Representability of the Root Functor and Counting Embeddings of a Finite Separable Extension

Let $k$$k$ be a field and $f(x)\in k[x]$$f(x)\in k[x]$ be a polynomial.

Let us define a functor $V_f:{\mathrm{Alg}}_k\to \mathrm{Set}$$V_f:\mathrm{Alg}_k\to\mathrm{Set}$, for a $k$$k$-algebra $L$$L$, $V_f(L)$$V_f(L)$ is the set of roots of $f(x)$$f(x)$ in $L$$L$.

Let $h:L\to H$$h:L\to H$ be a $k$$k$-algebra homomorphism, let $\alpha \in V_f(L)$$\alpha\in V_f(L)$, then $h(f(\alpha ))=f(h(\alpha ))=0$$h(f(\alpha))=f(h(\alpha))=0$, hence $h(\alpha )\in V_f(H)$$h(\alpha)\in V_f(H)$.

Thus it is a functor.

Proposition. $V_f(-)\cong {\mathrm{Hom}}_k(k[x]/(f(x)),-)$$V_f(-)\cong\mathrm{Hom}_{k}(k[x]/(f(x)),-)$.

Proof. Let $\alpha \in V_f(L)$$\alpha\in V_f(L)$, then we have an evaluation map ${\mathrm{ev}}_{\alpha }:k[x]\to L$$\mathrm{ev}_{\alpha}:k[x]\to L$ and ${\mathrm{ev}}_{\alpha }$$\mathrm{ev}_{\alpha}$ factors through $k[x]/(f(x))$$k[x]/(f(x))$ since $f(\alpha )=0$$f(\alpha)=0$. By the universal property of quotients, ${\mathrm{ev}}_{\alpha }$$\mathrm{ev}_{\alpha}$ uniquely corresponds to a $k$$k$-algebra homomorphism from $k[x]/(f(x))$$k[x]/(f(x))$ to $L$$L$.

Conversely, let $g:k[x]/(f(x))\to L$$g:k[x]/(f(x))\to L$ be a $k$$k$-algebra homomorphism, then we have $g\circ \pi :k[x]\to L$$g\circ\pi:k[x]\to L$, let $g\circ \pi (x)=\alpha$$g\circ\pi(x)=\alpha$, then $g\circ \pi ={\mathrm{ev}}_{\alpha }$$g\circ\pi=\mathrm{ev}_{\alpha}$. Also we have $f(\alpha )=f(g\circ \pi (x))=g(\pi (f(x)))=g(0)=0$$f(\alpha)=f(g\circ\pi(x))=g(\pi(f(x)))=g(0)=0$.

Hence we have ${\mathrm{\Phi }}_L:g⟼g(\pi (x))\in V_f(L)$$\Phi_L:g\longmapsto g(\pi(x))\in V_f(L)$, easy to see they are inverse of each other, and we can check the diagram of natural transformation.

$$\begin{array}{ccc}{\mathrm{Hom}}_k(k[x]/(f),L)& \stackrel{{\mathrm{\Phi }}_L}{\to }& V_f(L)\\ h\circ (-)\downarrow & & \downarrow h(-)& \\ {\mathrm{Hom}}_k(k[x]/(f),H)& \stackrel{{\mathrm{\Phi }}_H}{\to }& V_f(H)\end{array}$$

$◻$$\square$

Corollary. Let $k\subset K$$k\subset K$ be a finite separable extension, and $k_{\mathrm{sep}}$$k_{\mathrm{sep}}$ be the separable closure of $k$$k$. We have:

$$|{\mathrm{Hom}}_k(K,k_{\mathrm{sep}})|=[K:k]$$

Proof. By the primitive element theorem, we have $K=k[\theta ]\cong k[x]/(m_{\theta }(x))$$K=k[\theta]\cong k[x]/(m_{\theta}(x))$. Let $f(x)=m_{\theta }(x)$$f(x)=m_{\theta}(x)$.

Hence we have

$${\mathrm{Hom}}_k(K,k_{\mathrm{sep}})\cong V_f(k_{\mathrm{sep}})$$

Also, $|V_f(k_{\mathrm{sep}})|=\mathrm{deg}(f)=[K:k]$$|V_f(k_{\mathrm{sep}})|=\deg(f)=[K:k]$ implies $|{\mathrm{Hom}}_k(K,k_{\mathrm{sep}})|=[K:k]$$|\mathrm{Hom}_k(K,k_{\mathrm{sep}})|=[K:k]$. $◻$$\square$