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I would like to understand the calculation of higher Ext groups of a skyscraper sheaf $\mathcal{O}_p$ at $p$.

The calculation I have seen does this using a Koszul resolution. It starts out like this

"Assuming X is affine, local coordinates near p define a section s of \mathcal{O}^n (n = dim X) vanishing transversely at p. "

I know that for such a section we get a koszul resolution of $\mathcal{O}_p$ that will lead to the Ext groups. However, I fail to understand the quotation. More specifically: how is this section obtained and why is $p$ the zero locus?

Thanks in advance for your answers.


ps: the calculation is from

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up vote 8 down vote accepted

If $p\in X=Spec(A)$ is a regular point corresponding to the maximal ideal ${\frak m}\subset A$, you may assume (by definition of "regular point") that there exist functions $s_1,s_2,...,s_n\in A$ such that their germs $s_{1,p},s_{2,p},...,s_{n,p} \in \mathcal O_{X,p}=A_{\frak m}$ generate $\frak m$ .
The required section of $\mathcal O_{X}^n$ is then $(s_1,s_2,...,s_n)\in \Gamma(X,\mathcal O_{X}^n)=\Gamma(X,\mathcal O_{X})^n=A^n$.
"Transversally" refers to the fact that our sections locally generate the maximal ideal.

In the more classical language of complex manifolds, you would take a point $p$ of a holomorphic manifold and an open neighbourhood $X$ of $p$ on which you have local coordinates $z_1,z_2,...,z_n$ with $z_i(p)=0 $.
Consider the section $(z_1,z_2,...,z_n) $ of the trivial bundle $X\times \mathbb C^n$.
The zero set $z_1=z_2=...=z_n=0$ of that section is exactly $\lbrace p\rbrace$ and moreover the differentials $d_pz_1, d_pz_2,...,d_pz_n$ are linearly independant linear forms on $T_p(X)$ : this is the differential-geometric interpretation of transversality.

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Great, thanks a lot. – Carsten Nov 13 '11 at 14:15

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