Let $R=\frac{\mathbb{C}[x,y,z]}{\langle x^2-1, yz \rangle}$.
For convenience, I will write $x,y,z \in R$ instead of $\bar{x},\bar{y},\bar{z}$.
In $R[X,Y]$, take $A=(x+iy)X+yY$, $B=yX+(x-iy)Y$, $C=zX$.
We have: $\operatorname{Jac}(A,B)=A_XB_Y-A_YB_X=(x+iy)(x-iy)-yy=x^2-i^2y^2-y^2=x^2+y^2-y^2=x^2=1$ $\operatorname{Jac}(A,C)=A_XC_Y-A_YC_X=(x+iy)0-yz=-yz=0$.
Is it true that $C \notin R[A]$?
I think/hope that the answer is yes (though I may be wrong):
Otherwise, $C \in R[A]$ so there exists $h(t) \in R[t]$ such that $C=h(A)$. Since $\deg(C)=\deg(A)=1$, it follows that $h$ must be of degree $1$, so $h(t)=\lambda t + \mu$ for some $\lambda, \mu \in R$.
Therefore, $C=h(A)$ becomes $zX=\lambda A + \mu= \lambda ((x+iy)X+yY)+ \mu =\lambda(x+iy)X+\lambda yY+\mu$.
Then we must have: (i) $z=\lambda(x+iy)$; (ii) $0=\lambda y$; (iii) $0=\mu$.
Now, (ii) implies (if I am not wrong) that $\lambda=0$ or $\lambda=z$.
But each of these two options contradicts (i): The first option yields $z=\lambda(x+iy)=0(x+iy)=0$, and the second option yields $z=\lambda(x+iy)=z(x+iy)$, so $z(1-x-iy)=z-z(x+iy)=0$. (I think that $z(1-x-iy)=z(1-x)-izy=z(1-x)-i0=z(1-x)$ is not zero).
Am I missing something trivial?
Thank you very much!
Motivation: Actually, it seems (but again I may be wrong) that the two-dimensional Jacobian Conjecture in $R[X,Y]$, $R=\frac{\mathbb{C}[x,y,z]}{\langle x^2-1, yz \rangle}$, is false:
Just take $A=(x+iy)X+yY$, $B=yX+(x-iy)Y$. We have seen above that $\operatorname{Jac}(A,B)=1$.
It seems that $R[A,B] \subsetneq R[X,Y]$, so $(X,Y) \mapsto (A,B)$ is not an automorphism of $R[X,Y]$.
Moreover, if indeed $R[A,B] \subsetneq R[X,Y]$, then Proposition 1.1.12 implies that the two-dimensional Jacobian Conjecture in $\mathbb{C}[X,Y]$ is false..
It remains to show that $R[A,B] \subsetneq R[X,Y]$. I am not sure yet how to prove this (if this is true..). Observe that $zA=z(x+iy)X+zyY=z(x+iy)X$ and $zB=zyX+z(x-iy)Y=z(x-iy)Y$.
It seems impossible to obtain $X$ and $Y$; am I right?
Thank you very much again!