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?


3 Answers 3


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.

  • 2
    $\begingroup$ Indeed, my argument is incorrect (the explanation is given in the edit to my answer). $\endgroup$
    – Sasha
    Aug 10, 2022 at 9:15
  • $\begingroup$ I understand that this is outside scope but what about simply connected $X$? $\endgroup$ Aug 10, 2022 at 9:21
  • 1
    $\begingroup$ @მამუკა ჯიბლაძე: No difference, there are smooth simply connected surfaces in $\Bbb{P}^4$ which are not complete intersections. $\endgroup$
    – abx
    Aug 10, 2022 at 9:36

There is already a complete and correct answer. I am just writing another answer since I am having trouble finding the old answer mentioned in my comment above. Update. Thanks to user @MinseonShin for finding the old answer: A Bertini-type result for hypersurfaces containing a subvariety

The question in the original post is a special case of the following question (which I believe was asked in a previous post).

Question. Is every smooth proper closed subvariety $X$ of a smooth projective variety $Y$ contained in a smooth hypersurface $Z$ in $Y$?

Proposition. If there exists a hypersurface $Z$ in $Y$ that contains $X$ and is smooth at every point of $X$, then the total Chern class $c(N_{X/Y})$ equals $(1+[Z]|_X)\cup \alpha$ for the total Chern class $\alpha$ of a locally free sheaf of rank $\text{dim}(Y)-(1+\text{dim}(X))$.

Proof. If there is such a hypersurface $Z$, then there is a short exact sequence of locally free sheaves on $X$, $$0 \to N_{X/Z} \to N_{X/Y} \to N_{Z/Y}|_X \to 0.$$ Then, by the Whitney sum formula, the total Chern class $c(N_{X/Y})$ equals $(1+[Z]|_X)\cup c(N_{X/Z})$. QED

This fails in many cases. For instance, for the embedding of a $2$-plane $X$ in a smooth quadric hypersurface $Y$ in $\mathbb{P}^5$ (a Schubert subvariety of the Grassmannian $\text{Gr}(2,4)$, in other words), this gives, $$1+H+H^2 = (1+mH)(1+nH) = 1+(m+n)H +mnH^2,$$ for integers $m$ and $n$. Clearly this has no solution in integers, thus there is no hypersurface $Z$ in $Y$ that contains $X$ and is smooth at every point of $X$. (Note, the restriction map on Picard groups from $Y$ to $X$ is an isomorphism in this example.)


EDIT. The argument below is incorrect. Indeed, for smoothness of a divisor along $X$ one needs the zero locus of a section of $I_X/I_X^2(mH)$ to be empty (not just smooth), and this typically is impossible when the rank of this bundle (equal to $\mathrm{codim}(X)$) is less or equal than $\dim(X)$.

Yes, this follows from Bertini's Theorem.

Indeed, let us check that if $X \subset Y$ is an embedding of smooth projective varieties then there is a smooth divisor $D \subset Y$ containing $X$ (then we will proceed by induction). Indeed, let $I_X$ be the ideal of $X$ and let $H$ be an ample divisor class on $Y$. Then for $m \gg 0$ the sheaf $I_X(mH)$ is globally generated, hence by Bertini's Theorem on Y a general section of $I_X(mH)$ is a divisor smooth away from $X$.

On the other hand, for $m \gg 0$ we have $H^1(Y,I^2_X(mH)) = 0$, hence the morphism $$ H^0(Y,I_X(mH)) \to H^0(X,I_X/I_X^2(mH)) $$ is surjective, and the twisted conormal bundle $I_X/I_X^2(mH)$ is globally generated. Therefore, for a general its section (hence for general section of $I_X(mH)$) the zero locus on $X$ is also smooth (now by Bertini's Theorem on $X$), hence it is smooth everywhere.

Now, finally, we apply an inductive argument. First, we consider the embedding $X \subset X_k := \mathbb{P}^n$ and construct a smooth divisor $X_{k-1} \subset X_k$ containing $X$. Next we consider the embedding $X \subset X_{k-1}$ and construct a smooth divisor $X_{k-2} \subset X_{k-1}$ containing $X$. Iterating this procedure we construct the required chain.

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    $\begingroup$ I don't think that any smooth surface in $\mathbb{P}^4$ is contained in a smooth hypersurface, otherwise by Lefschetz theorem it would be a complete intersection? $\endgroup$
    – abx
    Aug 10, 2022 at 8:37
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    $\begingroup$ @abx: You are absolutely right, my argument is incorrect (see the edit). $\endgroup$
    – Sasha
    Aug 10, 2022 at 9:14
  • $\begingroup$ Nevertheless, this is an interesting error (and instructive to me as I failed to detect it and probably would have if I had been refereeing a paper). $\endgroup$
    – Gro-Tsen
    Aug 10, 2022 at 12:02
  • $\begingroup$ This has come up before on MO (I can find a link). One obstruction to existence of a smooth hypersurface of degree $m$ containing $X$ is the image of the top Chern class of the normal bundle of $X$ inside the quotient of the Chow ring of $X$ by the principal ideal generated by $mc_1(\mathcal{O}(1))$. $\endgroup$ Aug 10, 2022 at 14:55

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