# Quotient of category of f.g. modules by subcategory

Let $$\mathcal A$$ be the category of finitely generated modules over $$A[t]$$ and $$\mathcal B$$ be its subcategory of modules which is annihilated by some power of $$t$$. Then I want to show that quotient category $$\mathcal A/B$$ is equivalent to the category of finitely generated modules over $$A[t, t^{-1}]$$, where $$A$$ is Noetherian ring.

If $$M \in \mathcal B$$, suppose $$t^{n}.M = 0$$ in that case I can define the action of $$\frac{1}{1-t}$$ to be the action of $$1 + t + ... + t^{n-1}$$. But how will I define the action of $$\frac{1}{t}$$? Any help would be great.

This is only partial answer.It seems that you should consider the localization functor.

Denote $$\mathcal C$$ the category of finite generated $$A[t,t^{-1}]$$ modules.Consider the localization funtor $$L:\mathcal A\rightarrow \mathcal C$$ which maps $$M$$ to $$M_t$$.

https://stacks.math.columbia.edu/download/homology.pdf#nameddest=02MN lemma 9.7 is well-known:Let $$F:D\rightarrow E$$ be an exact functor between two $$Abelian$$ categories. If $$D_1$$ is $$Serre$$ subcateogry of $$D$$.Then $$D=KerF$$ iff the induced functor $$\bar F:D/D_1\rightarrow E$$ is faithful.

In your question $$\mathcal B=KerL$$, hence $$\mathcal{A/B}\rightarrow C$$ is faithful. Can we do better?

Consider the category of modules. Denote $$\hat{\mathcal A}$$ the category of $$A[t]$$ modules and $$\hat{\mathcal C}$$ the category of $$A[t,t^{-1}]$$ modules, consider the localization functor $$l:\hat{\mathcal A}\rightarrow \hat{\mathcal C}$$.Since the forgetful functor is fully faithful, we know $$\hat{\mathcal A}/\hat{\mathcal B}\rightarrow \hat{\mathcal C}$$ is equivalence (here we denote $$Kerl=\hat{\mathcal B}$$).http://www.numdam.org/article/BSMF_1962__90__323_0.pdf

Since $$\mathcal A\rightarrow \hat{\mathcal A}$$ and $$\mathcal C\rightarrow \hat{\mathcal C}$$ are fully faithful.We know：

$$\mathcal{A/B}\rightarrow C$$ is fully faithful if and only if the natural functor $$\mathcal{A/B}\rightarrow \hat{\mathcal A}/\hat{\mathcal B}$$ is fully faithful.

It is easy to check that $$\mathcal{A/B}\rightarrow \hat{\mathcal A}/\hat{\mathcal B}$$ is faithful. But it seems that it is not full. In fact, I can not prove this or give an example.