I guess that the answer to this question can be given using Gröbner basis among many other computational methods, but my goal is to see if there is a more elementary way of approaching this problem.

While trying to find a free graded resolution of $k[X_0,X_1, X_2,X_3]/(X_0,X_1)$, I noticed that there's a pretty general method that consists on noticing that we have an exact sequence $$S(-1)^2\rightarrow S\rightarrow S/(X_0,X_1)\rightarrow 0$$ where the first morphism is given by the matrix $(X_0\phantom{-}X_1)$ and then, since we want the sequence to be exact, we impose $X_0f+X_1g=0$, which implies that $X_0| g$ and $X_1| f$, hence $g=hX_0$ and $f=h'X_1$. If we substitute this in the equality $X_0f+X_1g=0$, we obtain that $h=-h'$, therefore we have another morphism $$S(-2)\rightarrow S(-1)^2$$ given by $h\mapsto (X_0h, -X_1 h)$. Since the only element in the kernel is $0$, we have an exact sequence $$0\rightarrow S(-2)\rightarrow S(-1)^2\rightarrow S\rightarrow S/(X_0,X_1)\rightarrow 0$$ therefore we have a resolution of the initial module.

My question is, can apply this method to solve other similar problems? For instance, I noticed that it worked perfectly in order to find a resolution for the twisted cubic or $k[X,Y]/(X^2,Y^2)$... but what happens for example in the case where $k[X,Y]/(X^2,XY,Y^3)$?

I am not even sure if there is a faster method actually.

  • $\begingroup$ The example $S=k[X,Y]$ you have above can be easily solved to get a resolution, $0\to S(-3)\oplus S(-4)\to S(-2)^2\oplus S(-3)\to S\to S/(X^2,XY,Y^3)\to 0$ exactly as in your first example. $\endgroup$ – Mohan Apr 2 '17 at 18:24
  • $\begingroup$ @Mohan Using the same method with the divisibility? Thanks. $\endgroup$ – Verónica Segura Draper Apr 2 '17 at 18:31
  • $\begingroup$ Yes, write down the relations and use divisiblity. $\endgroup$ – Mohan Apr 2 '17 at 19:16

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