A result of E. Formanek says: Let $k$ be a field of characteristic zero and let $R=k[x_1,\ldots,x_n]$ be the polynomial ring in $n$ variables. Let $F_1,\ldots,F_n \in R$ have invertible Jacobian in $R$ (namely, the Jacobian is a non-zero scalar). If there exists $F_{n+1} \in R$ such that $k[F_1,\ldots,F_n][F_{n+1}]=R$, then $k[F_1,\ldots,F_n]=R$.
Now let $I$ be the ideal of $R$ generated by $F_1,\ldots,F_n$: $I=\langle F_1\ldots,F_n \rangle= RF_1+\cdots+RF_n$.
Is it possible to prove the following claim by adjusting Formanek's proof:
If there exists $F_{n+1} \in R$ such that $k[F_1,\ldots,F_n][F_{n+1}]+I=R$, then $k[F_1,\ldots,F_n]=R$.
Of course, if $I \subseteq k[F_1,\ldots,F_n][F_{n+1}]$, then this is just Formanek's result.
Now also asked in MO. This is a relevant question.
Thank you very much!
Your condition $k[F_1,\ldots,F_n][F_{n+1}]+I=R$ is equivalent to $k[F_{n+1}]\to R/I$ is onto for a suitable $F_{n+1}\in R$. If $I=R$, there is nothing to prove (and Jacobian conjecture says this can not happen) and if not, the Jacobian condition says $R/I$ is a finite product of fields $L_1\times\cdots\times L_r$, where each $L_i$ is a finite extension of $k$. In characteristic zero, these are all simple extensions and then I will leave you to check that such an $F_{n+1}$ can be found.