Is there a non-homogeneous ideal whose radical is homogeneous? Let $I$ be an ideal, and let $\sqrt{I}$ denote its radical (intersection of all primes containing $I$).
One has that (1)(2): $I$ homogeneous $\implies $ $\sqrt{I}$ is homogeneous.

Question: Does one also have that: $\sqrt{I}$ homogeneous $\implies$ $I$ is homogeneous?
I.e., can anyone find a counterexample of a non-homogeneous ideal (in any graded ring) whose radical is homogeneous? (That would prove the negation of the contrapositive.)

Attempt: Given an ideal $J$, denote by $\mathfrak{h}(J)$ the "homogenization" of $J$, i.e. the ideal generated by the set of all homogeneous elements of $J$. In particular, $J$ is homogeneous $\iff J = \mathfrak{h}(J)$.
As shown in this answer, $\sqrt{\mathfrak{h}(I)} = \mathfrak{h}(\sqrt{I})$ always. If we assume that $\sqrt{I}$ is homogeneous, then as noted above, $\sqrt{I} = \mathfrak{h}(\sqrt{I})$, so in particular: $\sqrt{\mathfrak{h}(I)} = \sqrt{I}$.
So if $\sqrt{I}$ is homogeneous, then we can conclude that $\mathfrak{h}(I)$ and $I$ have the same radical -- however, this is not enough to conclude in general that $I$ and $\sqrt{I}$ are equal, from which it would follow, since $\sqrt{I}$ is homogeneous, that $I$ is homogeneous.
Not being able to go any further than this, and being able to think of non-equal non-radical ideals with the same radical (e.g. $\langle x^2 \rangle$ and $\langle x^3 \rangle$), I suspect that there is a counterexample.
Motivation: Since the product of homogeneous elements are again homogeneous, (at least for Noetherian rings) it is fairly easy to use their sets of generators (consisting only of homogeneous elements) to argue that the product of homogeneous ideals is again homogeneous. (Note)
I could find only one proof for the fact that the intersection of two homogeneous ideals is again homogeneous. However, if it were the case that $\sqrt{I}$ is homogeneous $\implies$ $I$ is homogeneous, then we could use the fact that the product of homogeneous ideals is homogeneous, and the relationship $\sqrt{IJ} = \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}$, to prove the result for all homogeneous ideals, not just radical ones.

Take $R=k[x,y]$, polynomial ring in two variables with the usual grading and let $I=(x+y^2, y^3)$. Then $I$ is not homogeneous, but $\sqrt{I}=(x,y)$ is.
