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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.

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