# Simple object of $k[X,Y]/(Y^2)$

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 $$\mathcal{C}$$?

• 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? – Marco Farinati Jul 6 at 23:29
• 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. – Ming Lu Jul 7 at 8:14
• 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 – Marco Farinati Jul 7 at 18:48
• With multilication by $Y$ as differential, and at some point, truncate by the image of the differential. – Marco Farinati Jul 7 at 18:49
• So, you are right that you have an exact category, but the kernels of multiplicative by $Y$ is not in that category – Marco Farinati Jul 7 at 18:50

For any $$\lambda\in k$$, $$A/(X-\lambda Y)$$ is another example, I think.
• But these modules have infinite projective dimension if $\lambda\neq0$. – Ming Lu Jul 7 at 8:15
• @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. – Jeremy Rickard Jul 7 at 8:24
• Another question: what is the Grothendieck group of $\mathcal{C}$? is it free of rank one? – Ming Lu Jul 8 at 4:11