Proposition 110.76.3. Let $A$ be a Noetherian integral domain. Let $K$ denote its field of fractions. Let $\text{Mod}_ A^{fg}$ denote the category of finitely generated $A$-modules and $\mathcal{T}^{fg}$ its Serre subcategory of finitely generated torsion modules. Then $\text{Mod}_ A^{fg}/\mathcal{T}^{fg}$ is canonically equivalent to the category of finite dimensional $K$-vector spaces.
Proof. The equivalence given in Proposition 110.76.2 restricts along the embedding $\text{Mod}_ A^{fg}/\mathcal{T}^{fg} \to \text{Mod}_ A/\mathcal{T}$ to an equivalence $\text{Mod}_ A^{fg}/\mathcal{T}^{fg} \to \text{Vect}_ K^{fd}$. The Noetherian assumption guarantees that $\text{Mod}_ A^{fg}$ is an abelian category (see More on Algebra, Section 15.53) and that the canonical functor $\text{Mod}_ A^{fg}/\mathcal{T}^{fg} \to \text{Mod}_ A/\mathcal{T}$ is full (else torsion submodules of finitely generated modules might not be objects of $\mathcal{T}^{fg}$). $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)