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Is there any known generalization of Jacobian conjecture which gives condition for $k[f_1, \ldots, f_m] = k[g_1, \ldots, g_m]$ where all $f_i$ and $g_i$ are functions over $x_1, \ldots, x_n$? Note that possibly, $n \neq m$.

Is there anything known even if we assume $n = m$?

One possibility is that the determinant of Jacobian for $f$ and $g$ (as in Jacobian conjecture) is same w.r.t. all $m$-tuples of $x_i$'s. This is stronger than Jacobian conjecture, but I cannot prove the other way round.

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    $\begingroup$ I fail to see what this has got to do with the Jacobian Conjecture, could you explain, please? $\endgroup$ – Ben Sep 14 '16 at 22:51
  • $\begingroup$ Well the jacobian conjecture gives the condition for $k[x_1,\ldots ,x_n] = k[f_1,\ldots ,f_n]$ $\endgroup$ – nikhil_vyas Sep 16 '16 at 8:32
  • $\begingroup$ I see, never thought about it that way, thank you. So, in geometric terms, what you're asking is related to the question when two regular maps $f,g\colon k^n\to k^m$ have isomorphic images..? $\endgroup$ – Ben Sep 16 '16 at 12:34
  • $\begingroup$ yeah, I think so. $\endgroup$ – nikhil_vyas Sep 16 '16 at 18:35
  • $\begingroup$ Perhaps arxiv.org/pdf/1601.01508v1.pdf is what you are looking for? (and other papers by the same authors). It seems to be in the spirit of what you have asked. $\endgroup$ – user237522 Jan 26 '17 at 2:29
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Let $m=n=2$. Same Jacobians (= determinants of Jacobi matrices. You probably meant up to multiple by a non-zero scalar) do not imply that the subrings are equal. For example, take $f_1=x, f_2=xy, g_1=x^2, g_2=\frac{1}{2}y$. The Jacobian of $f_1$ and $f_2$ is $x$, and also the Jacobian of $g_1$ and $g_2$ is $x$, but $k[x,xy] \neq k[x^2,\frac{1}{2}y]$.

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