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.


1 Answer 1


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.


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