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Let $X$ be a smooth projective variety and $Z$ is a closed subscheme in $X$ which is not a complete intersection in $X$. Assume the dimension of $X$ (resp. $Z$) is greater than $3$ (resp. $1$). Then,

1) For a general hyperplane $H$ in $X$, is it possible that $H.Z$ is complete intersection in $X$?

2) Suppose $X$ is embedded in $\mathbb{P}^n$. Fix a point $p \in \mathbb{P}^n$. Consider a projection map $\mbox{pr}:\mathbb{P}^n \backslash \{p\} \to \mathbb{P}^{n-1}$ from $p$. Denote by $X'$ (resp. $Z'$) the image of $X \backslash \{p\}$ (resp. $Z \backslash \{p\}$) in $\mathbb{P}^{n-1}$ under the morphism $\mbox{pr}$. Is it possible $Z'$ a complete intersection in $X'$?

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Yes for both questions. For (1) take $X = P^3$ and let $Z$ be a line with an embedded point. For (2) take $X = P^3$ and let $Z$ be a twisted rational cubic. Then $Z'$ is a nodal cubic curve on $X' = P^2$ which is a complete intersection.

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  • $\begingroup$ Thanks for the answer. Could you also give an example to the first question in the case $X$ is an hypersurface in $\mathbb{P}^{2N+1}$ and $Z$ is of codimension $N$ in $X$? Assume $N>1$. $\endgroup$
    – Jana
    Commented Nov 1, 2013 at 9:18
  • $\begingroup$ You can do the same trick by taking $Z_0$ to be a complete intersection in $X$ and taking $Z$ to be $Z_0$ with an embedded point so that it is no longer a complete intersection. Then $H\cap Z = H\cap Z_0$ is a complete intersection. $\endgroup$
    – Sasha
    Commented Nov 1, 2013 at 9:36

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