If $R_0 \not\cong S_0$ but $\exists n \in \mathbb N$ with $R_d \cong S_d$ for all $d \ge n$, then $\operatorname{Proj}R \cong \operatorname{Proj}S$?

Asked Sep 19 '21 at 17:50

Active Sep 19 '21 at 17:50

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Let $R=R_0 \oplus R_1 \oplus \cdots$ and $S=S_0 \oplus S_1 \oplus \cdots$ be finitely generated graded rings with $R_0 \not\cong S_0$. If there exists a positive integer $n$ such that $R_d \cong S_d$ for all $d \ge n$, then $\operatorname{Proj}R \cong \operatorname{Proj}S$?

I know the result is true if $R_0 \cong S_0$ since then the $d$th-Veronese subrings of $R$ and $S$ are isomorphic. Is this true if $R_0 \not\cong S_0$?

asked Sep 19 '21 at 17:50

user5826

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Two applications of the argument from the linked duplicate show that the answer is yes. – KReiser Sep 19 '21 at 18:00
@KReiser Very nice generalization. – user5826 Sep 19 '21 at 19:40

If $R_0 \not\cong S_0$ but $\exists n \in \mathbb N$ with $R_d \cong S_d$ for all $d \ge n$, then $\operatorname{Proj}R \cong \operatorname{Proj}S$?

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