From Orbits to Minimal Polynomials: A Categorical Proof

Proposition.

Let $K$$K$ be a field, and let $G\subseteq \mathrm{Aut}(K)$$G\subseteq \mathrm{Aut}(K)$ be a finite group. 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.

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$.

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

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

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.

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

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

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$.