Vtyjkyuk

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.

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).

Part (ii) has been answered by Magdiragdag. I am expanding on Part (i) from Magdiragdag's answer.

Consider the following function $phi:mathbbR^2tomathbbR^2$ sending $(x,y)mapsto big(F(x,y),G(x,y)big)$ for all $x,yinmathbbR$. Since $mathbbRbig[F(X,Y),G(X,Y)big]=mathbbR[X,Y]$, there exist $P(X,Y),Q(X,Y)inmathbbR[X,Y]$ such that
$$Pbig(F(X,Y),G(X,Y)big)=X$$
and
$$Qbig(F(X,Y),G(X,Y)big)=Y,.$$
This means the map $psi:mathbbR^2tomathbbR^2$ sending $(x,y)mapsto big(P(x,y),Q(x,y)big)$ for all $x,yinmathbbR$ is the inverse map of $phi$. This also shows that
$$Fbig(P(X,Y),Q(X,Y)big)=X$$
and
$$Gbig(P(X,Y),Q(X,Y)big)=Y,.$$
Now, let $tau:mathbbR[X,Y]tomathbbR[๊U,V]$ be the ring homomorphism extending the requirement that $1mapsto 1$, $Xmapsto P(U,V)$, and $Ymapsto Q(U,V)$. Then, the ideal $biglangle F(X,Y)bigrangle$ of $mathbbR[X,Y]$ is sent to the ideal $langle Urangle$ of $mathbbR[U,V]$. Since $mathbbRbig[P(U,V),Q(U,V)big]=mathbbR[U,V]$, we see that $tau$ is an isomorphism of rings. Hence, $tau$ induces an isomorphism
$$mathbbR[X,Y]/biglangle F(X,Y)bigranglecong mathbbR[U,V]/langle Urangle cong mathbbR[V],.$$
As a consequence, if $sigma:mathbbR[X,Y]tomathbbR[V]$ is a ring homomorphism extending $1mapsto 1$, $Xmapsto P(0,V)$, and $Ymapsto Q(0,V)$, then the kernel of $sigma$ is precisely $biglangle F(X,Y)bigrangle$.