Let $k$ be a field and $E$ be a vector bundle over $\mathbb{P}_{k}^{1}\times\mathbb{A}_{k}^{1}$, does it extend to $\mathbb{P}_{k}^{1}\times\mathbb{P}_{k}^{1}$?
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Yes, the point is that $\mathbb{P}_{k}^{1} \times \mathbb{P}_{k}^{1}$ is regular of dimension at most 2. Extend $E$ to a coherent sheaf $E'$ on $\mathbb{P}_{k}^{1} \times \mathbb{P}_{k}^{1}$, then take double dual $E'' := (E')^{\vee\vee}$ to get a reflexive sheaf; then $E''$ is flat by SP Tag 0B3N.
answered Dec 12, 2019 at 11:46
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$\begingroup$ and does it also hold for vector bundles over $\mathbb{P}^{n}\times\mathbb{A}^{1}$? $\endgroup$– prochetCommented Dec 12, 2019 at 13:44
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$\begingroup$ I don't know, and I would also like to know. $\endgroup$ Commented Dec 12, 2019 at 14:43
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2$\begingroup$ Re: @prochet's follow-up question. I doubt it, but an argument would have to be of global nature. There are moduli spaces of semistable reflexive sheaves on $\mathbb{P}^n$, and not all of these sheaves are locally free for $n>1$; if you take a curve $C$ in the moduli space whose intersection with the non-locally free locus is a finite set $S$, then your question with $C\setminus S$ instead of $\mathbb{A}^1$ will likely have a negative answer. But it is not obvious how to ensure $C\setminus S \simeq \mathbb{A}^1$. $\endgroup$ Commented Dec 12, 2019 at 20:12
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$\begingroup$ The point is precisely that I don't want to modify the $\mathbb{A}^{1}$-factor. $\endgroup$– prochetCommented Dec 12, 2019 at 20:58