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I have been trying to establish an isomorphic concerning graded rings, and there is a last step that I'm confused about.

Let $R$ be a $\Bbb{Z}$ - graded ring. Let $f$ be a homogeneous non-nilpotent element of degree $1$ in $R$. Let $R_f$ denote the localisation $S^{-1}R$ where $S = \{1,f,f^2 \ldots ...\}$. It is not hard to see that $R_f$ is also a graded ring and we denote by $(R_f)_0$ the degree zero component of $R_f$. Let $t$ be an indeterminate, I have established the following isomorphisms:

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The "?" on the bottom left corner of the diagram is something that I want to establish. Now firstly when I try to identify $\ker \pi \phi = \phi^{-1} (\ker \pi)$, I am curious to know if the ideal $(f-1)$ in $R_f$ is actually the extension of the ideal $(f-1)$ in $R$. Sorry for the bad notation, but perhaps I should call the one in $R_f$ say $(f-1)^e$ and the one in $R$ just $(f-1)$. Perhaps they are actually the same?

The dotted arrows from "?" to $R_f/(f-1)$ indicate some map that I am trying to establish. I believe such a map is an isomorphism, and I want to prove this using the first isomorphism theorem. The problem is I don't know if $\pi \phi$ is surjective so how can I conclude such an isomorphism exists? Perhaps it may not be true after all.

Thanks.

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    @BenjaminLim: Don't worry about it :) I think the key to getting a good screenshot is making the window with the math output full-screened, then zooming in as far as possible so the math is as large as possible while still fitting everything in on the screen, and then (after taking the screenshot) cropping the picture appropriately so only the math is left, and not the toolbars or whatever from the program one is using.2012-05-31

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Of cause, your idea is right: Passing to the quotient where $f$ is unity isn't affected by localizing at $f$ first. Anyway, localizing commutes with quotients (see comments above) and we have a commutative diagram $\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\sum}\right.} \begin{array}{c} R & \longrightarrow & R_f \\ \da{\;}& & \da{\;}\\ R/(f-1) & \xrightarrow{\varphi} & R_f/(f-1) \end{array},$ where $\varphi$ is the composition of canonical maps $R/(f-1)\xrightarrow{\psi} (R/(f-1))_f \;\tilde\to\; R_f/(f-1)$. To see that this is an isomorphism, you indeed could use the homomorphism theorems, but its easier.

The latter arrow is the canonical isomorphism and the former ($\psi$) is localizing. Hence, if we view $(R/(f-1))_f$ as $(R/(f-1))[u]/(fu-1)$, $\psi$ is just inclusion $R/(f-1)\hookrightarrow(R/(f-1))[u]$ and projection. On the other hand we have $(R/(f-1))[u]/(fu-1) = (R/(f-1))[u]/(u-1) \cong R/(f-1),$ where reading backwards, the isomorphism is nothing but inclusion $R/(f-1)\hookrightarrow(R/(f-1))[u]$ and projection. Hence there is the commuting diagram $\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\sum}\right.} \begin{array}{c} R/(f-1) & \xrightarrow{\psi} & (R/(f-1))_f \\ \;& \searrow & \da{\;}\\ \; & \; & (R/(f-1))[u]/(fu-1) \end{array}$ of $\searrow$ and $\downarrow$ being isomorphisms, such that $\psi$ is so, as desired.