# Bertini's theorem, divisors which are not smooth

Let $X$ be a non-singular projective variety over $\mathbb{C}$. Let $L$ be a very ample line bundle on $X$. Then it is base point free. So the Bertini's theorem says that the general element of $|L|$ is non-singular. If $\mathcal{D}\subset |L|$ is a linear system with basepoints. Then a general member $C\in\mathcal{D}$ is non-singular away from base-points. Then does it mean any element of $\mathcal{D}$ is not smooth as an element of $|L|$? Thanks in advance!

• I'm not quite sure what you're asking, as I don't know what you mean by $|D|$, and by "as an element of $|L|$." I'll make the observation that just because a linear system has base points doesn't force elements to be singular at them: Consider the linear system of lines in $\mathbb P^2$ that pass through a fixed point. Oct 27, 2015 at 8:42
• Sorry @John Brevik, I meant $\mathcal{D}$. I made the necessary edits. Ah yes that is similar to my situation. In an abelian surface I am considering the linear system of curves through the sixteen 2-torsion points. Can we say that a general member will be smooth? Oct 27, 2015 at 8:50
• Why do you say "not smooth as an element of $|L|$"? Do you mean is the corresponding divisor on $X$ not smooth?
– Cass
Oct 27, 2015 at 21:23
• Didn't fully read John's comment when I first posted, but yeah, as you can see, "as an element of $|L|$" is throwing people off.
– Cass
Oct 27, 2015 at 21:31
• Yes I just want to ask if the corresponding divisor will be smooth @Cass. Oct 28, 2015 at 1:32

Consider $\mathcal{O}(2)$ on $\mathbb{P}^2$. We have $|\mathcal{O}(2)|\cong\mathbb{P}(Span\{x^2,xy,xz,y^2,yz,z^2\})\cong\mathbb{P}^5$. Let $Z\subseteq |O(2)|$ be the subset of points $(a_0:...:a_5)$ in $\mathbb{P}^5$ such that the hypersurface $a_0x^2+...+a_5z^2=0$ contains the point $P=(1:0:0)$ and is not smooth there. By the Jacobi criterion, this is equivalent to the set of equations $a_0=a_1=a_2=0$. So $Z$ is a sub-linear system of $|\mathcal{O}(2)|$. But by definition, the divisor corresponding to any element of $Z$ is automatically nonsmooth at the base point $P$ of $Z$.