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

Question

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.

Answer 1

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.

Answer 2

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 $$\frac{x}{y}=\frac{x\cdot y^{deg(f)-1}}{y^{deg(f)}}=\psi\Big(\frac{x\cdot y^{deg(f)-1}/f^{deg(x)}}{y^{deg(f)}/f^{deg(x)}}\Big).$$