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Let $R=\mathbb{R}[X_1,\dots,X_n]$, and

$$\mathfrak{I}_d=\{ \text{ideals for which there is minimal generating system with $d$ elements} \}\setminus \{\text{ ideals generated by $d$ monomials}\}$$

be partially ordered by inclusions. Is there an algorithm or criterion that can determine if a given ideal $I$ is maximal in $\mathfrak{I}_2$. If an ideal is not maximal, then the algorithm should provide a list of all ideals in $\mathfrak{I}_2$ that contain $I$.

This question arose from my [previous question] about ideals 1 for which Pace Nielsen provided very nice examples: $$I_1:= \langle x_1 x_2, x_1(x_2^4+x_1^2)-x_2^3(x_1 x_2) \rangle = \langle x_1x_2,x_1^3 \rangle \subsetneq I_2:= \langle x_1x_2,x_2^4+x_1^2 \rangle . $$

In this case $I_1$ is not maximal but $I_2$ might be. For example, is $gcd(I)=1$ sufficient?

I am even more interested in the same question for subrings, let $$\mathfrak{S}_d=\{ \text{subrings of $R$ for which there is minimal generating system with $d$ elements} \}\setminus \{\text{ subrings generated by $d$ monomials}.\}$$

Given a subring $S$ how can we determine if it is maximal or what are larger subrings with the same number of minimal generators.

I placed the bounty for an answer dealing with the question about subrings.

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    $\begingroup$ I'm pretty sure that $I_2$ is not maximal, since it is contained in the 2-generated ideal $\langle x_1,x_2\rangle$. $\endgroup$ Commented Mar 10, 2015 at 23:02
  • $\begingroup$ @PaceNielsen: I always ask slightly wrong question. One could consider $\mathfrak{I}_d \setminus \{\text{ ideals generated by $d$ monomials}\}$ instead. $\endgroup$
    – warsaga
    Commented Mar 11, 2015 at 6:44
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    $\begingroup$ Your intermediate question about whether $\gcd(I)=1$ is sufficient has a negative answer. For instance $\langle x_3^2,x_1+x_2\rangle$ is not generated by monomials, and has trivial gcd, but it is not maximal with respect to those properties since it is properly contained in $\langle x_3,x_1+x_2\rangle$. The condition of not being generated by monomials is somewhat strange for a number of reasons. For example, it is not preserved under automorphisms of the ring. $\endgroup$ Commented Mar 16, 2015 at 2:58

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