Let $\pi:\mathcal{X} \to S$ be a flat, projective family of rational curves ($S$ is noetherian) over an algebraically closed field $k$. Assume $S$ is irreducible. Let $E$ be a locally-free sheaf on $\mathcal{X}$. Then, does there exists a decomposition $E \cong L_1 \oplus ... \oplus L_r$ of $E$ by $r$ invertible sheaves $L_i$ on $\mathcal{X}$? Furthermore, is $L_i \cong \mathcal{O}_{\mathcal{X}}(a_i)$ for some integer $a_i$, for each $i$?
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4$\begingroup$ Unfortunately no. Basic example: $S=\mathbb{A}^1$ with coordinate t, $\mathcal{X}= S\times \mathbb{P}^1$ with projection $\pi:\mathcal{X}\to \mathbb{P}^1$. Take the Euler sequence $0\to \mathcal{O}(-1) \to \mathcal{O}^2\to\mathcal{O}(1)\to 0$ on $\mathbb{P}^1$ and pull it back via $\pi$. Now take the push-out of the resulting sequence by $t$ acting on the left term, obtaining an extension $0\to \pi^*\mathcal{O}(-1) \to E\to \pi^* \mathcal{O}(1)\to 0$. The restriction of $E$ to the fiber over $0$ is $\mathcal{O}(-1)\oplus \mathcal{O}(1)$, and on the other fibers we have $E\cong \mathcal{O}^2$. $\endgroup$– Piotr AchingerCommented Aug 29, 2017 at 14:39
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4$\begingroup$ Just a comment; Piotr's example is the ``universal family" of extensions over $\text{Ext}^1(\mathcal{O}(1), \mathcal{O}(-1))$. $\endgroup$– Daniel LittCommented Aug 29, 2017 at 15:59
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