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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:

enter image description here

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.


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I've replaced your diagram with a version that should be a bit easier to read; please double-check that I didn't introduce any errors, and of course feel free to change back if you want. – Zev Chonoles May 27 '12 at 3:16
@ZevChonoles Thanks Zev for editing it, now it looks a lot better! I tried to export the figure out of latex so I just went print screen :D Do you think you can help me out with my question above? Thanks. – user38268 May 27 '12 at 3:20
In general, if $g\colon A \to B$ is a ring homomorphism and $(a)$ a principal ideal of $A$, then the extension of $(a)$ to $B$ is just $(g(a))$. You are definitely capable of proving this. – Dylan Moreland May 27 '12 at 4:09
The diagram sort of implies that you expect some object to go in the $?$ slot, but I don't think that's what you're asking. – Dylan Moreland May 27 '12 at 4:10
@BenjaminLim Then it seems to me that you know the answer to the question you ask right after the diagram! – Dylan Moreland May 27 '12 at 4:14
up vote 3 down vote accepted

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.

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