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Let $(R,\mathfrak{m},k)$ be a Noetherian local ring and $F.$ the Koszul complex of a minimal system of generators of $\mathfrak{m}$. Let $G.$ be the minimal free resolution of $k$. In which cases they are the same?

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up vote 7 down vote accepted

In general, the Koszul complex is not exact. It is if one is working with a regular sequence. This lack of exactness is the only obstacle to the Koszul complex being a minimal free resolution (because the matrix coefficients lie in the maximal ideal). If the local ring is regular, then a minimal system of generators will be a regular sequence and the Koszul complex will furnish a minimal free resolution of $k$.

If $R$ is not regular local, then $k$ has infinite Tor-dimension (this is essentially Serre's characterization of regular local rings as those noetherian local rings with finite global dimension), so the minimal free resolution of $k$ is infinite, and in particular is not the Koszul complex. (The other direction of Serre's theorem is essentially the observation that, for a regular local ring, $k$ has finite global dimension because of the Koszul complex, as above.)

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If we work with regular sequence the Koszul complex is exact, but why then it is the minimal free resolution? – Jacob Fox Jul 18 '11 at 18:06
@Jacob: If I'm not blundering here, the minimal free resolution $\{F^\bullet\}$ is characterized by the fact that $d_i(F^i) \subset \mathfrak{m} F^{i-1}$ (where $d_i$ is the differential). This is true for the Koszul complex by inspection, since one wedges forms with a vector with coefficients in the maximal ideal. – Akhil Mathew Jul 18 '11 at 18:18

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