Timeline for Algebraically closed ring extension
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13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 12, 2021 at 21:30 | answer | added | Ben C | timeline score: 1 | |
| Feb 12, 2021 at 20:51 | answer | added | Antoine Labelle | timeline score: 3 | |
| Feb 12, 2021 at 15:11 | comment | added | ofiz | @BenC I think your counterexample works. Thank you for pointing it out. I still wonder if it is always surjective | |
| Feb 11, 2021 at 23:41 | comment | added | ofiz | @BenC Thank you for the comment, Let me think about it | |
| Feb 11, 2021 at 23:23 | comment | added | Ben C | @ofiz I understand your terminology now. You are correct but now I modified the example by inverting $f = x - y$ so the fiber above zero loses the origin and becomes disconnected. Explicitly, let $A = B[x,y,z]/(xy - t, z(x- y) - 1)$. | |
| Feb 11, 2021 at 23:00 | comment | added | ofiz | By fibration I just mean surjective and connected fibres. It doesn't even need to be flat. I call a fibre bundle a morphism whose fibres are all isomorphic, but it is really rare. Sorry for the confusion. @BenC I still don't get your point, the fibres of $k[xy] \rightarrow k[x,y]$ are connected. The fibre over $0$ is just $xy = 0$, which is connected. | |
| Feb 11, 2021 at 19:52 | comment | added | Ben C | Exactly it's not a fibration. I think I can improve it to $B = k[t]$ and $B \to (B[x,y]/(xy - t))_{f}$ where $f = x - y$ so the fiber over $0$ is the coordinate axes in $\mathbb{A}^2$minus the origin which is disconnected by $V(x)$ and $V(y)$. | |
| Feb 11, 2021 at 19:36 | comment | added | Antoine Labelle | I think that Ben's point is that the fibers are not all the same so it's not a fibration, over t=0 it's two lines crossing while everywhere else it's an hyperbola | |
| Feb 11, 2021 at 19:24 | comment | added | ofiz | @BenC Do you mean to say that $t \rightarrow xy$? If so, the fibres are connected and it,s surjective. Perhaps you mean $A = B[x,y,t]/(xy - t)$? But in this case, I also think that the fibres are connected (at least over $\bar{k}$) | |
| Feb 11, 2021 at 19:21 | comment | added | ofiz | @AntoineLabelle Anything that looks like a product. If $C$ is an integral $k$-algebra, then something of the from $B \rightarrow B \otimes_k C = A$ would be algebraically closed. Otherwise $k[t] \rightarrow k[x,y]$ where $t \rightarrow xy$ is algebraically closed. This is a fibration by hyperbolae. Another one is the radial fibration $k[t] \rightarrow k[x,y,1/x]$ where $t \rightarrow y/x$. | |
| Feb 11, 2021 at 16:13 | comment | added | Ben C | Does $B = k[t]$ and $A = B[x,y]/(x y - t)$ give a counter example? | |
| Feb 11, 2021 at 16:00 | comment | added | Antoine Labelle | Do you have examples of such extensions where $A$ is not of the form $B[x_1,\cdots, x_n]$? | |
| Feb 11, 2021 at 11:35 | history | asked | ofiz | CC BY-SA 4.0 |