$\Bbb R[X,Y]/(F) \cong \Bbb R[Z]$ and $F_X G_Y - G_X F_Y \in \Bbb R^*$

Question

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.

Answer 1

Because ${\mathbb R}[F,G] = {\mathbb R}[X,Y]$, there are $P, Q \in {\mathbb R}[X,Y]$ such that $P(F,G) = X$ and $Q(F,G) = Y$. Then, by the multivariate chain rule, $$J_{P,Q}(F,G) \cdot J_{F,G}(X,Y) = J_{X,Y}(X,Y) = I.$$ (Here $J_{P,Q}(F,G)$ is the Jacobian matrix of $(P,Q)$ evaluated at $(F,G)$; note that you can view $(F,G)$ as a map from ${\mathbb R}^2$ to ${\mathbb R}^2$, as in Batominovski's answer). Now take determinants on both sides; this shows that $\det(J_{F,G}(X,Y)) = F_X G_Y - F_Y G_X$ is an invertible element of ${\mathbb R}[X,Y]$, i.e., an element of ${\mathbb R}^*$.

For item (i): ${\mathbb R}[X,Y]/(F) = {\mathbb R}[F,G]/(F) \cong {\mathbb R}[G]$. (The $\cong$ takes a bit of arguing; the point is that because ${\mathbb R}[F,G] = {\mathbb R}[X,Y]$, $F$ and $G$ behave as variables. This makes it possible to define a map ${\mathbb R}[F,G] \to {\mathbb R}[G]$ by mapping $H(F,G)$ to $H(G)$. The kernel of this map is $(F)$.)

Remarks:

• The (real) Jacobian Conjecture (in dimension 2) is the converse: if $F_X G_Y - F_Y G_X \in {\mathbb R}^*$, then ${\mathbb R}[X,Y] = {\mathbb R}[F,G]$.
• The field ${\mathbb R}$ is largely immaterial; it can be replaced by any field $k$. However, $F_X G_Y - F_Y G_X \in {\mathbb R}^*$ is a different condition than saying that the evaluation of $F_X G_Y - F_Y G_X$ is non-zero at every $(x,y) \in {\mathbb R}^*$. (For ${\mathbb C}$, those two conditions are equivalent).

Answer 2

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 $\big\langle F(X,Y)\big\rangle$.