Timeline for Does a semistable curve descend to a regular base?

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Jun 17, 2015 at 0:07	comment	added	Jason Starr		My formula was wrong. Let $p$ and $q$ be points such that $\underline{p}+2\underline{q}$ is equivalent to $2\underline{\tau(p)}+\underline{\tau(q)}$, i.e. $\underline{q} +3\underline{o}= \underline{p}+3\underline{\tau(o)}$, where $o$ is the group identity. Then let $D$ be $2\underline{p}+\underline{q}$ and let $D'$ be $\underline{\tau(p)}+2\underline{\tau(q)}$.
Jun 16, 2015 at 20:58	comment	added	Jason Starr		That works. Now I see that we can directly construct the local open affine covers. If $p$ is a closed point of $E$ such that $\underline{p}+\underline{\tau(p)} \sim 2\underline{\tau(\tau(p))}$ for your translation $\tau$, then the divisors $D=\underline{p} + 2\underline{\tau(p)}$ and $D'=\underline{\tau(p)}+2\underline{\tau(\tau(p))}$ are linearly equivalent. For an effective divsior $\mathcal{D}$ on $E\times T$ whose fibers over $t$, $t'$ are $D$, $D'$, the open complement of $\mathcal{D}$ maps to an open affine in $X$.
Jun 16, 2015 at 20:30	comment	added	Count Dracula		By webusers.imj-prg.fr/~daniel.ferrand/Conducteur.pdf
Jun 16, 2015 at 18:39	comment	added	Jason Starr		Aha, Count Dracula! Your example is definitely a valid example with algebraic spaces, but why is it a scheme?
Jan 30, 2015 at 14:47	history	answered	Count Dracula	CC BY-SA 3.0