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Feb 17, 2021 at 1:53 comment added LAPRAS yes. thank you very much for the correction.
Feb 16, 2021 at 13:19 comment added Evgeny Shinder Welcome to Mathoverlow! Regarding the first sentence of the question, why are you saying that a general element of a linear system is irreducible? Clearly if your linear system contains just one hypersurface, which is reducible, then it's false? Bertini's theorem will tell us that they are generically irreducible provided the system is sufficiently moving, e.g. basepoint free...
Feb 16, 2021 at 10:27 comment added Zach Teitler (Unique except when the quadric is reducible, but that doesn’t matter for the dimension count.)
Feb 16, 2021 at 8:04 comment added LAPRAS Thank you. I have not tried that way.
Feb 16, 2021 at 6:37 comment added Francesco Polizzi Did you try a dimension argument? Every reducible cubic is given by a unique pair (quadric, plane), hence the space of reducible cubics has dimension $9+3=12$, namely, codimension 7 in the projective space of all cubics...
Feb 16, 2021 at 2:41 history edited LAPRAS CC BY-SA 4.0
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Feb 16, 2021 at 2:30 history asked LAPRAS CC BY-SA 4.0