# Localization of a graded ring at degree zero

Let $S$ be a $Z^{\ge 0}$-graded ring and $f,g$ be two homogeneous elements of positive degree in $S$.

I wonder if $$(S_{fg})_0 \cong [(S_f)_0]_{g^{\text{deg} f}/f^{\text{deg} g}}$$

is true (and how to prove it).

Here $S_{fg}$ and $S_f$ are all localizations at elements. $(\cdot)_0$ means taking the degree zero component of $S$ and it is clear that ${g^{\text{deg} f}/f^{\text{deg} g}}$ is a degree zero element in $(S_f)_0$ and the RHS is the localization w.r.t. it.

• I assume that $S_{fg}$ and $S_f$ denote the localizations at the corresponding elements? Mar 15, 2018 at 20:19
• Also what does $[(S_f)_0]_{g^{\deg f}/f^{\deg g}}$ means? Mar 15, 2018 at 20:22
• Do you mean to assume $f$ and $g$ both have positive degree? This won't be true if $f$ has degree $0$, but maybe that's a degenerate case. Mar 15, 2018 at 20:23
• @Levent I added some explanations and hope it's clear now... Mar 15, 2018 at 20:29
• @EricWofsey Thank you! I have restricted it to this case Mar 15, 2018 at 20:30

## 1 Answer

Let $\deg f=a$ and $\deg g=b$. Let $R=S_f$, so $R_g\cong S_{fg}$. The inclusion map $R_0\to R$ induces a homomorphism $\varphi:(R_0)_{g^a/f^b}\to R_g$, since $g^a/f^b$ is a unit in $R_g$. First, I claim that $\varphi$ is injective. An element of $(R_0)_{g^a/f^b}$ has the form $\frac{r}{(g^a/f^b)^n}$ where $r\in R_0$, and $\varphi$ sends such an element to the fraction $\frac{rf^{bn}}{g^{an}}$ in $R_g$. This latter fraction is $0$ in $R_g$ iff there exists $m$ such that $g^mrf^{bn}=0$ in $R$. Since $f$ is a unit in $R$ and $am\geq m$, this implies $g^{am}f^{-bm}r=0$ as well, which implies $\frac{r}{(g^a/f^b)^n}=0$ in $(R_0)_{g^a/f^b}$.

So, $\varphi$ is injective. It remains to be shown that the image of $\varphi$ is exactly $(R_g)_0$. Clearly the image of $\varphi$ is contained in $(R_g)_0$. Conversely, consider an element $\frac{r}{g^n}\in (R_g)_0$, where $r\in R$ has degree $bn$. We can write $$\frac{r}{g^n}=\frac{(g^{an-n}f^{-bn}r)f^{bn}}{g^{an}}$$ and so $$\frac{r}{g^n}=\varphi\left(\frac{g^{an-n}f^{-bn}r}{(g^a/f^b)^n}\right)$$ (here we again use that $a>0$ so $an\geq n$). Thus $\frac{r}{g^n}$ is in the image of $\varphi$, and so the image of $\varphi$ is all of $(R_g)_0$.