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Jun 3, 2019 at 8:36 comment added user237522 @InfiniteLooper, thank you. So injectivity does not imply that the Jacobian is a non-zero scalar, but probably it must be at least a non-zero element of $k[x,x^{-1},y]$. (Please see math.stackexchange.com/questions/3147174/…).
Jun 2, 2019 at 15:41 comment added InfiniteLooper For $f(x) = x^2$, and $f(y) = y^2$ the Jacobian is $4xy$ vanishing at 0
Jun 2, 2019 at 14:43 history edited user237522 CC BY-SA 4.0
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Jun 2, 2019 at 14:28 history edited user237522 CC BY-SA 4.0
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Jun 2, 2019 at 14:21 history asked user237522 CC BY-SA 4.0