 0$$P'(\alpha) \neq 0$. Therefore, $(P^{\prime }(\alpha ))=L$$(P'(\alpha)) = L$ (since $L$$L$ is a field), so ${\mathrm{\Omega }}_{L/F}^1=L/(P^{\prime }(\alpha ))=0$$\Omega^1_{L/F} = L/(P'(\alpha)) = 0$.

• If $L/F$$L/F$ is not separable, then by the definition of inseparability, $P^{\prime }(\alpha )=0$$P'(\alpha) = 0$ (this happens only in characteristic $p>0$$p > 0$, where $P(X)$$P(X)$ has the form $Q(X^p)$$Q(X^p)$). Therefore, $(P^{\prime }(\alpha ))=(0)$$(P'(\alpha)) = (0)$, so ${\mathrm{\Omega }}_{L/F}^1=L/(0)=L\ne 0$$\Omega^1_{L/F} = L/(0) = L \neq 0$.

For a general finite field extension $L/F$$L/F$, we can use the primitive element theorem for field extensions: there exists $\alpha \in L$$\alpha \in L$ such that $L=F(\alpha )$$L = F(\alpha)$, so the above proof applies to the general case as well.

Through this computation, we establish the equivalence between the separability of field extensions and the vanishing of the Kähler differential module, which is the core reason for the correspondence between étale morphisms over a field and finite separable field extensions.

3 Concrete Examples and the Correspondence with Polynomial Roots

3.1 Detailed Analysis of Quadratic Field Extensions

Consider $F=\mathbb{Q}$$F = \mathbb{Q}$ and $L=\mathbb{Q}(\sqrt{2})$$L = \mathbb{Q}(\sqrt{2})$. The minimal polynomial of $\sqrt{2}$$\sqrt{2}$ is $P(X)=X^2-2$$P(X) = X^2 - 2$, with derivative $P^{\prime }(X)=2X$$P'(X) = 2X$.

Computation of the differential module: Evaluating at $\alpha =\sqrt{2}$$\alpha = \sqrt{2}$, we get $P^{\prime }(\sqrt{2})=2\sqrt{2}\ne 0$$P'(\sqrt{2}) = 2\sqrt{2} \neq 0$. Therefore, ${\mathrm{\Omega }}_{L/F}^1\cong L/(P^{\prime }(\alpha ))=L/(2\sqrt{2})=0$$\Omega^1_{L/F} \cong L/(P'(\alpha)) = L/(2\sqrt{2}) = 0$, so $L/F$$L/F$ is étale.

Correspondence with polynomial roots: $\mathrm{Spec}(L)=\mathrm{Spec}(\mathbb{Q}(\sqrt{2}))$$\operatorname{Spec}(L) = \operatorname{Spec}(\mathbb{Q}(\sqrt{2}))$ is a single point, and this point corresponds to the root $\sqrt{2}$$\sqrt{2}$ of the polynomial $X^2-2$$X^2 - 2$. From a geometric perspective, this point represents a solution to the algebraic equation $X^2-2=0$$X^2 - 2 = 0$.

Galois theory perspective: $L/\mathbb{Q}$$L/\mathbb{Q}$ is a Galois extension, with Galois group $\mathrm{Gal}(L/\mathbb{Q})\cong \mathbb{Z}/2\mathbb{Z}$$\operatorname{Gal}(L/\mathbb{Q}) \cong \mathbb{Z}/2\mathbb{Z}$, where the generator maps $\sqrt{2}$$\sqrt{2}$ to $-\sqrt{2}$$-\sqrt{2}$.

After base change to $\mathbb{C}$$\mathbb{C}$, we get $L{\otimes }_{\mathbb{Q}}\mathbb{C}\cong \mathbb{C}[X]/(X^2-2)\cong \mathbb{C}\times \mathbb{C}$$L \otimes_\mathbb{Q} \mathbb{C} \cong \mathbb{C}[X]/(X^2-2) \cong \mathbb{C} \times \mathbb{C}$ which corresponds to two points, representing $\sqrt{2}$$\sqrt{2}$ and $-\sqrt{2}$$-\sqrt{2}$.

3.2 Detailed Analysis of Inseparable Field Extensions in Characteristic $p$$p$

Consider a field $F={\mathbb{F}}_p(t)$$F = \mathbb{F}_p(t)$ of characteristic $p$$p$ and the extension $L=F(t^{1/p})={\mathbb{F}}_p(t^{1/p})$$L = F(t^{1/p}) = \mathbb{F}_p(t^{1/p})$.

The minimal polynomial of $t^{1/p}$$t^{1/p}$ is $P(X)=X^p-t$$P(X) = X^p - t$, with derivative $P^{\prime }(X)=p{X}^{p-1}=0$$P'(X) = pX^{p-1} = 0$ (in characteristic $p$$p$).

Computation of the differential module: Since $P^{\prime }(X)=0$$P'(X) = 0$, we have $P^{\prime }(t^{1/p})=0$$P'(t^{1/p}) = 0$. Therefore ${\mathrm{\Omega }}_{L/F}^1\cong L/(P^{\prime }(t^{1/p}))=L/(0)=L\ne 0$$\Omega^1_{L/F} \cong L/(P'(t^{1/p})) = L/(0) = L \neq 0$

So $L/F$$L/F$ is not étale, which is consistent with it being an inseparable extension.

Algebraic explanation: In characteristic $p$$p$, the polynomial $X^p-t$$X^p - t$ is irreducible, but it has no formal derivative ($P^{\prime }(X)=0$$P'(X) = 0$). This reflects the "pathological" nature of the extension—it is not separable.

Geometric explanation: Although $\mathrm{Spec}(L)$$\operatorname{Spec}(L)$ is still a single point, this point has "multiplicity $p$$p$", i.e., it is a "thick point" representing the degenerate case of a $p$$p$-fold root. Geometrically, this corresponds to a non-simple root, or a point with "ramification".

3.3 Detailed Analysis of Direct Products of Finite Field Extensions

Consider $B=\mathbb{Q}(\sqrt{2})\times \mathbb{Q}(\sqrt{3})$$B = \mathbb{Q}(\sqrt{2}) \times \mathbb{Q}(\sqrt{3})$.

Computation of the differential module: Since the Kähler differential is distributive over direct products, ${\mathrm{\Omega }}_{B/\mathbb{Q}}^1={\mathrm{\Omega }}_{\mathbb{Q}(\sqrt{2})/\mathbb{Q}}^1\oplus {\mathrm{\Omega }}_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}}^1$$\Omega^1_{B/\mathbb{Q}} = \Omega^1_{\mathbb{Q}(\sqrt{2})/\mathbb{Q}} \oplus \Omega^1_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}}$

We've already shown that ${\mathrm{\Omega }}_{\mathbb{Q}(\sqrt{2})/\mathbb{Q}}^1=0$$\Omega^1_{\mathbb{Q}(\sqrt{2})/\mathbb{Q}} = 0$ and ${\mathrm{\Omega }}_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}}^1=0$$\Omega^1_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}} = 0$ (since they are both separable extensions), so ${\mathrm{\Omega }}_{B/\mathbb{Q}}^1=0\oplus 0=0$$\Omega^1_{B/\mathbb{Q}} = 0 \oplus 0 = 0$

Therefore, $\mathrm{Spec}(B)\to \mathrm{Spec}(\mathbb{Q})$$\operatorname{Spec}(B) \to \operatorname{Spec}(\mathbb{Q})$ is a finite étale morphism.

Algebraic explanation: By the Chinese remainder theorem, $B\cong \mathbb{Q}[X]/((X^2-2)(X^2-3))\cong \mathbb{Q}[X]/(X^2-2)\times \mathbb{Q}[X]/(X^2-3)\cong \mathbb{Q}(\sqrt{2})\times \mathbb{Q}(\sqrt{3})$$B \cong \mathbb{Q}[X]/((X^2 - 2)(X^2 - 3)) \cong \mathbb{Q}[X]/(X^2 - 2) \times \mathbb{Q}[X]/(X^2 - 3) \cong \mathbb{Q}(\sqrt{2}) \times \mathbb{Q}(\sqrt{3})$

Geometric explanation: $\mathrm{Spec}(B)$$\operatorname{Spec}(B)$ consists of two discrete points, each corresponding to an irreducible factor of the polynomial $(X^2-2)(X^2-3)=X^4-5X^2+6$$(X^2 - 2)(X^2 - 3) = X^4 - 5X^2 + 6$, namely $X^2-2$$X^2 - 2$ and $X^2-3$$X^2 - 3$. The residue field at each point is $\mathbb{Q}(\sqrt{2})$$\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$$\mathbb{Q}(\sqrt{3})$, respectively.

After base change to $\mathbb{C}$$\mathbb{C}$, $B{\otimes }_{\mathbb{Q}}\mathbb{C}\cong \mathbb{C}[X]/((X^2-2)(X^2-3))\cong {\mathbb{C}}^4$$B \otimes_\mathbb{Q} \mathbb{C} \cong \mathbb{C}[X]/((X^2 - 2)(X^2 - 3)) \cong \mathbb{C}^4$ corresponds to four points, representing $\sqrt{2}$$\sqrt{2}$, $-\sqrt{2}$$-\sqrt{2}$, $\sqrt{3}$$\sqrt{3}$, and $-\sqrt{3}$$-\sqrt{3}$.

3.4 Detailed Analysis of Cubic Polynomials

Consider $F=\mathbb{Q}$$F = \mathbb{Q}$ and the polynomial $P(X)=X^3-3X-1$$P(X) = X^3 - 3X - 1$. This polynomial is irreducible over $\mathbb{Q}$$\mathbb{Q}$, but it has one real root and two complex conjugate roots in $\mathbb{R}$$\mathbb{R}$.

Computation of the differential module: Let $\alpha$$\alpha$ be a root of $P(X)$$P(X)$, then $P^{\prime }(X)=3X^2-3$$P'(X) = 3X^2 - 3$, so $P^{\prime }(\alpha )=3{\alpha }^2-3$$P'(\alpha) = 3\alpha^2 - 3$.

We can prove that $P^{\prime }(\alpha )\ne 0$$P'(\alpha) \neq 0$: if $P^{\prime }(\alpha )=0$$P'(\alpha) = 0$, then ${\alpha }^2=1$$\alpha^2 = 1$, but substituting into $P(X)$$P(X)$ gives $P(\pm 1)=\pm 1-3(\pm 1)-1=\mp 3-2\ne 0$$P(\pm 1) = \pm 1 - 3(\pm 1) - 1 = \mp 3 - 2 \neq 0$, a contradiction.

Therefore ${\mathrm{\Omega }}_{L/\mathbb{Q}}^1=0$$\Omega^1_{L/\mathbb{Q}} = 0$, where $L=\mathbb{Q}(\alpha )$$L = \mathbb{Q}(\alpha)$, so $L/\mathbb{Q}$$L/\mathbb{Q}$ is étale.

Geometric explanation: $\mathrm{Spec}(L)$$\operatorname{Spec}(L)$ has only one point, corresponding to the cubic field extension $L=\mathbb{Q}(\alpha )$$L = \mathbb{Q}(\alpha)$.

After base change to $\mathbb{R}$$\mathbb{R}$, $L{\otimes }_{\mathbb{Q}}\mathbb{R}\cong \mathbb{R}[X]/(P(X))\cong \mathbb{R}\times \mathbb{C}$$L \otimes_\mathbb{Q} \mathbb{R} \cong \mathbb{R}[X]/(P(X)) \cong \mathbb{R} \times \mathbb{C}$ because $P(X)$$P(X)$ has one real root and one pair of complex conjugate roots in $\mathbb{R}$$\mathbb{R}$.

After base change to $\mathbb{C}$$\mathbb{C}$, $L{\otimes }_{\mathbb{Q}}\mathbb{C}\cong \mathbb{C}[X]/(P(X))\cong {\mathbb{C}}^3$$L \otimes_\mathbb{Q} \mathbb{C} \cong \mathbb{C}[X]/(P(X)) \cong \mathbb{C}^3$ because $P(X)$$P(X)$ factors into three linear factors over $\mathbb{C}$$\mathbb{C}$.

This example demonstrates how base change can "split" a field extension, revealing its root structure.

3.5 Example of a Non-Simple Field Extension

Consider $F=\mathbb{Q}$$F = \mathbb{Q}$ and the algebra $B=\mathbb{Q}[X,Y]/(X^2-2,XY)$$B = \mathbb{Q}[X,Y]/(X^2 - 2, XY)$.

Structure analysis: Let $x$$x$ and $y$$y$ be the images of $X$$X$ and $Y$$Y$ in $B$$B$, respectively. Note that $x^2=2$$x^2 = 2$ and $xy=0$$xy = 0$.

We can represent $B$$B$ as $B=\mathbb{Q}[x,y]/(xy)$$B = \mathbb{Q}[x,y]/(xy)$ where $x^2=2$$x^2 = 2$. This is not a direct product of fields, because it contains zero divisors (since $xy=0$$xy = 0$ but $x\ne 0$$x \neq 0$ and $y\ne 0$$y \neq 0$).

Computation of the differential module: Since $x^2=2$$x^2 = 2$, we have $2x\cdot dx=0$$2x \cdot dx = 0$, i.e., $x\cdot dx=0$$x \cdot dx = 0$. Since $xy=0$$xy = 0$, we have $x\cdot dy+y\cdot dx=0$$x \cdot dy + y \cdot dx = 0$.

Since $x\ne 0$$x \neq 0$ and $x^2=2$$x^2 = 2$, $x$$x$ is invertible in $B$$B$. Therefore, from $x\cdot dx=0$$x \cdot dx = 0$, we get $dx=0$$dx = 0$. Then from $x\cdot dy+y\cdot dx=0$$x \cdot dy + y \cdot dx = 0$ and $dx=0$$dx = 0$, we get $x\cdot dy=0$$x \cdot dy = 0$, and by the invertibility of $x$$x$, we get $dy=0$$dy = 0$.

Therefore ${\mathrm{\Omega }}_{B/\mathbb{Q}}^1=0$$\Omega^1_{B/\mathbb{Q}} = 0$, so $B$$B$ is an étale algebra over $\mathbb{Q}$$\mathbb{Q}$.

Further analysis: We can decompose $B$$B$ as $B\cong \mathbb{Q}(\sqrt{2})\times \mathbb{Q}(\sqrt{2})/(y)$$B \cong \mathbb{Q}(\sqrt{2}) \times \mathbb{Q}(\sqrt{2})/(y)$. This is not a direct product of fields, but it is an extension construction based on $\mathbb{Q}(\sqrt{2})$$\mathbb{Q}(\sqrt{2})$.

This example illustrates the complexity of étale algebras, which are not necessarily direct products of field extensions but still maintain a certain "unramified" property.

4 Connection between Étale Morphisms and Galois Theory

4.1 The Absolute Galois Group

Before introducing the fiber functor, we first need to clearly define the concept of the absolute Galois group, which will play a central role throughout the theory.

Definition (Absolute Galois Group): Let $F$$F$ be a base field, and $F^{\text{sep}}$$F^{\text{sep}}$ a fixed separable closure (or a separably closed subfield within the algebraic closure). The absolute Galois group is denoted by $G_F=\mathrm{Gal}(F^{\text{sep}}/F)={\mathrm{Aut}}_F(F^{\text{sep}})$$G_F = \operatorname{Gal}(F^{\text{sep}}/F) = \operatorname{Aut}_F(F^{\text{sep}})$ which is the group of all automorphisms that map $F^{\text{sep}}$$F^{\text{sep}}$ to itself as an $F$$F$-algebra. It naturally forms a group with composition of mappings and inverse mappings as the group operations.

The absolute Galois group $G_F$$G_F$ is a topological group, with the topology known as the Krull topology: the basic open sets are given by $\mathrm{Gal}(E/F)$$\operatorname{Gal}(E/F)$ where $E/F$$E/F$ is a finite Galois extension within $F^{\text{sep}}$$F^{\text{sep}}$. Under this topology, $G_F$$G_F$ is a compact group.

Properties:

1. For any finite Galois extension $E$$E$ of $F$$F$, we have a natural surjective homomorphism $G_F\to \mathrm{Gal}(E/F)$$G_F \to \operatorname{Gal}(E/F)$, whose kernel is the closed subgroup consisting of all automorphisms that fix $E$$E$.

2. $G_F$$G_F$ is the inverse limit, in the categorical sense, of the inverse system of all finite Galois extensions of $F$$F$: $G_F=\underleftarrow{lim}\mathrm{Gal}(E/F)$$G_F = \lim_{\leftarrow} \operatorname{Gal}(E/F)$ where the limit ranges over all finite Galois extensions $E$$E$ of $F$$F$.

3. $G_F$$G_F$ completely characterizes the theory of algebraic extensions of $F$$F$: there is a one-to-one correspondence between all algebraic extensions of $F$$F$ and all closed subgroups of $G_F$$G_F$ (via the fixed field and subgroup relationship).

4. From a topological perspective, $G_F$$G_F$ is profinite, i.e., it is the inverse limit of finite groups.

Examples:

• For $F=\mathbb{Q}$$F = \mathbb{Q}$, the absolute Galois group $G_{\mathbb{Q}}$$G_{\mathbb{Q}}$ is a very complicated group containing rich arithmetic information. There is currently no complete explicit description of $G_{\mathbb{Q}}$$G_{\mathbb{Q}}$.

• For $F={\mathbb{F}}_q$$F = \mathbb{F}_q$ (a finite field), $G_F$$G_F$ is isomorphic to the completed integer group $\hat{\mathbb{Z}}$$\hat{\mathbb{Z}}$, generated by the Frobenius automorphism.

• For $F=\mathbb{C}$$F = \mathbb{C}$, $G_F$$G_F$ is the trivial group, because $\mathbb{C}$$\mathbb{C}$ is already algebraically closed.

 

4.2 The Fiber Functor: Geometric Meaning and Algebraic Correspondence

4.2.1 Geometric Perspective: Fibers and Polynomial Roots

The fiber functor is fundamentally a geometric concept that directly corresponds to the notion of "fiber" in algebraic geometry. Given a morphism $f:Y\to X$$f: Y \to X$ and a point $x\in X$$x \in X$, the fiber $f^{-1}(x)$$f^{-1}(x)$ is defined as the set of all points in $Y$$Y$ that map to $x$$x$. In scheme-theoretic terms, this corresponds to the fiber product: $f^{-1}(x)=Y{\times }_{X}x$$f^{-1}(x) = Y \times_X {x}$

In the context of fields, consider a morphism $f:\mathrm{Spec}(L)\to \mathrm{Spec}(F)$$f: \operatorname{Spec}(L) \to \operatorname{Spec}(F)$ induced by a field extension $L/F$$L/F$, and let $x:\mathrm{Spec}(K)\to \mathrm{Spec}(F)$$x: \operatorname{Spec}(K) \to \operatorname{Spec}(F)$ be a geometric point corresponding to an embedding $F\hookrightarrow K$$F \hookrightarrow K$. The fiber $f^{-1}(x)$$f^{-1}(x)$ can be expressed as: $f^{-1}(x)=\mathrm{Spec}(L){\times }_{\mathrm{Spec}(F)}\mathrm{Spec}(K)=\mathrm{Spec}(L{\otimes }_{F}K)$$f^{-1}(x) = \operatorname{Spec}(L) \times_{\operatorname{Spec}(F)} \operatorname{Spec}(K) = \operatorname{Spec}(L \otimes_F K)$

$$\begin{array}{ccc}\mathrm{Spec}(L{\otimes }_{F}K)& \stackrel{}{\to }& \mathrm{Spec}(L)\\ \downarrow & & \downarrow f& \\ \mathrm{Spec}(K)& \stackrel{x}{\to }& \mathrm{Spec}(F)\end{array}$$

This is precisely the geometric object that the fiber functor $F_x$$F_x$ aims to describe.

To illustrate this concept, consider a concrete example: the polynomial $P(X)=X^2-2$$P(X) = X^2 - 2$ defining the field extension $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$$\mathbb{Q}(\sqrt{2})/\mathbb{Q}$. From a geometric perspective:

• $\mathrm{Spec}(\mathbb{Q})$$\operatorname{Spec}(\mathbb{Q})$ is a single point $x$$x$

• $\mathrm{Spec}(\mathbb{Q}(\sqrt{2}))$$\operatorname{Spec}(\mathbb{Q}(\sqrt{2}))$ is also a single point $y$$y$

• The morphism $f:\mathrm{Spec}(\mathbb{Q}(\sqrt{2}))\to \mathrm{Spec}(\mathbb{Q})$$f: \operatorname{Spec}(\mathbb{Q}(\sqrt{2})) \to \operatorname{Spec}(\mathbb{Q})$ maps $y$$y$ to $x$$x$

For a geometric point $x^{\prime }:\mathrm{Spec}(\mathbb{C})\to \mathrm{Spec}(\mathbb{Q})$$x': \operatorname{Spec}(\mathbb{C}) \to \operatorname{Spec}(\mathbb{Q})$, the fiber is: $f^{-1}(x^{\prime })=\mathrm{Spec}(\mathbb{Q}(\sqrt{2})){\times }_{\mathrm{Spec}(\mathbb{Q})}\mathrm{Spec}(\mathbb{C})=\mathrm{Spec}(\mathbb{Q}(\sqrt{2}){\otimes }_{\mathbb{Q}}\mathbb{C})$$f^{-1}(x') = \operatorname{Spec}(\mathbb{Q}(\sqrt{2})) \times_{\operatorname{Spec}(\mathbb{Q})} \operatorname{Spec}(\mathbb{C}) = \operatorname{Spec}(\mathbb{Q}(\sqrt{2}) \otimes_{\mathbb{Q}} \mathbb{C})$

Computing this tensor product, we get: $\mathbb{Q}(\sqrt{2}){\otimes }_{\mathbb{Q}}\mathbb{C}\cong \mathbb{C}[X]/(X^2-2)\cong \mathbb{C}\times \mathbb{C}$$\mathbb{Q}(\sqrt{2}) \otimes_{\mathbb{Q}} \mathbb{C} \cong \mathbb{C}[X]/(X^2-2) \cong \mathbb{C} \times \mathbb{C}$

Therefore, $f^{-1}(x^{\prime })=\mathrm{Spec}(\mathbb{C}\times \mathbb{C})\cong {\alpha }_1,{\alpha }_2$$f^{-1}(x') = \operatorname{Spec}(\mathbb{C} \times \mathbb{C}) \cong {\alpha_1, \alpha_2}$, where ${\alpha }_1=\sqrt{2}$$\alpha_1 = \sqrt{2}$ and ${\alpha }_2=-\sqrt{2}$$\alpha_2 = -\sqrt{2}$ are the two roots of the polynomial $X^2-2$$X^2 - 2$ in $\mathbb{C}$$\mathbb{C}$.

Geometric interpretation: The single point $\mathrm{Spec}(\mathbb{Q}(\sqrt{2}))$$\operatorname{Spec}(\mathbb{Q}(\sqrt{2}))$ splits into two points in the fiber over the $\mathbb{C}$$\mathbb{C}$-point, precisely corresponding to the two roots of the polynomial $X^2-2=0$$X^2 - 2 = 0$. The fiber functor is the formal tool that captures this fiber structure.

4.2.2 The Fiber Functor: A Geometric Definition

The fiber functor can be most naturally understood in geometric terms, starting directly from the fiber of an étale morphism.

Definition (Fiber Functor - Geometric Approach): Let $F$$F$ be a field and $x:\mathrm{Spec}(K)\to \mathrm{Spec}(F)$$x: \operatorname{Spec}(K) \to \operatorname{Spec}(F)$ a geometric point corresponding to an embedding $F\hookrightarrow K$$F \hookrightarrow K$. For any finite étale $F$$F$-algebra $B$$B$ with the associated morphism $f:\mathrm{Spec}(B)\to \mathrm{Spec}(F)$$f: \operatorname{Spec}(B) \to \operatorname{Spec}(F)$, the fiber of $f$$f$ over $x$$x$ is:

$$F_x(B)=f^{-1}(x)=\mathrm{Spec}(B){\times }_{\mathrm{Spec}(F)}\mathrm{Spec}(K)=\mathrm{Spec}(B{\otimes }_{F}K)=\mathrm{Spec}(K^n)={\mathrm{Hom}}_K(K^n,K)$$

Proposition. $F_x(B)\cong {\mathrm{Hom}}_F(B,K)$$F_x(B)\cong\mathrm{Hom}_F(B,K)$.

Proof. We only need to prove

$${\mathrm{Hom}}_F(B,K)\cong {\mathrm{Hom}}_K(B{\otimes }_{F}K,K)$$

This is the adjoint between extension of scalar and restriction of scalar. See https://en.wikipedia.org/wiki/Change_of_rings $◻$$\square$

Hence we have

$$F_x(B)={\mathrm{Hom}}_F(\mathrm{Spec}K,\mathrm{Spec}B)\cong {\mathrm{Hom}}_F(B,K)$$

We define $F_x$$F_x$ to be the functor ${\mathrm{Hom}}_F(-,K):F$$\mathrm{Hom}_F(-,K):F$-$\mathrm{Alg}\to \mathrm{Set}$$\mathrm{Alg}\to\mathrm{Set}$.

It is a contravarient functor due to the algebra-geometry duality.

 

4.2.3 Structure of the Fiber and Correspondence with Polynomial Root

Let's analyze in more detail the structure of the fiber and its precise correspondence with polynomial roots. Given a field extension $L/F$$L/F$ and a geometric point $x:\mathrm{Spec}(K)\to \mathrm{Spec}(F)$$x: \operatorname{Spec}(K) \to \operatorname{Spec}(F)$, the structure of the fiber $f^{-1}(x)$$f^{-1}(x)$ is:

Theorem: If $L/F$$L/F$ is a finite separable field extension and $K$$K$ is a separate closure containing $F$$F$, then: $f^{-1}(x)=\mathrm{Spec}(L{\otimes }_{F}K)\cong \mathrm{Spec}(K^{[L:F]})$$f^{-1}(x) = \operatorname{Spec}(L \otimes_F K) \cong \operatorname{Spec}(K^{[L:F]})$ That is, the fiber consists of $[L:F]$$[L:F]$ $K$$K$-points.

Proof:

By the Primitive element theorem, every finite separate extension is simple, hence we could assume $L=F(\alpha )$$L = F(\alpha)$ where $\alpha$$\alpha$ has minimal polynomial $P(X)\in F[X]$$P(X) \in F[X]$, so $L\cong F[X]/(P(X))$$L \cong F[X]/(P(X))$. Computing the tensor product:

$$L{\otimes }_{F}K\cong F[X]/(P(X)){\otimes }_{F}K\cong K[X]/(P(X))$$

Since $K$$K$ is a separate closure containing $F$$F$, $P(X)$$P(X)$ factors completely into linear factors in $K$$K$: $P(X)=\prod _{i=1}^n(X-{\alpha }_i)$$P(X) = \prod_{i=1}^n (X - \alpha_i)$ where ${\alpha }_i\in K$$\alpha_i \in K$ are all the roots of $P(X)$$P(X)$ and $n=[L:F]$$n = [L:F]$.

By the Chinese Remainder Theorem: $K[X]/(P(X))\cong \prod _{i=1}^{n}K[X]/(X-{\alpha }_i)\cong K^n$$K[X]/(P(X)) \cong \prod_{i=1}^n K[X]/(X - \alpha_i) \cong K^n$

Therefore: $f^{-1}(x)=\mathrm{Spec}(L{\otimes }_{F}K)\cong \mathrm{Spec}(K^n)=\{{\mathfrak{p}}_1,{\mathfrak{p}}_2,...,{\mathfrak{p}}_n\}\cong \{{\pi }_1,{\pi }_2,\dots ,{\pi }_n\}$$f^{-1}(x) = \operatorname{Spec}(L \otimes_F K) \cong \operatorname{Spec}(K^n) =\{\mathfrak p_1,\mathfrak p_2,...,\mathfrak p_n\} \cong\{\pi_1, \pi_2, \ldots, \pi_n\}$

where each point ${\pi }_i$$\pi_i$ corresponds to a projection from $K[X]/(P(X))$$K[X]/(P(X))$ to $K$$K$, essentially corresponding to a root ${\alpha }_i$$\alpha_i$ of the polynomial $P(X)$$P(X)$ in $K$$K$. $◻$$\square$

Correspondence between fiber functor and fiber points: The fiber functor $F_x(L)$$F_x(L)$ precisely describes the $K$$K$-valued points of this fiber: $F_x(L)={\mathrm{Hom}}_F(L,K)={\phi }_1,{\phi }_2,\dots ,{\phi }_n$$F_x(L) = \operatorname{Hom}_F(L, K) = {\varphi_1, \varphi_2, \ldots, \varphi_n}$

Here each homomorphism ${\phi }_i:L\to K$$\varphi_i: L \to K$ is determined by ${\phi }_i(\alpha )={\alpha }_i$$\varphi_i(\alpha) = \alpha_i$, where ${\alpha }_i$$\alpha_i$ is a root of $P(X)$$P(X)$ in $K$$K$. Thus, the elements of $F_x(L)$$F_x(L)$ correspond one-to-one with the points in the fiber $f^{-1}(x)$$f^{-1}(x)$, and also one-to-one with the roots of the polynomial $P(X)$$P(X)$ in $K$$K$.

In this sense, the fiber functor $F_x$$F_x$ perfectly captures the geometric structure of the fiber, translating it into an algebraic structure (a set of homomorphisms).

4.3 The isomorphism between Étale Fundamental Group and Absolute Galois Group via Yoneda Lemma.

Before discussing the Galois theory correspondence, we need to formally introduce the concept of the étale fundamental group, which is a key construction that generalizes the topological fundamental group to algebraic geometry.

The Fundamental Group as the Automorphism Group of a Fiber Functor: Proof and Analysis

Introduction

In algebraic topology and algebraic geometry, the fundamental group has several equivalent definitions. This paper explores and proves a profound result: the topological fundamental group ${\pi }_1(X,x)$$\pi_1(X,x)$ can be equivalently defined as the automorphism group $\text{Aut}(F_x)$$\text{Aut}(F_x)$ of the fiber functor $F_x:\text{Cov}(X)\to \text{Sets}$$F_x: \text{Cov}(X) \to \text{Sets}$. This categorical perspective not only unifies the definitions of the topological fundamental group and the étale fundamental group but also reveals deep connections between the fundamental group and Galois theory.

Prerequisites

Remark. There is some more natural way to understand this fact, however, it will make this eassy become too long.

Covering Spaces and Fiber Functor

Given a topological space $X$$X$ and a basepoint $x\in X$$x \in X$, we define:

• $\text{Cov}(X)$$\text{Cov}(X)$ is the category of covering spaces of $X$$X$, where objects are covering maps $\pi :Y\to X$$\pi: Y \to X$, and morphisms are continuous maps that preserve the covering maps

• The fiber functor $F_x:\text{Cov}(X)\to \text{Sets}$$F_x: \text{Cov}(X) \to \text{Sets}$ maps a covering map $\pi :Y\to X$$\pi: Y \to X$ to its fiber ${\pi }^{-1}(x)$$\pi^{-1}(x)$ at $x$$x$

Fundamental Group Action

For a covering map $\pi :Y\to X$$\pi: Y \to X$ and a closed path $\gamma :[0,1]\to X$$\gamma: [0,1] \to X$ satisfying $\gamma (0)=\gamma (1)=x$$\gamma(0) = \gamma(1) = x$:

• For any $y\in {\pi }^{-1}(x)$$y \in \pi^{-1}(x)$, $\gamma$$\gamma$ can be uniquely lifted to a path $\tilde{\gamma }:[0,1]\to Y$$\tilde{\gamma}: [0,1] \to Y$ satisfying $\tilde{\gamma }(0)=y$$\tilde{\gamma}(0) = y$ and $\pi \circ \tilde{\gamma }=\gamma$$\pi \circ \tilde{\gamma} = \gamma$

• We define the action of $\gamma$$\gamma$ on $y$$y$ as: $\gamma .y=\tilde{\gamma }(1)$$\gamma.y = \tilde{\gamma}(1)$

• This defines a left action of ${\pi }_1(X,x)$$\pi_1(X,x)$ on ${\pi }^{-1}(x)$$\pi^{-1}(x)$

Main Theorem

Theorem: For a sufficiently nice topological space $X$$X$ (locally path-connected and semi-locally simply connected) and a basepoint $x\in X$$x \in X$, there exists a natural isomorphism: ${\pi }_1(X,x)\cong \text{Aut}(F_x)$$\pi_1(X,x) \cong \text{Aut}(F_x)$

Proof

We will construct an isomorphism $\mathrm{\Phi }:{\pi }_1(X,x)\to \text{Aut}(F_x)$$\Phi: \pi_1(X,x) \to \text{Aut}(F_x)$ and prove it is bijective.

Step 1: Constructing the Map $\mathrm{\Phi }$$\Phi$

For each element $[\gamma ]\in {\pi }_1(X,x)$$[\gamma] \in \pi_1(X,x)$ (where $\gamma$$\gamma$ is a closed path with endpoints at $x$$x$), we define a natural transformation $\mathrm{\Phi }([\gamma ])$$\Phi([\gamma])$ as follows:

For any covering $(Y,\pi )$$(Y,\pi)$ and any point $y\in {\pi }^{-1}(x)$$y \in \pi^{-1}(x)$, define: $\mathrm{\Phi }([\gamma ]{)}_{(Y,\pi )}(y)=\gamma .y$$\Phi([\gamma])_{(Y,\pi)}(y) = \gamma.y$

Step 2: Verifying $\mathrm{\Phi }([\gamma ])$$\Phi([\gamma])$ is a Natural Transformation

Given a morphism of covering spaces $f:(Y_1,{\pi }_1)\to (Y_2,{\pi }_2)$$f: (Y_1,\pi_1) \to (Y_2,\pi_2)$, we need to prove that the following diagram commutes:

$$\begin{array}{ccc}{F}_x(Y_1,{\pi }_1)& \stackrel{\mathrm{\Phi }([\gamma ]{)}_{(Y_1,{\pi }_1)}}{\to }& F_x(Y_1,{\pi }_1)\\ F_x(f)\downarrow & & \downarrow F_x(f)& \\ F_x(Y_2,{\pi }_2)& \underset{\mathrm{\Phi }([\gamma ]{)}_{(Y_2,{\pi }_2)}}{\overset{}{\to }}& F_x(Y_2,{\pi }_2)\end{array}$$

For any $y\in {\pi }_1^{-1}(x)$$y \in \pi_1^{-1}(x)$:

• Left side: $F_x(f)(\mathrm{\Phi }([\gamma ]{)}_{(Y_1,{\pi }_1)}(y))=f(\gamma .y)$$F_x(f)(\Phi([\gamma])_{(Y_1,\pi_1)}(y)) = f(\gamma.y)$

• Right side: $\mathrm{\Phi }([\gamma ]{)}_{(Y_2,{\pi }_2)}(F_x(f)(y))=\mathrm{\Phi }([\gamma ]{)}_{(Y_2,{\pi }_2)}(f(y))=\gamma .(f(y))$$\Phi([\gamma])_{(Y_2,\pi_2)}(F_x(f)(y)) = \Phi([\gamma])_{(Y_2,\pi_2)}(f(y)) = \gamma.(f(y))$

Since $f$$f$ is a morphism of covering spaces, it commutes with path lifting, thus $f(\gamma .y)=\gamma .(f(y))$$f(\gamma.y) = \gamma.(f(y))$. This proves that $\mathrm{\Phi }([\gamma ])$$\Phi([\gamma])$ is a natural transformation.

Step 3: Verifying $\mathrm{\Phi }$$\Phi$ is a Group Homomorphism

We need to prove:

1. $\mathrm{\Phi }([e_x])={\text{id}}_{F_x}$$\Phi([e_x]) = \text{id}_{F_x}$, where $e_x$$e_x$ is the constant path at $x$$x$

2. $\mathrm{\Phi }([\gamma ]\cdot [\delta ])=\mathrm{\Phi }([\gamma ])\circ \mathrm{\Phi }([\delta ])$$\Phi([\gamma] \cdot [\delta]) = \Phi([\gamma]) \circ \Phi([\delta])$

For the first point, clearly the lifting of the constant path preserves the starting point, thus $\mathrm{\Phi }([e_x]{)}_{(Y,\pi )}(y)=y$$\Phi([e_x])_{(Y,\pi)}(y) = y$ holds for all $y\in {\pi }^{-1}(x)$$y \in \pi^{-1}(x)$.

For the second point, consider the composite path $[\gamma ]\cdot [\delta ]$$[\gamma] \cdot [\delta]$ acting on a point $y\in {\pi }^{-1}(x)$$y \in \pi^{-1}(x)$:

• $\mathrm{\Phi }([\gamma ]\cdot [\delta ]{)}_{(Y,\pi )}(y)=(\gamma \cdot \delta ).y$$\Phi([\gamma] \cdot [\delta])_{(Y,\pi)}(y) = (\gamma \cdot \delta).y$

• $(\mathrm{\Phi }([\gamma ])\circ \mathrm{\Phi }([\delta ]){)}_{(Y,\pi )}(y)=\mathrm{\Phi }([\gamma ])\ast (Y,\pi )(\mathrm{\Phi }([\delta ]{)}_{(Y,\pi )}(y))=\mathrm{\Phi }([\gamma ]{)}_{(Y,\pi )}(\delta .y)=\gamma .(\delta .y)$$(\Phi([\gamma]) \circ \Phi([\delta]))_{(Y,\pi)}(y) = \Phi([\gamma])*{(Y,\pi)}(\Phi([\delta])_{(Y,\pi)}(y)) = \Phi([\gamma])_{(Y,\pi)}(\delta.y) = \gamma.(\delta.y)$

By the uniqueness of path lifting, $(\gamma \cdot \delta ).y=\gamma .(\delta .y)$$(\gamma \cdot \delta).y = \gamma.(\delta.y)$. Therefore $\mathrm{\Phi }$$\Phi$ is a group homomorphism.

Step 4: Proving $\mathrm{\Phi }$$\Phi$ is Injective

Suppose $\mathrm{\Phi }([\gamma ])={\text{id}}_{F_x}$$\Phi([\gamma]) = \text{id}_{F_x}$, we need to prove that $[\gamma ]=[e_x]$$[\gamma] = [e_x]$.

Consider the universal covering space $(\tilde{U},p)$$(\tilde{U},p)$ and a point $\tilde{u}$$\tilde{u}$ in its fiber $p^{-1}(x)$$p^{-1}(x)$. Since $\mathrm{\Phi }([\gamma ])=\text{id}\ast F_x$$\Phi([\gamma]) = \text{id}*{F_x}$, we have: $\mathrm{\Phi }([\gamma ])\ast (\tilde{U},p)(\tilde{u})=\tilde{u}$$\Phi([\gamma])*{(\tilde{U},p)}(\tilde{u}) = \tilde{u}$

This means $\gamma .\tilde{u}=\tilde{u}$$\gamma.\tilde{u} = \tilde{u}$. But in the universal covering space, only the trivial path class $[e_x]$$[e_x]$ maps any lifted point to itself. Therefore $[\gamma ]=[e_x]$$[\gamma] = [e_x]$, proving that $\mathrm{\Phi }$$\Phi$ is injective.

Step 5: Proving $\mathrm{\Phi }$$\Phi$ is Surjective

Given $\alpha \in \text{Aut}(F_x)$$\alpha \in \text{Aut}(F_x)$, we need to find $[\gamma ]\in {\pi }_1(X,x)$$[\gamma] \in \pi_1(X,x)$ such that $\mathrm{\Phi }([\gamma ])=\alpha$$\Phi([\gamma]) = \alpha$.

Key observation: Consider the universal covering space $(\tilde{U},p)$$(\tilde{U},p)$ and a chosen basepoint $\tilde{x}\in p^{-1}(x)$$\tilde{x} \in p^{-1}(x)$. There is a natural bijection between $p^{-1}(x)$$p^{-1}(x)$ and the fundamental group ${\pi }_1(X,x)$$\pi_1(X,x)$ (via endpoints of lifted paths starting from $\tilde{x}$$\tilde{x}$).

Thus, the natural transformation ${\alpha }_{(\tilde{U},p)}$$\alpha_{(\tilde{U},p)}$ maps $\tilde{x}$$\tilde{x}$ to some point ${\tilde{x}}^{\prime }\in p^{-1}(x)$$\tilde{x}' \in p^{-1}(x)$, corresponding to some $[\gamma ]\in {\pi }_1(X,x)$$[\gamma] \in \pi_1(X,x)$.

We can prove that this $[\gamma ]$$[\gamma]$ satisfies $\mathrm{\Phi }([\gamma ])=\alpha$$\Phi([\gamma]) = \alpha$. The proof involves verifying that for any covering space $(Y,\pi )$$(Y,\pi)$ and any $y\in {\pi }^{-1}(x)$$y \in \pi^{-1}(x)$, we have: $\mathrm{\Phi }([\gamma ]{)}_{(Y,\pi )}(y)={\alpha }_{(Y,\pi )}(y)$$\Phi([\gamma])_{(Y,\pi)}(y) = \alpha_{(Y,\pi)}(y)$

This verification uses:

1. Any covering space can be obtained from the universal covering space via quotient maps

2. Natural transformations must be compatible with these quotient maps

3. The definition of the fundamental group action

Therefore $\mathrm{\Phi }$$\Phi$ is surjective.

Conclusion

We have proven that $\mathrm{\Phi }:{\pi }_1(X,x)\to \text{Aut}(F_x)$$\Phi: \pi_1(X,x) \to \text{Aut}(F_x)$ is a group isomorphism, thus: ${\pi }_1(X,x)\cong \text{Aut}(F_x)$$\pi_1(X,x) \cong \text{Aut}(F_x)$

Étale Fundamental Group

Definition (Étale Fundamental Group): Let $X$$X$ be a connected Noetherian scheme, and $x:\mathrm{Spec}(\mathrm{\Omega })\to X$$x: \operatorname{Spec}(\Omega) \to X$ a geometric point (where $\mathrm{\Omega }$$\Omega$ is an algebraically closed field). The étale fundamental group of $X$$X$ at $x$$x$, denoted by ${\pi }_1^{\text{ét}}(X,x)$$\pi_1^{\text{ét}}(X, x)$, is defined as the automorphism group of the fiber functor $F_x$$F_x$: ${\pi }_1^{\text{ét}}(X,x)=\mathrm{Aut}(F_x)$$\pi_1^{\text{ét}}(X, x) = \operatorname{Aut}(F_x)$ .

Here $F_x:{\mathbf{FEt}}_X\to \mathbf{FinSet}$$F_x: \mathbf{FEt}_X \to \mathbf{FinSet}$ is the fiber functor from the category of finite étale coverings of $X$$X$ to the category of finite sets, defined as $F_x(Y)={\mathrm{Hom}}_X(x,Y)$$F_x(Y) = \operatorname{Hom}_X(x, Y)$, i.e., the set of all $X$$X$-morphisms from $x$$x$ to $Y$$Y$.

 

Topological structure: The étale fundamental group ${\pi }_1^{\text{ét}}(X,x)$$\pi_1^{\text{ét}}(X, x)$ naturally carries a topological structure, making it a compact topological group. Specifically, if $Y_i$${Y_i}$ is a cofinal system of finite étale Galois coverings of $X$$X$, then ${\pi }_1^{\text{ét}}(X,x)=\underleftarrow{lim}{\mathrm{Aut}}_X(Y_i)$$\pi_1^{\text{ét}}(X, x) = \lim_{\leftarrow} \operatorname{Aut}_X(Y_i)$ is the inverse limit of finite groups, and therefore has a natural compact topology (given by the limit construction).

Analogy with the topological fundamental group: In complex analysis, the topological fundamental group ${\pi }_1(M,p)$$\pi_1(M, p)$ of a connected complex manifold $M$$M$ can be viewed as the automorphism group acting on the universal covering space $\tilde{M}$$\tilde{M}$ of $M$$M$. Similarly, the étale fundamental group ${\pi }_1^{\text{ét}}(X,x)$$\pi_1^{\text{ét}}(X, x)$ can be viewed as the automorphism group of the "algebraic universal covering," except that in algebraic geometry, this "universal covering" usually does not exist as a scheme, but as an inverse system.

Main properties:

If $X$$X$ is a smooth projective curve over an algebraically closed field $k$$k$, then ${\pi }_1^{\text{ét}}(X,x)$$\pi_1^{\text{ét}}(X, x)$ is the completion of the (classical) topological fundamental group of $X$$X$.

The functor $F_x$$F_x$ establishes a category equivalence ${\mathbf{FEt}}_X\simeq {\mathbf{FinSet}}^{{\pi }_1^{\text{ét}}(X,x)}$$\mathbf{FEt}_X \simeq \mathbf{FinSet}^{\pi_1^{\text{ét}}(X, x)}$ i.e., the category of finite étale coverings of $X$$X$ is equivalent to the category of finite sets with a ${\pi }_1^{\text{ét}}(X,x)$$\pi_1^{\text{ét}}(X, x)$-action. We will prove it soon.

If $f:Y\to X$$f: Y \to X$ is a morphism of connected schemes, $y$$y$ is a geometric point of $Y$$Y$, and $f(y)=x$$f(y) = x$, then there is a natural homomorphism ${\pi }_1^{\text{ét}}(Y,y)\to {\pi }_1^{\text{ét}}(X,x)$$\pi_1^{\text{ét}}(Y, y) \to \pi_1^{\text{ét}}(X, x)$ corresponding to the base change functor $f^{\ast }:{\mathbf{FEt}}_X\to {\mathbf{FEt}}_Y$$f^*: \mathbf{FEt}_X \to \mathbf{FEt}_Y$.

The case of fields: For the spectrum of a field $X=\mathrm{Spec}(F)$$X = \operatorname{Spec}(F)$, if we choose a geometric point $x:\mathrm{Spec}(K)\to \mathrm{Spec}(F)$$x: \operatorname{Spec}(K) \to \operatorname{Spec}(F)$ where $K=F^{\mathrm{sep}}$$K=F^{\mathrm{sep}}$ (corresponding to the inclusion map $F\to K$$F \to K$), then the étale fundamental group

$${\pi }_1^{\text{ét}}(\mathrm{Spec}(F),x)\cong \mathrm{Gal}(F^{\text{sep}}/F)=G_F$$

is precisely the absolute Galois group of $F$$F$!

The proof is easy. Notice that $F_x={\mathrm{Hom}}_F(-,K)$$F_x=\mathrm{Hom}_F(-,K)$, by Yonde lemma, $\mathrm{Aut}(F_x)\cong {\mathrm{Aut}}_F(K)=\mathrm{Gal}(K/F)=G_F$$\mathrm{Aut}(F_x)\cong\mathrm{Aut}_F(K)=\mathrm{Gal}(K/F)=G_F$.

In this case, the category equivalence ${\mathbf{FEt}}_{\mathrm{Spec}(F)}\simeq {\mathbf{FinSet}}^{G_F}$$\mathbf{FEt}_{\operatorname{Spec}(F)} \simeq \mathbf{FinSet}^{G_F}$ is the categorical form of Galois theory.

4.4 Galois Theory Correspondence

When we choose $K=F^{\text{sep}}$$K = F^{\text{sep}}$ (the separable closure of $F$$F$), we obtain the classical Galois theory correspondence:

• Finite étale coverings $\mathrm{Spec}(L)\to \mathrm{Spec}(F)$$\operatorname{Spec}(L) \to \operatorname{Spec}(F)$ correspond to finite separable field extensions $L/F$$L/F$.

• The fiber functor $F_x(L)={\mathrm{Hom}}_F(L,F^{\text{sep}})$$F_x(L) = \operatorname{Hom}_F(L, F^{\text{sep}})$ gives the set of all embeddings of $L$$L$ into $F^{\text{sep}}$$F^{\text{sep}}$, equipped with the natural action of $G_F$$G_F$.

• The étale fundamental group ${\pi }_1^{\text{ét}}(\mathrm{Spec}(F),x)=\mathrm{Aut}(F_x)$$\pi_1^{\text{ét}}(\operatorname{Spec}(F), x) = \operatorname{Aut}(F_x)$ is precisely the absolute Galois group $G_F=\mathrm{Gal}(F^{\text{sep}}/F)$$G_F = \operatorname{Gal}(F^{\text{sep}}/F)$.

This correspondence allows Galois theory to be viewed as a special case of the étale fundamental group theory for the point $\mathrm{Spec}(F)$$\operatorname{Spec}(F)$.

4.4.1 Concrete Examples of Galois Correspondence with Polynomial Roots

Let's illustrate the relationship between polynomial roots and étale coverings through several concrete Galois extension examples:

Example 1: Cyclic Quartic Extension

Consider $F=\mathbb{Q}$$F = \mathbb{Q}$ and the polynomial $P(X)=X^4-2$$P(X) = X^4 - 2$. Let $L=\mathbb{Q}(2^{1/4})$$L = \mathbb{Q}(2^{1/4})$. Then $L/\mathbb{Q}$$L/\mathbb{Q}$ is a cyclic Galois extension of degree 4.

From the perspective of polynomial roots:

• The four roots of $P(X)$$P(X)$ in $\mathbb{C}$$\mathbb{C}$ are ${\alpha }_1=2^{1/4}$$\alpha_1 = 2^{1/4}$, ${\alpha }_2=i\cdot 2^{1/4}$$\alpha_2 = i \cdot 2^{1/4}$, ${\alpha }_3=-2^{1/4}$$\alpha_3 = -2^{1/4}$, ${\alpha }_4=-i\cdot 2^{1/4}$$\alpha_4 = -i \cdot 2^{1/4}$

• These four roots correspond to the four points of $\mathrm{Spec}(L{\otimes }_{\mathbb{Q}}\mathbb{C})$$\operatorname{Spec}(L \otimes_\mathbb{Q} \mathbb{C})$

• The Galois group $\mathrm{Gal}(L/\mathbb{Q})\cong \mathbb{Z}/4\mathbb{Z}$$\operatorname{Gal}(L/\mathbb{Q}) \cong \mathbb{Z}/4\mathbb{Z}$ acts on these four roots, with the generator mapping ${\alpha }_j$$\alpha_j$ to ${\alpha }_{j+1}$$\alpha_{j+1}$ (with cyclic indexing)

Computation of Kähler differentials: $P^{\prime }(X)=4X^3$$P'(X) = 4X^3$, so $P^{\prime }(2^{1/4})=4\cdot (2^{1/4}{)}^3=4\cdot 2^{3/4}\ne 0$$P'(2^{1/4}) = 4 \cdot (2^{1/4})^3 = 4 \cdot 2^{3/4} \neq 0$. Therefore ${\mathrm{\Omega }}_{L/\mathbb{Q}}^1=0$$\Omega^1_{L/\mathbb{Q}} = 0$, which is consistent with the fact that $L/\mathbb{Q}$$L/\mathbb{Q}$ is étale.

From the perspective of the fiber functor:

• $F_x(L)={\mathrm{Hom}}_{\mathbb{Q}}(L,{\mathbb{Q}}^{\text{sep}})$$F_x(L) = \operatorname{Hom}_\mathbb{Q}(L, \mathbb{Q}^{\text{sep}})$ is precisely the set of four embeddings, each uniquely determined by which root $2^{1/4}$$2^{1/4}$ maps to

• The absolute Galois group $\mathrm{Gal}({\mathbb{Q}}^{\text{sep}}/\mathbb{Q})$$\operatorname{Gal}(\mathbb{Q}^{\text{sep}}/\mathbb{Q})$ acts on $F_x(L)$$F_x(L)$ by permuting these roots

• $\mathrm{Gal}(L/\mathbb{Q})$$\operatorname{Gal}(L/\mathbb{Q})$ is isomorphic to $\mathrm{Gal}({\mathbb{Q}}^{\text{sep}}/\mathbb{Q})/\mathrm{Gal}({\mathbb{Q}}^{\text{sep}}/L)$$\operatorname{Gal}(\mathbb{Q}^{\text{sep}}/\mathbb{Q})/\operatorname{Gal}(\mathbb{Q}^{\text{sep}}/L)$

Geometric interpretation: $\mathrm{Spec}(L)$$\operatorname{Spec}(L)$ is a single point, but this point "contains" information about all four roots of $P(X)$$P(X)$. When base changed to $\mathbb{C}$$\mathbb{C}$, this single point "splits" into four points, each corresponding to a root. The action of the Galois group reflects the symmetry of these roots.

Example 2: Non-Galois Cubic Extension

Consider $F=\mathbb{Q}$$F = \mathbb{Q}$ and the polynomial $P(X)=X^3-2$$P(X) = X^3 - 2$. Let $L=\mathbb{Q}(2^{1/3})$$L = \mathbb{Q}(2^{1/3})$. This is a non-Galois extension.

From the perspective of polynomial roots:

• The three roots of $P(X)$$P(X)$ in $\mathbb{C}$$\mathbb{C}$ are ${\alpha }_1=2^{1/3}$$\alpha_1 = 2^{1/3}$, ${\alpha }_2=2^{1/3}{e}^{2\pi i/3}$$\alpha_2 = 2^{1/3}e^{2\pi i/3}$, ${\alpha }_3=2^{1/3}{e}^{4\pi i/3}$$\alpha_3 = 2^{1/3}e^{4\pi i/3}$

• These three roots correspond to the three points of $\mathrm{Spec}(L{\otimes }_{\mathbb{Q}}\mathbb{C})$$\operatorname{Spec}(L \otimes_\mathbb{Q} \mathbb{C})$

• Since $L/\mathbb{Q}$$L/\mathbb{Q}$ is not Galois, there is no Galois group acting directly on these roots

Computation of Kähler differentials: $P^{\prime }(X)=3X^2$$P'(X) = 3X^2$, so $P^{\prime }(2^{1/3})=3\cdot (2^{1/3}{)}^2=3\cdot 2^{2/3}\ne 0$$P'(2^{1/3}) = 3 \cdot (2^{1/3})^2 = 3 \cdot 2^{2/3} \neq 0$. Therefore ${\mathrm{\Omega }}_{L/\mathbb{Q}}^1=0$$\Omega^1_{L/\mathbb{Q}} = 0$, so $L/\mathbb{Q}$$L/\mathbb{Q}$ is étale, even though it is not a Galois extension.

Galois closure: The Galois closure is $E=\mathbb{Q}(2^{1/3},\omega )$$E = \mathbb{Q}(2^{1/3}, \omega)$, where $\omega =e^{2\pi i/3}$$\omega = e^{2\pi i/3}$.

• $\mathrm{Gal}(E/\mathbb{Q})\cong S_3$$\operatorname{Gal}(E/\mathbb{Q}) \cong S_3$ (the symmetric group of order 3) acts on the three roots

• The fiber functor $F_x(E)$$F_x(E)$ gives a set of 6 embeddings, corresponding to the 6 elements of $S_3$$S_3$

• The fiber functor $F_x(L)$$F_x(L)$ still has 3 elements, but lacks the complete Galois action structure

Geometric interpretation: $\mathrm{Spec}(L)$$\operatorname{Spec}(L)$ is a single point, representing $2^{1/3}$$2^{1/3}$, but when base changed to $\mathbb{C}$$\mathbb{C}$, we get three points, representing the three cube roots. Since $L/\mathbb{Q}$$L/\mathbb{Q}$ is not Galois, these three points do not have complete symmetry. To obtain full symmetry, we need to consider the Galois closure $E$$E$.

Example 3: Direct Product of Square Root Extensions

Consider $B=\mathbb{Q}(\sqrt{2})\times \mathbb{Q}(\sqrt{3})\times \mathbb{Q}(\sqrt{5})$$B = \mathbb{Q}(\sqrt{2}) \times \mathbb{Q}(\sqrt{3}) \times \mathbb{Q}(\sqrt{5})$, which corresponds to three points in the scheme $\mathrm{Spec}(B)$$\operatorname{Spec}(B)$.

From the polynomial perspective: This corresponds to the set of roots of the polynomial $(X^2-2)(X^2-3)(X^2-5)=0$$(X^2 - 2)(X^2 - 3)(X^2 - 5) = 0$:

• The first point corresponds to $\sqrt{2}$$\sqrt{2}$ and $-\sqrt{2}$$-\sqrt{2}$ (indistinguishable in $\mathbb{Q}(\sqrt{2})$$\mathbb{Q}(\sqrt{2})$)

• The second point corresponds to $\sqrt{3}$$\sqrt{3}$ and $-\sqrt{3}$$-\sqrt{3}$ (indistinguishable in $\mathbb{Q}(\sqrt{3})$$\mathbb{Q}(\sqrt{3})$)

• The third point corresponds to $\sqrt{5}$$\sqrt{5}$ and $-\sqrt{5}$$-\sqrt{5}$ (indistinguishable in $\mathbb{Q}(\sqrt{5})$$\mathbb{Q}(\sqrt{5})$)

Computation of Kähler differentials: Since the Kähler differential is distributive over direct products, and each component is a separable extension, we have ${\mathrm{\Omega }}_{B/\mathbb{Q}}^1={\mathrm{\Omega }}_{\mathbb{Q}(\sqrt{2})/\mathbb{Q}}^1\oplus {\mathrm{\Omega }}_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}}^1\oplus {\mathrm{\Omega }}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}^1=0$$\Omega^1_{B/\mathbb{Q}} = \Omega^1_{\mathbb{Q}(\sqrt{2})/\mathbb{Q}} \oplus \Omega^1_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}} \oplus \Omega^1_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}} = 0$ Therefore, $B$$B$ is étale over $\mathbb{Q}$$\mathbb{Q}$.

Base change to $\mathbb{C}$$\mathbb{C}$: $B{\otimes }_{\mathbb{Q}}\mathbb{C}\cong {\mathbb{C}}^6$$B \otimes_\mathbb{Q} \mathbb{C} \cong \mathbb{C}^6$ corresponding to six discrete points, each representing a root of the above equation.

From the perspective of the fiber functor: $F_x(B)$$F_x(B)$ is a set with 6 elements, corresponding to embeddings of the three fields: $F_x(B)=F_x(\mathbb{Q}(\sqrt{2}))\bigsqcup F_x(\mathbb{Q}(\sqrt{3}))\bigsqcup F_x(\mathbb{Q}(\sqrt{5}))$$F_x(B) = F_x(\mathbb{Q}(\sqrt{2})) \sqcup F_x(\mathbb{Q}(\sqrt{3})) \sqcup F_x(\mathbb{Q}(\sqrt{5}))$ Each $F_x(\mathbb{Q}(\sqrt{d}))$$F_x(\mathbb{Q}(\sqrt{d}))$ contains two elements, corresponding to embeddings mapping $\sqrt{d}$$\sqrt{d}$ to either $\sqrt{d}$$\sqrt{d}$ or $-\sqrt{d}$$-\sqrt{d}$.

Galois theory interpretation: If we consider the extension $E=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$$E = \mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})$ containing all three square roots, then $E/\mathbb{Q}$$E/\mathbb{Q}$ is a Galois extension, with Galois group $\mathrm{Gal}(E/\mathbb{Q})\cong (\mathbb{Z}/2\mathbb{Z}{)}^3$$\operatorname{Gal}(E/\mathbb{Q}) \cong (\mathbb{Z}/2\mathbb{Z})^3$. We can view $B$$B$ as the direct product of certain intermediate fields of $E$$E$, which correspond to certain subgroups of $\mathrm{Gal}(E/\mathbb{Q})$$\operatorname{Gal}(E/\mathbb{Q})$.

This example clearly demonstrates how the language of schemes can be used to uniformly handle multiple different field extensions, and their correspondence with polynomial roots.

 

5.3 Fiber Functor and Category Equivalence

5.3.1 Construction and Intuitive Understanding of the Fiber Functor

Given a field $F$$F$, we aim to establish a profound connection between finite étale coverings and representations of the absolute Galois group. This connection is realized through the fiber functor, defined by choosing a geometric point $x:\mathrm{Spec}(F^{\text{sep}})\to \mathrm{Spec}(F)$$x: \operatorname{Spec}(F^{\text{sep}}) \to \operatorname{Spec}(F)$, where $F^{\text{sep}}$$F^{\text{sep}}$ is the separable closure of $F$$F$.

First, let's recall two key categories:

• ${\mathbf{FEt}}_F$$\mathbf{FEt}_F$: The category of finite étale $F$$F$-algebras (or equivalently, finite étale coverings of $\mathrm{Spec}(F)$$\operatorname{Spec}(F)$)

• ${\mathbf{FinSet}}^G$$\mathbf{FinSet}^G$: The category of finite $G$$G$-sets, where $G=\mathrm{Gal}(F^{\text{sep}}/F)$$G = \operatorname{Gal}(F^{\text{sep}}/F)$ is the absolute Galois group

The fiber functor $F_x:{\mathbf{FEt}}_F\to {\mathbf{FinSet}}^G$$F_x: \mathbf{FEt}_F \to \mathbf{FinSet}^G$ is defined as follows:

• Object mapping: For any finite étale covering $f:\mathrm{Spec}(B)\to \mathrm{Spec}(F)$$f: \operatorname{Spec}(B) \to \operatorname{Spec}(F)$, where $B$$B$ is a finite étale $F$$F$-algebra, the functor $F_x$$F_x$ gives: $F_x(B)={\mathrm{Hom}}_F(B,F^{\text{sep}})$$F_x(B) = \operatorname{Hom}_F(B, F^{\text{sep}})$

  This represents the set of all $F$$F$-algebra homomorphisms from $B$$B$ to $F^{\text{sep}}$$F^{\text{sep}}$. Intuitively, it can be understood as the "fiber" of $\mathrm{Spec}(B)$$\operatorname{Spec}(B)$ at the geometric point $x$$x$, comprising all different ways to embed $B$$B$ into $F^{\text{sep}}$$F^{\text{sep}}$.

• Morphism mapping: Given a morphism $\varphi :\mathrm{Spec}(B_1)\to \mathrm{Spec}(B_2)$$\phi: \operatorname{Spec}(B_1) \to \operatorname{Spec}(B_2)$ (corresponding to an $F$$F$-algebra homomorphism ${\varphi }^{\ast }:B_2\to B_1$$\phi^*: B_2 \to B_1$), the functor $F_x$$F_x$ maps it to: $F_x(\varphi ):{\mathrm{Hom}}_F(B_2,F^{\text{sep}})\to {\mathrm{Hom}}_F(B_1,F^{\text{sep}}),\psi \mapsto \psi \circ {\varphi }^{\ast }$$F_x(\phi): \operatorname{Hom}_F(B_2, F^{\text{sep}}) \to \operatorname{Hom}_F(B_1, F^{\text{sep}}),\quad \psi \mapsto \psi \circ \phi^*$

Key Property: Galois Action

The group $G=\mathrm{Gal}(F^{\text{sep}}/F)$$G = \operatorname{Gal}(F^{\text{sep}}/F)$ acts naturally on $F_x(B)={\mathrm{Hom}}_F(B,F^{\text{sep}})$$F_x(B) = \operatorname{Hom}_F(B, F^{\text{sep}})$: for any $\sigma \in G$$\sigma \in G$ and $\psi \in {\mathrm{Hom}}_F(B,F^{\text{sep}})$$\psi \in \operatorname{Hom}_F(B, F^{\text{sep}})$, we define $\sigma \cdot \psi =\sigma \circ \psi$$\sigma \cdot \psi = \sigma \circ \psi$. This action transforms $F_x(B)$$F_x(B)$ into a $G$$G$-set, ensuring that $F_x$$F_x$ is indeed a functor mapping to ${\mathbf{FinSet}}^G$$\mathbf{FinSet}^G$.

5.3.2 Category Equivalence Theorem and Its Rigorous Proo2

We now state the core result in category theory:

Theorem (Category Equivalence): The fiber functor $F_x:\mathrm{Cov}(F)\cong {\mathbf{FEt}}_F^{\mathrm{op}}\to {\mathbf{FinSet}}^G$$F_x: \mathrm{Cov}(F)\cong\mathbf{FEt}_F^\mathrm{op} \to \mathbf{FinSet}^G$ establishes a category equivalence $\mathrm{Cov}(F)\simeq {\mathbf{FinSet}}^G$$\mathrm{Cov}(F) \simeq \mathbf{FinSet}^G$.

This profound theorem, first proposed by Grothendieck, transforms the study of étale coverings into the study of representation theory of the absolute Galois group, pioneering modern research in algebraic fundamental groups and étale cohomology theory.

To prove this category equivalence, we will construct an adjoint functor $G_x:({\mathbf{FinSet}}^G{)}^{\mathrm{op}}\to {\mathbf{FEt}}_F$$G_x: (\mathbf{FinSet}^G)^{\mathrm{op}} \to \mathbf{FEt}_F$ and demonstrate that $F_x$$F_x$ and $G_x$$G_x$ are mutual quasi-inverses, i.e., there exist natural isomorphisms $F_x\circ G_x\cong {\text{Id}}_{{\mathbf{FinSet}}^G}$$F_x \circ G_x \cong \text{Id}_{\mathbf{FinSet}^G}$ and $G_x\circ F_x\cong {\text{Id}}_{{\mathbf{FEt}}_F}$$G_x \circ F_x \cong \text{Id}_{\mathbf{FEt}_F}$.

1. Construction of the Inverse Functor $G_x$$G_x$

For any finite $G$$G$-set $S$$S$, we need to construct a finite étale $F$$F$-algebra $G_x(S)$$G_x(S)$. The steps are as follows:

(a) First, according to the structure theorem for finite $G$$G$-sets, $S$$S$ can be decomposed into a disjoint union of orbits: $S=\underset{i=1}{\overset{r}{⨆}}G/H_i$$S = \bigsqcup_{i=1}^r G/H_i$.

(b) For each subgroup $H_i$$H_i$, consider the fixed field $F_i=(F^{\text{sep}}{)}^{H_i}$$F_i = (F^{\text{sep}})^{H_i}$, which is the subfield of $F^{\text{sep}}$$F^{\text{sep}}$ consisting of elements fixed by $H_i$$H_i$.

(c) Define $G_x(S)=\prod _{i=1}^r{F}_i$$G_x(S) = \prod_{i=1}^r F_i$, which is a finite étale $F$$F$-algebra.

For a morphism $\alpha :S_1\to S_2$$\alpha: S_1 \to S_2$ between $G$$G$-sets, we define $G_x(\alpha ):G_x(S_2)\to G_x(S_1)$$G_x(\alpha): G_x(S_2) \to G_x(S_1)$ as follows:

Decompose $S_1$$S_1$ and $S_2$$S_2$ into orbits: $S_1=\underset{i=1}{\overset{n}{⨆}}G/H_i$$S_1 = \bigsqcup_{i=1}^n G/H_i$ and $S_2=\underset{j=1}{\overset{m}{⨆}}G/K_j$$S_2 = \bigsqcup_{j=1}^m G/K_j$. Each $G$$G$-equivariant map $\alpha$$\alpha$ must map certain $G/H_i$$G/H_i$ to certain $G/K_j$$G/K_j$, and such a mapping exists if and only if $H_i\supseteq K_j$$H_i \supseteq K_j$ (i.e., $K_j$$K_j$ is a subgroup of $H_i$$H_i$), corresponding to an embedding from the fixed field $F_{K_j}$$F_{K_j}$ to $F_{H_i}$$F_{H_i}$. Based on the specific structure of $\alpha$$\alpha$, we can construct the corresponding algebra homomorphism $G_x(\alpha ):\prod _{j=1}^m{F}_{K_j}\to \prod _{i=1}^n{F}_{H_i}$$G_x(\alpha): \prod_{j=1}^m F_{K_j} \to \prod_{i=1}^n F_{H_i}$.

2. Proving $F_x\circ G_x\cong {\text{Id}}_{{\mathbf{FinSet}}^G}$$F_x \circ G_x \cong \text{Id}_{\mathbf{FinSet}^G}$

For any finite $G$$G$-set $S=\underset{i=1}{\overset{r}{⨆}}G/H_i$$S = \bigsqcup_{i=1}^r G/H_i$, we need to prove that $F_x(G_x(S))\cong S$$F_x(G_x(S)) \cong S$ as $G$$G$-sets.

By definition, $G_x(S)=\prod _{i=1}^r{F}_i$$G_x(S) = \prod_{i=1}^r F_i$, where $F_i=(F^{\text{sep}}{)}^{H_i}$$F_i = (F^{\text{sep}})^{H_i}$.

Computing $F_x(G_x(S))$$F_x(G_x(S))$: $F_x(G_x(S))={\mathrm{Hom}}_F(\prod _{i=1}^r{F}_i,F^{\text{sep}})$$F_x(G_x(S)) = \operatorname{Hom}_F\left(\prod_{i=1}^r F_i, F^{\text{sep}}\right)$

For finite direct products, we have an isomorphism: ${\mathrm{Hom}}_F(\prod _{i=1}^r{F}_i,F^{\text{sep}})\cong \underset{i=1}{\overset{r}{⨆}}{\mathrm{Hom}}_F(F_i,F^{\text{sep}})$$\operatorname{Hom}_F\left(\prod_{i=1}^r F_i, F^{\text{sep}}\right) \cong \bigsqcup_{i=1}^r \operatorname{Hom}_F(F_i, F^{\text{sep}})$

Now we analyze each ${\mathrm{Hom}}_F(F_i,F^{\text{sep}})$$\operatorname{Hom}_F(F_i, F^{\text{sep}})$. Since $F_i/F$$F_i/F$ is a finite separable extension, any $F$$F$-algebra homomorphism $\psi :F_i\to F^{\text{sep}}$$\psi: F_i \to F^{\text{sep}}$ must be an embedding (injection).

Key observation: There is a natural $G$$G$-equivariant bijection between $\mathrm{Hom}\ast F(F_i,F^{\text{sep}})$$\operatorname{Hom}*F(F_i, F^{\text{sep}})$ and $G/H_i$$G/H_i$. Specifically, fixing an embedding ${\iota }_i:F_i\hookrightarrow F^{\text{sep}}$$\iota_i: F_i \hookrightarrow F^{\text{sep}}$, all other embeddings can be expressed as $\sigma \circ {\iota }_i$$\sigma \circ \iota_i$, where $\sigma \in G$$\sigma \in G$. Furthermore, ${\sigma }_1\circ {\iota }_i={\sigma }_2\circ {\iota }_i$$\sigma_1 \circ \iota_i = \sigma_2 \circ \iota_i$ if and only if ${\sigma }_1{|}_{F_i}={\sigma }_2{|}_{F_i}$$\sigma_1|_{F_i} = \sigma_2|_{F_i}$, which is equivalent to ${\sigma }_1^{-1}{\sigma }_2\in H_i$$\sigma_1^{-1}\sigma_2 \in H_i$, meaning that ${\sigma }_1$$\sigma_1$ and ${\sigma }_2$$\sigma_2$ represent the same coset in the quotient set $G/H_i$$G/H_i$.

Therefore, we have a $G$$G$-set isomorphism: ${\mathrm{Hom}}_F(F_i,F^{\text{sep}})\cong G/H_i$$\operatorname{Hom}_F(F_i, F^{\text{sep}}) \cong G/H_i$, which leads to: $F_x(G_x(S))\cong \underset{i=1}{\overset{r}{⨆}}G/H_i\cong S$$F_x(G_x(S)) \cong \bigsqcup_{i=1}^r G/H_i \cong S$

This isomorphism is natural, and one can verify that for morphisms between $G$$G$-sets, the diagrams commute, establishing a natural isomorphism $F_x\circ G_x\cong {\text{Id}}_{{\mathbf{FinSet}}^G}$$F_x \circ G_x \cong \text{Id}_{\mathbf{FinSet}^G}$.

3. Proving $G_x\circ F_x\cong {\text{Id}}_{{\mathbf{FEt}}_F}$$G_x \circ F_x \cong \text{Id}_{\mathbf{FEt}_F}$

For any finite étale $F$$F$-algebra $B$$B$, we need to prove that $G_x(F_x(B))\cong B$$G_x(F_x(B)) \cong B$.

First, we can decompose $B$$B$ into a direct product of fields: $B=\prod _{i=1}^n{E}_i$$B = \prod_{i=1}^n E_i$, where each $E_i$$E_i$ is a finite separable extension of $F$$F$.

From our previous analysis, we know: $F_x(B)={\mathrm{Hom}}_F(B,F^{\text{sep}})\cong \underset{i=1}{\overset{n}{⨆}}{\mathrm{Hom}}_F(E_i,F^{\text{sep}})\cong \underset{i=1}{\overset{n}{⨆}}G/H_i$$F_x(B) = \operatorname{Hom}_F(B, F^{\text{sep}}) \cong \bigsqcup_{i=1}^n \operatorname{Hom}_F(E_i, F^{\text{sep}}) \cong \bigsqcup_{i=1}^n G/H_i$

where $H_i=\mathrm{Gal}(F^{\text{sep}}/E_i)$$H_i = \operatorname{Gal}(F^{\text{sep}}/E_i)$.

By the definition of $G_x$$G_x$, we have: $G_x(F_x(B))=G_x\left(\underset{i=1}{\overset{n}{⨆}}G/H_i\right)=\prod _{i=1}^n(F^{\text{sep}}{)}^{H_i}=\prod _{i=1}^n{E}_i=B$$G_x(F_x(B)) = G_x\left(\bigsqcup_{i=1}^n G/H_i\right) = \prod_{i=1}^n (F^{\text{sep}})^{H_i} = \prod_{i=1}^n E_i = B$

Here, we use the important result from Galois theory: $(F^{\text{sep}}{)}^{H_i}=E_i$$(F^{\text{sep}})^{H_i} = E_i$, i.e., the elements of $F^{\text{sep}}$$F^{\text{sep}}$ fixed by $H_i$$H_i$ precisely form the subfield $E_i$$E_i$.

Similarly, this isomorphism is also natural, and one can verify that for morphisms between finite étale $F$$F$-algebras, the corresponding diagrams commute, establishing a natural isomorphism $G_x\circ F_x\cong {\text{Id}}_{{\mathbf{FEt}}_F}$$G_x \circ F_x \cong \text{Id}_{\mathbf{FEt}_F}$.

4. Constructive Understanding of Category Equivalence

The above proof not only demonstrates the equivalence between ${\mathbf{FEt}}_F$$\mathbf{FEt}_F$ and ${\mathbf{FinSet}}^G$$\mathbf{FinSet}^G$ but also provides a concrete implementation of this equivalence through the explicit construction of mutually inverse functors. This constructive proof reveals the deep intrinsic connection between finite étale coverings and Galois representations:

• Each finite étale $F$$F$-algebra is essentially fully determined by the Galois action on its "fiber."

• Conversely, each finite Galois representation can be "geometrized" into a finite étale covering.

This bidirectional correspondence provides a powerful method, enabling us to transform algebraic problems into topological ones, or representational problems into geometric ones, and vice versa.

5.4 From the duality to classical Galois Correspondence

We know that ${\mathbf{FEt}}_F$$\mathbf{FEt}_F$ is the finite product complement of $L/F$$L/F$ where $L/F$$L/F$ is finite separate extension and ${\mathbf{FinSet}}^G$$\mathbf{FinSet}^G$ is the finite coproduct complement. Hence we have the duality between this two full subcategories.

$$L/F⟷G_F/H$$

Hence we get a correspondence between finite intermediate fields between $F\subset L\subset F^{\mathrm{sep}}$$F\subset L\subset F^{\mathrm{sep}}$ and finite index subgroup of $G_F$$G_F$.

$$F\subseteq L\subseteq F^{\mathrm{sep}}⟷\mathrm{Gal}(F^{\mathrm{sep}}/F)\supseteq \mathrm{Gal}(F^{\mathrm{sep}}/L)\supseteq \mathrm{Gal}(F^{\mathrm{sep}}/F^{\mathrm{sep}})$$

We could restrict it to the intermediate field $F\subseteq M\subseteq L$$F\subseteq M\subseteq L$ and get the correspondence to

$$\mathrm{Gal}(F^{\mathrm{sep}}/F)\supseteq \mathrm{Gal}(F^{\mathrm{sep}}/M)\supseteq \mathrm{Gal}(F^{\mathrm{sep}}/L)$$

 

Normal Extensions and Normal Subgroups

Proposition. The extension $L/F$$L/F$ is a normal extension if and only if $\mathrm{Gal}(F^{\mathrm{sep}}/L)$$\mathrm{Gal}(F^{\mathrm{sep}}/L)$ is a normal subgroup of $\mathrm{Gal}(F^{\mathrm{sep}}/F)$$\mathrm{Gal}(F^{\mathrm{sep}}/F)$.

Proposition. The extension $L/F$$L/F$ is a normal extension if and only if $\mathrm{Gal}(F^{\mathrm{sep}}/L)$$\mathrm{Gal}(F^{\mathrm{sep}}/L)$ is a normal subgroup of $\mathrm{Gal}(F^{\mathrm{sep}}/F)$$\mathrm{Gal}(F^{\mathrm{sep}}/F)$.

Proof. Let $H=\mathrm{Gal}(F^{\mathrm{sep}}/L)$$H = \mathrm{Gal}(F^{\mathrm{sep}}/L)$.

($\Rightarrow$$\Rightarrow$) Suppose $L/F$$L/F$ is normal. By definition, this means that $L$$L$ is the splitting field of a family of polynomials in $F[x]$$F[x]$, or equivalently, for any $F$$F$-embedding $\sigma :L\to F^{\mathrm{sep}}$$\sigma: L \rightarrow F^{\mathrm{sep}}$, we have $\sigma (L)=L$$\sigma(L) = L$.

For any $\tau \in \mathrm{Gal}(F^{\mathrm{sep}}/F)$$\tau \in \mathrm{Gal}(F^{\mathrm{sep}}/F)$ and $h\in H$$h \in H$, we need to show $\tau h{\tau }^{-1}\in H$$\tau h \tau^{-1} \in H$.

For any $a\in L$$a \in L$, we want to show $(\tau h{\tau }^{-1})(a)=a$$(\tau h \tau^{-1})(a) = a$.

Now, ${\tau }^{-1}$$\tau^{-1}$ is an $F$$F$-automorphism of $F^{\mathrm{sep}}$$F^{\mathrm{sep}}$, and when restricted to $L$$L$, it gives an $F$$F$-embedding of $L$$L$ into $F^{\mathrm{sep}}$$F^{\mathrm{sep}}$. Since $L/F$$L/F$ is normal, any $F$$F$-embedding of $L$$L$ maps $L$$L$ to itself, so ${\tau }^{-1}(L)\subseteq L$$\tau^{-1}(L) \subseteq L$. Therefore, for any $a\in L$$a \in L$, we have ${\tau }^{-1}(a)\in L$$\tau^{-1}(a) \in L$.

Since ${\tau }^{-1}(a)\in L$$\tau^{-1}(a) \in L$ and $h\in H=\mathrm{Gal}(F^{\mathrm{sep}}/L)$$h \in H = \mathrm{Gal}(F^{\mathrm{sep}}/L)$, we have $h({\tau }^{-1}(a))={\tau }^{-1}(a)$$h(\tau^{-1}(a)) = \tau^{-1}(a)$.

Therefore $(\tau h{\tau }^{-1})(a)=\tau (h({\tau }^{-1}(a)))=\tau ({\tau }^{-1}(a))=a$$(\tau h \tau^{-1})(a) = \tau(h(\tau^{-1}(a))) = \tau(\tau^{-1}(a)) = a$, which means $\tau h{\tau }^{-1}\in H$$\tau h \tau^{-1} \in H$.

($\Leftarrow$$\Leftarrow$) Conversely, suppose $H$$H$ is normal in $\mathrm{Gal}(F^{\mathrm{sep}}/F)$$\mathrm{Gal}(F^{\mathrm{sep}}/F)$. Then for any $\sigma \in \mathrm{Gal}(F^{\mathrm{sep}}/F)$$\sigma \in \mathrm{Gal}(F^{\mathrm{sep}}/F)$ and any $a\in L$$a \in L$, we need to show $\sigma (a)\in L$$\sigma(a) \in L$.

For any $h\in H$$h \in H$, we have ${\sigma }^{-1}h\sigma \in H$$\sigma^{-1}h\sigma \in H$ since $H$$H$ is normal. This means $({\sigma }^{-1}h\sigma )(a)=a$$(\sigma^{-1}h\sigma)(a) = a$ for all $a\in L$$a \in L$. Therefore, $h(\sigma (a))=\sigma (a)$$h(\sigma(a)) = \sigma(a)$ for all $h\in H$$h \in H$, which implies $\sigma (a)\in L$$\sigma(a) \in L$. Thus, $L/F$$L/F$ is normal.

Proposition. If $L/F$$L/F$ is a normal extension, then $H=\mathrm{Gal}(F^{\mathrm{sep}}/L)$$H = \mathrm{Gal}(F^{\mathrm{sep}}/L)$ is precisely the kernel of the natural homomorphism

$$\phi :\mathrm{Gal}(F^{\mathrm{sep}}/F)\to \mathrm{Gal}(L/F)$$

Where $\phi (\sigma )=\sigma {|}_L$$\varphi(\sigma) = \sigma|_L$ (the restriction of $\sigma$$\sigma$ to $L$$L$).

Proof. First, let's verify that $\phi$$\varphi$ is a homomorphism. For any $\sigma ,\tau \in \mathrm{Gal}(F^{\mathrm{sep}}/F)$$\sigma, \tau \in \mathrm{Gal}(F^{\mathrm{sep}}/F)$ and $a\in L$$a \in L$:

$$\phi (\sigma \cdot \tau )(a)=(\sigma \cdot \tau ){|}_L(a)=\sigma (\tau (a))=\sigma {|}_L(\tau {|}_L(a))=(\phi (\sigma )\cdot \phi (\tau ))(a)$$

Now we need to determine the kernel of $\phi$$\varphi$:

$$\ker (\phi )=\{\sigma \in \mathrm{Gal}(F^{\mathrm{sep}}/F)\mid \phi (\sigma )={\mathrm{id}}_L\}=\{\sigma \in \mathrm{Gal}(F^{\mathrm{sep}}/F)\mid \sigma {|}_L={\mathrm{id}}_L\}=\{\sigma \in \mathrm{Gal}(F^{\mathrm{sep}}/F)\mid \sigma (a)=a\text{ for all }a\in L\}$$

But this is precisely the definition of $\mathrm{Gal}(F^{\mathrm{sep}}/L)$$\mathrm{Gal}(F^{\mathrm{sep}}/L)$, which is the group of $F$$F$-automorphisms of $F^{\mathrm{sep}}$$F^{\mathrm{sep}}$ that fix $L$$L$ pointwise.

Therefore, $H=\mathrm{Gal}(F^{\mathrm{sep}}/L)=\ker (\phi )$$H = \mathrm{Gal}(F^{\mathrm{sep}}/L) = \ker(\varphi)$. Now we need to show that $\phi$$\varphi$ is surjective, i.e., $\mathrm{Im}(\phi )=\mathrm{Gal}(L/F)$$\mathrm{Im}(\varphi) = \mathrm{Gal}(L/F)$.

Take any $\tau \in \mathrm{Gal}(L/F)$$\tau \in \mathrm{Gal}(L/F)$. Since $L/F$$L/F$ is normal and $F^{\mathrm{sep}}$$F^{\mathrm{sep}}$ is a separable closure of $F$$F$, the automorphism $\tau$$\tau$ can be extended to an $F$$F$-automorphism $\tilde{\tau }$$\tilde{\tau}$ of $F^{\mathrm{sep}}$$F^{\mathrm{sep}}$. That is, there exists $\tilde{\tau }\in \mathrm{Gal}(F^{\mathrm{sep}}/F)$$\tilde{\tau} \in \mathrm{Gal}(F^{\mathrm{sep}}/F)$ such that $\tilde{\tau }{|}_L=\tau$$\tilde{\tau}|_L = \tau$.

Therefore, $\phi (\tilde{\tau })=\tilde{\tau }{|}_L=\tau$$\varphi(\tilde{\tau}) = \tilde{\tau}|_L = \tau$, which means $\tau \in \mathrm{Im}(\phi )$$\tau \in \mathrm{Im}(\varphi)$. Since $\tau$$\tau$ was arbitrary, we have $\mathrm{Gal}(L/F)\subseteq \mathrm{Im}(\phi )$$\mathrm{Gal}(L/F) \subseteq \mathrm{Im}(\varphi)$.

Also, for any $\sigma \in \mathrm{Gal}(F^{\mathrm{sep}}/F)$$\sigma \in \mathrm{Gal}(F^{\mathrm{sep}}/F)$, since $L/F$$L/F$ is normal, we have $\sigma (L)=L$$\sigma(L) = L$. This means $\sigma {|}_L$$\sigma|_L$ is an $F$$F$-automorphism of $L$$L$, i.e., $\sigma {|}_L\in \mathrm{Gal}(L/F)$$\sigma|_L \in \mathrm{Gal}(L/F)$. Thus, $\mathrm{Im}(\phi )\subseteq \mathrm{Gal}(L/F)$$\mathrm{Im}(\varphi) \subseteq \mathrm{Gal}(L/F)$.

Combining these inclusions, we get $\mathrm{Im}(\phi )=\mathrm{Gal}(L/F)$$\mathrm{Im}(\varphi) = \mathrm{Gal}(L/F)$, which shows that $\phi$$\varphi$ is surjective.

By the First Isomorphism Theorem, we have:

$$\mathrm{Gal}(F^{\mathrm{sep}}/F)/\ker (\phi )\cong \mathrm{Im}(\phi )$$

Therefore, $\mathrm{Gal}(F^{\mathrm{sep}}/F)/H\cong \mathrm{Gal}(L/F)$$\mathrm{Gal}(F^{\mathrm{sep}}/F)/H \cong \mathrm{Gal}(L/F)$. Hence, if $H=\mathrm{Gal}(F^{\mathrm{sep}}/L)$$H=\mathrm{Gal}(F^{\mathrm{sep}}/L)$ is a normal subgroup, then

$$\mathrm{Gal}(F^{\mathrm{sep}}/F)/H\cong \mathrm{Gal}(L/F)$$

Classical Galois Correspondence

Assume that $L/F$$L/F$ is a normal extension, then we have:

$$F\subseteq M\subseteq L⟷\mathrm{Gal}(F^{\mathrm{sep}}/F)\supseteq \mathrm{Gal}(F^{\mathrm{sep}}/M)\supseteq \mathrm{Gal}(F^{\mathrm{sep}}/L)$$

Taking the quotient by $\mathrm{Gal}(F^{\mathrm{sep}}/L)$$\mathrm{Gal}(F^{\mathrm{sep}}/L)$ we get

$$F\subseteq M\subseteq L⟷\mathrm{Gal}(L/F)\supseteq \mathrm{Gal}(L/M)\supseteq \{e\}$$

Degree and Galois Group Order Relationship

Proposition. Let $E$$E$ be an intermediate field $F\subseteq E\subseteq L$$F \subseteq E \subseteq L$. Then $[E:F]=|F_x(E)|$$[E:F] = |F_x(E)|$.

Proof. Since $E/F$$E/F$ is a finite separable extension, by the Primitive Element Theorem, there exists $\alpha \in E$$\alpha \in E$ such that $E=F(\alpha )$$E = F(\alpha)$. Let the minimal polynomial of $\alpha$$\alpha$ over $F[x]$$F[x]$ be: $f(x)=\underset{\alpha ,F}{min}(x)$$f(x) = \min_{\alpha,F}(x)$, $\mathrm{deg}f=n$$\deg f = n$.

Separability guarantees that $f$$f$ has exactly $n$$n$ distinct roots ${\alpha }_1,\dots ,{\alpha }_n$$\alpha_1, \dots, \alpha_n$ in the separable closure $(F^{\mathrm{sep}})$$(F^{\mathrm{sep}})$.

For any $F$$F$-embedding $\phi :E\to F^{\mathrm{sep}}$$\varphi \colon E \to F^{\mathrm{sep}}$, it is completely determined by $\phi (\alpha )$$\varphi(\alpha)$, which must be one of the roots ${\alpha }_i$$\alpha_i$ mentioned above.

Conversely, for each root ${\alpha }_i$$\alpha_i$ we define: ${\phi }_i:F(\alpha )⟶F^{\mathrm{sep}},{\phi }_i(\alpha )={\alpha }_i$$\varphi_i \colon F(\alpha) \longrightarrow F^{\mathrm{sep}}, \quad \varphi_i(\alpha) = \alpha_i$

We can verify that this is a field homomorphism. This gives us a one-to-one correspondence: $\phi \in {\mathrm{Hom}}_F(E,F^{\mathrm{sep}})⟷{\alpha }_1,\dots ,{\alpha }_n$${\varphi \in \mathrm{Hom}_F(E, F^{\mathrm{sep}})} \longleftrightarrow {\alpha_1, \dots, \alpha_n}$

Therefore: $|{\mathrm{Hom}}_F(E,F^{\mathrm{sep}})|=n=\mathrm{deg}f=[F(\alpha ):F]=[E:F]$$\left|\mathrm{Hom}_F(E, F^{\mathrm{sep}})\right| = n = \deg f = [F(\alpha):F] = [E:F]$

Proposition. Let $L/F$$L/F$ be a Galois extension, and $F\subseteq E\subseteq L$$F\subseteq E\subseteq L$. Then $|F_x(E)|=[\mathrm{Gal}(L/F):\mathrm{Gal}(E/F)]$$|F_x(E)|=[\mathrm{Gal}(L/F):\mathrm{Gal}(E/F)]$

Proof. Since $|F_x(E)|=[\mathrm{Gal}(F^{\mathrm{sep}}/F):\mathrm{Gal}(F^{\mathrm{sep}}/E)]=[\mathrm{Gal}(F^{\mathrm{sep}}/F)/\mathrm{Gal}(F^{\mathrm{sep}}/L):\mathrm{Gal}(F^{\mathrm{sep}}/E)/\mathrm{Gal}(F^{\mathrm{sep}}/L)]$$|F_x(E)|=[\mathrm{Gal}(F^{\mathrm{sep}}/F):\mathrm{Gal}(F^{\mathrm{sep}}/E)]=[\mathrm{Gal}(F^{\mathrm{sep}}/F)/\mathrm{Gal}(F^{\mathrm{sep}}/L):\mathrm{Gal}(F^{\mathrm{sep}}/E)/\mathrm{Gal}(F^{\mathrm{sep}}/L)]$.

But

$$[\mathrm{Gal}(F^{\mathrm{sep}}/F)/\mathrm{Gal}(F^{\mathrm{sep}}/L):\mathrm{Gal}(F^{\mathrm{sep}}/E)/\mathrm{Gal}(F^{\mathrm{sep}}/L)]=|F_x(E)|=[\mathrm{Gal}(L/F):\mathrm{Gal}(E/F)]◻$$

Grothendieck's insight is not limited to the case of fields; it extends to general schemes:

Theorem (SGA1): Let $X$$X$ be a connected Noetherian scheme, and $x:\mathrm{Spec}(\mathrm{\Omega })\to X$$x: \operatorname{Spec}(\Omega) \to X$ a geometric point. Then there is a category equivalence ${\mathbf{FEt}}_X\simeq {\mathbf{FinSet}}^{{\pi }_1^{\text{ét}}(X,x)}$$\mathbf{FEt}_X \simeq \mathbf{FinSet}^{\pi_1^\text{ét}(X,x)}$ where ${\mathbf{FEt}}_X$$\mathbf{FEt}_X$ is the category of finite étale coverings of $X$$X$, and ${\pi }_1^{\text{ét}}(X,x)$$\pi_1^\text{ét}(X,x)$ is the étale fundamental group of $X$$X$ at $x$$x$.

This theorem generalizes the topological analogue of the fundamental group of Riemann surfaces to algebraic geometry, providing a powerful tool for studying arithmetic surfaces and number theory. In particular, when $X=\mathrm{Spec}(F)$$X = \operatorname{Spec}(F)$ is the spectrum of a field, we have ${\pi }_1^{\text{ét}}(X,x)=\mathrm{Gal}(F^{\text{sep}}/F)$$\pi_1^\text{ét}(X,x) = \operatorname{Gal}(F^\text{sep}/F)$, recovering the result discussed earlier.

6 Conclusion

The correspondence between étale morphisms and finite separable field extensions reveals a profound connection between algebraic geometry and Galois theory:

1. Finite étale algebras over a field $F$$F$ are precisely finite products of finite separable field extensions of $F$$F$.

2. The separability of field extensions is equivalent to the vanishing of the Kähler differential module, which elucidates the essence of separability from a perspective of differential geometry.

3. There is a natural correspondence between polynomial roots and points of schemes, and through base change to an algebraically closed field, the "potential" root structure in field extensions can be revealed.

4. The category equivalence ${\mathbf{FEt}}_F\simeq {\mathbf{FinSet}}^{\mathrm{Gal}(F^{\text{sep}}/F)}$$\mathbf{FEt}_F \simeq \mathbf{FinSet}^{\operatorname{Gal}(F^{\text{sep}}/F)}$ elevates Galois theory to a categorical level, unifying the perspectives of algebra, geometry, and topology.

5. The theory of the étale fundamental group unifies traditional Galois theory with the fundamental group theory of algebraic topology, pioneering an important direction in modern arithmetic geometry.

This theoretical framework not only deepens our understanding of field extensions but also provides powerful geometric insights into arithmetic problems over number fields, function fields, and finite fields, forming the foundation for the intersection of modern number theory and algebraic geometry.

Appendix

1 Étale Morphisms in Affine Schemes

1.1 Definition of Étale Morphisms

Let $A$$A$ and $B$$B$ be commutative rings, and $f:X^{\prime }=\mathrm{Spec}(B)\to X=\mathrm{Spec}(A)$$f: X' = \operatorname{Spec}(B) \to X = \operatorname{Spec}(A)$ be the corresponding morphism of affine schemes. We say that $f$$f$ is an étale morphism (or $X^{\prime }$$X'$ is an étale covering of $X$$X$) if it satisfies the following three conditions:

1. Finite type: $B$$B$ is a finitely generated $A$$A$-algebra, i.e., there exist elements $b_1,b_2,\dots ,b_n$$b_1, b_2, \ldots, b_n$ of $A$$A$ such that $B=A[b_1,b_2,\dots ,b_n]$$B = A[b_1, b_2, \ldots, b_n]$.

2. Flat: $B$$B$ is a flat $A$$A$-module, i.e., the functor $(-){\otimes }_{A}B$$(-) \otimes_A B$ is exact.

3. Unramified: The Kähler differential module ${\mathrm{\Omega }}_{B/A}^1=0$$\Omega^1_{B/A} = 0$.

If, in addition, $B$$B$ is a finitely generated $A$$A$-module (not just a finitely generated $A$$A$-algebra), then we say that $f$$f$ is a finite étale morphism.

1.2 Detailed Explanation of Flatness

Flatness is a central concept in algebraic geometry that ensures the "uniformity" of a morphism. Specifically:

Definition (Flat Module): Let $A$$A$ be a ring and $M$$M$ an $A$$A$-module. We say that $M$$M$ is flat if the functor $-{\otimes }_{A}M:A\text{-Mod}\to A\text{-Mod}$$- \otimes_A M: A\text{-Mod} \to A\text{-Mod}$ is exact, i.e., for any injection $N_1\to N_2$$N_1 \to N_2$, the induced map $N_1{\otimes }_{A}M\to N_2{\otimes }_{A}M$$N_1 \otimes_A M \to N_2 \otimes_A M$ is also an injection.

There are several equivalent conditions for flatness:

1. For any ideal $I\subset A$$I \subset A$, the natural map $I{\otimes }_{A}M\to A{\otimes }_{A}M\cong M$$I \otimes_A M \to A \otimes_A M \cong M$ is injective.

2. For any finitely generated ideal $I=(a_1,\dots ,a_n)\subset A$$I = (a_1, \ldots, a_n) \subset A$, if $\sum _{i=1}^n{a}_i{m}_i=0$$\sum_{i=1}^n a_i m_i = 0$ for some $m_i\in M$$m_i \in M$, then there exist elements $m_{ij}\in M$$m_{ij} \in M$ such that $m_i=\sum _{j=1}^n{a}_j{m}_{ij}$$m_i = \sum_{j=1}^n a_j m_{ij}$ and $m_{ij}=m_{ji}$$m_{ij} = m_{ji}$.

3. All localizations $M_{\mathfrak{p}}$$M_{\mathfrak{p}}$ of $M$$M$ are free $A_{\mathfrak{p}}$$A_{\mathfrak{p}}$-modules for all prime ideals $\mathfrak{p}\subset A$$\mathfrak{p} \subset A$.

Examples:

• Free modules and projective modules are always flat.

• Localizations $S^{-1}A$$S^{-1}A$ for any multiplicative subset $S\subset A$$S \subset A$ are flat $A$$A$-modules.

• Any vector space over a field $K$$K$ is a flat $K$$K$-module.

• Over the ring $\mathbb{Z}$$\mathbb{Z}$, $\mathbb{Q}$$\mathbb{Q}$ is flat, but $\mathbb{Z}/n\mathbb{Z}$$\mathbb{Z}/n\mathbb{Z}$ for $n>1$$n > 1$ is not flat.

Geometric interpretation: If $f:X^{\prime }\to X$$f: X' \to X$ is flat, then the fibers of $X^{\prime }$$X'$ have "uniform" dimension. In particular, for a fixed point $x\in X$$x \in X$, the dimension of the fiber $f^{-1}(x)$$f^{-1}(x)$ does not suddenly "jump".

1.3 Kähler Differential Module and the Unramified Condition

The Kähler differential module ${\mathrm{\Omega }}_{B/A}^1$$\Omega^1_{B/A}$ is precisely defined through the following universal property:

Definition (Kähler Differential Module): Let $A\to B$$A \to B$ be a ring homomorphism. The Kähler differential module ${\mathrm{\Omega }}_{B/A}^1$$\Omega^1_{B/A}$ is a $B$$B$-module satisfying the following universal property: there exists an $A$$A$-linear derivation (also called a differential map) $d:B\to {\mathrm{\Omega }}_{B/A}^1$$d: B \to \Omega^1_{B/A}$ satisfying the Leibniz rule $d(b{b}^{\prime })=b\cdot d(b^{\prime })+b^{\prime }\cdot d(b)$$d(bb') = b \cdot d(b') + b' \cdot d(b)$, such that for any $B$$B$-module $M$$M$ with an $A$$A$-linear derivation $\delta :B\to M$$\delta: B \to M$, there exists a unique $B$$B$-linear map $\varphi :{\mathrm{\Omega }}_{B/A}^1\to M$$\phi: \Omega^1_{B/A} \to M$ such that $\delta =\varphi \circ d$$\delta = \phi \circ d$.

Construction method: The Kähler differential module can be explicitly constructed as follows:

1. Consider $F=B{\otimes }_{A}B$$F = B \otimes_A B$.

2. Let $I$$I$ be the kernel of $\mu :B{\otimes }_{A}B\to B$$\mu:B\otimes _A B\to B$, where $\mu$$\mu$ is the multiplication map for the $A$$A$-algebra $B$$B$.

3. Then ${\mathrm{\Omega }}_{B/A}^1=I/I^2$$\Omega^1_{B/A} = I/I^2$, with the natural differential map given by $d(b)=1\otimes b-b\otimes 1\mathrm{mod}{I}^2$$d(b) = 1 \otimes b - b \otimes 1 \mod I^2$.

   The $B$$B$-Module structure of $I/I^2$$I/I^2$ is given by $b\cdot x=1\otimes b\cdot x=b\otimes 1\cdot x$$b\cdot x=1\otimes b\cdot x=b\otimes 1\cdot x$. The equation holds due to the fact that

   $$(1\otimes b-b\otimes 1)\in I⟹(1\otimes b-b\otimes 1)\cdot x\in I^2$$

    

Remark. $d:B\to I/I^2$$d:B\to I/I^2$ satisfies Leibniz Law becasue

$$\begin{array}{rl}d(b{b}^{\prime })& =[1\otimes (b{b}^{\prime })-(b{b}^{\prime })\otimes 1]\\ & =[(1\otimes b)(1\otimes b^{\prime })-(b\otimes 1)(b^{\prime }\otimes 1)]\end{array}$$

Notice that we have

$$AB-CD=A(B-D)+(A-C)D,$$

Thus

$$[(1\otimes b)(1\otimes b^{\prime })-(b\otimes 1)(b^{\prime }\otimes 1)]=(1\otimes b)(1\otimes b^{\prime }-b^{\prime }\otimes 1)+(1\otimes b-b\otimes 1)(b^{\prime }\otimes 1)$$

The $B$$B$-Module structure on $I/I^2$$I/I^2$ tells us that it is equal to $b(1\otimes b^{\prime }-b^{\prime }\otimes 1)+(1\otimes b-b\otimes 1)b^{\prime }=bd(b^{\prime })+d(b)b^{\prime }$$b(1\otimes b'-b'\otimes 1)+(1\otimes b-b\otimes1)b'=bd(b')+d(b)b'$

Remark. If you feel confused about the definition of $\mu$$\mu$, you shoud read the definition of monoid object in $A$$A$-Module.

Using the diagram of monoid object in $A$$A$-Module it is not hard to see that we have for a $A$$A$-algebra homomorphism

$$\phi :B_1\to B_2$$

We have

$$\phi :I_{B_1}\to I_{B_2},I_{B_1},I_{B_1}^2\to I_{B_2}^2⟹{\mathrm{\Omega }}_{(\phi )/A}^1:{\mathrm{\Omega }}_{B_1/A}^1\to {\mathrm{\Omega }}_{B_1/A}^1$$

That is the functorial property of ${\mathrm{\Omega }}_{(-)/A}^1$$\Omega^1_{(-)/A}$.

${\mathrm{\Omega }}_{(-)/A}^1$$\Omega^1_{(-)/A}$ as left adjoint functor.

Definition. Let $A$$A$-$\mathrm{Alg}/B$$\mathrm{Alg}/B$ be the slice category of $A$$A$ algebra over $B$$B$. Define a functor

$$\begin{array}{rl}\mathrm{\Phi }:B\text{-Mod}& ⟶A\text{-Alg}/B,\\ M& ⟼B\oplus M,(m\cdot m^{\prime }=0),\end{array}$$

The multiplication in $B\oplus M$$B\oplus M$ is given by:

$$(b+m)\cdot (b^{\prime }+m^{\prime })=b{b}^{\prime }+b{m}^{\prime }+b^{\prime }m$$

Then we have the following adjoint:

$$\begin{array}{ccc}{\mathrm{Hom}}_{A\text{-Alg}/B}(C,B\oplus M)& \stackrel{\simeq }{\to }& {\mathrm{Hom}}_{B\text{-Mod}}({\mathrm{\Omega }}_{C/A},M),\end{array}$$

Proof see appendix.

1.4 Geometric Interpretation of Étale Morphisms over General Commutative Rings

Proof of the Adjunction for Kähler Differentials

This document proves the fundamental adjunction relationship for Kähler differentials:

$\begin{array}{ccc}\mathrm{Hom}\ast A\text{-Alg}/B(C,;B\oplus M)& \stackrel{;\simeq ;}{\to }& \mathrm{Hom}\ast B\text{-Mod}({\mathrm{\Omega }}_{C/A},;M)\end{array}$$\begin{CD} \mathrm{Hom}*{A\text{-Alg}/B}\bigl(C,;B\oplus M\bigr) @>;\simeq;>> \mathrm{Hom}*{B\text{-Mod}}\bigl(\Omega_{C/A},;M\bigr) \end{CD}$

Prerequisites

Trivial Extensions

For a ring $B$$B$ and $B$$B$-module $M$$M$, the trivial extension $B\oplus M$$B \oplus M$ is:

• Set: $(b,m)\mid b\in B,m\in M$${(b,m) \mid b \in B, m \in M}$

• Addition: $(b_1,m_1)+(b_2,m_2)=(b_1+b_2,m_1+m_2)$$(b_1,m_1)+(b_2,m_2)=(b_1+b_2,m_1+m_2)$

• Multiplication: $(b_1,m_1)\cdot (b_2,m_2)=(b_1{b}_2,b_1{m}_2+b_2{m}_1)$$(b_1,m_1)\cdot(b_2,m_2)=(b_1b_2,b_1m_2+b_2m_1)$

The projection $\pi :B\oplus M\to B$$\pi: B \oplus M \to B$ is given by $\pi (b,m)=b$$\pi(b,m)=b$.

Comma Category $A\text{-Alg}/B$$A\text{-Alg}/B$

For rings $A$$A$ and $B$$B$ with $B$$B$ an $A$$A$-algebra:

• Objects: $A$$A$-algebra homomorphisms $f:C\to B$$f: C \to B$

• Morphisms: Commutative triangles of $A$$A$-algebra homomorphisms

Kähler Differentials

For an $A$$A$-algebra $C$$C$, the Kähler differential module ${\mathrm{\Omega }}_{C/A}$$\Omega_{C/A}$ is:

• A $C$$C$-module

• Equipped with a universal $A$$A$-derivation $d:C\to {\mathrm{\Omega }}_{C/A}$$d: C \to \Omega_{C/A}$

• Universal in that any $A$$A$-derivation $\delta :C\to N$$\delta: C \to N$ uniquely factors through $d$$d$

Proof of the Adjunction

We establish a bijection between $\mathrm{Hom}\ast A\text{-Alg}/B(C,B\oplus M)$$\mathrm{Hom}*{A\text{-Alg}/B}(C,B\oplus M)$ and $\mathrm{Hom}\ast B\text{-Mod}({\mathrm{\Omega }}_{C/A},M)$$\mathrm{Hom}*{B\text{-Mod}}(\Omega_{C/A},M)$.

Forward Direction: $\mathrm{\Phi }:{\mathrm{Hom}}_{A\text{-Alg}/B}(C,B\oplus M)\to {\mathrm{Hom}}_{B\text{-Mod}}({\mathrm{\Omega }}_{C/A},M)$$\Phi: \mathrm{Hom}_{A\text{-Alg}/B}(C,B\oplus M) \to \mathrm{Hom}_{B\text{-Mod}}(\Omega_{C/A},M)$

Let $g:C\to B\oplus M$$g: C \to B \oplus M$ be a morphism in $A\text{-Alg}/B$$A\text{-Alg}/B$, meaning $\pi \circ g=f$$\pi \circ g = f$ where $f:C\to B$$f: C \to B$ is given.

1. Write $g(c)=(f(c),h(c))$$g(c) = (f(c), h(c))$ for some function $h:C\to M$$h: C \to M$.

2. Since $g$$g$ is an $A$$A$-algebra homomorphism:

   $$g(c_1\cdot c_2)=g(c_1)\cdot g(c_2)$$

   Expanding:

   $$(f(c_1{c}_2),h(c_1{c}_2))=(f(c_1)f(c_2),f(c_1)h(c_2)+f(c_2)h(c_1))$$

    

3. Comparing second components:

   $$h(c_1{c}_2)=f(c_1)h(c_2)+f(c_2)h(c_1)$$

   This shows $h$$h$ is a "twisted derivation."

4. Define $\delta :C\to M$$\delta: C \to M$ by $\delta (c)=h(c)$$\delta(c) = h(c)$, which is an $A$$A$-derivation when $M$$M$ is viewed as a $C$$C$-module via $f$$f$.

5. By the universal property of ${\mathrm{\Omega }}_{C/A}$$\Omega_{C/A}$, there exists a unique $C$$C$-module homomorphism $\varphi :{\mathrm{\Omega }}_{C/A}\to M$$\phi: \Omega_{C/A} \to M$ with $\varphi \circ d=\delta$$\phi \circ d = \delta$.

6. View $\varphi$$\phi$ as a $B$$B$-module homomorphism, where ${\mathrm{\Omega }}_{C/A}$$\Omega_{C/A}$ is considered as a $B$$B$-module via $f:C\to B$$f: C \to B$.

7. Define $\mathrm{\Phi }(g)=\varphi$$\Phi(g) = \phi$.

Reverse Direction: $\mathrm{\Psi }:{\mathrm{Hom}}_{B\text{-Mod}}({\mathrm{\Omega }}_{C/A},M)\to {\mathrm{Hom}}_{A\text{-Alg}/B}(C,B\oplus M)$$\Psi: \mathrm{Hom}_{B\text{-Mod}}(\Omega_{C/A},M) \to \mathrm{Hom}_{A\text{-Alg}/B}(C,B\oplus M)$

Let $\varphi :{\mathrm{\Omega }}_{C/A}\to M$$\phi: \Omega_{C/A} \to M$ be a $B$$B$-module homomorphism.

1. Compose with $d$$d$ to get $\delta =\varphi \circ d:C\to M$$\delta = \phi \circ d: C \to M$, an $A$$A$-derivation.

2. Define $h(c)=\delta (c)$$h(c) = \delta(c)$.

3. Construct $g:C\to B\oplus M$$g: C \to B \oplus M$ by $g(c)=(f(c),h(c))$$g(c) = (f(c), h(c))$.

4. Verify $g$$g$ is an $A$$A$-algebra homomorphism:

   For multiplication:

   $g(c_1{c}_2)=(f(c_1{c}_2),h(c_1{c}_2))$$g(c_1c_2) = (f(c_1c_2), h(c_1c_2))$

   $=(f(c_1)f(c_2),f(c_1)h(c_2)+f(c_2)h(c_1))$$= (f(c_1)f(c_2), f(c_1)h(c_2) + f(c_2)h(c_1))$

   $=g(c_1)\cdot g(c_2)$$= g(c_1) \cdot g(c_2)$

5. Since $\pi \circ g=f$$\pi \circ g = f$, we have $g\in {\mathrm{Hom}}_{A\text{-Alg}/B}(C,B\oplus M)$$g \in \mathrm{Hom}_{A\text{-Alg}/B}(C, B\oplus M)$.

6. Define $\mathrm{\Psi }(\varphi )=g$$\Psi(\phi) = g$.

Verification of Inverse Relationship

Claim 1: $\mathrm{\Phi }\circ \mathrm{\Psi }=\mathrm{id}$$\Phi \circ \Psi = \mathrm{id}$

Given $\varphi :{\mathrm{\Omega }}_{C/A}\to M$$\phi: \Omega_{C/A} \to M$, we constructed:

• $\delta =\varphi \circ d$$\delta = \phi \circ d$

• $g(c)=(f(c),\delta (c))=(f(c),\varphi (d(c)))$$g(c) = (f(c), \delta(c)) = (f(c), \phi(d(c)))$

Applying $\mathrm{\Phi }$$\Phi$ to $g$$g$ recovers $\varphi$$\phi$ since the derivation part of $g$$g$ is precisely $\varphi \circ d$$\phi \circ d$.

Claim 2: $\mathrm{\Psi }\circ \mathrm{\Phi }=\mathrm{id}$$\Psi \circ \Phi = \mathrm{id}$

Given $g:C\to B\oplus M$$g: C \to B \oplus M$ with $g(c)=(f(c),h(c))$$g(c) = (f(c), h(c))$, we constructed:

• $\varphi :{\mathrm{\Omega }}_{C/A}\to M$$\phi: \Omega_{C/A} \to M$ with $\varphi \circ d=h$$\phi \circ d = h$

Applying $\mathrm{\Psi }$$\Psi$ to $\varphi$$\phi$ recovers $g$$g$ since:

$\mathrm{\Psi }(\varphi )(c)=(f(c),\varphi (d(c)))=(f(c),h(c))=g(c)$$\Psi(\phi)(c) = (f(c), \phi(d(c))) = (f(c), h(c)) = g(c)$

Naturality

The bijection is natural with respect to morphisms in both categories, meaning the appropriate diagrams commute.

Implications

This adjunction has several important consequences:

1. Preservation of Colimits: As a left adjoint, ${\mathrm{\Omega }}_{-/A}$$\Omega_{-/A}$ preserves all colimits.

2. Representability: Kähler differentials represent the derivation functor.

3. Deformation Theory: The trivial extension $B\oplus M$$B \oplus M$ relates to first-order deformations.

4. Geometric Interpretation: In algebraic geometry, this adjunction connects vector fields (derivations) to differential forms.

Conclusion

We have proven:

$$\begin{array}{ccc}{\mathrm{Hom}}_{A\text{-Alg}/B}(C,;B\oplus M)& \stackrel{\simeq }{\to }& {\mathrm{Hom}}_{B\text{-Mod}}({\mathrm{\Omega }}_{C/A},;M)\end{array}$$

This establishes that ${\mathrm{\Omega }}_{-/A}:A\text{-Alg}/B\to B\text{-Mod}$$\Omega_{-/A}: A\text{-Alg}/B \to B\text{-Mod}$ is left adjoint to the trivial extension functor $B\oplus -:B\text{-Mod}\to A\text{-Alg}/B$$B\oplus-: B\text{-Mod} \to A\text{-Alg}/B$.

1.4 Geometric Interpretation of Étale Morphisms over General Commutative Rings

Before focusing on the special case of fields, let's understand étale morphisms over general commutative rings from a geometric perspective.

For a general commutative ring $A$$A$, $\mathrm{Spec}(A)$$\operatorname{Spec}(A)$ is a scheme that may contain multiple points. Given an étale morphism $f:\mathrm{Spec}(B)\to \mathrm{Spec}(A)$$f: \operatorname{Spec}(B) \to \operatorname{Spec}(A)$, we can understand its global structure by studying its local behavior at each point.

Theorem (Local Structure Theorem): Let $f:\mathrm{Spec}(B)\to \mathrm{Spec}(A)$$f: \operatorname{Spec}(B) \to \operatorname{Spec}(A)$ be an étale morphism, $\mathfrak{q}$$\mathfrak{q}$ a prime ideal in $B$$B$, and $\mathfrak{p}=f^{-1}(\mathfrak{q})=\mathfrak{q}\cap A$$\mathfrak{p} = f^{-1}(\mathfrak{q}) = \mathfrak{q} \cap A$ the corresponding prime ideal in $A$$A$. Then there exists an element $a\notin \mathfrak{p}$$a \not\in \mathfrak{p}$ of $A$$A$, elements $b_1,\dots ,b_n$$b_1, \ldots, b_n$ of $B$$B$, and polynomials $P_1(X),\dots ,P_n(X)$$P_1(X), \ldots, P_n(X)$ over $A_a$$A_a$ such that:

1. $B_a\cong A_a[X_1,\dots ,X_n]/(P_1(X_1),\dots ,P_n(X_n))$$B_a \cong A_a[X_1, \ldots, X_n]/(P_1(X_1), \ldots, P_n(X_n))$

2. The derivatives $\partial P_i/\partial X_i$$\partial P_i/\partial X_i$ are units in the quotient ring $B_a$$B_a$ (i.e., each $P_i$$P_i$ is "separable" with respect to $X_i$$X_i$ after modding out by the other variables)

This theorem indicates that étale morphisms locally look like "solutions to separable polynomial equations in multiple variables." In particular, if we consider the case over the complex field, étale morphisms correspond to local homeomorphisms in the analytic topology.

Fiber structure: For a point $x$$x$ in the scheme $X=\mathrm{Spec}(A)$$X = \operatorname{Spec}(A)$ (corresponding to a prime ideal $\mathfrak{p}\subset A$$\mathfrak{p} \subset A$), the fiber of an étale morphism $f:Y=\mathrm{Spec}(B)\to X$$f: Y = \operatorname{Spec}(B) \to X$ at $x$$x$ is $f^{-1}(x)=Y{\times }_X\mathrm{Spec}(\kappa (x))=\mathrm{Spec}(B{\otimes }_A\kappa (x))$$f^{-1}(x) = Y \times_X \operatorname{Spec}(\kappa(x)) = \operatorname{Spec}(B \otimes_A \kappa(x))$ where $\kappa (x)=A_{\mathfrak{p}}/\mathfrak{p}{A}_{\mathfrak{p}}$$\kappa(x) = A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$ is the residue field at point $x$$x$.

By the unramified condition, we know that $B{\otimes }_A\kappa (x)\cong \prod _{i=1}^r{L}_i$$B \otimes_A \kappa(x) \cong \prod_{i=1}^r L_i$ where each $L_i$$L_i$ is a finite separable field extension of $\kappa (x)$$\kappa(x)$.

This means that the fiber of an étale morphism is a separated set of finitely many points, each corresponding to a field extension: $f^{-1}(x)={\text{point}}_1,\dots ,{\text{point}}_r$$f^{-1}(x) = {\text{point}_1, \ldots, \text{point}_r}$ where the residue field of ${\text{point}}_i$$\text{point}_i$ is $L_i$$L_i$.

Branching behavior: When considering maps on curves, étale property ensures there are no "branch points." For example, for the parabola $y^2=x$$y^2 = x$ in the complex plane, there is a branch point at $x=0$$x = 0$ (where the slope becomes infinite). This mapping is not étale at $x=0$$x = 0$ because the Kähler differential is non-zero at that point. In contrast, the mapping $y^2=x(x-1)$$y^2 = x(x-1)$ is étale at all points except $x=0$$x = 0$ and $x=1$$x = 1$.

Comparison with complex analysis: In complex analysis, a holomorphic function $f:U\to \mathbb{C}$$f: U \to \mathbb{C}$ is a local homeomorphism at point $z_0$$z_0$ if and only if $f^{\prime }(z_0)\ne 0$$f'(z_0) \neq 0$. This is completely analogous to the condition ${\mathrm{\Omega }}_{B/A}^1=0$$\Omega^1_{B/A} = 0$ (i.e., the "algebraic derivative" does not vanish) for étale morphisms.

Analogy between topology/analysis and algebra/schemes:

Topological/Analytic Concept	Algebraic/Scheme Concept
Local homeomorphism	Étale morphism
Non-vanishing derivative	Kähler differential module is zero
Branch point	Non-étale point
Covering space	Étale covering

Through this analogy, many concepts from complex analysis and topology (such as covering spaces, fundamental groups, etc.) can be generalized to the setting of algebraic geometry, leading to the development of important tools like étale cohomology theory and the theory of the étale fundamental group.