# A Bertini-type result for hypersurfaces containing a subvariety

Let $P$ be a smooth projective variety of dimension $4$ and let $Z$ be an irreducible subvariety of dimension $2$ ($Z$ is not necessarily smooth, but you can assume it).

Is there a smooth, irreducible ample hypersurface $H \subseteq P$ such that $H$ contains $Z$?

I first tried to mimic Hartshorne's argument for Bertini theorem, but I could only prove the case when $Z$ is smooth of dimension $\leq 1$.

Then I tried to blowup $Z$ and take a general hypersurface on $\textrm{Bl}_Z(X)$. But I don't know how to control the regularity and ampleness for the pushforward of this hypersurface.

For non-smooth $Z$ the answer is in general no.

Take $P=\mathbb{P}^4$ and let $Z \subset \mathbb P^4$ be a surface with a non-normal double point $p$ (i.e. a singularity locally analytically isomorphic to the one given by two planes intersecting in a single point, it is no difficult to construct irreducible examples, for instance taking general projections from smooth surfaces in $\mathbb{P}^5$).

Then there is no smooth hypersurface $H$ containing $Z$, since the singularity $p$ has embedding dimension $4$.

Added. Here is another counterexample, showing that the answer is in general no even if $Z$ smooth.

Let $P=\mathbb{P}^4$ and $Z \subset \mathbb{P}^4$ be a smooth surface which is not a complete intersection (for instance, an abelian surface). If $H$ is any hypersurface containing $X$, by the Lefschetz hyperplane theorem the induced restriction map between Picard groups $$\textrm{Pic} \, \mathbb{P}^4 \longrightarrow \ \textrm{Pic} \, H$$ is an isomorphism. Then, if $X$ were a Cartier divisor in $H$, it would be a complete intersection there, hence a complete intersection in $\mathbb{P}^4$, contradiction.

This shows that $Z$ is not a Cartier divisor inside $H$, hence $H$ is necessarily singular at some point of $Z$.

• Even for smooth $Z$, the answer is still no. Let $S$ be a K3 surface that contains a smooth rational curve $C$. Let $P$, resp. $Z$, be the Hilbert scheme parameterizing $0$-dimensional, length $2$ closed subschemes of $S$, resp. of $C$. By Fujiki / Beauville, $P$ is a hyperKaehler manifold. Of course $Z$ is $\mathbb{P}^2$. The normal bundle of $Z$ in $P$ is $\Omega_Z$. Since this is not an extension of two invertible sheaves (as I have explained here before -- use Whitney sum), there is no divisor in $P$ containing $Z$ that is smooth at every point of $Z$. – Jason Starr Dec 16 '15 at 11:10
• @JasonStarr: another counterexample in the smooth case is provided by a surface $Z \subset \mathbb{P}^4$ which is not a complete intersection. I added this in the answer. – Francesco Polizzi Dec 16 '15 at 12:08
• @FrancescoPolizzi I rethink about your second example, and found the following thing which I cannot explain: Suppose we blowup $Z$ and obtain a new variety $\tilde P$. Choose a general ample divisor $\tilde H$ in $\tilde P$, it will cut the exceptional divisor horizontally, hence its image $H$ in $P$ should contain $Z$. Then you find a Cartier divisor containing $Z$. The problem I guess may come from $Z$ is a complete intersection in $H$ may not imply it is a c.i. in $P$. – Li Yutong Dec 17 '15 at 7:47
• Also, if the above argument works (i.e. blowup, intersect then pushforward), then any subvariety will be contained as a hypersurface of some ambient variety (but the ambient variety may not smooth as your first example showed). – Li Yutong Dec 17 '15 at 8:02
• @LiYutong: I do not completely follow your first comment. Of course there exist Cartier divisors in $P$ containing $Z$, every divisor $H \subset P$ is Cartier, since $P$ is smooth. The point is that $Z$ is not Cartier inside $H$, hence $H$ is necessarily a singular (Cartier) divisor in $P$. This means that for any choice of your "horizontal" divisor $\tilde H \subset \tilde P$, the pushforward $H:=\pi_* \tilde H$ inside $P$ will never be smooth. – Francesco Polizzi Dec 17 '15 at 9:11