For ideal $m$ maximal and principal, there's no ideal between $m^2$ and $m$. Prove that this can be false when $m$ is not principal or maximal.

Question

Prove that for ideal $m$ maximal and principal, there's no ideal $I$ such that $m^2 \subsetneq I \subsetneq m$. Show that this can be false when $m$ is not principal or maximal.

Suppose $\mathfrak m=(a)$, and $a\notin I$. Let's show that $I\subseteq\mathfrak m^2$. Pick $x\in I$. Then $x=ay$, $y\in R$. If $y\in\mathfrak m$, then $y=az$ and thus $x=a^2z\in\mathfrak m^2$. Otherwise, $\mathfrak m+(y)=R$, so $1=am+yn$. Then $a=a^2m+ayn$, so $a=a^2m+xn\in I$, a contradiction.
This proves the first part but I don't know how to do the second.

Answer 1

Can you find an ideal $I$ such that $(x^2,xy,y^2)=(x,y)^2\subsetneq I\subsetneq (x,y)$? (The ideal $\mathfrak m=(x,y)$ is maximal in $K[x,y]$, but not principal.)

The same question for $36\mathbb Z\subsetneq I\subsetneq 6\mathbb Z$. (The ideal $6\mathbb Z$ is principal, but not maximal in $\mathbb Z$.)