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I am trying to find out the kernel (syzygy) in a free resolution.

Here $R = K[x,y,z]$ where $K$ is a field. I am trying to resolve the ideal $M=(x,y,z)$. I have the following resolution.

$ \phi_0 : R \rightarrow R/M$ by natural map. I get kernel as $(x,y,z)$. Now I need to map $R^3$ to $R$.

$\phi_1 : R^3 \rightarrow R$ by mapping the 3 generators of $R^3$ to $x, y$ and $z$ respectively. Now I get that

$\begin{bmatrix} 0 &z &-y \\-z &0 &x \\y &-x &0 \end{bmatrix}$ will be in the kernel. But I am not being able to show that the kernel is itself generated by this.

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1 Answer 1

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So, $(f_1,f_2,f_3)\in\ker\phi_1$ iff $xf_1+yf_2+zf_3=0$. Then $xf_1\in(y,z)$, so $f_1\in(y,z)$. This shows that $f_1=yu_1-zv_1$. Do the same for $f_2,f_3$ and find what you want.

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  • $\begingroup$ Will this argument work for this statement ? $xf_1\in(y,z) \implies f_1\in(y,z)$. Agument : as $x$ is independent of $y$ and $z$ thus $f_1\in(y,z)$ $\endgroup$
    – Abhi80
    Oct 15, 2015 at 20:29
  • $\begingroup$ @Abhi80 I'd say "because the ideal $(y,z)$ is prime and $x\notin(y,z)$". $\endgroup$
    – user26857
    Oct 15, 2015 at 20:59
  • $\begingroup$ what next? Please complete the resolution. $\endgroup$
    – user118827
    Apr 15, 2017 at 7:27

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