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S May 2, 2019 at 17:03 history bounty ended CommunityBot
S May 2, 2019 at 17:03 history notice removed user138661
May 2, 2019 at 16:21 answer added user140149 timeline score: 5
S Apr 27, 2019 at 3:22 history bounty started CommunityBot
S Apr 27, 2019 at 3:22 history notice added user138661 Draw attention
S Apr 25, 2019 at 10:01 history bounty ended CommunityBot
S Apr 25, 2019 at 10:01 history notice removed CommunityBot
Apr 21, 2019 at 18:19 comment added user138661 Nobody has yet given an answer to the first question with the target separated (if you can prove that there can not be such example in dimension 1, please write it up as an answer, I would gladly upvote that).
Apr 21, 2019 at 18:19 comment added user138661 @TabesBridges you can not have a bijective morphism between integral separated schemes of finite type over $\mathbb{C}$ with target normal and source connected that is not an isomorphism (as shown in the link in the bounty description), so results about normal varieties should be useless. I do not understand what you mean by "curves are hopeless". The answer to the second question is no whether you are talking about 1-dimensional schemes or not (as is shown in the second link in the answer).
Apr 20, 2019 at 19:12 comment added Tabes Bridges As others have said, curves are hopeless in the Zariski topology. But in higher dimensions you might be interested in the recent Lieblich-Olsson reconstruction theorem (arxiv.org/pdf/1902.04668.pdf). This is very much in the normal projective variety setting, but in that case they show that the Zariski topological space together with the data of the rational equivalence relation on divisors is enough to determine the underlying abstract scheme. This suggests that your question probably requires additional hypotheses.
Apr 17, 2019 at 16:31 comment added David E Speyer An idea that I don't have time to pursue: Let $E$ be a supersingular elliptic curve over $\mathbb{F}_p$. For each $\alpha \in H^1(E, \mathcal{O})$, there is an affine bundle $X_{\alpha}$ over $E$, whose corresponding line bundle is trivial. There is a Frobenius map $E_{\alpha} \to E_{\alpha^p}$ which is a homeomorphism on the topological spaces. Since $E$ is supersingular, $\alpha^p$ is $0$ in $H^1(E, \mathcal{O})$, so we get a map $A_{\alpha} \to A_0$. I know that $A_0 \cong A \times \mathbb{A}^1$ is not affine; is $A_{\alpha}$ affine?
S Apr 17, 2019 at 8:50 history bounty started CommunityBot
S Apr 17, 2019 at 8:50 history notice added user137767 Draw attention
Apr 13, 2019 at 21:35 history edited user137767 CC BY-SA 4.0
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Apr 13, 2019 at 8:23 history became hot network question
Apr 13, 2019 at 8:18 history edited YCor
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Apr 13, 2019 at 3:19 answer added Julian Rosen timeline score: 17
Apr 13, 2019 at 3:08 comment added Kapil No. At this point one can use that $X$ is affine to describe $U_i$ in terms of the co-ordinate ring of $X$. Anyway, it is not a completely worked out idea.
Apr 13, 2019 at 3:07 comment added user137767 @Kapil maybe I am missing something but do you prove that $X$ is affine (which is an assumption)?
Apr 13, 2019 at 3:04 comment added Kapil The following argument may lead to a proof when $Y$ is quasi-projective. If $Y$ is quasi-projective, then it is completely described by "patching" of affine open sets: an affine cover $\{U_i\}$ such that $U_i\cap U_j$ is affine. It then follows from the quoted result that $f^{-1}(U_i)$ and $f^{-1}(U_i\cap U_j)$ are affine. So $X$ is defined by the same patching.
Apr 13, 2019 at 2:40 comment added Julian Rosen The answer to the second question is no: over an infinite field, $\mathbb{A}^1$ and $\mathbb{P}^1$ are homeomorphic.
Apr 13, 2019 at 2:11 history asked user137767 CC BY-SA 4.0