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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.

Thanks.

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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.2012-05-27
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    @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.2012-05-27
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    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.2012-05-27
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    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.2012-05-27
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    @DylanMoreland Regarding your first comment I know that already, however I'm not sure if it's related to what I'm asking. Also, your second comment is very relevant to my question, definitely I would like to know what goes in the slot there! It's what I kinda trying to figure out, my guess is that it is $R/(f-1)$.2012-05-27
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    @BenjaminLim Then it seems to me that you know the answer to the question you ask right after the diagram!2012-05-27
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    @DylanMoreland Hmmm it seems the question I asked after the diagram is kinda of the "reverse" of the fact you stated though....2012-05-27
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    @DylanMoreland You know what I'm stupid my first question is trivial. Although I am very curious to know what goes in that slot there......2012-05-27
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    I think $R/(f - 1)$ is a good guess. There are many reasons for this. But let's just try to prove that it works. What is the map $R/(f - 1)R \to R_f/(f - 1)R_f$?2012-05-27
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    @DylanMoreland If it is helpful, my goal at the end is to establish the isomorphism in Ex.2.17 (b) of Eisenbud. From the diagram above, I already have $(R_f)_0[t,t^{-1}]/(t-1)$ being isomorphic to $(R_f)_0$ so the "?" in the diagram is the last bit to complete the proof.2012-05-27
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    @DylanMoreland I believe I have proven the isomorphism. I will post an answer tomorrow.2012-05-27
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    What is left to proof if you consider the canonical isomorphism $R_f = R[u]/(uf-1)$? (If I'm not mistaken this is also an Eisenbud-exercise.) In this manner its evident that $R_f/(f-1) = (R/(f-1))_f = R/(f-1)$ where the first isomorphism holds by the general commutativity of localization with ideal quotients.2012-05-27
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    @BenA. Ah I think I see what you mean, the first one comes from the general fact that $S^{-1}(M/N) \cong S^{-1}M/ S^{-1}N$. If you post an answer, I would be glad to accept it!2012-05-27
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    @BenjaminLim: I didn't do anything automatic with your diagram - I just coded it by hand on my computer and took a screenshot of my screen, so I don't want this to be an expected service :) I think the diagrams in Ben's answer are sufficiently readable to not require the use of a picture, whereas the diagram in your question could not be easily done with MathJax.2012-05-29
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    @ZevChonoles Sorry I did not mean to say it like that, however my diagram too was done out of a screen shot but somehow did not come out as nice as yours!2012-05-29
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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.