If $I\leq K[X_0,\dots,X_n]$ (for $K$ a field, let's say algebraically closed) is an ideal whose radical is homogeneous, is it always the case that $I$ is homogeneous?
I'm trying to understand projective varieties, which have been defined as follows:
For $V$ an $n+1$ dimensional vector space, $\pi:V\setminus\{0\} \rightarrow \mathbb{P}V$ the map sending a point to the line through that point, a subset $Z$ of $\mathbb{P}V$ is said to be Zariski closed if $\widetilde{Z}:=\pi^{-1}(Z)$ is Zariski closed in $V$ (i.e. we use the quotient topology for $\mathbb{P}V$).
Now I know that if some subset $\widetilde{Z}$ of $V$ is $K^\times$ invariant then $\pi^{-1}(\pi(\widetilde{Z}))=\widetilde{Z}$, so if $\widetilde{Z}=Z(I)$ for $I$ a homogeneous ideal then $\widetilde{Z}$ is $K^\times$ invariant and so $\pi(\widetilde{Z}\setminus\{0\})$ is a projective variety.
I also (think I) know that if $Z$ is a projective variety and $K$ is infinite, then $I(\widetilde{Z})$ is homogeneous. In other words, if $\widetilde{Z}=Z(I)$ then $\sqrt{I}$ is homogeneous.
Putting these together, projective varieties in $\mathbb{P}V$ correspond exactly to affine varieties in $V$ which can be written as the zero set of a homogeneous ideal (when we work in a field that allows the use of Nullstellensatz). However, if we define an affine variety in terms of a non-homogeneous variety and don't know it's radical, we don't know from this whether or not it corresponds to a projective variety, unless the answer to the question in the post is yes, which is why I'm asking it.
Consider the ideal $I=(x+y^2,x^2)$, whose radical is $\sqrt{I}=(x,y)$. $I$ is not homogeneous, yet $\sqrt{I}$ is.