Let $k$ be a field, $R$ a commutative $k$-algebra and $p,q \in R$.
Denote $I=Rp+Rq=\langle p,q \rangle_{R}$ (= the ideal of $R$ generated by $p$ and $q$) and $J=k[p,q]p+k[p,q]q=\langle p,q \rangle_{k[p,q]}$ (= the ideal of $k[p,q]$ generated by $p$ and $q$). Clearly, $J \subseteq I$.
The answer to this question proves the following:
$I \subseteq k[p,q]$ if and only if $J=I$.
Example (appears in the above mentioned answer): It is not difficult to see that for $R=k[x]$, $p=x^2$, $q=x^3$, we have $I \subseteq k[p,q]$ and $J=I$.
Question: Could one find a sufficient condition on the ring extension $k[p,q] \subseteq R$ that guarantees that $I=J$? For example, if $k[p,q] \subseteq R$ is etale (flat+unramified), then $I=J$?
Edit: What if we further assume that $J= I \cap k[p,q]$? Clearly, $J \subseteq I \cap k[p,q]$. Does this help?
Denote $S=k[p,q]$. Obviously, $J$ is a maximal ideal of $S$ (since its quotient is a field). So with the further assumption we get that $i : S \to R$ the inclusion map, $i^{-1}(I)=I \cap k[p,q]=J$, which is a maximal ideal. But this does not imply that $I$ is maximal. Should such I be at least a prime ideal? (probably not).
Thank you very much!