Let $p:=[0,0,...,0,1] \in \mathbb{P}^n$ the point whose all the coordinates are zero except for the $n$-th. This defines a linear projection map $\phi:\mathbb{P}^n-p \to \mathbb{P}^{n-1}$, given by $[x\_0,...,x\_n] \mapsto [x\_0,...,x\_{n-1}]$. Let $X$ be a local complete intersection subscheme in $\mathbb{P}^{n-1}$. Is it true that the scheme theoretic closure of $\phi^{-1}(X)$ in $\mathbb{P}^n$ is a local complete intersection subscheme in $\mathbb{P}^n$? If not, is there any known condition on $X$ under which we have an affirmitive answer to the question? (Assume the underlying field is the field of complex numbers and $n \ge 3$)


**EDIT**: If I understand correctly, $\overline{\phi^{-1}(X)}$ is locally complete intersection at all points in $\phi^{-1}(X)$. It only remains to prove it is locally complete intersection at $p$ (if $p \in \overline{\phi^{-1}(X)}$). Since $p$ is of codimension at least $2$ Hartshorne's Ex. III.$3.5$ tells us that the ideal sheaf of $\overline{\phi^{-1}(X)}$ in $\mathbb{P}^n$ is simply the pushforward of the ideal sheaf of $\phi^{-1}(X)$ in $\mathbb{P}^n\backslash p$ under the open immersion $i:\mathbb{P}^n\backslash p \to \mathbb{P}^n$. Finally, we just need to show that the stalk at $p$ of the ideal sheaf of $\overline{\phi^{-1}(X)}$ is generated by a regular sequence. The question is then, when is this possible?



Linear projection from a point and local complete intersection
The ideal of $\overline{\phi^{-1}(X)}$ in an affine neighborhood of $p$ is just the graded ideal of $X$, viewed as a usual ideal. So I think for any locally complete intersection that is not a complete intersection you will be in trouble.


Let $m$ be the maximal ideal at $p$. If it is a locally complete intersection, the dimension of $I/mI$ should be the codinension of $X$. But a basis of $I/mI$ gives a set of generators of the graded ideal of $X$, making it a flobal complete intersection.


The twisted cubic curve should provide a suitable counterexample.


