Integral Closure as a Filtered Colimit and Its Applications

Let $A\subset C$$A\subset C$ be a commutative ring extension, the integral closure of $A$$A$ in $C$$C$, denoted as $B$$B$. We would like to define $B$$B$ as a small filtered colimit in $\mathsf{Ring}$$\mathsf{Ring}$, and as a corollary, we will get the following fact:

$$\boxed{{\displaystyle S^{-1}B\text{ is the integral closure of }{S}^{-1}A\text{ in }{S}^{-1}C}}$$

This follows from $S^{-1}A\otimes -$$S^{-1}A\otimes-$ is the left adjoint functor, hence preserves colimits. Let $G$$G$ be a finite group and consider $G\text{-}{\mathsf{Alg}}_A:=\mathrm{Func}(B(G),{\mathsf{Alg}}_A)$$G\text{-}\mathsf{Alg}_A:=\mathrm{Func}(B(G),\mathsf{Alg}_A)$. Now let $B$$B$ be a $G\text{-}$$G\text{-}$algebra ($G$$G$ acts on $C$$C$ via $A\text{-}$$A\text{-}$algebra automorphisms). Then we have:

$$\boxed{{\displaystyle B^G\text{ is the integral closure of }A\text{ in }{C}^G}}$$

This follows from $B^G$$B^G$ is the limit of $B$$B$ as a $G$$G$-algebra (or if you want, you could write it as the equalizer of all the $g:B\to B$$g:B\to B$) and there is a natural isomorphism exchanging filtered colimits and finite limits:

$$\boxed{{\displaystyle \underset{i\in \mathcal{I}}{\underrightarrow{\lim }}\underset{j\in \mathcal{J}}{\underleftarrow{\lim }}F(i,j)\cong \underset{j\in \mathcal{J}}{\underleftarrow{\lim }}\underset{i\in \mathcal{I}}{\underrightarrow{\lim }}F(i,j)}}$$

Here $\mathcal{I}$$\mathcal{I}$ is the filtered category and $\mathcal{J}$$\mathcal{J}$ is the small category.

Proposition. Let $A\subset C$$A\subset C$ be a ring extension, the integral closure of $A$$A$ in $C$$C$, denoted as $B$$B$, then $B$$B$ is a small filtered colimit.

Let us build the filtered system in $\mathsf{Ring}$$\mathsf{Ring}$ as follows: Let $\mathcal{I}$$\mathcal{I}$ be the following filtered category, the objects are all the $E$$E$ such that $A\subset E\subset C$$A\subset E\subset C$ and $E$$E$ is a finitely generated $A\text{-}$$A\text{-}$module.

Remark. $E$$E$ is finite generated $A$$A$-module implies that $E\subset B$$E\subset B$.

The morphisms are inclusion maps. Let $F:\mathcal{I}\hookrightarrow \mathsf{Ring}$$F:\mathcal{I}\hookrightarrow\mathsf{Ring}$ be the embedding functor. It is easy to see that $\mathcal{I}$$\mathcal{I}$ is a small filtered category.

Now we prove that $B$$B$ is the colimit of $F$$F$, i.e.,

$$\underset{i\in \mathcal{I}}{\underrightarrow{\lim }}F(i)=B$$

Let $D$$D$ be an $A\text{-}$$A\text{-}$algebra and $({\phi }_E{)}_{E\in \mathrm{Ob}(\mathcal{I})}$$(\varphi_E)_{E\in\mathrm{Ob}(\mathcal{I})}$ be the natural transformation(the cocone) from $F$$F$ to $\mathrm{\Delta }(D)$$\Delta(D)$. Define $\psi :B\to D$$\psi:B\to D$ by $\psi (x)={\phi }_E(x)$$\psi(x)=\varphi_E(x)$ if $x\in E$$x\in E$. We always could find such $E$$E$, for example, let $E=A[x]$$E=A[x]$. This is well-defined since $({\phi }_E{)}_{E\in \mathrm{Ob}(\mathcal{I})}$$(\varphi_E)_{E\in\mathrm{Ob}(\mathcal{I})}$ is compatible. Also, this is a ring homomorphism since for $x,y\in B$$x,y\in B$, let $E=A[x,y]$$E=A[x,y]$, then ${\phi }_E$$\varphi_E$ is a ring homomorphism. This morphism is unique, since if there is another $f:B$$f:B$ such that $f{|}_E={\phi }_E$$f|_E=\varphi_E$, we have $f=\psi$$f=\psi$. Hence we prove that ${\underrightarrow{\lim }}_{i\in \mathcal{I}}F(i)=B$$\varinjlim_{i\in \mathcal{I}} F(i)=B$. $◻$$\square$