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May 23, 2020 at 8:36 comment added A.Skutin So if I understand it correctly, from Jacobian conjecture for $F = \mathbb{C}(x)$ there exist polynomials $R, T\in\mathbb{C}^{[2]}$ with $0\not= R(P, Q) = T(P, Q)\in\mathbb{C}[x]$.
May 22, 2020 at 18:36 comment added Mohan For the last question (for simplicity, take $F=\mathbb{C}$), I suppose you know that the these can be split into two cases, one in which you get that Jacobian to be zero and the other non-zero. The latter leads you to the notorious Jacobian conjecture and the former can be completely characterized.
May 22, 2020 at 10:37 comment added YCor For any scalar ring, on the $L$-algebra $L[y,z]$, the usual product along with $\{P,Q\}=\partial_yP\partial_zQ-\partial_zP\partial_yQ$ form a commutative Poisson algebra. That is, $\{\cdot,\cdot\}$ is a Lie bracket, and is a derivation (for the multiplication) with respect to each variable. I don't know to which extent this can answer the question, but at least it's a very well-studied kind of structure. See Wikipedia: Poisson algebra.
May 22, 2020 at 8:49 history edited A.Skutin CC BY-SA 4.0
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May 22, 2020 at 8:40 comment added A.Skutin It is not additive. Right now I am also interested in any not trivial examples. By trivial I mean $P, Q\in\mathbb{C}[x, y]$, or $P, Q\in\mathbb{C}[x, z]$, also $(P, Q) = (f(x)\cdot y, g(x)\cdot z)$.
May 22, 2020 at 8:17 comment added YCor @ZachTeitler is right. Actually, for every $(P,Q)$ we can write $(P,Q)=(P,P)+(0,Q-P)$, and both pairs $(P,P)$ and $(0,Q-P)$ satisfy the equation, but not $(P,Q)$ in general.
May 22, 2020 at 8:13 comment added YCor To start with you could ask the question in $L[y,z]$ for $L$ a field of characteristic zero, then specify to $L=\mathbf{C}(x)$, and then intersect with $A[y,z]$ with $A=\mathbf{C}[x]$.
May 22, 2020 at 8:12 comment added Zach Teitler Is that set of pairs closed under addition? That’s a nonlinear differential operator...
May 22, 2020 at 7:52 comment added A.Skutin These pairs form a $\mathbb{C}[x]$-submodule in $\mathbb{C}[x, y, z]^2$. For example, it obviously holds if $P$ or $Q$ in $\mathbb{C}[x]$.
May 22, 2020 at 7:48 comment added HenrikRüping The pair of such polynomials forms a set? I don't know what this question aims at.
May 22, 2020 at 7:38 history asked A.Skutin CC BY-SA 4.0