# Isomorphism between localizations of graded ring $S_{(P)} \cong [S_{(f)}]_{PS_f \cap S_{(f)}}$

I know that if $$S$$ is a graded ring, and $$f$$ is a homogeneous element of positive degree, then there is a bijection between the homogeneous prime ideals of the localization $$S_f$$ and the prime ideals of $$S_{(f)}$$, the subring of $$S_f$$ comprising the homogeneous elements of degree $$0$$ such as from this MSE question. How can I prove that this bijection gives rise to an isomorphism of the following rings: for a homogenous prime ideal $$P$$ of $$S$$ with $$f \notin P$$, there is an isomorphism

$$S_{(P)} \cong [S_{(f)}]_{P S_f \cap S_{(f)}}.$$

Is this true? If yes, what is this isomorphism? In other words, how do I lift the bijection between prime ideals mentioned above, to elements in the prime ideals related by this bijection. Here $$S_{(P)}$$ is the subring of degree $$0$$ elements of the localization $$S_{P}$$ like always.

• See also Stacks 00JR which proves the result you seek without any extra hypotheses but leaves much of the verification to the reader. Commented Mar 30, 2021 at 16:11

I suspect you may need that $$S$$ is generated in degree 1, but perhaps I'm missing a way to fix the following argument.

Since $$P$$ is a homogeneous prime and $$S$$ is generated in degree 1, there is an element $$\lambda$$ of degree one such that $$\lambda\not\in P$$. Let $$\phi : S_{(P)}\to [S_{(f)}]_{PS_f\cap S_{(f)}}$$ be defined by $$\phi\left(\frac{s}{r}\right)=\frac{s\lambda^k/f^n}{r\lambda^k/f^n},$$ where $$n$$ is large enough that $$n\deg f \ge \deg s$$ and $$k$$ is such that $$\deg s + k = n \deg f$$. To see that this is well defined, observe that if $$\frac{s}{r} = \frac{s'}{r'}$$, then there is $$h\not\in P$$ such that $$h(r's-rs')=0$$, and for appropriate choices of $$\ell,j,k,n,m,o$$, we have $$\frac{h\lambda^\ell}{f^o}\left(\frac{r'\lambda^j}{f^m}\frac{s\lambda^k}{f^n}-\frac{s'\lambda^j}{f^m}\frac{r\lambda^k}{f^n}\right)=\frac{\lambda^{\ell+j+k}}{f^{o+m+n}}(h(r's-s'r))=0.$$ Thus $$\phi(s/r)=\phi(s'/r')$$.

To see that $$\phi$$ is the desired isomorphism, note $$PS_f$$ is the set of elements $$a$$ of $$S_f$$ such that $$af^n\in P$$ for some $$n$$. Thus if $$s/f^m\not\in PS_f$$ with $$s\in S_{m\deg f}$$, and if $$r\in S_{n\deg f}$$, and if $$\ell=\max\{n,m\}$$ we can define the inverse by $$\psi\left(\frac{r/f^n}{s/f^m}\right)=\frac{rf^{\ell-n}}{sf^{\ell-m}},$$ which works since we know $$sf^{\ell-m}\not\in P$$, since $$s/f^m \not\in PS_f$$, and both numerator and denominator have degree $$\ell \deg f$$.

It shouldn't be hard to verify that $$\psi$$ is also well defined and $$\phi$$ and $$\psi$$ are inverses.

• Thanks for the answer @jgon. I believe the extra hypothesis is not necessary, but I am not able to prove it otherwise. So I will wait a couple more days to see if I get any more answers before accepting your excellent answer. Commented Nov 22, 2018 at 5:32

Here I believe is a proof without assuming S is generated in degree 1. For $$P\subset S$$ homogenous with $$f\notin P$$ let $$\varphi(P):=PS_f\cap S_{(f)}$$. Define $$\psi: \Big(S_{(f)}\Big)_{\varphi(P)}\to S_{(P)}$$ by $$\frac{a/f^n}{b/f^m}\mapsto \frac{af^m}{bf^n}$$.

To see that $$\psi$$ is injective suppose $$\frac{af^m}{bf^n}=0$$ in $$S_{(P)}$$. Then $$\frac{af^m}{bf^n}=0$$ in $$S_{P}$$ so there is an $$h\notin P$$ such that $$haf^m=0$$. We may assume $$h$$ is homogeneous because if it's not we replace it by one of its nonzero homogeneous components. But then $$\frac{h^{deg(f)}f^m}{f^{deg(h)+m}}\notin \varphi(P)$$ and $$\frac{h^{deg(f)}f^m}{f^{deg(h)+m}}\cdot \frac{a}{f^n}=0$$ in $$S_{(f)}$$. Therefore $$\frac{a/f^n}{b/f^m}=0$$ in $$(S_{(f)})_{\varphi(P)}$$ and $$\psi$$ is injective.

To se that $$\psi$$ is surjective let $$\frac{x}{y}\in S_{(P)}$$ and note that $$deg(x)=deg(y)$$ and $$y\notin P$$. Then $$\frac{x}{y}=\frac{x\cdot y^{deg(f)-1}}{y^{deg(f)}}=\psi\Big(\frac{x\cdot y^{deg(f)-1}/f^{deg(x)}}{y^{deg(f)}/f^{deg(x)}}\Big).$$