# local versus graded free resolutions

I'm currently trying to learn about syzygies. Most material is written in the context of graded rings and/or graded modules but I'm interested in a specific question about local rings. Hence I need to translate results from one context to another.

To ask a precise question: suppose you have a finitely generated module $$M$$ over a (noetherian) local ring $$(A,\mathfrak{m},k)$$. In this case one can also speak about the associated graded objects: $$N=gr_{\mathfrak{m}}M$$ over $$G=gr_{\mathfrak{m}}A$$. Suppose you know the (minimal) free resolutions $$\{F_s=A^{\oplus\beta_s}\}$$ of $$M$$ and $$\{H_i=\oplus_j G(-j)^{\beta_{i,j}}\}$$ of $$N$$. Here the $$\beta_{i,j}$$ are graded Betti numbers which represent the (minimal) number of generators for $$H_i$$ in degree $$j$$ and (I would like to say (see (2), below)) the $$\beta_{s}$$ are the total(?) Betti numbers.

1) How do these two free resolutions relate?

2) Is it true - at least - something like the formulae $$\beta_s=\sum_j\beta_{s,j}$$?

I would also appreciate if someone could indicate a (well-known?) reference doing this kind of translations. Thank you.

1) Your question is somewhat broad, but in some sense the general answer is "they don't", or at least that the relationship is not well understood in general, and I think constructing a minimal graded free resolution of $$\operatorname{gr}_{\mathfrak{m}}(M)$$ is quite a hard problem. Given a minimal free resolution $$F_{\bullet}$$ of $$M$$ over a local ring, one can construct an associated graded complex $$\operatorname{gr}_{\mathfrak{m}} (F_{\bullet})$$ of free modules, but it is rarely acyclic. In fact, $$\operatorname{gr}_{\mathfrak{m}}(F_{\bullet})$$ is a minimal graded free resolution of $$\operatorname{gr}_{\mathfrak{m}}(M)$$ if and only if $$M$$ is Koszul (this is sometimes taken as the definition of Koszul). See this paper for a reference on Koszul modules.
2) No. For example, let $$S=k[\![x,y,z]\!]$$ with $$\mathfrak{n}=(x,y,z)$$ and consider the $$S$$-module $$R=S/I$$, where $$I$$ is the ideal $$I=(xz-y^3,yz-x^4,z^2-x^3y^2)$$; one can prove $$R \cong k[\![t^4,t^5,t^{11}]\!]$$. This is a classical example of a domain of dimension $$1$$ whose associated graded ring is not Cohen-Macaulay, and it can be found, for instance, as Example 1.3 in these notes. Concretely, one can show that $$\operatorname{gr}_{\mathfrak{n}}(R) \cong k[x,y,z]/(xz,yz,z^2,y^4)$$. We observe that $$z$$ is a nonzero socle element of $$\operatorname{gr}_{\mathfrak{n}}(R)$$, so $$\operatorname{gr}_{\mathfrak{n}}(R)$$ has depth $$0$$. In particular, the Auslander-Buchsbaum formula tells us that the minimal graded $$\operatorname{gr}_{\mathfrak{n}}(S)$$-free resolution of $$\operatorname{gr}_{\mathfrak{n}}(R)$$ does not even have the same length as the minimal $$S$$-free resolution of $$M$$. That is, these modules have different projective dimensions.