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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!

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    $\begingroup$ 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. $\endgroup$ Oct 27, 2015 at 8:42
  • $\begingroup$ 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? $\endgroup$ Oct 27, 2015 at 8:50
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    $\begingroup$ Why do you say "not smooth as an element of $|L|$"? Do you mean is the corresponding divisor on $X$ not smooth? $\endgroup$
    – Cass
    Oct 27, 2015 at 21:23
  • $\begingroup$ 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. $\endgroup$
    – Cass
    Oct 27, 2015 at 21:31
  • $\begingroup$ Yes I just want to ask if the corresponding divisor will be smooth @Cass. $\endgroup$ Oct 28, 2015 at 1:32

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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 $Z$.

Combining this with John's example in the comments, you can have sublinear systems with base points for which the general element is smooth, and you can have sublinear systems with base points for which the general element is nonsmooth.

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