11
votes
Accepted
8
votes
Original proof of Hilbert's syzygy theorem
Actually, there is an English translation of Hilbert's
"Über die Theorie der algebraischen Formen" (Mathematische Annalen 36, 473--530, 1890), where the theorem is in Part III of that five-...
5
votes
Syzygies of projective varieties
If you want examples you could try to play around with Macaulay2 and compute syzygies for (almost) every possible example of varieties that you can produce.
4
votes
Accepted
Relative version of Hilbert syzygy theorem
The answers to the first two questions readily follow from Proposition 7.5.2 in [J. C. McConnell and J. C. Robson, "Noncommutative Noetherian rings", AMS, 1987]. Actually, they prove a much ...
3
votes
Accepted
Degrees of syzygies of points in $\mathbb P^2$
I asked David Eisenbud and was told about Exercises 12 and 13 of Chapter 3 in his book "Geometry of Syzygies". Putting together, they show that for a generic set of $n$ points $X$ in $\mathbb P^2$, ...
3
votes
Accepted
Constructing a free resolution of a $\mathbb Z[x_1,\dotsc,x_n]$-module using a related free resolution of a $\mathbb Q[x_1,\dotsc,x_n]$-module
Let $\mathfrak{a}:=(X_1-2X_2,X_1-2X_3,X_1)$ as ideal of $R$. Then the Koszul complex of the mentioned generating set of $\mathfrak{a}$ is not acyclic, because $X_1-2X_2,X_1-2X_3,X_1$ is not a regular ...
2
votes
For a local complete intersection ring $(R,\mathfrak m)$ with $\mathfrak m^3=0\ne \mathfrak m^2$, $\mathfrak m$ can be generated by two elements
I an writing this as an answer. It is indeed the case that the graded ring of a local Artinian complete intersection ring can fail to be a local Artinian complete intersection ring. Here is a direct ...
Community wiki
2
votes
Stability of syzygy bundles of smooth curves
I think what you look for is in Butler's paper [3], Theorem 1.2, where he proves semistability of the syzygy bundle for any semistable vector bundle $\cal{E}$ over a curve $C$ of slope $\mu(\cal{E})\...
1
vote
Relative version of Hilbert syzygy theorem
I contacted some people privately and was suggested the following reference, which answers both the projective and injective dimension questions. The flat dimension question surely can be dealt with ...
1
vote
Relative version of Hilbert syzygy theorem
I think that I can answer the first one of the three questions (which I guess is the simplest one), about the flat dimension. The idea is to reduce the problem to the case when $R$ is a field, using ...
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