When square of an ideal is the square of a maximal ideal in a polynomial ring Question (1) Let $J$ be an ideal in $\mathbb C[X,Y]$ such that $J^2=(X,Y)^2$. Then is it true that $J=(X,Y)$ ?
Question (2) Let $J$ be an ideal in $\mathbb C[X,Y,Z]$ such that $J^2=(X,Y,Z)^2$. Then is it true that $J=(X,Y,Z)$ ?
If the answers to any one is no, then a further question would does some extra condition on $J$ (like being homogeneous or monomial, or some bound on $\mu (J)$) force $J$ to be equal to the corresponding maximal ideal ? 
For a similar question, see https://math.stackexchange.com/questions/3101911/ideal-in-local-domain-whose-square-equals-the-square-of-the-maximal-ideal
Answer to any part of the questions is very appreciated.
Thanks in advance.
 A: (1): Yes. First note that $J\subseteq (X,Y)$. If $a\in J$ but $a\not \in (X,Y)$, then $a^2 \in J^2 = (X,Y)^2 \subseteq (X,Y)$. Since $(X,Y)$ is prime this implies that $a\in (X,Y)$. 
Now if you look at $\mathbb{C}[X,Y]/(X,Y)^2$, note that this is a 3-dimensional complex vector space with basis $1,X,Y$. The ideal corresponding to $J$ in this ring, if it is not equal to $(X,Y)$, must have dimension one as a vector space. Call the generator $aX+bY$. Then $J = (aX+bY) + (X,Y)^2$. But it is clear that in this case $J^2 \ne (X,Y)^2$. 
(2): Yes. Same argument shows $J\subseteq (X,Y,Z)$, and you only have to argue that the corresponding ideal in the quotient cannot have dimension two. Call generators $f,g$. Then $J = (f,g) + (X,Y,Z)^2$, where $f,g$ have degree $1$. Now $J$ is homogeneous and its degree $1$ component has dimension 2 and dimension $0$ component has dimension $0$. So the degree $2$ component of $J^2$ has dimension at most $4$, hence $J^2 \ne (X,Y,Z)^2$. 
Note that the argument in (2) can be generalised (and (1) is really a baby case)! 
