It is known that in characteristic zero, in a linear system (not necessarily complete) in $\mathbb{P}^3$ a general element is irreducible. Now my question is the following: Suppsoe we have a subspace $V \subset H^0(\mathbb{P}^3, \mathcal{O}(3))$ of dimension $\ge 3$. Then is it possible always to find an element in $V$ which is reducible ?
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3$\begingroup$ 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... $\endgroup$– Francesco PolizziCommented Feb 16, 2021 at 6:37
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$\begingroup$ Thank you. I have not tried that way. $\endgroup$– LAPRASCommented Feb 16, 2021 at 8:04
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$\begingroup$ (Unique except when the quadric is reducible, but that doesn’t matter for the dimension count.) $\endgroup$– Zach TeitlerCommented Feb 16, 2021 at 10:27
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1$\begingroup$ 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... $\endgroup$– Evgeny ShinderCommented Feb 16, 2021 at 13:19
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$\begingroup$ yes. thank you very much for the correction. $\endgroup$– LAPRASCommented Feb 17, 2021 at 1:53
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