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 $V$? 



Irreducible divisors containing an arbitrary closed set
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.


