Let $X$ be a very general surface of degree $\ge 5$ in $\mathbb{P}^3$ and $ Y$ is arbitrary irreducible cubic hypersurface. Is $X \cap Y$ reduced ?
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Yes (of course, $Y$ should be reduced). Since $X$ is very general we may assume $\operatorname{Pic}(X)=\mathbb{Z}\cdot [\mathscr{O}_X(1)]$. If the divisor $Y_{|X}$ on $X$ is not reduced, it is of the form $2H+H'$, where $H$ and $H'$ are hyperplane sections of $X$ (possibly equal). But since the restriction map $H^0(\mathbb{P}^3,\mathscr{O}_{\mathbb{P}^3}(3))\rightarrow H^0(X,\mathscr{O}_X(3))$ is injective, this implies that $Y$ is non reduced.
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$\begingroup$ Do we need $Y$ to be reduced or irreducible is enough ? $\endgroup$ Commented Mar 2, 2020 at 4:55
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1$\begingroup$ If $Y$ is not reduced, there is no chance that $Y\cap X$ is reduced. $\endgroup$– abxCommented Mar 2, 2020 at 5:27