Timeline for Degree of a generically finite map
Current License: CC BY-SA 3.0
15 events
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| Nov 9, 2012 at 19:13 | history | edited | Ongaro Nyang' | CC BY-SA 3.0 |
edited title
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| Nov 9, 2012 at 12:13 | comment | added | Ongaro Nyang' | I have checked finiteness genericall by calculating the inverse image at a point. | |
| Nov 9, 2012 at 11:54 | comment | added | Daniel Loughran | So you actually have a rational map from projective space to itself and you want to know the degree of this map? As Felipe asks, do you know that if map is generically finite? The degree will not be defined if every fibre is infinite. | |
| Nov 9, 2012 at 10:52 | comment | added | Ongaro Nyang' | @Daniel,that was just an example to answer "defined by polynomials". To me such a map is a morphism if the polynomials have no common root and I get an induced map $\mathbb P^{n−1}\longrightarrow \mathbb P^{n−1}$. | |
| Nov 9, 2012 at 9:50 | comment | added | Daniel Loughran | @Ongaro: The example you have written down is indeed a morphism (it is defined everywhere). For a rational map not to be a morphism it needs to contain expressions which are rational functions (i.e. quotients of polynomials). What does your rational map look like? Is it made up of rational functions or polynomials? | |
| Nov 9, 2012 at 7:55 | comment | added | Ongaro Nyang' | Example $n=3,$ $f$ consinder something like $(a,b,c)\longmapsto(ab-2c^2,7ac, c^2+8ab).$ | |
| Nov 9, 2012 at 6:54 | comment | added | Laurent Moret-Bailly | If $f$ is not a morphism, what does "defined by polynomials" mean? | |
| Nov 8, 2012 at 22:52 | answer | added | Margaret Friedland | timeline score: 3 | |
| Nov 8, 2012 at 22:23 | comment | added | Ongaro Nyang' | @Felipe for $\ker f\neq 0$ I only wanted to say $f$ is not a morphism, fixed it now. | |
| Nov 8, 2012 at 22:16 | history | edited | Ongaro Nyang' | CC BY-SA 3.0 |
deleted 2 characters in body; added 12 characters in body
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| Nov 8, 2012 at 22:06 | comment | added | Ongaro Nyang' | @ Felipe it seems like that is true if $f$ is a morphism never true in general. | |
| Nov 8, 2012 at 19:47 | comment | added | Felipe Voloch | If the map is generically finite, I think the degree is $m^n$. It certainly is that for "most" $f$. | |
| Nov 8, 2012 at 18:53 | history | edited | Ongaro Nyang' | CC BY-SA 3.0 |
added 11 characters in body
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| Nov 8, 2012 at 18:15 | comment | added | Felipe Voloch | What do you mean by $\ker f$? | |
| Nov 8, 2012 at 17:54 | history | asked | Ongaro Nyang' | CC BY-SA 3.0 |