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Let $A=\mathbb{Q}[x,y,z]$ and let $M=A/(x,y)$,$N=A/(x,y)^3$ be two $A$-modules. I am supposed to compute a free resolution for the two modules respectively and then compute $\text{Tor}^{A}_i(M,N)$ using the two different resolutions. I am struggling with the resolutions, this is what I have done so far:

For $M$, I have found this resolution $$0 \to A \xrightarrow{\cdot (-y,x)} A \oplus A \xrightarrow{(x,y)} A \to A/(x,y) \to 0$$.

For the next part I am struggling more, I tried to use the same method, i.e $$\cdots \to A \xrightarrow{g} A^4 \xrightarrow{f} A \to A/(x,y)^3 \to 0$$ where $f$ is given by the map $(x^3,x^2y, xy^2,y^3)$ (multiplication componentwise). But I noticed that the kernel of $f$ was to difficult to describe, as of this I tried to find a resolution of the form $$((x^2), (x^2) \oplus (xy) \oplus (y^2)) \xrightarrow{(x,y)} A \to A/(x,y)^3 \to 0$$ because we have $(x,y)^3=(x^3,x^2y,xy^2,y^3)$ and the direct sum of free modules is free so indeed the component is free. However the kernels still difficult to describe and I feel that this should be easier, I asked a similar question before but did not receive an answer, this is why I am looking for some more help.

Alon Amit
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1 Answers1

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I don't see that the kernel of the map $f$ is too difficult to describe. It consists of the polynomial quadruples $(\phi_1,\phi_2,\phi_3,\phi_4)$ with $x^3\phi_1+x^2y\phi_2+xy^2\phi_3+y^3\phi_4=0$. If I am not mistaken the kernel is free on the three generators $(-y,x,0,0)$, $(0,-y,x,0)$ and $(0,0,-y,x)$.

Angina Seng
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  • But I need to have a explicit description of those quadruples in order to describe the image of $g$. How do you motivate the part with the three generators? I don't see this. –  Oct 22 '17 at 14:00
  • Don't you see that $\phi_4$ must be divisible by $x$? Then taking away a multiple of $(0,0,-y,x)$ gives you a kernel element $(\psi_1,\psi_2,\psi_3,0)$ where $x^2\psi_1+xy\psi_2+y^2\psi_3=0$ etc. – Angina Seng Oct 22 '17 at 14:04
  • Yes I do see that $\phi_4$ is divisible by $x$, but I don't get from where you removal the multiple? Sorry but I really can't see your reasoning, although it is correct for sure. Would you mind updating your answer with some details? –  Oct 22 '17 at 16:14