Hartshorne Example II.7.17.3

Question

Let $X$ be a smooth projective variety over $\mathbb{C}$ of dimension $n$. Let $L$ be a line bundle on $X$, and let $V\subset H^0(X,L)$ be a subspace of sections. Suppose that $s_1,..,s_{l+1}$ is a basis of $V$, and that $s_i$ correspond to the divisors $Z_i$ on $X$.

a) Then the base locus of the linear system $|V|$ is $Z=\cap_{i=1}^l Z_i$, isn't it?

So the we get a rational map $\phi: X-\! -\!\rightarrow \mathbb{P}^l$, which is regular on the complement of $Z$. We can blow up $X$ along $Z$ to get $Y$, so that $\phi$ extends to a morphism $\tilde{\phi}:Y\rightarrow\mathbb{P}^l$. Let $\pi:Y\rightarrow X$ be the blow up map.

Now in Algebraic geometry, Hartshorne, Chapter 2, Section 7, Example 7.17.3, he says that the sections $\pi^*s_i$ of $\pi^*L$ generate an invertible coherent subsheaf $L'$ of $L$.

b) How do we see this? Is this invertible subsheaf $\pi^*L\otimes\mathcal{O}_Y(-E)$ where $E$ is the exceptional divisor? I think this line bundle is base point free. How to show this?

c) Also, will $Y$ be smooth?

Thank you!

Answer 1

a) is correct, and it should be really understood in terms of schemes and not just a closed set. So, The sections give a map $\mathcal{O}_X^{l+1}\to L$ and its image is $I_Z\otimes L$. Blowing up $I_Z$, we get $\pi:Y\to X$ and $\pi^*I_Z\mathcal{O}_Y$ defines the exceptional divisor $E$, which is Cartier. By right exactness of tensor products, thus we get a surjection $\mathcal{O}_Y^{l+1}\to \mathcal{O}_Y(-E)\otimes\pi^*L$. As you can see c) may or may not be true, depending on the scheme $Z$.

Answer 2

As usual, it is helpful to examine the affine local picture, from which the global picture is glued together. So let $X = \operatorname{Spec}A$ and $\mathscr{L}$ an invertible sheaf on $X$. Let $s_0,\cdots,s_n$ be global sections of $\mathscr{L}$ and let $\mathscr{F}$ be the subsheaf of $\mathscr{L}$ generated by them. With $U$ an open in $X$ on which $\mathscr{L}$ is trivial, we have an isomorphism of $\mathscr{O}_U$-modules $\psi_U: \mathscr{L}|_U \rightarrow \mathscr{O}_U$. Now $\psi_U(\mathscr{F}|_U)$ is an ideal sheaf of $\mathscr{O}_U$, which does not depend on the choice of isomorphism $\psi_U$. Doing this for an open cover of $X$ we see that $\mathscr{F}$ is isomorphic as an $\mathscr{O}_X$-module to an ideal sheaf $\mathscr{J}$ of $X$. Now $\mathscr{J}=\tilde{I}$, where $I=(a_0,\dots,a_n)$ is the ideal of $A$ inducing $\mathscr{J}$, with the $a_i$'s corresponding to the $s_i$'s. The blow-up of $X$ along the closed subscheme defined by $\mathscr{J}$ is $\tilde{X} = \operatorname{Proj} S$, with $S = A \oplus I \oplus I^2 \oplus \cdots$. Let $\pi: \tilde{X} \rightarrow X$ be the blow-up morphism. Then $\pi^* \mathscr{J} = \tilde{J}$, where $J$ is the ideal of $S$ defined as $J = I \oplus I^2 \oplus I^3 \oplus \cdots$; this can be checked on the standard affine opens $D(a_i)$ of $\operatorname{Proj}S$, where $a_i$ is viewed as an element of $S$ in degree $1$. We see that $\pi^* \mathscr{J}$ is already a subsheaf of $\mathscr{O}_{\tilde{X}}$, so that it coincides with $\pi^{-1} \mathscr{J} \cdot \mathscr{O}_{\tilde{X}}$; here recall Caution II.7.12.2 in Hartshorne. By Proposition II.7.13, $\pi^* \mathscr{J}$ is an invertible ideal sheaf. Hence $\pi^* \mathscr{F}$ is an invertible coherent sheaf. It is generated by the $\pi^* s_i$'s because $\mathscr{F}$ is generated by the $s_i$'s; we have a surjection $\mathscr{O}_X^{n+1} \rightarrow \mathscr{F} \rightarrow 0$, which remains a surjection upon pullback by $\pi$, because $\pi^*$ is right-exact.