Suppose $X\rightarrow Z$ is a projective smooth morphism. Let $0\in Z$ be a closed point, $X_0$ the corresponding fiber. Suppose $H^1(X_0,\mathcal{O})=H^2(X_0,\mathcal{O})=0$, then a line bundle $L$ on $X_0$ can be extended (after an 'etale cover) to a line bundle $\mathcal{L}$ on $X$ by Professor Totaro's result in "Jumping of the nef cone for Fano varieties".

My question is, if $L=\mathcal{O}_{X_0}(D_0)$ for a divisor $D_0$ on $X_0$, then can we extend the divisor $D_0$ (after an 'etale cover) to a divisor $D$ on $X$? And if $D_0$ is effective, can we choose $D$ to be also effective?

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    $\begingroup$ Certainly Totaro uses that result, but it goes back (at least) to Kodaira and Spencer. $\endgroup$ Dec 8, 2023 at 13:07
  • $\begingroup$ Thank you for pointing out this issue, I didn't know about it before. $\endgroup$ Dec 10, 2023 at 4:00

1 Answer 1


The statement for divisors is true. Indeed, up to an étale cover, we may suppose that $Z$ is affine and that $\mathscr L$ is a lift of $L$ to $X$. Take $\mathscr M$ on $X$ very ample relative to $Z$, and set $M = \mathscr M|_{X_0}$. By Serre vanishing, there exists $n \gg 0$ such that $$H^1\big(X,\mathscr I \otimes \mathscr L^{\otimes i} \otimes \mathscr M^{\otimes n}\big) = 0$$ for $i \in \{0,1\}$, where $\mathscr I$ is the ideal sheaf of $X_0 \hookrightarrow X$. In particular, the maps $$H^0(X,\mathscr L^{\otimes i} \otimes \mathscr M^{\otimes n}) \to H^0(X_0,L^{\otimes i} \otimes M^{\otimes n})$$ are surjective for $i \in \{0,1\}$, so if $D_0 \in \lvert L\rvert$ and $E_0 \in \lvert M^{\otimes n} \rvert$, then both $D_0 + E_0$ and $E_0$ can be lifted to divisors on $X$.

The statement for effective divisors is not true, by the following (well-known) example.

Example. Let $Z = \mathbf A^1$ with variable $t$ and set $Y = \mathbf P^1 \times Z$, with its structure map $f \colon Y \to Z$. Consider the group $\operatorname{Ext}^1_{\mathbf P^1}(\mathcal O_{\mathbf P^1}(1),\mathcal O_{\mathbf P^1}(-1)) \cong k$, and let $\alpha$ be a generator. Then the element $\alpha t \in \operatorname{Ext}^1_Y(\mathcal O_f(1),\mathcal O_f(-1))$ gives a family of extensions $$0 \to \mathcal O_{\mathbf P^1}(-1) \to \mathscr E_t \to \mathcal O_{\mathbf P^1}(1) \to 0$$ over $Z$ that is non-split for $t \neq 0$ and split for $t = 0$. In particular, $$\mathscr E_t \cong \begin{cases} \mathcal O_{\mathbf P^1}(-1) \oplus \mathcal O_{\mathbf P^1}(1), & t = 0, \\ \mathcal O_{\mathbf P^1} \oplus \mathcal O_{\mathbf P^1}, & t \neq 0. \end{cases}$$ This gives a family $X := \mathbf P_Y(\mathscr E) \to Z$ of Hirzebruch surfaces such that $$X_t \cong \begin{cases} F_2, & t = 0, \\ F_0 = \mathbf P^1 \times \mathbf P^1, & t \neq 0. \end{cases}$$ (There are also other ways to construct this degeneration of $F_n$ to $F_{n-2}$.)

Then all fibres satisfy $H^1(X_t,\mathcal O_{X_t}) = H^2(X_t,\mathcal O_{X_t}) = 0$, and the Picard groups are all free of rank $2$, generated by $g^*\mathcal O_f(1)$ and $\mathcal O_g(1)$, where $g \colon X \to Y$ is the projection onto the constant family $Y = \mathbf P^1 \times Z$.

But $X_0 \cong F_2$ contains an irreducible effective divisor $C$ with $C^2 = -2$, whereas all effective divisors $D$ in $X_t$ for $t \neq 0$ have $D^2 \geq 0$. So in particular, $C$ cannot lift to an effective divisor on $X$ (up to an étale cover $Z' \to Z$ around $0$).

(Footnote: in the smooth case, the theorem of Totaro is a well-known result from deformation theory, going back many decades. The contribution is the specific generalisation to a certain mildly singular setup.)

  • $\begingroup$ Thank you for your kind reply! May I ask under what circumstances does the statement about effective divisors hold? $\endgroup$ Dec 10, 2023 at 4:09
  • $\begingroup$ There is no easy criterion. You need to know that $H^0(X,\mathscr L) \to H^0(X_0,\mathscr L|_{X_0})$ is surjective for every line bundle $\mathscr L$ on $X$. Without any hypothesis on $\mathscr L$, it seems pointless to try to obtain this surjectivity from vanishing theorems. $\endgroup$ Dec 10, 2023 at 16:44

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