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Assume that the algebraically independent polynomials $f, g\in\mathbb{C}[x, y]$ are such that the Jacobian matrix $\text{Jac}_{x, y}^{f, g}\in\mathbb{C}\setminus\{0\}$.

Is it true that $\mathbb{C}[x, y] = \mathbb{C}[f, g]+g\cdot\mathbb{C}[x, y]$?

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    $\begingroup$ Do you have some reason to suspect this? This is equivalent to the Jacobian conjecture. $\endgroup$ – Mohan Jul 21 at 17:44
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    $\begingroup$ Do you have proof of its equivalence to JC? $\endgroup$ – Algorithm Jul 21 at 17:52
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    $\begingroup$ If JC is true, the above equality is obvious. If the above equality is true in general, consider $g_i\in\mathbb{C}[x_1,\ldots, x_{n+1}]$ with $g_i=f_i, i\leq n$ and $g_{n+1}=x_{n+1}$ and apply the equality for the $g_i$s. $\endgroup$ – Mohan Jul 21 at 18:00
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    $\begingroup$ So You know that if $Q(n)$ is my question and $JC(n)$ is the Jacobian conjecture, then $JC(n)=>Q(n)=>JC(n-1)$. Assume we have proven $JC(i)$ for $i<n$ then can $Q(n)$ be proved? $\endgroup$ – Algorithm Jul 21 at 18:16
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    $\begingroup$ As I know Jacobian conjecture is open even for $n=2$, so I edited the question to the case $n=2$. $\endgroup$ – Algorithm Jul 21 at 18:40
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This is still is equivalent to JC.

Your equality says, $\mathbb{C}[x,y]=\mathbb{C}[f,g]+g\mathbb{C}[x,y]$, the last term is equal to $\mathbb{C}[f]+g\mathbb{C}[f,g]+g\mathbb{C}[x,y]=\mathbb{C}[f]+g\mathbb{C}[x,y]$, since $g\mathbb{C}[f,g]\subset g\mathbb{C}[x,y]$. This says, the map $\mathbb{C}[f]\to \mathbb{C}[x,y]/g\mathbb{C}[x,y]$ is onto and then it is clear that this is an isomorphism. Then, $g=0$ is an embedded line in $\mathbb{C}^2$ and by Abhyankar-Moh, is a co-ordinate line after an automorphism. Then, it is easy to verify that $\mathbb{C}[f,g]=\mathbb{C}[x,y]$.

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