# Inverse limit of graded rings

Let $(I,\le)$ be a directed set and let $(\rho^{\beta\alpha}: R^\beta \to R^\alpha)_{\alpha \le \beta}$ be an $I$-directed system of $\mathbb{Z}$-graded rings whose multiplication is denoted by

$$\mu_{nm}^\alpha: R_n^\alpha \otimes R_m^\alpha \to R_{n+m}^\alpha.$$

For fixed integers $n,m$ we have the inverse limit of abelian groups $\hat{R}_n = \varprojlim R_n^\alpha$ and $\mu_{nm}^\alpha$ induces a linear map $\hat{\mu}_{nm}: \hat{R}_n \otimes \hat{R}_m \to \hat{R}_k\; (k=n+m)$.

Question: Is it possible to describe the inverse limit $\hat{R} := \varprojlim R^\alpha$ by $\hat{R_n}$ and $\hat{\mu}_{nm}$ ? Or is there another nice description ?

Note: The inverse limit is taken in the category of rings (not in the category of graded rings)!

Such a description exists, if the rings are $\mathbb{N}$-graded: For, then $\hat{R} = \prod_{n \ge 0} \hat{R}_n$ with the usual multiplication

$$x \cdot y = \big(\sum_{i+j=n} \hat{\mu}_{ij}(x_i,y_j)\big)_{n \ge 0}.$$

However, I wasn't able to figure out something similar in the $\mathbb{Z}$-graded case. By writing $x_n = (x_n^\alpha)_\alpha \in \hat{R}_n$ we have

$$\hat{R} = \lbrace (x_n) \in \prod_n \hat{R}_n \mid \forall \alpha: x_n^\alpha = 0 \text{ for almost every }n\rbrace$$ as abelian group. But I hope for a description that doesn't resort to an explicit construction of the inverse limits $\hat{R}_n$.

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Does your problem come from some geometric situation, like some formal projective space, at is it purely algebraic! I do not know whether this really matters, but it simplifies the generalized picture... – Filippo Alberto Edoardo Sep 28 '12 at 0:08
No, there is no geometric background. Originally I was interested in the inverse limit taken in the graded category - it's $\bigoplus_n \hat{R}_n$ (with the multiplication described in the $\mathbb{N}$-graded case). Then I started wondering what the inverse limit as abstract rings might be. – Ralph Sep 28 '12 at 0:37