Timeline for When two k-varieties with the same underlying topological spaces isomorphic?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 23, 2010 at 23:07 | comment | added | Anweshi | @Pete. It is easy to see that the two are indeed the same. Sorry for the stupid question and thanks for clarifying. | |
| Jan 23, 2010 at 22:51 | comment | added | Pete L. Clark | @Anweshi: $k[x^2,x^3]$ is the first thing you say it is, and it is isomorphic to the second thing you say it is. Hence the convention. | |
| Jan 23, 2010 at 22:48 | comment | added | Anweshi | So $k[x^2, x^3]$ is not the sub-$k$-algebra of $k[x]$ generated by $x^2$ and $x^3$, it is rather $k[x,y]/(x^2 - y^3)$. Why is such a convention used? I do not understand. | |
| Jan 23, 2010 at 22:04 | comment | added | Kevin Buzzard | Anweshi: he didn't mean that, he meant x^2=y^2 and x^3=y^3 implies x=y. | |
| Jan 23, 2010 at 21:38 | comment | added | Pete L. Clark | But in this case, isn't the fiber over the singular point non-reduced? | |
| Jan 23, 2010 at 21:24 | comment | added | Anweshi | Pedantic interruption: zero and one have the same cube and square. | |
| Jan 23, 2010 at 21:15 | history | answered | Ben Webster♦ | CC BY-SA 2.5 |