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The two-dimensional Jacobian Conjecture over $\mathbb{C}$ says the following: Let $p,q \in \mathbb{C}[x,y]$ satisfy $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in \mathbb{C}-\{0\}$. Then $\mathbb{C}[p,q]=\mathbb{C}[x,y]$.

Let $\beta: \mathbb{C}[x,y] \to \mathbb{C}[x,y]$ be the following involution (= automorphism of order two): $x \mapsto x, y \mapsto -y$.

Then we can write: $p=s_1+k_1$ and $q=s_2+k_2$, where $s_1,s_2$ are symmetric w.r.t. $\beta$ and $k_1,k_2$ are skew-symmetric w.r.t. $\beta$.

It is well-know and was proved in several ways that $\mathbb{C}[p,q] \subseteq \mathbb{C}[x,y]$ is flat.

Question. Is $\mathbb{C}[s_1,k_1,s_2,k_2] \subseteq \mathbb{C}[x,y]$ flat?

I am not sure if one of the known proofs for flatness of $\mathbb{C}[p,q] \subseteq \mathbb{C}[x,y]$ can be adjusted here.

I have also asked the above question in MSE.

Any hints and comments are welcome! Thank you.

Edit: The one-dimensional case does not seem to help solve my above question, since if $f: \mathbb{C}[x] \to \mathbb{C}[x]$, $x \mapsto p$, has an invertible Jacobian $p_x \in \mathbb{C}-\{0\}$, then $p=\lambda x + \mu$, for some $0 \neq \lambda,\mu \in \mathbb{C}$. Then, $s_1=\mu, k_1=\lambda x$ ($\beta: x \mapsto -x$), so $\mathbb{C}[s_1,k_1]=\mathbb{C}[x] \subseteq \mathbb{C}[x]$ is flat.

If we do not require an invertible Jacobian, then of course there are counterexamples, the easiest: $p=x^2+x^3$, so $s_1=x^2, k_1=x^3$, and then $\mathbb{C}[s_1,k_1]=\mathbb{C}[x^2,x^3] \subseteq \mathbb{C}[x]$ is not flat.

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  • $\begingroup$ Isn't it true that the 2-dimensional Jacobian conjecture implies the full Jacobian conjecture? $\endgroup$
    – markvs
    Commented Mar 14, 2021 at 19:36
  • $\begingroup$ If I am not wrong, no, the $2$-dimensional JC does not imply the generalized JC (= $n$-dimensional JC, for all $n$); at least I have not seen such an implication. But if for a fixed $M$, the $M$-dimensional JC holds, then it holds for all $m \leq M$. $\endgroup$
    – user237522
    Commented Mar 14, 2021 at 20:06
  • $\begingroup$ @dodd: not that I know. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Mar 14, 2021 at 20:07
  • $\begingroup$ Yes, I meant old results by Gary Meisters, but my recollection was wrong. $\endgroup$
    – markvs
    Commented Mar 14, 2021 at 20:35
  • $\begingroup$ @dodd, thank you very much for trying to help (I will try to find relevant results of Meisters). If you have any ideas about the above question-- you are welcome! $\endgroup$
    – user237522
    Commented Mar 14, 2021 at 20:59

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