# How to prove that a linear system is *not* base point free?

Assume I have a line bundle $\mathcal{L}$ on a projective, nonsingular variety $X$ over, say, an algebraically closed field (in fact, you may assume $\mathbb{C}$). Given global sections $s_0,\ldots,s_k\in\mathcal{L}(X)$, how to prove that they do not generate $\mathcal{L}$ globally?

I'd think that this is best done by contradiction, but I can't think of any theorems that strongly rely on the assumption that a line bundle is globally generated by certain particular sections. Note that $\mathcal{L}$ may very well be globally generated (in fact, in my case, it most certainly is) - but I want to prove that certain sections do not generate it globally. So, what I am basically asking for, are results that include a base-point free linear system in their assumptions, so I can use that to produce a contradiction. My first impule was to consider the induced morphism $\phi:X\to\mathbb{P}^k$, but that's already where I get stuck.

If it helps, the case $X=\mathbb{P}^n$ and $\mathcal{L}=\mathcal{O}_X(1)$ is already interesting to me, but it'd (of course) be better if the strategy works even in the more general setting.

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## 1 Answer

1) Let's analyze the situation for $X=\mathbb P(V)=\mathbb P(k^{n+1})$ and $\mathcal L=\mathcal O_{X}(1)$ .
The sections of the vector space $\Gamma(X,\mathcal L)=V^*$ generate the line bundle $\mathcal L$ .
However I claim that the sections of any strict subspace $H\subsetneq \Gamma(X,\mathcal L)=V^*$ do not generate $\mathcal L$.
Indeed , we may assume that $H\subsetneq V^*$ is a hyperplane.
It consists of the linear forms $a_0x_0+...+a_nx_n \in V^*$ satisfying $p_0a_0+...+p_na_n=0$ for some $(p_0,\cdots,p_n)\neq 0\in V$ .
Then it is tautological that all sections $s\in H$ vanish at $(p_0:\cdots:p_n)\in X$.

r) Let's analyze the situation for $X=\mathbb P(V)=\mathbb P(k^{n+1})$ and $\mathcal L^r=\mathcal O_{X}(r)\quad (r\geq 2)$
I claim that there exists a hyperplane $H\subsetneq \Gamma(X,\mathcal L^r)=S^r(V^*)=(k[x_0,\cdots,x_n])_r$ whose sections still generate $\mathcal L^r$.
A hyperplane $H_p\subsetneq (k[x_0,\cdots,x_n])_r$ consists of those homogeneous polynomials $\Sigma_{|I|=r}a_Ix^I\in (k[x_0,\cdots,x_n])_r$ of degree $r$ satisfying the relation $\Sigma_{|I|=r}p_Ia_I=0$ for some fixed family $p=(p_I)_{|I|=r}\in \mathbb P^N=\mathbb P(S^rV)$.

Given an arbitrary point $b=(b_0:\cdots :b_n)\in \mathbb P^n$, we must show that there exists a family $(a_I)_{|I|=r}$ such that $\Sigma_{|I|=r}p_Ia_I=0$ but $\Sigma_{|I|=r}a_Ib^I\neq 0$: indeed that family will define a section $s=\Sigma a_Ix^I\in H_p$ not vanishing at $b$.

For that it suffices to take $p\in \mathbb P^N \setminus v(\mathbb P^n)$, i.e. outside of the image of the Veronese embedding $$v=v_r:\mathbb P^n=\mathbb P(V)\to \mathbb P^N=\mathbb P(S^rV):(c_0:\cdots:c_n)\mapsto (\cdots:c^I:\cdots)\quad (|I|=r)$$

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