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I am interested in (as explicit as possible) descriptions of non-Gorenstein integral projective curves. Most of the literature on singular curves appears to be focused around the Gorenstein case, with a notable exception being the paper of Kleiman, The Canonical Model of a Singular Curve, and a handful of others. Any other references would be appreciated. Specifically I am interested in knowing what has been understood about their derived categories of coherent sheaves (or the subcategory of perfect complexes), so I would very interested in any developments along those lines.

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I don't think there is a reasonably short explicit description of non-Gorenstein curves. They could be described as lacking some of those conditions that define Gorenstein curves.

In any case, here is a simple example: Let $C$ be the union of three coordinate axis in $\mathbb A^3$. Then $C$ is not Gorenstein. It is probably a good exercise for you to check that this is true.

What I find fascinating about this is that the union of three lines contained in $\mathbb A^2$ is a Gorenstein curve, so this shows that being Gorenstein or not could depend on very subtle details.

Of course, $C$ can be projected to a plane and if the projection is general, then it will be one-to-one on $C$ and an isomorphism on each component to their respective images, however, it is not an isomorphism on $C$. This follows from the fact that the image of $C$ is Gorenstein, or more directly from the fact that the Zariski tangent space of $C$ at its singular point is $3$-dimensional while that of its image is $2$-dimensional, so they can't be isomorphic.

(In case you are worried that this example is reducible, then think about how to get this singularity on an irreducible curve).

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    $\begingroup$ Thanks for your answer. I would say that an integral curve in $\mathbb{A}^3$ with a singularity at the origin analytically isomorphic to $C$ above would be non-Gorenstein, yes? If no one is able to say anything about the derived categories of such curves (or provide any other references) in a bit, I will accept your answer. $\endgroup$
    – DKS
    Commented Apr 30, 2019 at 15:39
  • $\begingroup$ Yes, that's right. $\endgroup$ Commented Apr 30, 2019 at 19:12

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