TITLE: Embedding of a complex submanifold in projective space
QUESTION [5 upvotes]: Suppose you have a projective manifold $M$, a very ample bundle $\scr L$ and  a transverse holomorphic section $s \in H^0(\scr L)$. Then the zero set of $s$ is a complex submanifold $S_M$. 
Can we have a embedding of the the projective manifold $M$ in some projective space such that image of $S_M$ will not be contained
in a hyperplane? 
You can assume $M$ has complex dimension $2$.

REPLY [7 votes]: For $M=P^2$ the projective plane and $L=S_M$ a line in $P^2$, there are simple examples of projective embeddings of $P^2$ where $L$ does not lie in a hyperplane.
To see this consider e.g. a generic 5-dimensional linear system $D$ in the 20-dimensional space of quintic curves on P^2.  Then $D$ provides an embedding of P^2 into $P^5$, and $D$ is disjoint from the 14-dimensional space of reducible quintics that contain $L$ (which is isomorphic to the space of quartic curves), because $5+14 < 20$.  Therefore no hyperplane section (no divisor in $D$) contains $L$. 
(Here $Pic(M)=Z$ so this contradicts the comment by abx; the point is the linear system embedding $M$ here is not be complete.)
The same reasoning applies to any divisor $S$ of any projective variety $M$ in $P^n$ with $dim(M)>1$.  To see this pick a finite set $T$ in $S$ with $|T| > 2 dim(M)+1$.  Then for $d$ sufficiently large, the space of divisors of degree $d$ on $M$ passing through $T$ has codimension $|T|$ in the space of all divisors of degree $d$.  (This is elementary.)  A generic linear system $D$ of degree $d$ and dimension $N= 2 dim(M)+1$ will then contain no such divisors, and hence no divisors containing $S$; and it will embed $M$ in $P^N$ since it gives a generic projection of the Veronese embedding $M$ of degree $d$.