From Orbits to Minimal Polynomials and Artin's Lemma: A Categorical Proof

Proposition.

Let $K$$K$ be a field, and let $G\subseteq \mathrm{Aut}(K)$$G\subseteq \mathrm{Aut}(K)$ be a subgroup. For $a\in K$$a\in K$, let $\mathcal{O}=G\cdot a$$\mathcal O=G\cdot a$ be its orbit. Write $F=K^G$$F=K^G$ for the fixed field, and put $H={\mathrm{Stab}}_G(a)$$H=\mathrm{Stab}_G(a)$.

If $\mathcal{O}$$\mathcal O$ is finite, then $q(x)=\prod _{b\in \mathcal{O}}(x-b)$$q(x)=\prod_{b\in\mathcal O}(x-b)$ is the minimal polynomial of $a$$a$ over $F$$F$.

Also, we have

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

Proof.

Since $G$$G$ permutes $\mathcal{O}$$\mathcal O$, we have

$$\sigma q(x)=\prod _{b\in \mathcal{O}}(x-\sigma b)=q(x)$$

for every $\sigma \in G$$\sigma\in G$. Thus $q(x)\in F[x]$$q(x)\in F[x]$, and the coefficientwise action of $G$$G$ on $K[x]$$K[x]$ descends to $K[x]/(q(x))$$K[x]/(q(x))$.

We first show that $K[x]/(q(x))\cong [\mathcal{O},K]$$K[x]/(q(x))\cong[\mathcal O,K]$ as ring objects in the category of $G$$G$-sets, where $[\mathcal{O},K]$$[\mathcal O,K]$ is the internal hom, equipped with pointwise ring operations.

By the Chinese remainder theorem, there is an isomorphism of rings

$$K[x]/(q(x))\cong \prod _{b\in \mathcal{O}}K\cong [\mathcal{O},K].$$

The isomorphism is given by $[p]\mapsto f_p$$[p]\mapsto f_p$, where $f_p(b)=p(b)$$f_p(b)=p(b)$. Now we show that the actions agree.

The action on polynomials is given by

$$(\sigma p)(t)=\sum _i\sigma (c_i)t^i,p(t)=\sum _i{c}_i{t}^i,$$

while the action on the internal hom is given by

$$(\sigma \cdot f)(b)=\sigma (f({\sigma }^{-1}b)).$$

Thus

$$\begin{array}{rl}\sigma (p({\sigma }^{-1}b))& =\sigma (\sum _i{c}_i({\sigma }^{-1}b{)}^i)\\ & =\sum _i\sigma (c_i)(\sigma ({\sigma }^{-1}b){)}^i\\ & =\sum _i\sigma (c_i)b^i\\ & =(\sigma p)(b).\end{array}$$

Hence the ring isomorphism is $G$$G$-equivariant.

Now we consider the $G$$G$-invariant parts of $K[x]/(q(x))\cong [\mathcal{O},K]$$K[x]/(q(x))\cong[\mathcal O,K]$.

First, we have a natural ring isomorphism

$$F[x]/(q(x))\cong (K[x]/(q(x)){)}^G.$$

Indeed, every class in $K[x]/(q(x))$$K[x]/(q(x))$ has a unique representative $r$$r$ of degree less than $\mathrm{deg}q$$\deg q$. If the class is $G$$G$-invariant, then $[\sigma r]=[r]$$[\sigma r]=[r]$ for every $\sigma \in G$$\sigma\in G$. By uniqueness of the representative, $\sigma r=r$$\sigma r=r$, so all coefficients of $r$$r$ lie in $F$$F$. Conversely, any polynomial with coefficients in $F$$F$ represents a $G$$G$-invariant class.

Taking the global sections functor

$${\mathrm{Hom}}_G(1,-):G\text{-}\mathbf{Set}⟶\mathbf{Set},$$

we therefore obtain ring isomorphisms

$$\begin{array}{rl}F[x]/(q(x))& \cong (K[x]/(q(x)){)}^G\\ & \cong {\mathrm{Hom}}_G(1,K[x]/(q(x)))\\ & \cong {\mathrm{Hom}}_G(1,[\mathcal{O},K])\\ & \cong {\mathrm{Hom}}_G(1\times \mathcal{O},K)\\ & \cong {\mathrm{Hom}}_G(\mathcal{O},K).\end{array}$$

What we need is $F[x]/(q(x))\cong {\mathrm{Hom}}_G(\mathcal{O},K)$$F[x]/(q(x))\cong\mathrm{Hom}_G(\mathcal O,K)$. If we can prove that ${\mathrm{Hom}}_G(\mathcal{O},K)$$\mathrm{Hom}_G(\mathcal O,K)$ is a field, then we know that $q(x)$$q(x)$ is irreducible.

We have two proofs here.

The first proof is easy: notice that $\mathcal{O}\cong G/H$$\mathcal O\cong G/H$, where $H={\mathrm{Stab}}_G(a)$$H=\mathrm{Stab}_G(a)$. Hence

$${\mathrm{Hom}}_G(\mathcal{O},K)\cong {\mathrm{Hom}}_G(G/H,K)\cong K^H,$$

where the last isomorphism is given by evaluation at the coset $H$$H$. Since $K^H$$K^H$ is a field, ${\mathrm{Hom}}_G(\mathcal{O},K)$$\mathrm{Hom}_G(\mathcal O,K)$ is a field.

Thus we have $F[a]\cong K^H,[K^H:K^G]=\mathrm{deg}q=|G/H|=[G:H]$$F[a]\cong K^H,[K^H:K^G]=\deg q=|G/H|=[G:H]$.

Here is the second proof.

We prove that ${\mathrm{Hom}}_G(\mathcal{O},-)$$\mathrm{Hom}_G(\mathcal O,-)$ preserves field objects.

Readers should be familiar with extensive categories, connected objects in an extensive category, and the idea of a sketch.

For details, see the links below.

Connected objects in extensive categories

Sketch — nLab

Now let us introduce a sketch for fields.

A sketch for fields (finite limits and finite coproducts).

Start with the finite-product sketch for commutative unital rings: a sort $R$$R$, operations

$$+,\cdot :R\times R⟶R,-:R⟶R,z,e:1⟶R,$$

and the usual ring identities. Here $1$$1$ is terminal, and $z,e$$z,e$ represent zero and one.

Adjoin an object $U$$U$ and arrows $\iota ,\nu :U\to R$$\iota,\nu:U\to R$, subject to the following two requirements.

Specified finite limit.

The arrow

$$⟨\iota ,\nu ⟩:U⟶R\times R$$

is the equalizer of

$$\cdot ,e\circ {!}_{R\times R}:R\times R\rightrightarrows R.$$

Thus $U$$U$ is the object of pairs $(x,y)$$(x,y)$ satisfying $xy=1$$xy=1$. Since multiplicative inverses are unique, $\iota$$\iota$ identifies $U$$U$ with the subobject of invertible elements.

Specified finite coproduct.

The cocone

$$1\stackrel{z}{\to }R\stackrel{\iota }{\leftarrow }U$$

is a coproduct cocone. Equivalently, the canonical map

$$[z,\iota ]:1\bigsqcup U\stackrel{\sim }{\to }R$$

is an isomorphism.

In $\mathbf{Set}$$\mathbf{Set}$, the models of this sketch are precisely fields: every element is uniquely either zero or invertible. Model morphisms are unital field homomorphisms.

Now let us prove that ${\mathrm{Hom}}_G(\mathcal{O},K)$$\mathrm{Hom}_G(\mathcal O,K)$ is a field.

Since $G$$G$ acts on $K$$K$ by field automorphisms, $K$$K$ is a model of this sketch in $G\text{-}\mathbf{Set}$$G\text{-}\mathbf{Set}$. Indeed, take

$$U_K=\{(u,v)\in K\times K:uv=1\}$$

with the diagonal $G$$G$-action. The required equalizer is computed on underlying sets, and the map

$$[z,\iota ]:1\bigsqcup U_K\stackrel{\sim }{\to }K$$

is a $G$$G$-equivariant isomorphism.

Since $\mathcal{O}=G\cdot a$$\mathcal O=G\cdot a$, it is nonempty and transitive, hence connected in $G\text{-}\mathbf{Set}$$G\text{-}\mathbf{Set}$.

The representable functor ${\mathrm{Hom}}_G(\mathcal{O},-)$$\mathrm{Hom}_G(\mathcal O,-)$ preserves limits, and connectedness implies that it preserves finite coproducts. Thus it preserves all the structure specified by the field sketch, so ${\mathrm{Hom}}_G(\mathcal{O},K)$$\mathrm{Hom}_G(\mathcal O,K)$ is a field.

In either proof, $F[x]/(q(x))$$F[x]/(q(x))$ is therefore a field, so $q(x)$$q(x)$ is irreducible over $F$$F$. Since $q(x)$$q(x)$ is monic and $q(a)=0$$q(a)=0$, it is the minimal polynomial of $a$$a$ over $F$$F$.

$◻$$\square$

---

Corollary (Artin's fixed-field theorem). 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

$$[K:F]=|G|.$$

Proof. By the preceding result, for every $a\in K$$a\in K$, the minimal polynomial of $a$$a$ over $F$$F$ is

$$q(x)=\prod _{b\in G\cdot a}(x-b).$$

Thus every element of $K$$K$ is separable over $F$$F$, and

$$\begin{array}{}\text{(1)}& [F(a):F]=|G\cdot a|\le |G|.\end{array}$$

Choose $a\in K$$a\in K$ such that $d=[F(a):F]$$d=[F(a):F]$ is maximal. Such a choice exists because these degrees are positive integers bounded by $|G|$$|G|$.

For any $b\in K$$b\in K$, the extension $F(a,b)/F$$F(a,b)/F$ is finite and separable. By the primitive element theorem, there is some $c\in F(a,b)$$c\in F(a,b)$ such that

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

By maximality,

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

Hence $F(a,b)=F(a)$$F(a,b)=F(a)$, so $b\in F(a)$$b\in F(a)$. Since $b$$b$ was arbitrary, $K=F(a)$$K=F(a)$.

Now put $H={\mathrm{Stab}}_G(a)$$H=\operatorname{Stab}_G(a)$. Every element of $H$$H$ fixes both $F$$F$ and $a$$a$, hence fixes $K=F(a)$$K=F(a)$. Therefore $H=\{1\}$$H=\{1\}$, and (1) gives

$$[K:F]=[F(a):F]=|G\cdot a|=[G:H]=|G|.$$

$◻$$\square$

Remark. The maximal-degree argument is the classical one in Lang, Algebra, revised third edition, Chapter VI, §1, Lemma 1.7, used there to prove Artin's theorem (Theorem 1.8). The primitive element theorem is understood here in a form established independently of Artin's theorem.