Let $X ⊆ \mathbb{P}^n$ be a smooth projective variety (over $\mathbb{C}$). I think we can find a chain of irreducible varieties $X = X\_0 ⊆ X\_1 ⊆ X\_2 ⊆ \cdots ⊆ X\_k = \mathbb{P}^n$ whose dimension increases by one at every step by writing $X = \mathcal{V}(f\_1, \dots, f\_n)$ and dropping some of the $f\_i$ until the dimension of the irreducible component containing $X$ increases, and then proceeding by induction on $k = \operatorname{codim}(X)$.


Is it possible to a chain where all of the $X\_i$ are smooth?



Is every smooth projective variety contained in a chain of smooth projective varieties of increasing dimension?
Suppose that $\operatorname{dim}(X)>1$ and that such a chain exists. Since $\operatorname{Pic}(\mathbf{P}^n)\simeq \mathbf{Z}$, the variety $X\_{k-1}$ is an ample divisor in $\mathbf{P}^n$, and hence by the Lefschetz hyperplane theorem we have $\operatorname{Pic}(X\_{k-1})\simeq \mathbf{Z}$. So $X\_{k-2}$ is an ample divisor on $X\_{k-1}$, and again its Picard group is $\mathbf{Z}$. By induction, we obtain that each $X\_{i-1}$ is an ample divisor on $X\_{i}$, and then by hyperplane Lefschetz for $\pi\_1$ we obtain that $\pi\_1(X)\simeq \pi\_1(\mathbf{P}^n)$ is the trivial group. So to conclude, an abelian variety of dimension at least two embedded in $\mathbf{P}^n$ gives a counterexample. (N.B. There exist abelian surfaces in $\mathbf{P}^4$, constructed by Horrocks and Mumford).


**Edit.** Of course this contradicts Sasha's answer posted roughly at the same time. I am puzzled as to where the mistake is.


