# For any ideal $I$ in a graded ring, $V_+(I)=V_+(J)$ for a specific homogeneous $J$

Let $$S=\bigoplus_{n\geq 0}S_n$$ be a graded ring and $$S_+:=\bigoplus_{n\geq 1}S_n$$. We define $$\text{Proj}(S)$$ as the set of homogeneous prime ideals of $$S$$ not containing $$S_+$$ and, for any ideal $$I\subset S$$, let $$V_+(I):=\{\mathfrak{p}\in\text{Proj}(S)\mid \mathfrak{p}\supset I\}$$.

We define homogeneous ideals associated to $$I$$: \begin{align*} I_*&:=\bigoplus_{n\geq 0}I\cap S_n\\ I^*&:=\langle g\in S\mid g\text{ is the homogeneous component of some }f\in I\rangle \end{align*}

[a "homogeneous component" of $$f$$ is any of the $$f_i$$ in the decomposition $$f=f_0+...+f_n$$ with $$f_i\in S_i$$]

I'm trying to prove the following:

$$V_+(I_*)=V_+(I^*)=V_+(I)$$

I've already verified that $$I_*,I^*$$ are indeed homogeneous. Furthemore, I've proven that:

(i) $$I_*\subset I\subset I^*$$;

(ii) $$I_*$$ is maximal among the homogeneous ideals contained in $$I$$;

(iii) $$I^*$$ is minimal among the homogeneous ideals containing $$I$$.

(iv) If $$I$$ is homogeneous, $$I_*=I^*=I$$

Having that, we immediately see that $$V_+(I^*)\subset V_+(I)\subset V_+(I_*)$$. Now, if $$\mathfrak{p}\in\text{Proj}(S)$$ with $$\mathfrak{p}\supset I$$ then $$\mathfrak{p}\supset I^*$$ by minimality of $$I^*$$, which proves $$V_+(I)\subset V_+(I^*)$$, therefore $$V_+(I^*)=V_+(I)$$.

But I'm having trouble proving $$V_+(I_*)\subset V_+(I)$$, since the maximality of $$I_*$$ doesn't help me here.