Let $I,J \subset S=k[x_0,...,x_n]$ be homogeneous ideals. How can I show that $\operatorname{Proj}S/\bar{I}=\operatorname{Proj}S/\bar{J}$ iff $\bar{I}=\bar{J}$?
(Here $\operatorname{Proj}R$ is the set of all homogeneous prime ideals $p$ in a graded ring $R$ such that $R^+=R \backslash R^0 \nsubseteq p$; the saturation of $I$, $\bar{I}:=\{s \in S:x_i^ms\in I$ for some $m$ and all $i\}$.)
This is needed in showing the homogeneous coordinate ring of $X=\operatorname{Proj}S/I$ is well defined by $S(X):=S/\bar{I}$. Any help is appreciated!
Let $\mathbb{P}^n=\operatorname{Proj} S$, $X=\operatorname{Proj} S/I$ and $Y=\operatorname{Proj} S/J$: we denote the corresponding ideal sheaves by $\mathcal{I}_X,\mathcal{I}_Y$. Then it is easy to see that $$ X=Y \text{ as subschemes} \iff \mathcal{I}_X=\mathcal{I}_Y \iff \bigoplus_{q=0}^{\infty} H^0(\mathbb{P}^n,\mathcal{I}_X(q))=\bigoplus_{q=0}^{\infty}H^0(\mathbb{P}^n,\mathcal{I}_Y(q)) $$ To conclude it is enough to show the following:
Lemma: In the above notation, we have $$\bigoplus_{q=0}^{\infty}H^0(\mathbb{P}^n,\mathcal{I}_X(q))=\overline{I}$$
Proof: Denote by $U_i=\text{Spec} (S_{X_i})_0$ for $i=0,\dots,n$ the standard affine open cover of $\mathbb{P}^n$. Then by definition $(\mathcal{I}_X(q))_{|U_i}$ corresponds to the ideal $(I_{X_i})_q$. Now we prove the two inclusions:
$\subseteq$ : suppose that $f\in H^0(\mathbb{P}^n,\mathcal{I}_X(q))$. Then this means that for any $i$ $\frac{f}{1} \in (I_{X_i})_q$ , i.e.there exists $N_i\geq 0$ and $g_i \in I_{N_i+q}$ such that $\frac{f}{1}=\frac{g_i}{X_i^N}$, which is the same as saying that $X_i^{N_i}f \in I$. Taking $m=\max{N_i}$ we see that $fX_i^{m}\in I$ for all $i$, so that $f\in \overline{I}$
$\supseteq$: let $f\in \overline{I}$ be homogeneous of degree $q$. Then there is an $m$ such that $f\cdot X_i^m \in I_{q+m}$ for all $i$. This means that $\frac{f}{1}=\frac{f\cdot X_i^m}{X_i^m} \in (I_{X_i})_q$ for all $q$, and then it follows from the definition that $f\in H^0(\mathbb{P}^n,\mathcal{I}_X(q))$. Since $\overline{I}$ is homogeneous, we are done.