I'm given the ideal $I = \langle x+y,x^2y^2 \rangle$ as an example of an ideal whose radical $\sqrt{I} = \langle x,y \rangle$ is monomial even $I $ is not monomial itself.
I'm trying to fill in the details of this example.
On
There are a few important things that weren't stated in the helpful answers.
The key to solve this exercise is to find a minimal Groebner basis of the ideal with the help of Buchberger algorithm. Then, we know that the membership problem into an ideal can be solved by dividing the polynomial by the Groebner basis.
The minimal Groebner basis I computed was $\{x+y,y^4\}$. From here it is easy to see that $x \notin I$ (as @LordSharktheUnknown pointed out in his answer). But in fact, if the ideal was monomial then $x,y \in I$ should be in the ideal as @user26857 pointed out. This can be reasoned from a minimal basis formed by monomials (one writes $x+y = \sum c_k x^{(a_k,b_k)}$ and deduces that the right hand side should only contain $x,y$ as summands because minimality forbids any summand to be expressed in term of the rest of the sum).
Finally, one can check what power of $x,y$ is in the ideal (as @LordSharktheUnkown pointed out, but with the help of Groebner basis). In my case is $x^4,y^4 \in I$. I think that this could help to solve others questions such as this one.
For the computations I used sage, in particular function sage.rings.polynomial.toy_buchberger.spol(f, g) for knowing what I was doing.
1: You need to prove $x\notin I$.
2: you need both $x+y$, $x^2y^2\in\left<x,y\right>$ (easy) and also $x^n\in I$ and $y^n\in I$ for some $n$.