Irreducible divisors containing an arbitrary closed set

Question

Let $X$ be a normal complex projective variety, let $V$ be a closed subset of $X$ (possibly reducible), and let $I_V$ be its ideal sheaf (consider the reduced scheme structure for example).

If $A$ is an ample divisor on $X$, then $\mathcal{O}_X(mA)\otimes I_V$ is globally generated if $m \gg 0$ by one of the definitions of ampleness. This implies in particular that there exists a divisor $A'\in |mA|$ such that $A'$ contains $V$ in its support. In fact there is much more than one divisor with this property.

In general there is no reason why $A'$ should be irreducible. In particular if $V$ contains at least two prime divisors, it is clear that every divisor containing $V$ will be reducible.

My question is the following:

Suppse $\dim V\leq \dim X-2$. Can I find, for $m \gg 0$, an irreducible divisor $A'\in |mA|$ containing $V$?

Answer 1

EDIT : Olivier Wittenberg pointed out to me that a positive answer to the question follows from Theorem 1 of [Altman-Kleiman, Bertini theorems for hypersurface sections containing a subscheme]. I keep below my previous answer (whose argument is different, but more complicated).

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The answer to your question is positive. Let $X$ be an integral projective variety of dimension $n\geq 2$ (over an algebraically closed field) and $V\subset X$ a subvariety of codimension $\geq 2$. Let $A$ be ample on $X$ : up to replacing $A$ by a multiple, we may assume that $X\subset\mathbb{P}^N$ and $A=\mathcal{O}_X(1)$.

Let $\mathcal{H}_e$ be the projective space of degree $e$ hypersurfaces in $\mathbb{P}^N$, $\mathcal{F}_e^{igr}(X)\subset\mathcal{H}_e$ the subset consisting of hypersurfaces whose intersections with $X$ are not irreducible generically reduced of codimension $1$ in $X$, and $\mathcal{G}_e(V)\subset\mathcal{H}_e$ the subset consisting of hypersurfaces containing $V$.

The exact sequence $0\to\mathcal{I}_V(e)\to\mathcal{O}_{\mathbb{P}^N}(e)\to\mathcal{O}_V(e)\to 0$ shows that, when $ e\gg 0$, the codimension of $\mathcal{G}_e(V)$ in $\mathcal{H}_e$ is a polynomial of degree $\leq n-2$ in $e$ (the Hilbert polynomial of $V$). On the other hand, Théorème 0.4 of arXiv:0911.1118 shows that, when $e\geq 2$, the codimension of $\mathcal{F}_e^{igr}(X)$ in $\mathcal{H}_e$ is $\geq \binom{e+n-1}{n-1}-n$, that is at least a polynomial of degree $n-1$ in $e$. As a consequence, if $e \gg 0$, $\mathcal{G}_e(V)$ is not included in $\mathcal{F}_e^{igr}(X)$. A hypersurface in $\mathcal{G}_e(V)$ but not in $\mathcal{F}_e^{igr}(X)$ induces on $X$ the divisor you are looking for.

Note that if $X$ is moreover $S_2$ (for instance, normal), then this divisor is itself integral.

Answer 2

EDIT 18/1 to make it clearer and somehow address the singular case:

First assume $X$ nonsingular. Denote $|mA-V|$ the linear system of divisors in $|mA|$ containing $V$. For $m\gg 0$, the base locus of $|mA-V|$ is exactly $V$ (blowing up the components of $V$ you can transform $|mA-V|$ into a base-point free system). So by Bertini (in characteristic zero), either you have an irreducible $A'$ or the image of $X$ by the corresponding map $f$ to projective space is ${\mathbb P}^1$. As diverietti says in the comments, the exact sequence in cohomology of $0 \rightarrow \mathcal{I}_V(mA) \rightarrow \mathcal{O}_X(mA) \rightarrow \mathcal{O}_V(mA)\rightarrow 0$ and Serre vanishing show that $\dim |mA-V|$ grows like $m^{\dim X}$. This being larger than 2 does not guarantee that the image of $f$ is not $\mathbb{P}^1$ (you could have $H^0(\mathcal{I}_V(mA))=H^0(f^*(\mathcal{O}_{\mathbb{P}^1}(k))$) but if that were the case, the rate of growth of $\dim |tmA-V|$ shows that for $t \gg 0$, $H^0(\mathcal{I}_V(tmA))$ strictly contains $H^0(f^*(\mathcal{O}_{\mathbb{P}^1}(tk))$ and so for some $t$ the system is not composed with a pencil.

In the singular case, let $\pi:Y \rightarrow X$ be a resolution of singularities such that $E=\pi^{-1}(\operatorname{Sing}(X))$ is a divisor, and assume that (*) for each component $V_i$ of $V$ there is an irreducible $\tilde V_i\subset Y$ of codimension at least 2 with $\pi(\tilde V_i)=V_i$. Let $\tilde A$ be an ample divisor of the form $\pi^*(kA)-D$ where $D$ is some divisor supported at $E$. The proof for the nonsingular case gives an irreducible divisor in $|m\tilde A-\tilde V|$ whose image in $X$ is irreducible and lies in $|kmA-V|$.

(*) is always true if no component of $V$ is contained in the singular locus, and I guess it is also true for components contained in the singular locus but don't know of a quick proof. On the other hand, Olivier's proof does not need characteristic zero, so unless someone asks, I won't try to polish this one further.