Categorification of Artin's Lemma

Theorem (Categorification of Artin's Lemma). Let $K$$K$ be a field, let $G\le \mathrm{Aut}(K)$$G\leq\operatorname{Aut}(K)$ be a finite subgroup, and put $F=K^G$$F=K^G$. Then the canonical map

$$\mathrm{\Phi }:K{\otimes }_{F}K\stackrel{\sim }{\to }\prod _{\sigma \in G}K,u\otimes v⟼(u\sigma (v){)}_{\sigma \in G},$$

is an isomorphism of $K$$K$-algebras. Here the source is a $K$$K$-algebra through the first tensor factor, and the target has the diagonal $K$$K$-algebra structure.

Taking $K$$K$-dimensions on both sides gives

$$[K:F]={\dim }_K(K{\otimes }_{F}K)={\dim }_K\left(\prod _{\sigma \in G}K\right)=|G|.$$

Proof. Since every element of $G$$G$ fixes $F$$F$, the displayed formula defines a $K$$K$-algebra homomorphism $\mathrm{\Phi }$$\Phi$. We show that it is an isomorphism.

We use the results established in From Orbits to Minimal Polynomials: A Categorical Proof. For any $a\in K$$a\in K$, write

$$O=G\cdot a,H={\mathrm{Stab}}_G(a),q(x)=\prod _{b\in O}(x-b).$$

$$\begin{array}{}\text{(1)}& F(a)=K^H,[F(a):F]=\mathrm{deg}q=|O|=[G:H]\le |G|.\end{array}$$

We now show that $K/F$$K/F$ is a simple extension. By (1), the degrees $[F(a):F]$$[F(a):F]$, as $a$$a$ ranges over $K$$K$, are positive integers bounded above by $|G|$$|G|$. Choose $a\in K$$a\in K$ such that

$$d=[F(a):F]$$

is maximal, and retain the notation $O$$O$, $H$$H$, and $q$$q$ for this choice of $a$$a$.

Let $b\in K$$b\in K$. Both $a$$a$ and $b$$b$ are separable over $F$$F$, so $F(a,b)/F$$F(a,b)/F$ is a finite separable extension. By the primitive element theorem, there exists $c\in F(a,b)$$c\in F(a,b)$ such that

$$F(a,b)=F(c).$$

The maximality of $d$$d$ gives

$$d=[F(a):F]\le [F(a,b):F]=[F(c):F]\le d.$$

Thus $F(a,b)=F(a)$$F(a,b)=F(a)$, and hence $b\in F(a)$$b\in F(a)$. Since $b$$b$ was arbitrary,

$$K=F(a).$$

In particular, $K/F$$K/F$ is finite and separable.

If $\sigma \in H$$\sigma\in H$, then $\sigma$$\sigma$ fixes both $F$$F$ and $a$$a$, and therefore fixes every element of $K=F(a)$$K=F(a)$. Hence

$$H=\{1\}.$$

It follows that

$$G\stackrel{\sim }{\to }O,\sigma ⟼\sigma (a),$$

is a bijection. We may therefore write

$$q(x)=\prod _{\sigma \in G}(x-\sigma (a)),$$

with distinct linear factors.

Using $K\cong F[x]/(q)$$K\cong F[x]/(q)$, extension of scalars, and the Chinese remainder theorem, we obtain $K$$K$-algebra isomorphisms

$$\begin{array}{rl}K{\otimes }_{F}K& \cong K{\otimes }_F(F[x]/(q))\\ & \cong K[x]/(q)\\ & \cong \prod _{\sigma \in G}K.\end{array}$$

The last isomorphism is evaluation at the roots $\sigma (a)$$\sigma(a)$.

It remains to identify this composite with $\mathrm{\Phi }$$\Phi$. Since $a$$a$ is algebraic over $F$$F$, every $v\in K=F(a)$$v\in K=F(a)$ can be written as $v=p(a)$$v=p(a)$ for some $p\in F[x]$$p\in F[x]$. Under the composite above,

$$u\otimes p(a)⟼u\otimes [p(x)]⟼[up(x)]⟼(up(\sigma (a)){)}_{\sigma \in G}.$$

Since $\sigma$$\sigma$ fixes the coefficients of $p$$p$,

$$p(\sigma (a))=\sigma (p(a))=\sigma (v).$$

Thus the composite sends

$$u\otimes v⟼(u\sigma (v){)}_{\sigma \in G},$$

which is precisely $\mathrm{\Phi }$$\Phi$. Therefore $\mathrm{\Phi }$$\Phi$ is an isomorphism.

Finally, extension of scalars preserves the dimension of a vector space, so

$${\dim }_K(K{\otimes }_{F}K)={\dim }_{F}K=[K:F].$$

The stated degree formula follows. $◻$$\square$

Remark. The primitive element theorem used here is taken with an elementary proof independent of Artin's lemma and the Galois correspondence. The choice of $a$$a$ is only used to prove that $\mathrm{\Phi }$$\Phi$ is an isomorphism; the map $\mathrm{\Phi }$$\Phi$ itself does not depend on this choice.

Corollary. With the notation of the theorem, for every subgroup $H\le G$$H\leq G$,

$$\mathrm{Gal}(K/K^H)=H.$$

Proof. Put $E=K^H$$E=K^H$. Applying the theorem to $H$$H$ gives a $K$$K$-algebra isomorphism

$${\mathrm{\Phi }}_H:K{\otimes }_{E}K\stackrel{\sim }{\to }\prod _{h\in H}K,u\otimes v⟼(uh(v){)}_{h\in H}.$$

By the extension-of-scalars adjunction and the connectedness of $\mathrm{Spec}K$$\operatorname{Spec}K$, we obtain

$$\begin{array}{rl}{\mathrm{Hom}}_{E\text{-alg}}(K,K)& \cong {\mathrm{Hom}}_{K\text{-alg}}(K{\otimes }_{E}K,K)\\ & \cong {\mathrm{Hom}}_{K\text{-alg}}(\prod _{h\in H}K,K)\\ & \cong \coprod _{h\in H}{\mathrm{Hom}}_{K\text{-alg}}(K,K)\\ & \cong H.\end{array}$$

Here the third bijection follows because a $K$$K$-morphism from the connected scheme $\mathrm{Spec}K$$\operatorname{Spec}K$ to the finite disjoint union

$$\coprod _{h\in H}\mathrm{Spec}K$$

factors through a unique component. Each component contributes the unique $K$$K$-algebra endomorphism ${\mathrm{id}}_K$$\operatorname{id}_K$.

The element indexed by $h$$h$ corresponds to the coordinate projection ${\pi }_h$$\pi_h$, and hence to

$$v⟼{\pi }_h({\mathrm{\Phi }}_H(1\otimes v))=h(v).$$

Thus the resulting bijection

$$H\stackrel{\sim }{\to }{\mathrm{Hom}}_{E\text{-alg}}(K,K)$$

is the natural inclusion. Since every element of $H$$H$ is an automorphism,

$$\mathrm{Gal}(K/K^H)=H.$$

$◻$$\square$