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Most books on commutative algebra explain Grobner bases in the non graded case and minimal free resolutions in the local case. I like projective geometry and want to compute the minimal free resolution of a coherent sheaf on projective space. What follows is my understanding of how this should work, along with some points at which I am confused:

We are given a coherent sheaf $ \mathscr{F}$ on $ \mathbb{P}^n$ via generators and relations. That is, we have an exact sequence $$ \oplus \mathscr{O}(-b^2_{p}) \to \oplus \mathscr{O}(-b^1_{p}) \to \mathscr{F} \to 0 $$ with $ b^2_{p} \geq b^1_{q} $ for each $ p,q$. The first map in this sequence is given by a matrix $ (F_{ij})$ of homogeneous polynomials such that $ F_{ij}$ has degree $ b^2_j - b^1_i$. For the resolution which we are going to compute to be minimal we want $b_p^2 > b^1_q$ for each $p,q$. This brings us to our first question:

Question 1: How can we modify the presentation so that $ b^2_p > b_q^1$?

Lets continue assuming that $ b_p^2 > b_q^1 $. Write $ S$ for the homogeneous coordinate ring of $ \mathbb{P}^n$. Then we have a map $ (F_{ij}) : \oplus S \to \oplus S $. Suppose that we run Schreyer's Algorithm as explained on page 338 of Eisenbud's book. This gives us a matrix $(G_{ij}) : \oplus S \to \oplus S $ such that $$ \oplus S \xrightarrow{(G_{ij})} \oplus S \xrightarrow{(F_{ij})} \oplus S $$ Is exact.

Question 2: If we run the version of Schreyer's Algorithm from Eisenbud's book, then will it be true that the $ G_{ij}$ are homogeneous? Also, is there any way to modify the algorithm to ensure that the $G_{ij}$ all have positive degree?

Suppose that now we have produced a free resolution $ \oplus S \to \oplus S \to \dots \to \oplus S $ where the matrix entries are all homogeneous with positive degree. By modifying the gradings on all the copies of $S$, we do not destroy exactness, so we have an exact sequence of graded maps $$ \oplus S (-b^d_p) \to \dots \to \oplus S(-b^2_p) \to \oplus S(-b^1_p) $$ where $b_p^i > b_q^{i-1}$ for all $p,q$. Applying the graded $ \tilde{} $ functor gives us a free resolution of $ \mathscr{F}$.

Question 3: Is the overall picture correct?

One of the main things I am worried about if forgetting the grading when we actually do the Grobner bases computation, but if we try and carry it along then things get really messy (at least for a human). OK, I guess that is it. Thanks for reading all that!

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Write $\mathscr{F} = \tilde{M}$ for a f.g. graded module $M$ over the polynomial ring. If you apply the tilde operation to the minimal graded free resolution of $M$, you'll get a "minimal" locally free resolution of $\mathscr{F}$.

As for your specific questions, it looks like you have the correct picture in mind. For Question 1, you can't always have the inequality you want (for example take $\mathscr{F}$ to be the direct sum of $\mathcal{O}(-1)$ and the cokernel of a linear map $\mathcal{O}(-1)\to \mathcal{O}$). In any case, minimality of a map isn't about inequalities on degrees: it's about not having any nonzero maps of the form $\mathcal{O}(d) \to \mathcal{O}(d)$.

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I don't think the conlusion right above Question 1 is correct. Let $S = k[x,y,z]$ and $I = (z^2,zy,zx^3-y^4)$. Then by the Hilbert-Burch theorem a graded minimal free resolution of $I$ is given by the following presentation matrix $$ \begin{pmatrix} y & -x^3 \\ z & y^3 \\ 0 & z. \end{pmatrix} $$ By analyzing the matrix you can see that the shifts are $-2,-2,-4$ and $-3, -5$. In particular $b^2_1 = 3$ and $b^1_3 = 4$.

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