If $L$ is a very ample line bundle over a smooth complex projective surface $X$ and $s\_0, \dots, s\_n$ is a basis of the global sections of $L$, is there some choice of $i,j$ such that the pencil generated by $s\_i$ and $s\_j$ does not have a curve in its base locus?


I know that for generic choices for the basis it is rather trivial, but I have to work with a fixed basis from which I don't have much information.


A simple example would be a basis of monomials for the global sections of $\mathcal{O}\_{\mathbb{P}^2}(n)$. There are many monomials sharing a common factor, but one can easily choose two relatively prime.



Pencils in very ample linear systems without curve in its base locus
Here is a more precise formulation of my suggestion above. First of all, I do not believe that you can find such a surface in $\mathbb{P}^3$. The base locus curves must be coordinate lines. In order to get the behavior you describe, you will need the entire coordinate tetrahedron to be contained in the surface. But then the vertices where three coordinate lines intersect would be singular points of your surface. Thus, the first place to look is in $\mathbb{P}^4$.


Let $[x\_0,x\_1,x\_2,x\_3,x\_4]$ be the standard homogeneous coordinates on $\mathbb{P}^4$. For every $0\leq i < j \leq 4$, denote by $\Pi\_{\{i,j\} }$ the common zero locus $\text{Zero}(x\_i,x\_j)$. This is a $2$-plane. For every $0\leq i < j < k \leq 4$, denote by $L\_{\{i,j,k\} }$ the common zero locus $\text{Zero}(x\_i,x\_j,x\_k)$. This is a line. Inside each $\Pi\_{\{i,j\} }$, let $B\_{\{i,j\} }$ be a line distinct from each $L\_{\{i,j,k\} }$ and such that the points of intersection $p\_{\{i,j\};k}\in L\_{\{i,j,k\} }\cap B\_{\{i,j\} }$ satisfy $(p\_{\{i,j\} ;k},p\_{\{j,k\} ;i},p\_{\{k,i\} ;j})$ are three distinct points of $L\_{\{i,j,k\} }$. Let $B$ be the curve $\cup\_{i,j} B\_{\{i,j\} }$. This is a disjoint union of $10$ lines. 


Denote by $\mathcal{I}\_{B/\mathbb{P}^4}$ the ideal sheaf of $B$ in $\mathbb{P}^4$. By Serre vanishing, there exists an integer $d\_0$ (which we could bound above, if it were important) such that for every integer $d\geq d\_0$, both $h^1(\mathbb{P}^4,\mathcal{I}\_{B/\mathbb{P}^4}(d))$ and $h^1(\mathbb{P}^4,\mathcal{I}^2\_{B/\mathbb{P}^4}(d))$ equal zero. Thus, the following restriction map is surjective, $$r\_d:V\_d\to W\_d,$$
$$H^0(\mathbb{P}^4,\mathcal{I}\_{B/\mathbb{P}^4}(d)) \to H^0(\mathbb{P}^4,\left( \mathcal{I}\_{B/\mathbb{P}^4}/\mathcal{I}^2\_{B/\mathbb{P}^4} \right)(d)). $$ Of course $\mathcal{I}\_{B/\mathbb{P}^4}/\mathcal{I}^2\_{B/\mathbb{P}^4}$ is the pushforward from $B$ of a locally free sheaf of rank $3$. For every line $M = B\_{\{ i,j \} }$, the locally free sheaf on $M$ is $N\_{M/\mathbb{P}^4}^{\vee} \cong \mathcal{O}\_M(-1)^{\oplus 3}$, the conormal sheaf of $M$ in $\mathbb{P}^4$. Thus, each section of $W\_d$ can be interpreted as a collection of sheaf homomorphisms $N\_{M/\mathbb{P}^4}\to \mathcal{O}(d)|\_M$, one for each of the $10$ lines in $B$. 


For integers $d,e\geq d\_0$, a pair $(s,t) \in W\_d\times W\_e$ can be interpreted as a collection of sheaf homomorphisms $$\phi\_{s,t,N}:N\_{M/\mathbb{P}^4}\to \mathcal{O}(d)|\_M \oplus \mathcal{O}(e)|\_M,$$ one for each of the $10$ lines in $B$. The surjectivity above implies that every such $10$-tuple of sheaf homomorphisms arises from a pair in $V\_d\times V\_e$. For a general such sheaf homomorphism $\phi$ on $M$, the ~~cokernel~~ kernel is locally free of rank $1$ and the homormophism is surjective. For instance, if we choose homogeneous coordinates $y\_0,y\_1$ on $M$, then for $d,e\geq 1$, the following homomorphism has this property, $$\phi:\mathcal{O}(1)^{\oplus 3} \to \mathcal{O}(d)\oplus \mathcal{O}(e), \ \ \ \left[ \begin{array}{rrr} y\_0^{d-1} & y\_1^{d-1} & 0 \\ 0 & y\_0^{e-1} & y\_1^{e-1} \end{array} \right] .$$ 


Because of this, for a general choice $(\widetilde{s},\widetilde{t})$ in $V\_d\times V\_e$, interpreting $\widetilde{s}$ and $\widetilde{t}$ as global sections of $\mathcal{O}\_{\mathbb{P}^4}(d)$, resp. $\mathcal{O}\_{\mathbb{P}^4}(e)$, the common zero locus $X$ of $\widetilde{s}$ and $\widetilde{t}$ contains $B$, and is everywhere a smooth surface along $B$. By Bertini's theorem, $X$ is also everywhere a smooth surface away from the base locus $B$. Thus, $X$ is a smooth complete intersection surface in $\mathbb{P}^4$ that contains $B$. In particular, because $h^1$ and $h^2$ of invertible sheaves on $\mathbb{P}^4$ are zero, it follows that the following restriction map is an isomorphism, $$H^0(\mathbb{P}^4,\mathcal{O}(1)) \to H^0(X,\mathcal{O}(1)|\_X).$$ Thus, the standard basis $x\_0,\dots,x\_4$ for $H^0(\mathbb{P}^4,\mathcal{O}(1))$ maps to a basis for $H^0(X,\mathcal{O}(1)|\_X)$. By construction, for every $0\leq i<j \leq 4$, the intersection of $X$ with the common zero locus $\text{Zero}(x\_i,x\_j)$ contains a line $B\_{\{i,j\} }$.


