A question on linear projection of a smooth projective variety

Let $$X$$ be a smooth, projective $$\mathbb{C}$$-variety of dimension $$n$$. Fix a closed point $$x \in X$$ and an embedding of $$X$$ in $$\mathbb{P}^m$$ for some integer $$m$$. For a given $$d$$, denote by $$\sigma_d : \mathbb{P}^m \to \mathbb{P}^{N_d}$$ the $$d$$-tuple embedding. My question is: for $$d \gg 0$$, does there exist a linear subspace $$L_x \subset \mathbb{P}^{N_d}$$ (depending on $$x$$) of dimension $$N_d-n-2$$, not intersecting $$\sigma_d(X)$$ such that for the linear projection from $$L_x$$ (sometimes called projection with centre $$L$$): $$\pi_{L_x} : \sigma_d(X) \to \mathbb{P}^{n+1}$$ we have $$\pi_{L_x}^{-1}(\pi_{L_x}(\sigma_d(x)))=\sigma_d(x)$$ i.e., the preimage of $$\sigma_d(x)$$ is only $$\sigma_d(x)$$ for the chosen closed point $$x$$? Any hint/reference is most welcome.

EDIT Note that, the choice of $$L_x$$ depends on the choice of $$x$$.

• Do you want the equality $\pi_L^{-1}(\pi_L(\sigma_d(x)))=\sigma_d(x)$ to be set-theoretical or scheme-theoretical? Apr 6 at 5:33
• @Sasha I would prefer the equality to be scheme-theoretic, but set-theoretic equality is also OK. From the work of Joel Roberts (Theorem 1 of "Generic projections of algebraic varieties") it seems that a scheme-theoretic equality is possible for a generic point of $X$. But, I am not sure a similar result exists for all points of $X$ i.e., if my question has a positive answer. Apr 6 at 8:11
• For the scheme-theoretic equality the answer is negative, see my answer. Apr 6 at 8:58

It is definitely not possible to expect the existence of a linear projection $$\pi_L$$ such that the equality $$\pi_L^{-1}(\pi_L(\sigma_d(x)))=\sigma_d(x)\tag{*}$$ holds scheme-theoretically for each point $$x \in X$$. Indeed, this equality means that $$\pi_L$$ defines an isomorphism of $$X$$ onto its image in $$\mathbb{P}^{n+1}$$, which is thus a smooth hypersurface. But for $$n \ge 3$$ a smooth hypersurface in $$\mathbb{P}^{n+1}$$ has Picard group isomorphic to $$\mathbb{Z}$$, while the Picard group of $$X$$ could be arbitrary.
On the other hand, for a given point $$x \in X$$ it is easy to find a subspace $$L$$ such that $$(*)$$ holds. Indeed, the closure of the union of lines joining $$x$$ with other points of $$X$$ has dimension at most $$n + 1$$, and any $$L$$ disjoint from this union works.
• Perhaps it is not clear from the question, I do not expect $\pi_L^{-1}(\pi_L(\sigma_d(x))=\sigma_d(x)$ for each point $x$ and fixed $L$. Of course, I do not expect $\pi_L$ to be an isomorphism. My question was, given an $x$, can I find an $L$ such that $\pi_L^{-1}(\pi_L(\sigma_d(x))=\sigma_d(x)$ . For different $x$, I will need to choose different $L$. I have edited the question to make this point slightly clearer. Apr 6 at 10:23