Timeline for If $F(x_1,0,\ldots,0)=(x_1,0,\ldots,0)$, then $F$ is bijective?
Current License: CC BY-SA 3.0
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| Apr 22, 2018 at 19:44 | comment | added | user237522 | Thank you. (I had in mind a much more complicated argument then your argument). | |
| Apr 22, 2018 at 18:52 | comment | added | Mohan | Let $f,g\in k[x,y]$ be such that their Jacobian is a non-zero constant. Let $k[x,y,z]\to k[x,y,z]$ be defined as $(f,g,z)$. Then $z(0,0,z)=z$, and their Jacobian is still a non-zero constant. So, if this map is bijective so is the original map given by $f,g$. Thus 2 variable Jacobian conjecture would be true. Similarly for any $n$. | |
| Apr 22, 2018 at 18:07 | history | edited | user237522 | CC BY-SA 3.0 |
added 4 characters in body
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| Apr 22, 2018 at 18:02 | history | asked | user237522 | CC BY-SA 3.0 |