The following is explained in Hartshorne, chapter 2.7. I will be considering varieties instead of schemes. Let $X$ be a variety over $k$, and let $L$ be a line bundle on $X$. Let $s_0,\dots,s_n$ be global sections of $L$. This determines a morphism $\phi:X\setminus B\to\mathbb{P}^n$, where $B$ is the base locus of the linear system. Then Hartshorne claims that there is a natural way to extend this to a morphism $\widetilde{\phi}:\widetilde{X}\to\mathbb{P}^n$, where $\pi:\widetilde{X}\to X$ is the blow up along $B$. He claims that this follows from the fact that the $\pi^*s_i$ generate a line bundle on $\widetilde{X}$. My question is: how can this be true? If we are just pulling back the sections of $L$, then these should vanish on the exceptional divisor. So how can they generate an invertible sheaf on $\widetilde{X}$? Edit: as was pointed out below, vanishing of the sections does not imply vanishing of the germs, so this is fine. However, I still do not see how the morphism to projective space is meant to be defined if the sections still all vanish along the exceptional divisor, regardless of whether they generate the stalk.

  • $\begingroup$ The proof tells you why: by II.7.13a, the pullback of that sheaf is invertible and it's defined to be generated by the $s_i$. $\endgroup$
    – KReiser
    Commented Jul 4, 2022 at 21:21
  • $\begingroup$ Thanks for the response, maybe I should have been a bit clearer. I meant to ask how I should unify this fact, with the fact that $\pi^*s_i$ all vanish on the exceptional divisor. Am I missing something there? $\endgroup$
    – user569579
    Commented Jul 4, 2022 at 21:26
  • $\begingroup$ $x\in k[x]$ generates an invertible submodule but vanishes on a point, right? Same thing here. $\endgroup$
    – KReiser
    Commented Jul 4, 2022 at 23:39
  • $\begingroup$ True, the germ doesn't have to vanish. But then, it still is true that $\pi^*s_i$ all vanish on the exceptional divisor. Since the morphism to $\mathbb{P}^n$ is $(s_0(x):\dots:s_n(x))$, this still seems problematic to me. $\endgroup$
    – user569579
    Commented Jul 5, 2022 at 7:33
  • $\begingroup$ At the naive level: it's true that $\pi^{*} s_i$ all vanish on the exceptional divisor, but that being of codim $1$, it's defined by an equation. So simplify al of the $\pi^{*} s_i$ by that factor. $\endgroup$
    – orangeskid
    Commented Jul 5, 2022 at 14:05

1 Answer 1


sorry for the late answer! I hope it helps a bit: For simplicity, I'll explain the idea for a smooth surface $S$ and the blow-up of a point but the general version works just like that: Instead of working with a line bundle $L$ on $S$ its better to take its (complete) linear system $|L|=\Gamma$ defined by the (equivalence class of) global sections of $L$.
Imagine you have a birational map $\phi_\Gamma:S \dashrightarrow \mathbf{P}^n$ determined by a linear system $\Gamma$ on $S$. Assume $p$ is a base point of $\Gamma$, so $\phi$ is not defined at $p$.Then blowing up $p$ gives a smooth surface $f: \tilde{S} \rightarrow S$ with exceptional divisor $E$. So now you want to get a new linear system $\Gamma'$ by pullback on $\tilde{S}$, which you get by $$f^*\Gamma = \Gamma' + kE$$ where $k$ is the multiplicity of $\Gamma$ at $p$.
So you get $\Gamma'$ by pulling back the curves (or sections if you prefer that language) from $\Gamma$ and then getting rid of $kE$ which is a fixed curve of the pullback linear system. Algebraically, you do exactly what @orangeskid mentioned in the comments. $kE$ is a divisor on $\tilde{S}$ and defined by a single polynomial and appears in every local equation for the members of $f^*\Gamma$ and subtracting $kE$ from this is factoring out this equation from every element of $f^*\Gamma$.
To get back the line bundle $L'$ on $\tilde{S}$, take any divisor (=curve) $D \in \Gamma'$ and consider $\mathcal{O}(D)$.

  • 1
    $\begingroup$ Thank you, that makes sense. Seems like there is slightly more to do than what Hartshorne lets on. $\endgroup$
    – user569579
    Commented Jul 7, 2022 at 9:38
  • $\begingroup$ glad you like it - you're welcome! $\endgroup$
    – Simonsays
    Commented Jul 7, 2022 at 11:35

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