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Based on arXiv:math/9802041v1, there is a definition for $NC$-filtration and $NC$-completion of an associated algebra over the complex numbers:

Let $R$ be an associative algebra and $R^{\rm Lie} = (R, [a,b] = ab - ba)$ be $R$ regarded as a ${\rm Lie}$ algebra. By definition the following hold:

(a) The NC-filtration of $R$ is the decreasing filtration $\{F^dR\}_{d\ge 0}$ where $F^dR$ is the two-sided ideal $$F^dR = \sum_m \sum_{i_1+\cdots+ i_m -m = d} R \cdot R^{\rm Lie}_{i_1} \cdot R \cdot{\kern 1.5pt}{\cdots}{\kern 1.5pt}\cdot R\cdot R^{\rm Lie}_{i_m}\cdot R.$$

(b) For an associative algebra $R$ its NC-completion is the algebra $$R_{[ ab ]} = \lim_{\leftarrow} R/F^dR.$$ My question is: How can we generalize this definition to a Noetherian $R$-algebra $A$, when $R$ is a commutative Noetherian ring?

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  • $\begingroup$ I suggest to not call $A$'s base ring $R$ but $k$ or something similar, because $(A,k)$ plays the role of $(R,\mathbb{C})$ at the beginning. Besides from the confusing notation, can't you apply the definition of $F^dR$ and $R_{[ab]}$ literarly to $A$ ? $\endgroup$
    – Todd Leason
    Commented Apr 7, 2016 at 10:26
  • $\begingroup$ First, I think your suggestion about the base ring to be an algebraically closed field is interesting. Second, for your question, it's not necessary. $\endgroup$
    – user900000
    Commented Apr 7, 2016 at 11:35
  • $\begingroup$ (1) I didn't suggest an algebraically closed field as base ring, I just suggest to rename the latter $R$ by $k$ or something similar to not mix it with the former $R$. (2) Could you please elaborate why you think the definitions you wrote down for $F^dR$ and $R_{[ab]}$ don't apply to $A$ in general ? $\endgroup$
    – Todd Leason
    Commented Apr 7, 2016 at 11:44

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