Let $X$ be an irreducible variety. Is there some necessary condition on a hyperplane $H$ such $X\cap H$ is reducible? Also, suppose that $H\cap X$ is reducible, i.e., $H\cap X=Y_1\cup Y_2 \cup \cdots \cup Y_k$, I am interested in the case when there exists a $Y_i$ such that the spam $Y_i$ is not $H$ but a subspace of codimension 2, i.e., $H\cap H'$. When can that happen? If $Y_i \subset H\cap H'\cap X$, is it possible there is another component of $Y_1\cup Y_2 \cup \cdots \cup Y_k$ in $H\cap H'$?
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$\begingroup$ All you want can happen for plane curves, so do you want to consider only higher dimensions? $\endgroup$– MohanCommented Nov 14, 2016 at 13:56
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$\begingroup$ Yes, only higher dimensions. $\endgroup$– user46071Commented Nov 14, 2016 at 14:39
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$\begingroup$ What about a one dimensional conic in 3 space degenerating into 2 skew lines ? Shouldn't the total space of that work ? $\endgroup$– mehCommented Nov 14, 2016 at 16:15
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