Proving R[X]/I[X] ≅ (R/I)[X] via Adjoint Functors

For the reason we consider $\mathrm{Rng}$$\mathrm{Rng}$ rather than $\mathrm{Ring}$$\mathrm{Ring}$, click here.

Let us define a functor $\mathcal{X}$$\mathcal X$ from $\mathrm{Rng}$$\mathrm{Rng}$ to ${\mathrm{Rng}}_{\ast }$$\mathrm{Rng}_*$, whcih is the category of rng with base point.

$\mathcal{X}(R):=(R[X],X)$$\mathcal X(R):=(R[X],X)$, and $\begin{array}{c}\mathcal{X}(f)\left(\sum _{i=0}^n{a}_i{X}^i\right)=\sum _{i=0}^{n}f(a_i)X^i\end{array}$$\mathcal X(f)\left(\sum_{i=0}^n a_iX^i\right)=\sum_{i=0}^n f(a_i)X^i\\$.

Then we claim that $\mathcal{X}$$\mathcal X$ is the left adjoint of the forgetful functor from ${\mathrm{Rng}}_{\ast }$$\mathrm{Rng}_*$ to $\mathrm{Rng}$$\mathrm{Rng}$

$$\begin{array}{}\text{(1)}& {\mathrm{Hom}}_{{\mathrm{Rng}}_{\ast }}(\mathcal{X}(R),(S,s))={\mathrm{Hom}}_{{\mathrm{Rng}}_{\ast }}((R[X],X),(S,s))\cong {\mathrm{Hom}}_{\mathrm{Rng}}(R,S)\end{array}$$

Proof. Let $f:R\to S$$f:R \to S$ be a rng homomorphism, then $\varphi (f)={\mathrm{ev}}_s\circ X(f)\in {\mathrm{Hom}}_{{\mathrm{Rng}}_{\ast }}((R[X],X),(S,s))$$\phi(f)=\mathrm{ev}_s\circ X(f)\in \mathrm{Hom_{{Rng}_{*}}}((R[X],X),(S,s))$

For any morphism $g\in {\mathrm{Hom}}_{{\mathrm{Rng}}_{\ast }}((R[X],X),(S,s))$$g\in \mathrm{Hom_{{Rng}_{*}}}((R[X],X),(S,s))$, $\psi (g)=g{|}_R\in {\mathrm{Hom}}_{\mathrm{Rng}}(R,S)$$\psi(g)=g|_R\in\mathrm{Hom}_{\mathrm{Rng}}(R,S)$.

Easy to see that $\psi \circ \varphi =\mathrm{id}$$\psi\circ\phi=\mathrm{id}$ and $\varphi \circ \psi =\mathrm{id}$$\phi\circ \psi=\mathrm{id}$, and check this gives you the natural isomorphism. $◻$$\square$

Corollary. $\mathcal{X}$$\mathcal X$ will preserve colimit. In particular, coequalizer.

In particular, we have $R[X]/(I[X])\cong (R/I)[X]$$R[X]/(I[X])\cong (R/I)[X]$.