Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  fpqc May 27 '12 at 3:20
1  
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
1  
@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
show 11 more comments

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.