Timeline for How to show a set of polynomials is algebraically independent?
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
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| Oct 10, 2010 at 14:26 | vote | accept | Jon | ||
| Oct 9, 2010 at 12:47 | comment | added | BCnrd | Martin, in char. 0 the map $f:\mathbf{A}^m \rightarrow \mathbf{A}^n$ is smooth on a dense open in the source if and only if it is dominant, and such generic smoothness is equivalent to generic injectivity of the induced map $f^{\ast}(\Omega^1_{\mathbf{A}^n/k}) \rightarrow \Omega^1_{\mathbf{A}^m/k}$. But at the generic point, such injectivity can be checked between $n$th exterior powers. | |
| Oct 9, 2010 at 11:06 | comment | added | Martin Brandenburg | Can you give a proof? | |
| Oct 8, 2010 at 21:37 | history | answered | Jorge Vitório Pereira | CC BY-SA 2.5 |