# A noneffective descent datum: isomorphism not satisfying the cocycle condition

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 $$\pi^{\ast}\mathcal{E} \simeq \mathcal{E}'$$?

Thoughts:

• 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 $$\mathcal{E}' \simeq \mathcal{O}_{S'}$$.

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 $$X(\mathbb{R})=\emptyset$$.