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Say, you have an ideal $I$ of a polynomial ring $R = K\lbrack X_1,\ldots,X_n \rbrack$ over an algebraically closed field $K$ (you can assume $K = \mathbb{C}$). What does a minimal free resolution of $R/I$ as an $R$-module tell me about the variety defined by $I$? Anything at all?

I know that this question is very similar but the answers were not satisfying for me as they essentially reduce to graded settings. I really do not want to assume that $I$ is homogenous. In this case a minimal free resolution is not unique and so the "Betti numbers" (i.e., the ranks of the terms in a minimal free resolution) are not unique. Are they still interesting?

To get something unique I could still do the following two things:

  1. Localize $R/I$ in some point. Then minimal free resolutions and the Betti numbers are unique. But what do they tell me about $R/I$? How do they relate to the non-localized ones?

  2. I could homogenize $I$ (i.e., consider the projective closure of the variety). Again minimal free resolutions and the Betti numbers are unique. But what do they tell me about $R/I$? How do they relate to the non-projectivized ones?

I know there is a bunch of literature on free resolutions and syzygies (book by Eisenbud for example) but usually everything is restricted to the graded setting.

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  • $\begingroup$ what do these Betti numbers tell you in the graded setting? $\endgroup$ – pro Jun 5 '15 at 8:36
  • $\begingroup$ Quote from Eisenbud's book: "Hilbert originally studied free resolutions because their discrete invariants, the graded Betti numbers, determine the Hilbert function. But the graded Betti numbers contain more information than the Hilbert function. [...]" $\endgroup$ – user74608 Jun 5 '15 at 8:42
  • $\begingroup$ thanks. (additional comment: in general having a free resolution is useful to compute Tor and thus intersection numbers with other subvarieties) $\endgroup$ – pro Jun 5 '15 at 9:34
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    $\begingroup$ The length of the resolution is an important invariant. If the length of the resolution equals the height of the ideal, then the quotient ring is Cohen-Macaulay. In particular, for Cohen-Macaulay quotient rings of codimension 2, this is the basis for the Hilbert-Schaps(-Burch) theorem. $\endgroup$ – Jason Starr Jun 5 '15 at 12:12

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