Timeline for Isomorphic schemes over DVR
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 27, 2016 at 19:34 | comment | added | David Benjamin Lim | @nfdc23 True, the fiber dimension jumps. However all is not lost. The families $t^2x^3 + ty^3 + z^3 = 0$ and $t^5x^3 + t^4y^3 + z^3 = 0$ are flat as they are families of hypersurfaces. The special fibers are both $z^3 = 0$ and generic fibers are isomorphic via $(x,y,z) \mapsto (x/t,y/t,z)$. | |
| Nov 27, 2016 at 18:12 | comment | added | HeinrichD | Well, your comment precisely answers the question as given. When people come to mathoverflow sites they tend to read the answers first, and it is not so good if valuable answers are hidden in the comments. Luckily, good comments get upvoted many times so that people also consider reading them. Also notice that there is no possibility to easily find again comments hidden in answers. Comments have smaller font size and cannot be edited later on. It doesn't really matter if the answer is optimal or not, usually every answer is valuable for many people. | |
| Nov 27, 2016 at 18:05 | comment | added | nfdc23 | @HeinrichD: I consider an "answer" to be something more substantial than can fit in the size of comment box (like labeling a result as a Proposition vs. Theorem when writing a paper); just a personal preference (with which others can reasonably disagree). In addition, I didn't consider my suggestion to be particularly optimal (best are $S$ and $S'$ that are smooth, proper, and geometrically connected, as were provided later in examples suggested by others). | |
| Nov 27, 2016 at 14:42 | comment | added | nfdc23 | @BenLim: Your suggested families are not flat over the dvr, and the OP requested for $S$ and $S'$ to be flat over the dvr. | |
| Nov 27, 2016 at 9:46 | answer | added | Jason Starr | timeline score: 3 | |
| Nov 27, 2016 at 8:45 | comment | added | David Benjamin Lim | Here is an idea. Take the families $t^3x^3 + t^2y^3 + tz^3 = 0$ and $t^6x^3 + t^5y^3 + t^4z^3 = 0$ in $\Bbb{P}^2 \times \Bbb{C}[[t]]$. When we specialize at $t = 0$ we get all of $\Bbb{P}^2$. The generic fibers are isomorphic via $(x,y,z) \mapsto (x/t, y/t, z/t)$. However I do not know how to show that these two families are not the same. | |
| Nov 27, 2016 at 8:31 | answer | added | Philip Engel | timeline score: 3 | |
| Nov 27, 2016 at 8:15 | comment | added | HeinrichD | @nfdc23: This is an answer, right? Why posting it as a comment? | |
| Nov 27, 2016 at 7:42 | answer | added | R. van Dobben de Bruyn | timeline score: 8 | |
| Nov 27, 2016 at 7:33 | comment | added | nfdc23 | What is the motivation for this question? Let $A$ be a dvr with fraction field $F$, and $K/F$ a finite separable extension of degree $d > 1$ with no non-trivial autmorphism, and suppose the maximal ideal of $A$ is totally split in the integral closure $B$ of $A$ in $K$. (Algebraic curves give many such.) Let $S$ and $S'$ be complements of distinct closed points in ${\rm{Spec}}(B)$. Their generic fibers are uniquely $F$-isomorphic since ${\rm{Aut}}(K/F)=1$, so $S$ and $S'$ are not $A$-isomorphic (missing different closed points from ${\rm{Spec}}(B)$) but their special fibers are isomorphic. | |
| Nov 27, 2016 at 7:28 | comment | added | Kevin Casto | @R.vanDobbendeBruyn Here's a handwavy example of what I have in mind: take a surface with two intersecting -1 curves $C, C'$, such that the blowdowns $X, X'$ of each are not isomorphic. The images of the non-blown-down curve ($C'$ in $X$, $C$ in $X'$) should be isomorphic, so embed them in $\mathbb{A}^n$ such that $\pi^{-1}(0) = C' = C = (\pi')^{-1}(0)$, where $\pi, \pi'$ are projection to first coord. $X, X'$ are already isomorphic away from $C, C'$, so we don't lose anything passing to localization. | |
| Nov 27, 2016 at 7:15 | comment | added | R. van Dobben de Bruyn | @KevinCasto: how do you make sure that this localisation doesn't accidentally make them isomorphic? | |
| Nov 27, 2016 at 5:49 | comment | added | Kevin Casto | On a more down-to-earth level, it seems like the following should work: take two projective varieties that are birational but not isomorphic. Take affine opens, and embed them in affine space so that they have the same points with (say) first coordinate 0. Consider projection to the first coordinate as a map to $\mathbb{A}^1$, making them schemes over $\mathbb{C}[x]$, and localize at (x). Then the generic fibers are the same since they're birational, and the special fibers are the same since they have the same points over 0. | |
| Nov 27, 2016 at 5:42 | comment | added | მამუკა ჯიბლაძე | Try to google around for Artin glu(e)ing | |
| Nov 27, 2016 at 2:32 | review | First posts | |||
| Nov 27, 2016 at 2:39 | |||||
| Nov 27, 2016 at 2:31 | history | asked | Skidoo | CC BY-SA 3.0 |