From Grothendieck’s Galois Theory to the Finite Galois Correspondence

Definition–Proposition (Finite étale algebras).

Let $F$ be a field, and let $A$ be a finite-dimensional commutative $F$-algebra. Write $n={\dim }_{F}A$. The following conditions are equivalent:

1. $A$ is isomorphic, as an $F$-algebra, to a finite product

   $$A\cong E_1\times \cdots \times E_r,$$

   where each $E_i/F$ is a finite separable field extension.

2. For an algebraic closure $\bar{F}$ of $F$, there is an isomorphism of $\bar{F}$-algebras

   $$\bar{F}{\otimes }_{F}A\cong {\bar{F}}^n.$$

3. $A$ is geometrically reduced over $F$: for every field extension $L/F$, the algebra

   $$L{\otimes }_{F}A$$

   is reduced, meaning that it has no nonzero nilpotent elements.

4. There exists a finite Galois extension $L/F$ such that

   $$L{\otimes }_{F}A\cong L^n$$

   as $L$-algebras.

An algebra satisfying these conditions is called a finite étale $F$-algebra. Equivalently, the morphism

$$\mathrm{Spec}(A)⟶\mathrm{Spec}(F)$$

is finite étale.

Here $L^n$ denotes the product of $n$ copies of $L$, with coordinatewise operations. We allow the zero algebra, corresponding to the empty product.

Definition (Splitting).

Let $K/F$ be a field extension. A finite étale $F$-algebra $A$ is split by $K$ if

$$K{\otimes }_{F}A\cong K^{{\dim }_{F}A}$$

as $K$-algebras. It is called split over $F$ if $A\cong F^n$.

Example.

For a nonconstant polynomial $f\in F[t]$, the algebra $F[t]/(f)$ is finite étale over $F$ if and only if

$$gcd(f,f^{\prime })=1,$$

equivalently, $f$ has no repeated roots in an algebraic closure of $F$. In this case, it is split by $K$ if and only if $f$ splits completely over $K$.

Theorem.

Let $F\hookrightarrow K$ be a finite Galois extension, and let $G=\mathrm{Gal}(K/F)$. Then there is an equivalence of categories ${\mathsf{Et}}_F(K{)}^{\mathsf{op}}\simeq G\text{-FinSet}$. Here ${\mathsf{Et}}_F(K)$ is the category of finite étale $F$-algebras split by $K$, with $F$-algebra homomorphisms as morphisms.

Remark. $G\text{-FinSet}$ is a Boolean topos!

Proof.

Notice that $K$ is an internal commutative ring in $G\text{-Set}$, hence $\mathcal{A}:={\mathrm{Hom}}_G(-,K)$ is a functor from $G{\text{-FinSet}}^{\mathrm{op}}$ to $\mathsf{CRing}$. In fact it takes values in $F$-algebras: since $K^G=F$, the constant functions with values in $F$ are $G$-equivariant, so each ${\mathrm{Hom}}_G(X,K)$ is an $F$-algebra, and precomposition with a $G$-map is an $F$-algebra homomorphism.

We know that every $G\text{-Set}$ has the orbit decomposition: $X\cong \coprod _{i\in I}{\mathcal{O}}_i$ and each orbit $\mathcal{O}$ is isomorphic to $G/H$ for some subgroup $H$.

Hence we have

$${\mathrm{Hom}}_G(X,K)\cong {\mathrm{Hom}}_G(\coprod {\mathcal{O}}_i,K)\cong \prod _{i\in I}{\mathrm{Hom}}_G({\mathcal{O}}_i,K)\cong \prod _{i\in I}{\mathrm{Hom}}_G(G/H_i,K)\cong \prod _{i\in I}{K}^{H_i}$$

Moreover, each $K^{H_i}$ is split by $K$. Indeed, write $K^{H_i}=F[a_i]$. Since $K/F$ is normal and separable, the minimal polynomial of $a_i$ over $F$ splits into distinct linear factors over $K$. The Chinese remainder theorem therefore gives

$$K{\otimes }_F{K}^{H_i}\cong K^{[K^{H_i}:F]}.$$

Consequently, since $K{\otimes }_F-$ commutes with finite products, $\mathcal{A}(X)={\mathrm{Hom}}_G(X,K)$ belongs to ${\mathsf{Et}}_F(K)$. In particular, taking $X=G/\{1\}$ shows that $K\in {\mathsf{Et}}_F(K)$.

Conversely, define

$$\mathcal{X}:={\mathrm{Hom}}_{{\mathsf{Et}}_F(K)}(-,K)={\mathrm{Hom}}_{F\text{-alg}}(-,K):{\mathsf{Et}}_F(K{)}^{\mathsf{op}}\to G\text{-FinSet},$$

which makes sense since $K\in {\mathsf{Et}}_F(K)$. It is clear that we define a functor to $G\text{-Set}$ by $g\cdot \phi :=g\circ \phi .$ But we still need to show that ${\mathrm{Hom}}_{{\mathsf{Et}}_F(K)}(A,K)$ is finite.

Let $A\in {\mathsf{Et}}_F(K)$ and put $n={\dim }_{F}A$. Since $A$ is split by $K$, extension of scalars gives

$${\mathrm{Hom}}_{F\text{-alg}}(A,K)\cong {\mathrm{Hom}}_{K\text{-alg}}(K{\otimes }_{F}A,K)\cong {\mathrm{Hom}}_{K\text{-alg}}(K^n,K).$$

Every $K$-algebra homomorphism $K^n\to K$ is a coordinate projection: the standard orthogonal idempotents must map to $0$ or $1$, with exactly one mapping to $1$. Hence

$$|{\mathrm{Hom}}_{F\text{-alg}}(A,K)|={\dim }_{F}A,$$

so this set is finite.

Let us prove $\mathcal{X}\circ \mathcal{A}\cong {\mathrm{id}}_{G\text{-FinSet}}$ and $\mathcal{A}\circ \mathcal{X}\cong {\mathrm{id}}_{{\mathsf{Et}}_F(K)}$.

For a finite $G\text{-Set}$ $X$, define ${\eta }_X:X\to \mathcal{X}(\mathcal{A}(X)),x\mapsto {\mathrm{ev}}_x$.

The map ${\eta }_X$ is $G$-equivariant, since for $u\in \mathcal{A}(X)$

$${\mathrm{ev}}_{gx}(u)=u(gx)=g(u(x))=(g\circ {\mathrm{ev}}_x)(u),$$

and it is natural in $X$: for a $G$-map $f:X\to Y$ and $u\in \mathcal{A}(Y)$,

$$(\mathcal{X}\mathcal{A}(f)({\mathrm{ev}}_x))(u)={\mathrm{ev}}_x(u\circ f)=u(f(x))={\mathrm{ev}}_{f(x)}(u),$$

that is, $\mathcal{X}\mathcal{A}(f)\circ {\eta }_X={\eta }_Y\circ f$.

To reduce to orbits, we also need that $\mathcal{X}$ turns finite products into coproducts.

Lemma. For $A,B\in {\mathsf{Et}}_F(K)$, the map

$$\mathcal{X}(A)\bigsqcup \mathcal{X}(B)⟶\mathcal{X}(A\times B),\phi ⟼\phi \circ {\mathrm{pr}}_A,\psi ⟼\psi \circ {\mathrm{pr}}_B,$$

is a $G$-equivariant bijection.

Proof. Equivariance is clear, and the map is injective on each summand because the projections are surjective. Put $e=(1,0)$. A homomorphism of the form $\phi \circ {\mathrm{pr}}_A$ sends $e$ to $1$, while one of the form $\psi \circ {\mathrm{pr}}_B$ sends $e$ to $0$, so the images of the two summands are disjoint. Now let $\chi :A\times B\to K$ be an $F$-algebra homomorphism. Then $\chi (e)$ is an idempotent of the field $K$, so $\chi (e)\in \{0,1\}$. If $\chi (e)=1$, then for all $b\in B$,

$$\chi (0,b)=\chi (e)\chi (0,b)=\chi (e\cdot (0,b))=\chi (0)=0,$$

so $\chi =\phi \circ {\mathrm{pr}}_A$ with $\phi (a)=\chi (a,0)$. If $\chi (e)=0$, the same argument with $1-e=(0,1)$ gives $\chi =\psi \circ {\mathrm{pr}}_B$. $◻$

Now let $X=Y\bigsqcup Z$ with inclusions $i_Y,i_Z$. Restriction gives $\mathcal{A}(X)\cong \mathcal{A}(Y)\times \mathcal{A}(Z)$, under which $\mathcal{A}(i_Y)$ and $\mathcal{A}(i_Z)$ become the two projections. By naturality, ${\eta }_X{|}_Y=\mathcal{X}\mathcal{A}(i_Y)\circ {\eta }_Y$ and ${\eta }_X{|}_Z=\mathcal{X}\mathcal{A}(i_Z)\circ {\eta }_Z$, so by the lemma ${\eta }_X$ is identified with ${\eta }_Y\bigsqcup {\eta }_Z$. In particular, ${\eta }_X$ is bijective as soon as ${\eta }_Y$ and ${\eta }_Z$ are. Moreover, ${\eta }_{\varnothing }$ is trivially bijective, since $\mathcal{A}(\varnothing )=0$ and $\mathcal{X}(0)=\varnothing$, and by naturality $\eta$ is compatible with any isomorphism $\mathcal{O}\cong G/H$.

By the orbit decomposition, it is therefore enough to consider $X=G/H$, then ${\eta }_{G/H}:G/H\to {\mathrm{Hom}}_{F\text{-alg}}(K^H,K)$ is given by $gH⟼g{|}_{K^H}$.

Let $B={\mathrm{Hom}}_G(G/H,K)$. The isomorphism

$$\alpha :B\stackrel{\sim }{\to }{K}^H,f⟼f(H)$$

has inverse $b\mapsto f_b$, where $f_b(sH)=s(b)$. By precomposition with ${\alpha }^{-1}$, it induces an identification

$${\mathrm{Hom}}_{F\text{-alg}}(B,K)\stackrel{\sim }{\to }{\mathrm{Hom}}_{F\text{-alg}}(K^H,K),\lambda ⟼\lambda \circ {\alpha }^{-1}.$$

Under this identification, the evaluation homomorphism ${\mathrm{ev}}_{gH}$ corresponds to ${\mathrm{ev}}_{gH}\circ {\alpha }^{-1}$. For every $b\in K^H$,

$$({\mathrm{ev}}_{gH}\circ {\alpha }^{-1})(b)={\mathrm{ev}}_{gH}(f_b)=f_b(gH)=g(b).$$

Thus the evaluation map ${\eta }_{G/H}$ is identified with

$$G/H⟶{\mathrm{Hom}}_{F\text{-alg}}(K^H,K),gH⟼g{|}_{K^H}.$$

Since $K/F$ is Galois, the hypotheses of the following proposition hold for $G=\mathrm{Gal}(K/F)$, so this map is bijective. Therefore ${\eta }_{G/H}$, and by the reduction above every ${\eta }_X$, is an isomorphism of $G$-sets.

Proposition. Let $K/F$ be a finite separable extension, and let $G\le {\mathrm{Aut}}_F(K)$ be a finite group with $K^G=F$. For every subgroup $H\le G$, the map

$$\rho :G/H⟶{\mathrm{Hom}}_{F\text{-alg}}(K^H,K),gH⟼g{|}_{K^H}$$

is a $G$-equivariant bijection. Moreover, $[K^H:F]=[G:H]$.

Proof. Write $E=K^H$. The map is well-defined because $H$ fixes $E$ pointwise.

Choose $\theta$ with $K=F(\theta )$ and set

$$r(t)=\prod _{h\in H}(t-h(\theta ))\in E[t].$$

If $\tau \in {\mathrm{Aut}}_E(K)$, then

$$r(\tau (\theta ))=\tau (r(\theta ))=0.$$

Hence $\tau (\theta )=h(\theta )$ for some $h\in H$. Since $\theta$ generates $K/F$, we have $\tau =h$. Thus ${\mathrm{Aut}}_E(K)=H$, and consequently

$$g{|}_E=g^{\prime }{|}_E⟺g^{-1}{g}^{\prime }\in H⟺gH=g^{\prime }H.$$

This proves injectivity.

For surjectivity, choose $a$ with $E=F(a)$, let $\mathcal{O}=G\cdot a$, and put

$$q(t)=\prod _{b\in \mathcal{O}}(t-b).$$

Since $G$ permutes $\mathcal{O}$, we have $q\in K^G[t]=F[t]$. For any $F$-algebra homomorphism $\phi :E\to K$,

$$q(\phi (a))=\phi (q(a))=0.$$

Thus $\phi (a)=g(a)$ for some $g\in G$. Since $E=F(a)$, this gives $\phi =g{|}_E$, proving surjectivity. Finally,

$$\rho (sgH)=s\circ \rho (gH),$$

so $\rho$ is $G$-equivariant.

Moreover, ${\mathrm{Stab}}_G(a)=H$: since $E=F(a)$, an element $g\in G$ fixes $a$ if and only if it fixes $E$ pointwise, i.e. if and only if $g\in G\cap {\mathrm{Aut}}_E(K)=H$. Equivariant CRT and the uniqueness of representatives of degree less than $\mathrm{deg}q$ give

$$F[t]/(q)\cong (K[t]/(q){)}^G\cong {\mathrm{Hom}}_G(\mathcal{O},K)\cong K^H.$$

The composite is $[p]\mapsto p(a)$. Taking $F$-dimensions therefore yields

$$[K^H:F]=\mathrm{deg}q=|\mathcal{O}|=[G:H].◻$$

Proposition. For every $A\in {\mathsf{Et}}_F(K)$, the evaluation map

$${\epsilon }_A:A⟶{\mathrm{Hom}}_G({\mathrm{Hom}}_{F\text{-alg}}(A,K),K),a⟼(\phi \mapsto \phi (a)),$$

is an $F$-algebra isomorphism, natural in $A$.

Proof. Write

$$X_A={\mathrm{Hom}}_{F\text{-alg}}(A,K),g\cdot \phi =g\circ \phi .$$

For each $a\in A$, the function $\hat{a}:X_A\to K$ defined by $\hat{a}(\phi )=\phi (a)$ is $G$-equivariant, since

$$\hat{a}(g\cdot \phi )=(g\circ \phi )(a)=g(\hat{a}(\phi )).$$

Because the algebra operations on ${\mathrm{Hom}}_G(X_A,K)$ are pointwise, the assignment $a\mapsto \hat{a}$ defines an $F$-algebra homomorphism ${\epsilon }_A$.

We shall obtain ${\epsilon }_A$ from the following chain of isomorphisms:

$$A\stackrel{\sim }{\to }(K{\otimes }_{F}A{)}^G\stackrel{\sim }{\to }\mathrm{Map}(X_A,K{)}^G={\mathrm{Hom}}_G(X_A,K).$$

Consider the $K$-algebra homomorphism

$${\mathrm{\Theta }}_A:K{\otimes }_{F}A⟶\mathrm{Map}(X_A,K),c\otimes a⟼(\phi \mapsto c\phi (a)).$$

Extension of scalars gives a bijection

$$X_A\stackrel{\sim }{\to }{\mathrm{Hom}}_{K\text{-alg}}(K{\otimes }_{F}A,K),\phi ⟼\tilde{\phi },$$

where

$$\tilde{\phi }(c\otimes a)=c\phi (a).$$

Its inverse sends a $K$-algebra homomorphism $\lambda$ to the map $a\mapsto \lambda (1\otimes a)$.

Since $A$ is split by $K$, there is a $K$-algebra isomorphism

$$K{\otimes }_{F}A\cong K^n,n={\dim }_{F}A.$$

Every $K$-algebra homomorphism $K^n\to K$ is a coordinate projection: the standard orthogonal idempotents must map to $0$ or $1$, with exactly one mapping to $1$. Thus, under a splitting isomorphism, the maps $\tilde{\phi }$ are precisely the $n$ coordinate projections. Consequently,

$${\mathrm{\Theta }}_A(z)(\phi )=\tilde{\phi }(z)$$

records all the coordinates of $z$, so ${\mathrm{\Theta }}_A$ is an isomorphism.

Equip its source and target with the actions

$$g\cdot (c\otimes a)=g(c)\otimes a,(g\cdot u)(\phi )=g(u(g^{-1}\cdot \phi )).$$

For $c\in K$, $a\in A$, and $\phi \in X_A$, we have

$$\begin{array}{rl}(g\cdot {\mathrm{\Theta }}_A(c\otimes a))(\phi )& =g(c(g^{-1}\circ \phi )(a))\\ & =g(c)\phi (a)\\ & ={\mathrm{\Theta }}_A(g(c)\otimes a)(\phi ).\end{array}$$

Hence ${\mathrm{\Theta }}_A$ is $G$-equivariant and restricts to an $F$-algebra isomorphism

$$(K{\otimes }_{F}A{)}^G\stackrel{\sim }{\to }\mathrm{Map}(X_A,K{)}^G.$$

To identify the left-hand side, choose an $F$-basis $e_1,\dots ,e_n$ of $A$. Every element of $K{\otimes }_{F}A$ has a unique expression

$$z=\sum _{i=1}^n{c}_i\otimes e_i.$$

Since $G$ acts only on the coefficients,

$$z\in (K{\otimes }_{F}A{)}^G⟺g(c_i)=c_i\text{ for all }g\in G\text{ and all }i⟺c_i\in K^G=F\text{ for all }i.$$

Such an element is precisely

$$z=1\otimes \left(\sum _{i=1}^n{c}_i{e}_i\right).$$

Thus $a\mapsto 1\otimes a$ identifies $A$ with $(K{\otimes }_{F}A{)}^G$.

On the right-hand side, a function $u:X_A\to K$ is fixed by $G$ exactly when

$$u(g\cdot \phi )=g(u(\phi ))$$

for every $g\in G$ and $\phi \in X_A$. Therefore

$$\mathrm{Map}(X_A,K{)}^G={\mathrm{Hom}}_G(X_A,K).$$

Combining these identifications yields an isomorphism

$$A\stackrel{\sim }{\to }(K{\otimes }_{F}A{)}^G\stackrel{\sim }{\to }{\mathrm{Hom}}_G(X_A,K).$$

On elements, this composite is

$$a⟼1\otimes a⟼(\phi \mapsto \phi (a)),$$

so it is exactly ${\epsilon }_A$.

Finally, let $v:A\to B$ be an $F$-algebra homomorphism. The induced map $X_B\to X_A$ sends $\psi$ to $\psi \circ v$. For every $a\in A$ and $\psi \in X_B$,

$${\epsilon }_A(a)(\psi \circ v)=\psi (v(a))={\epsilon }_B(v(a))(\psi ).$$

This proves that ${\epsilon }_A$ is natural in $A$. $◻$

End of the proof of the Theorem. We have shown that

$$\eta :{\mathrm{id}}_{G\text{-FinSet}}⟹\mathcal{X}\circ \mathcal{A}\text{and}\epsilon :{\mathrm{id}}_{{\mathsf{Et}}_F(K)}⟹\mathcal{A}\circ \mathcal{X}$$

are natural isomorphisms. Hence $\mathcal{A}$ and $\mathcal{X}$ are mutually quasi-inverse, and

$${\mathsf{Et}}_F(K{)}^{\mathsf{op}}\simeq G\text{-FinSet}.◼$$

Corollary (Fundamental theorem of Galois theory). Let $K/F$ be a finite Galois extension, and let $G=\mathrm{Gal}(K/F)$.

1. The assignments

   $$H⟼K^H,E⟼\mathrm{Gal}(K/E)$$

   are mutually inverse, inclusion-reversing bijections between subgroups of $G$ and intermediate fields of $K/F$. Moreover,

   $$[K^H:F]=[G:H],[K:K^H]=|H|.$$

2. For every $H\le G$, restriction induces a group isomorphism

   $$N_G(H)/H\stackrel{\sim }{\to }{\mathrm{Aut}}_F(K^H),gH⟼g{|}_{K^H}.$$

   Consequently, $K^H/F$ is Galois if and only if $H⊴G$, in which case

   $$\mathrm{Gal}(K^H/F)\cong G/H.$$

Proof. (1) Under the equivalence

$${\mathsf{Et}}_F(K{)}^{\mathrm{op}}\simeq G\text{-}\mathsf{FinSet},$$

the algebra $K$ corresponds to $G$ with its left regular action. Thus subobjects of $K$ correspond to equivariant quotients of $G$.

Every quotient $q:G\twoheadrightarrow X$ makes $X$ transitive and specifies a point $x=q(1)$. Setting $H={\mathrm{Stab}}_G(x)$, the isomorphism

$$G/H\stackrel{\sim }{\to }X,gH⟼gx$$

identifies $q$ with the standard quotient $q_H:G\twoheadrightarrow G/H$. Hence quotients of $G$, up to isomorphism commuting with the quotient maps, are classified by subgroups of $G$.

Applying $\mathcal{A}$ to $q_H$ gives

$$q_H^{\ast }:{\mathrm{Hom}}_G(G/H,K)⟶{\mathrm{Hom}}_G(G,K).$$

Evaluation at $H$ and $1$, respectively, identifies this map with the inclusion

$$K^H\hookrightarrow K,$$

since $(f\circ q_H)(1)=f(H)$.

Conversely, an intermediate field $i:E\hookrightarrow K$ corresponds to the restriction quotient

$$G\twoheadrightarrow \mathcal{X}(E),g⟼g\circ i.$$

Its distinguished point is $i$, whose stabilizer is precisely $\mathrm{Gal}(K/E)$. These constructions are inverse by the equivalence, proving the claimed bijection.

For $H_1\subseteq H_2$, the factorization

$$G\twoheadrightarrow G/H_1\twoheadrightarrow G/H_2$$

corresponds to

$$K^{H_2}\hookrightarrow K^{H_1}\hookrightarrow K,$$

so the bijection reverses inclusion. Finally,

$$[K^H:F]=|\mathcal{X}(K^H)|=|G/H|=[G:H],$$

and the tower formula gives $[K:K^H]=|H|$.

(2) Put $E=K^H$. For every $g\in G$,

$$g(E)=K^{gH{g}^{-1}}.$$

By (1), $g(E)=E$ precisely when $g\in N_G(H)$. Since every $F$-embedding $E\to K$ is a restriction of an element of $G$, restriction therefore gives a surjective homomorphism

$$N_G(H)⟶{\mathrm{Aut}}_F(E)$$

with kernel $H$, proving the asserted isomorphism.

As $E/F$ is separable and $K/F$ is normal, $E/F$ is Galois precisely when every $F$-embedding $E\to K$ has image $E$. By the preceding argument, this is equivalent to $N_G(H)=G$, or $H⊴G$. $◻$

Remark. Fields in ${\mathsf{Et}}_F(K)$ correspond to transitive $G$-sets. Specifying an embedding $E\hookrightarrow K$ specifies a point of the corresponding $G$-set, and hence a particular stabilizer subgroup. Without the chosen point, the subgroup is determined only up to conjugacy.

Remark (Naturality in $K$)

Fix $F$, and consider finite Galois extensions $K/F$ together with $F$-embeddings $\iota :K\to K^{\prime }$ between them. Write ${\mathsf{Et}}_F$ for the category of all finite étale $F$-algebras, and ${\mathcal{X}}_K={\mathrm{Hom}}_{F\text{-alg}}(-,K)$.

(a) Both sides are functorial in $K$

On the algebra side, let $A\in {\mathsf{Et}}_F(K)$ and $n={\dim }_{F}A$. Regarding $K^{\prime }$ as a $K$-algebra via $\iota$, we have

$$K^{\prime }{\otimes }_{F}A\cong K^{\prime }{\otimes }_K(K{\otimes }_{F}A)\cong K^{\prime }{\otimes }_K{K}^n\cong K^{\prime n}.$$

Hence ${\mathsf{Et}}_F(K)\subseteq {\mathsf{Et}}_F(K^{\prime })$ as full subcategories of ${\mathsf{Et}}_F$. So $K\mapsto {\mathsf{Et}}_F(K)$ is covariant, and the inclusion depends only on the existence of $\iota$, not on $\iota$ itself.

On the group side, $\iota (K)/F$ is normal, so every $\sigma \in \mathrm{Gal}(K^{\prime }/F)$ maps $\iota (K)$ onto itself. This gives a homomorphism

$${\pi }_{\iota }:\mathrm{Gal}(K^{\prime }/F)⟶\mathrm{Gal}(K/F),\sigma ⟼{\iota }^{-1}\circ \sigma \circ \iota .$$

It is surjective. Given $\tau \in \mathrm{Gal}(K/F)$, we have $\iota (K)=K^{\prime H}$ for $H={\mathrm{Aut}}_{\iota (K)}(K^{\prime })$ by the Corollary. By the first Proposition applied to $K^{\prime }/F$, the $F$-embedding $\iota \tau {\iota }^{-1}:\iota (K)\to K^{\prime }$ equals $\sigma {|}_{\iota (K)}$ for some $\sigma \in \mathrm{Gal}(K^{\prime }/F)$, and then ${\pi }_{\iota }(\sigma )=\tau$.

For $K\stackrel{\iota }{\to }{K}^{\prime }\stackrel{{\iota }^{\prime }}{\to }{K}^{″}$ we have ${\pi }_{{\iota }^{\prime }\iota }={\pi }_{\iota }\circ {\pi }_{{\iota }^{\prime }}$, so $K\mapsto \mathrm{Gal}(K/F)$ is contravariant. The functor $G\mapsto G\text{-FinSet}$, given by restricting the action along homomorphisms, is also contravariant. Composing the two gives a covariant functor $K\mapsto \mathrm{Gal}(K/F)\text{-FinSet}$, whose transition functors are the inflations ${\pi }_{\iota }^{\ast }$. Since ${\pi }_{\iota }$ is surjective, ${\pi }_{\iota }^{\ast }$ is fully faithful. This matches the full inclusion on the algebra side.

(b) Compatibility with the equivalence

For $A\in {\mathsf{Et}}_F(K)$, consider the map

$${\iota }_{\ast }:{\pi }_{\iota }^{\ast }{\mathcal{X}}_K(A)⟶{\mathcal{X}}_{K^{\prime }}(A),\phi ⟼\iota \circ \phi .$$

It is injective because $\iota$ is, and both sides have ${\dim }_{F}A$ elements, so it is a bijection. It is $\mathrm{Gal}(K^{\prime }/F)$-equivariant:

$$\sigma \circ (\iota \circ \phi )=\iota \circ ({\iota }^{-1}\sigma \iota )\circ \phi =\iota \circ ({\pi }_{\iota }(\sigma )\cdot \phi ).$$

It is natural in $A$, since $\iota \circ (\psi \circ v)=(\iota \circ \psi )\circ v$. Hence the square

$$\begin{array}{ccc}{\mathsf{Et}}_F(K{)}^{\mathsf{op}}& \stackrel{{\mathcal{X}}_K}{\to }& \mathrm{Gal}(K/F)\text{-FinSet}\\ \downarrow \subseteq & & \downarrow {\pi }_{\iota }^{\ast }\\ {\mathsf{Et}}_F(K^{\prime }{)}^{\mathsf{op}}& \stackrel{{\mathcal{X}}_{K^{\prime }}}{\to }& \mathrm{Gal}(K^{\prime }/F)\text{-FinSet}\end{array}$$

commutes up to the natural isomorphism ${\iota }_{\ast }$. These isomorphisms are compatible with composition and identities:

$$({\iota }^{\prime }\circ \iota {)}_{\ast }={\iota }_{\ast }^{\prime }\circ {\iota }_{\ast },({\mathrm{id}}_K{)}_{\ast }=\mathrm{id}.$$

(c) Only pseudonaturality

Any other $F$-embedding $K\to K^{\prime }$ has the form $\iota \circ \tau$ with $\tau \in \mathrm{Gal}(K/F)$. Indeed, both embeddings have the same image, namely the subfield of $K^{\prime }$ generated by the roots of the minimal polynomial of a primitive element of $K$. Then

$${\pi }_{\iota \tau }=c_{\tau }\circ {\pi }_{\iota },c_{\tau }(g)={\tau }^{-1}g\tau .$$

On the algebra side, $\iota$ and $\iota \tau$ induce the same inclusion ${\mathsf{Et}}_F(K)\subseteq {\mathsf{Et}}_F(K^{\prime })$. On the group side, ${\pi }_{\iota }^{\ast }$ and ${\pi }_{\iota \tau }^{\ast }$ are different functors unless $\tau$ is central. They are only naturally isomorphic, via

$${\pi }_{\iota \tau }^{\ast }X\stackrel{\sim }{\to }{\pi }_{\iota }^{\ast }X,x⟼\tau x,$$

and correspondingly $(\iota \tau {)}_{\ast }={\iota }_{\ast }\circ (\tau \cdot -)$.

Thus the right-hand functor sees the choice of embedding while the left-hand one does not, and the discrepancy is absorbed by coherent natural isomorphisms. In other words, $K\mapsto {\mathcal{X}}_K$ is a pseudonatural equivalence between two functors into the 2-category $\mathsf{Cat}$, rather than a strictly natural one.

(d) Fixing a separable closure

Now fix a separable closure $F^{\mathrm{sep}}$, and use only subfields $K\subseteq F^{\mathrm{sep}}$, with inclusions as morphisms. The indexing category becomes a poset, and the ambiguity in (c) disappears.

Moreover, for $A\in {\mathsf{Et}}_F(K)$, every $F$-algebra homomorphism $A\to F^{\mathrm{sep}}$ lands in $K$, since ${\mathrm{Hom}}_{F\text{-alg}}(A,K)$ and ${\mathrm{Hom}}_{F\text{-alg}}(A,F^{\mathrm{sep}})$ both have ${\dim }_{F}A$ elements. So all the sets ${\mathcal{X}}_K(A)$ are identified with ${\mathrm{Hom}}_{F\text{-alg}}(A,F^{\mathrm{sep}})$, and each ${\iota }_{\ast }$ becomes the identity.

Passing to the union over all $K$, one obtains Grothendieck's form of Galois theory:

$${\mathsf{Et}}_F^{\mathsf{op}}\simeq {\mathrm{\Gamma }}_F{\text{-FinSet}}_{\mathrm{cts}},A⟼{\mathrm{Hom}}_{F\text{-alg}}(A,F^{\mathrm{sep}}).$$

Here ${\mathrm{\Gamma }}_F=\mathrm{Gal}(F^{\mathrm{sep}}/F)={\underleftarrow{\lim }}_K\mathrm{Gal}(K/F)$, and the right-hand side is the category of finite sets with a continuous ${\mathrm{\Gamma }}_F$-action.