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Let $S$ be a graded ring and $f,g\in S$ be homogeneous of positive degree such that $D_+(g)\subset D_+(f)$, then I wonder if it is true that $S_{(g)}=S_{(fg)}$

I think this result should be true but I did not seem to find the proof. By this question we can assume $fr=g^n$ for some $r\in S$ homogeneous.

We can define $\phi:S_{(g)}\to S_{(fg)}$, $$ \frac{h^{\deg(g)}}{g^{\deg(h)}}\mapsto \frac{f^{\deg(h)}h^{\deg(g)}}{(fg)^{\deg(h)}} $$ and its inverse $S_{(fg)}\to S_{(g)}$, $$ \frac{h^{\deg(f)+\deg(g)}}{(fg)^{\deg(h)}}\mapsto \frac{h^{\deg(f)+\deg(g)}r^{\deg(h)}}{g^{(n+1)\deg(h)}} $$ extending both numerator and denominator by $r^{\deg(h)}$. I know this site is not for checking correctness of proof. But I want to know if I am in right direction. Thanks.

Sammy Black
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user11695417
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    By $D_+(f)$, do you mean the set of homogeneous prime ideals not containing $f$? – walkar Dec 02 '24 at 15:04
  • $D_+(f)$ the set of homogeneous prime ideals not containing $f$. I think the author also assumes $f$ to be of positive degree even he did not write it. If $f$ is homogeneous of positive degree. Then $D_+(f)=Proj(S)\cap D(f)$ – user11695417 Dec 02 '24 at 16:35
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    I believe your proof is correct then. You can avoid constructing the isomorphism directly by using the universal property of (homogeneous) localization. – walkar Dec 02 '24 at 17:40
  • Thank you for contributing to answer my question:) – user11695417 Dec 02 '24 at 17:50

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