Let $k$ be a field. Let $A=k[X,Y]/(Y^2)$ be the quotient of polynomial ring $k[X,Y]$. Let $\mathcal{C}$ be the category of finite-dimensional $A$-modules $M$ with the action of $X$ nilpotent, and of finite projective dimension. In fact, for any $M\in\mathcal{C}$, we have $\mathrm{proj.dim}_A M\leq 1$. Then $\mathcal{C}$ is an exact category.

Let $U=k[Y]/(Y^2)$ be the quotient of $A$ by $(X)$. Then $U$ can be viewed as an $A$-module, and $U$ has finite projective dimension and is a simple object in $\mathcal{C}$. Does $\mathcal{C}$ have any other simple object? What is the Grothendieck group of $\mathcal{C}$?

  • $\begingroup$ Is it an exact category? I assume the Hom is the $A$-linear maps. The kernels of multiplicative by $Y$ seems to me that has infinite projective dimensión. Am I wrong? $\endgroup$ – Marco Farinati Jul 6 at 23:29
  • $\begingroup$ It is an exact category, because the modules of finite projective dimensions are closed under taking extensions, similar for nilpotent modules. the kernel of multiplicative by Y has finite projective dimension. $\endgroup$ – Ming Lu Jul 7 at 8:14
  • $\begingroup$ I got confussed with the extensión property, because the cokernels of multiplicative by y is not of finite projective dimension. But the kernels of multiplicación by y is on infinite dimension. Just consider the periodic complex with $k[Y]/Y^2$ evewrywhwre $\endgroup$ – Marco Farinati Jul 7 at 18:48
  • $\begingroup$ With multilication by $Y$ as differential, and at some point, truncate by the image of the differential. $\endgroup$ – Marco Farinati Jul 7 at 18:49
  • $\begingroup$ So, you are right that you have an exact category, but the kernels of multiplicative by $Y$ is not in that category $\endgroup$ – Marco Farinati Jul 7 at 18:50

For any $\lambda\in k$, $A/(X-\lambda Y)$ is another example, I think.

  • $\begingroup$ But these modules have infinite projective dimension if $\lambda\neq0$. $\endgroup$ – Ming Lu Jul 7 at 8:15
  • $\begingroup$ @user142751 No, they are the cokernels of the injective maps $A\to A$ given by multiplication by $X-\lambda Y$, so they all have projective dimension one. $\endgroup$ – Jeremy Rickard Jul 7 at 8:24
  • $\begingroup$ Thank you very much for your detailed explaination. $\endgroup$ – Ming Lu Jul 8 at 0:32
  • $\begingroup$ Another question: what is the Grothendieck group of $\mathcal{C}$? is it free of rank one? $\endgroup$ – Ming Lu Jul 8 at 4:11
  • $\begingroup$ @user142751 I don’t know, sorry. $\endgroup$ – Jeremy Rickard Jul 8 at 7:38

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