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.

  • 1
    $\begingroup$ I fail to see what this has got to do with the Jacobian Conjecture, could you explain, please? $\endgroup$
    – Ben
    Sep 14, 2016 at 22:51
  • $\begingroup$ Well the jacobian conjecture gives the condition for $k[x_1,\ldots ,x_n] = k[f_1,\ldots ,f_n]$ $\endgroup$ Sep 16, 2016 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, 2016 at 12:34
  • $\begingroup$ yeah, I think so. $\endgroup$ Sep 16, 2016 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, 2017 at 2:29

1 Answer 1


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 $k[x,xy] \neq k[x^2,\frac{1}{2}y]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.