Let $A$ be the non-commutative ring given by $$ A=\mathbb{C}\langle x,y,z \rangle /(xy=ayx,yz=bzy,zx=cxz) $$ for some $a,b,c\in \mathbb{C}$. What is the localization $A_{(x)}$ of A with respect to the (two-sided) ideal $(x)$? If it can be defined, is it graded ring? In general what condition is required to localize a ring?

I think of $A$ as a non-commutative $\mathbb{P}^2$ and wonder whether or not we can study it by local patch.

Thank you in advance.

I should mention this; my main problem is I don't know about the definition of localization. Moreover even if it is defined I am not sure whether this technique is useful or not. I would like to conclude for example smoothness of the non-commutative $\mathbb{P}^2$ or a hypersurface in it. I hope to confirm this by checking it locally.

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    $\begingroup$ Regarding localization in the non-commutative setting, you might want to consider this MO answer which gives some intuition as to why localization can be problematic in the non-commutative setting. Regards, $\endgroup$ – Matt E Aug 2 '12 at 2:45
  • $\begingroup$ Thanks for the link. I still think that my example is one of the easiest case and we can say something by localization. $\endgroup$ – Michel Aug 2 '12 at 4:08

This isn't really an answer to your question (except perhaps the last part), but: Wikipedia claims that the localization of a noncommutative ring $R$ with respect to some subset $S$ does not always exist. This, I think, comes from using the wrong definition of localization. The definition that seems natural to me is the following.

Definition: The localization $S^{-1} R$ of a ring $R$ with respect to a subset $S$ is the universal ring equipped with a morphism $\phi : R \to S^{-1} R$ such that $\phi(s)$ is invertible for every $s \in S$.

The localization in this sense always exists. It can be constructed explicitly as consisting of sums of formal symbols $r_1 s_1^{-1} r_2 s_2^{-1} ... $ where $r_i \in R, s_i \in S$ quotiented by suitable relations. The real question is whether one can say anything reasonable about the localization in general (in particular determine whether it is nonzero)...

The concern that Wikipedia actually has seems to be with whether every element of the localization can be written $r s_1^{-1} s_2^{-1} ...$ (that is, whether we need to alternate elements of $R$). My understanding is that this is addressed by the Ore condition. For more general notions of noncommutative localization see for example the nLab.

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  • $\begingroup$ Thanks for your response. You are right: The localization in your sense always exists. The thing is I am not sure if this is useful or not (I should have clarified this point in my post). For example, I would like to conclude for example smoothness or dimension by localizing my ring (this is common technique in commutative case). I will edit my question a bit. $\endgroup$ – Michel Aug 2 '12 at 1:47
  • $\begingroup$ I agree that it is far from clear whether this is a useful thing to do in general. $\endgroup$ – Qiaochu Yuan Aug 2 '12 at 1:52
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    $\begingroup$ Dear Qiaochu, Regarding whether localization has a reasonable meaning in the non-commutative setting, you may want to look at this MO answer of mine. Regards, $\endgroup$ – Matt E Aug 2 '12 at 2:45

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