Structural Properties of Finitely Generated Modules and Their Applications in Integrality Theory over Noetherian Rings - An Elements Free Approach

Definition. Let R be a ring, an R-module M is called finitely generated if there exists a $n\in \mathbb{N}$$n\in\mathbb N$ and a surjective map $\pi :R^n\to M$$\pi:R^n\to M$.

Definition. Let R be a ring and $R$$R$-${\mathbf{Mod}}^f$$\mathbf{Mod}^f$ be the category of finitely generated R-modules.

Proposition. The quotient module of a finitely generated module is still finitely generated.

Proof.

$$R^n\stackrel{\pi }{\to }M\stackrel{{\pi }^{\prime }}{\to }M/N◻$$

Proposition. The direct sum of finitely generated R-modules is still finitely generated.

Proof. Consider

$$f:R^m\twoheadrightarrow M,g:R^n\twoheadrightarrow N$$

Then we have the surjective map (since $\oplus$$\oplus$ is a bifunctor)

$$f\oplus g:R^{m+n}\cong R^m\oplus R^n\to M\oplus N◻$$

Proposition. The tensor product of finitely generated R-modules is still finitely generated.

Proof. Consider

$$R^m\twoheadrightarrow M,R^n\twoheadrightarrow N$$

Consider the exact sequence and the right exact functor $M{\otimes }_R-$$M\otimes_R-$

$$R^n⟶N⟶0$$

We have

$$M^n\cong M{\otimes }_R{R}^n⟶M{\otimes }_{R}N⟶0$$

Hence $M{\otimes }_{R}N$$M\otimes_R N$ is still finitely generated. $◻$$\square$

If R is not a Noetherian ring, then the submodule of a finitely generated R-module may not be finitely generated. For example, view $R[x_1,x_2,x_3...]$$R[x_1,x_2,x_3...]$ as a module over itself, then it is finitely generated, but $I=(x_1,x_2,x_3...)$$I=(x_1,x_2,x_3...)$ is not.

Proposition. Let R be a Noetherian ring, then the submodule of a finitely generated module is still finitely generated.

Proof. Let $f:R^m\twoheadrightarrow M$$f:R^m\twoheadrightarrow M$ be the surjective map, and $\iota :N\hookrightarrow M$$\iota:N\hookrightarrow M$ be the inclusion map. Consider the pullback, or $N^{\prime }=f^{-1}(N)\subseteq R^m$$N'=f^{-1}(N)\subseteq R^m$, it is a finitely generated R-module. Hence we have $f{|}_{N^{\prime }}\twoheadrightarrow N$$f|_{N'}\twoheadrightarrow N$ $◻$$\square$

Corollary. $R$$R$-${\mathbf{Mod}}^f$$\mathbf{Mod}^f$ is an abelian category.

Proposition. Assume that A is a Noetherian ring. Let $A\subseteq B$$A\subseteq B$ be a ring extension, then $b_1,...,b_n\in B$$b_1,...,b_n\in B$ are integral over A iff $A[b_1,...,b_n]$$A[b_1,...,b_n]$ is a finitely generated A-module.

Proof. $\Rightarrow$$\Rightarrow$ We know that $b_i$$b_i$ is integral over A iff $A[b_i]$$A[b_i]$ is a finitely generated A-module. Notice that we have

$$\pi :A[b_1]{\otimes }_A...{\otimes }_{A}A[b_n]\twoheadrightarrow A[b_1,...,b_n],x_1\otimes ...\otimes x_n⟼x_1...x_n$$

$\Leftarrow$$\Leftarrow$ Since A is a Noetherian ring, then $A[b_i]$$A[b_i]$ as a submodule of a finitely generated module is still finitely generated. $◻$$\square$