Let $S,S'$ be schemes, let $\pi : S' \to S$ be a morphism which is faithfully flat and locally of finite presentation, set $S'' := S' \times_{S} S'$ and $S''' := S' \times_{S} S' \times_{S} S'$ with projections $p_{1},p_{2} : S'' \to S'$ and $p_{12},p_{13},p_{23} : S''' \to S''$. Let $\mathcal{E}'$ be a vector bundle on $S'$ such that there is an isomorphism $\varphi : p_{1}^{\ast}\mathcal{E}' \to p_{2}^{\ast}\mathcal{E}'$ of $\mathcal{O}_{S''}$-modules. What's an example of such $S,S',\pi,\mathcal{E}'$ such that $\mathcal{E}'$ does not descend to $S$, i.e. there does not exist a vector bundle $\mathcal{E}$ on $S$ and an $\mathcal{O}_{S'}$-module isomorphism $\pi^{\ast}\mathcal{E} \simeq \mathcal{E}'$?


  • Such $\mathcal{E}$ exists if and only if there exists some $\psi \in \operatorname{Aut}_{\mathcal{O}_{S''}}(p_{1}^{\ast}\mathcal{E}')$ such that the cocycle condition $p_{23}^{\ast}(\varphi \circ \psi) \circ p_{12}^{\ast}(\varphi \circ \psi) = p_{13}^{\ast}(\varphi \circ \psi)$ is satisfied.
  • If $\pi$ admits a section $\sigma : S \to S'$, then we can take $\mathcal{E} = \sigma^{\ast}\mathcal{E}'$ and pull back $\varphi$ via an induced section $S' \to S''$.
  • If $S = \operatorname{Spec} k$ and $S'$ is a smooth projective geometrically integral $k$-scheme and $\mathcal{E}'$ is a line bundle, then $\operatorname{Pic}(S') \times \operatorname{Pic}(S') \to \operatorname{Pic}(S'')$ is injective so the existence of $\varphi$ implies that $\mathcal{E}' \simeq \mathcal{O}_{S'}$.

1 Answer 1


This already fails for line bundles on smooth projective curves: let $X$ be 'the' pointless conic over $\mathbb{R}$, given by the closed subscheme of $\mathbb{P}^2_{\mathbb{R}}$ cut out by $X^2+Y^2+Z^2 = 0$. It is smooth, projective, geometrically integral over $\mathbb{R}$, and $X(\mathbb{R})= \emptyset$. In your notation, we let $\pi\colon S' \rightarrow S$ be the fppf morphism $X_{\mathbb{C}} \rightarrow X$. In this case fppf descent translates into Galois descent.

Since $X_{\mathbb{C}} \simeq \mathbb{P}^1_{\mathbb{C}}$, the isomorphism class of a line bundle on $X_{\mathbb{C}}$ is determined by its degree. Since the action of $Gal(\mathbb{C}|\mathbb{R})$ preserves degrees of divisors, we have that $L \simeq c^*L$ if $c$ denotes complex conjugation $X_{\mathbb{C}} \rightarrow X_{\mathbb{C}}$. Therefore a line bundle $L$ on $X_{\mathbb{C}}$ of degree $1$ gives an example you're looking for. Indeed we have just seen that $c^* L\simeq L$, and if there were a line bundle $M$ on $X_{\mathbb{R}}$ complexifying to $L$, it would need to have degree $1$ hence (by Riemann-Roch) be effective, which is impossible since $X(\mathbb{R})=\emptyset$.

  • $\begingroup$ Thanks, I see this should work for any nontrivial Brauer-Severi variety over any field. $\endgroup$ Jan 17, 2021 at 16:18

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