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Jul 27, 2019 at 4:27 comment added abx @Eoin: No, the computation of the canonical bundle does not necessitate any assumption on the hypersurfaces.
Nov 14, 2018 at 16:48 history edited Zach Teitler CC BY-SA 4.0
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Jul 24, 2018 at 4:30 comment added abx No, it is still valid for a singular curve $X$ if you have in mind the arithmetic genus, that is $\dim H^1(X,\mathcal{O}_X)$. For singular curves there is also the geometric genus, that is, the genus of the normalization of $X$; it is equal to the arithmetic genus minus the contribution of the singularities.
Jul 23, 2018 at 14:53 comment added Alm Does the above-given equation for the genus hold only for polynomials that do not have singularity?
Sep 5, 2017 at 18:40 comment added roy smith The answer of abx of course also reveals that the canonical bundle of a c.i. curve of genus ≥ 2 is very ample, hence no hyperelliptic curve of any genus ≥ 2 can occur as a c.i. This is just to illustrate explicitly the remark of abx that even among the allowable genera, exceptions exist.
Feb 9, 2016 at 18:45 comment added pbelmans There is now an entry in the OEIS giving the numbers up to a certain cutoff, see oeis.org/A266322.
Dec 27, 2015 at 17:22 comment added abx You are absolutely right. Sorry I didn't pay too much attention to the precise numbers -- I just wanted to indicate the principle.
Dec 27, 2015 at 13:41 comment added pbelmans Shouldn't 6 also be in that list, using a planar curve of degree 5? I also think 12 should not be in the list, whilst 13 should (using (3,2,2)) and so does 15 (as a curve of degree 7).
Aug 30, 2014 at 5:59 comment added abx Yes. Sorry I went too fast!
Aug 30, 2014 at 4:26 comment added Noam D. Elkies You write "$g=1,\ldots,5,9,10,12,16,\ldots$", but $g=2$ does not occur (while $g=0$ of course does).
Aug 29, 2014 at 20:20 comment added Robin Thanks! I feel a little foolish to have not realized (2)...
Aug 29, 2014 at 20:16 vote accept Robin
Aug 29, 2014 at 20:14 history answered abx CC BY-SA 3.0