TITLE: $\Bbb R[X,Y]/(F) \cong \Bbb R[Z]$ and $F_X G_Y - G_X F_Y \in \Bbb R^*$
QUESTION [6 upvotes]: Hope this isn't a duplicate.
I was trying to solve the following problem : 
Let $F,G \in \Bbb R[X,Y]$ satisfy $\Bbb R[F,G]= \Bbb R[X,Y]$. Prove that :
(i) $\Bbb R[X,Y]/(F) \cong \Bbb R[Z]$ , for some $Z$ and (ii) $F_X G_Y - G_X F_Y \in \Bbb R^*$ (where $F_X$ etc. denote partial derivatives) 
I don't have any idea regarding the problem.

REPLY [3 votes]: Part (ii) has been answered by Magdiragdag.  I am expanding on Part (i) from Magdiragdag's answer.
Consider the following function $\phi:\mathbb{R}^2\to\mathbb{R}^2$ sending $(x,y)\mapsto \big(F(x,y),G(x,y)\big)$ for all $x,y\in\mathbb{R}$.  Since $\mathbb{R}\big[F(X,Y),G(X,Y)\big]=\mathbb{R}[X,Y]$, there exist $P(X,Y),Q(X,Y)\in\mathbb{R}[X,Y]$ such that
$$P\big(F(X,Y),G(X,Y)\big)=X$$
and
$$Q\big(F(X,Y),G(X,Y)\big)=Y\,.$$
This means the map $\psi:\mathbb{R}^2\to\mathbb{R}^2$ sending $(x,y)\mapsto \big(P(x,y),Q(x,y)\big)$ for all $x,y\in\mathbb{R}$ is the inverse map of $\phi$.  This also shows that
$$F\big(P(X,Y),Q(X,Y)\big)=X$$
and
$$G\big(P(X,Y),Q(X,Y)\big)=Y\,.$$
Now, let $\tau:\mathbb{R}[X,Y]\to\mathbb{R}[๊U,V]$ be the ring homomorphism extending the requirement that $1\mapsto 1$, $X\mapsto P(U,V)$, and $Y\mapsto Q(U,V)$.  Then, the ideal $\big\langle F(X,Y)\big\rangle$ of $\mathbb{R}[X,Y]$ is sent to the ideal $\langle U\rangle$ of $\mathbb{R}[U,V]$.  Since $\mathbb{R}\big[P(U,V),Q(U,V)\big]=\mathbb{R}[U,V]$, we see that $\tau$ is an isomorphism of rings.  Hence, $\tau$ induces an isomorphism
$$\mathbb{R}[X,Y]/\big\langle F(X,Y)\big\rangle\cong \mathbb{R}[U,V]/\langle U\rangle \cong \mathbb{R}[V]\,.$$
As a consequence, if $\sigma:\mathbb{R}[X,Y]\to\mathbb{R}[V]$ is a ring homomorphism extending $1\mapsto 1$, $X\mapsto P(0,V)$, and $Y\mapsto Q(0,V)$, then the kernel of $\sigma$ is precisely $\big\langle F(X,Y)\big\rangle$.