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Let $R$ be a commutative unital ring and $g\in R[X]$ a polynomial with the property that $g(0)$ is a unit in $R$ and $g(1)=1$. Is there any possible way to understand either $$R[X]/g$$ or $$g^{-1}R[X]$$ better? Here $g^{-1}R[X]$ is the localised ring for the multiplicative set $\{g^n\}$. I can't find a way to incorporate the extra conditions on $g$. It would be favorable if the rings were expressable using the ring $R[X,X^{-1}]$ or the ideal $X(X-1)R[X]$. Also any graded ring with a simple expression in degree zero is much appreciated.

Background: Im working on some exact sequences in the K-theory of rings. The two rings above are part of some localisation sequence and understanding either one of them would help me simplifying things.

Edit: I tried to go for the basic commutative algebra version of it, since I expected it to be easy. What I am actually trying to show is that if $S$ is the set of all those $g$ whit the property as described above, then there is an exact sequence (assume $R$ regular, or take homotopy K-theory) $$0\to K_i(R)\to K_i(S^{-1}(R[X]))\to K_i(R)\to 0$$ where the first map comes from the natural inclusion and the second map should be something like the difference of the two possible evaluation maps (as described in @Martin's answer). There are some reasons to believe that this is true (coming from a way more abstract setting), but it still looks strange to me. Moreover we should be able to work with one $g$ at a time and then take the colimit. With this edit I assume this question should go to MO as well -_-

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    Concerning your edit: Do you know the localization theory developed in Bass' book on Algebraic K-Theory? Anyway, the question should fit better to MO.2012-05-30
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    I haven't read it in Bass back but am pretty sure that I read more modern versions which refer to Bass. The problem is, that I need to work the other way round, i.e. the usual localisation theory wants to comupte K-thy of the localisation by some sequence. I want to use the very same sequence to compute something else and therefore need the K-theory of the localisation... But maybe it doesn't work.2012-05-30

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