Let $X$ be a nonsingular projective algebraic variety over $\mathbb{C}$. Let $L$ be a base point free line bundle. Bertini's theorem asserts that general divisors in the complete linear system $|L|$ are nonsingular.


An example is $X=\mathbb{P}^n$, and $L=\mathcal{O}(m)$, for $m\ge 2$. The singular divisors of $|L|$ always forms a hypersurface (codimension =1).


My question is:


Can singular divisors form an algebraic set of arbitrary codimension in $|L|$ for $(X, L)$?


Thanks.



singular divisors in a complete linear system
The answer is yes. Here is an example generalising the one of Serge and giving arbitrary codimension. 


Consider line bundle $L=O(1)\boxtimes O(1)$ on $\mathbb CP^1\times \mathbb CP^n$. A section of such a bundle has shape $$\sum\_{i=0,j=0}^{i=1,j=n}a\_{ij}x\_iy\_j=0,$$ i.e. $|L|=\mathbb C^{2n+2}$. A section is singular iff the matrix $a\_{ij}$ has rank $1$. This happen in codimension $n$. If the rank of $a\_{ij}$ is two the section is smooth and is a $\mathbb CP^{n-1}$ bundle over $\mathbb P^1$.


