Let $\mathbb{F}$ be an algebraically closed field and let $I\subsetneq\mathbb{F}[x_1,\ldots,x_n]$ be a radical ideal. We can send $I$ to a homogeneous radical ideal in three steps:
$$I \mapsto V(I) \mapsto \mathrm{Cone}(V(I)) \mapsto I(\mathrm{Cone}(V(I))).$$
Here $V(I)\subseteq\mathbf{F}^n$ is the zero set of the ideal $I$, $\mathrm{Cone}(V(I))\subseteq\mathbf{F}^n$ is the cone over this set, i.e., $$\mathrm{Cone}(V(I))=\{\lambda\mathbf{p}: \mathbf{p}\in V(I), \lambda\in\mathbb{F}\},$$ and $I(\mathrm{Cone}(V(I)))$ is the ideal of polynomials that vanish on this cone. (This final ideal is homogeneous since it is the ideal of a conical set and it is radical by the Nullstellensatz.) Let us define $\mathrm{Cone}(I):=I(\mathrm{Cone}(V(I)))$.
Question: How is $\mathrm{Cone}(I)$ described in terms of $I$? If $I=(f_1,\ldots,f_m)$ then is it true that $\mathrm{Cone}(I)$ is generated by the homogeneous parts of the polynomials $f_i$?
Best Answer
The correspondence between ideals and closed subsets reverses inclusion. Therefore, the expected ideal should not be the ideal $I^+$ generated by homogeneous parts of generators of $I$, as such ideal would contain $I$, but rather $$\mathrm{Cone}(I)=I^{-}:=(\{f \in I\; |\;f \text{ is homogeneous}\}) \subseteq I.$$ Indeed, as $I^-$ is the biggest homogeneous ideal contained in $I$, $V(I^-)$ is the smallest conical subset containing $V(I)$, which is $\mathrm{Cone}(V(I))$.
(If you prefer to talk strictly about varieties, i.e. $I$ is assumed to be a radical ideal, note that then $I^-$ is radical too, so it works out.)