# The global (weak) dimension of formal power series rings

Given a commutative ring $R$, what are relations between w.gldim$(R)$ and w.gldim$(R[[x]])$ (gldim$(R)$ and gldim$(R[[x]])$)?

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is this homework? –  Fernando Muro Feb 18 '12 at 14:49
I like the question since $R$ is rather general (ok, the commutivity condition could be dropped, but that wouldn't essentially effect the results stated in my answer). In particular I would be interested in knowing the global dimension if $R$ is not Noetherian. –  Ralph Feb 18 '12 at 15:06

Let $R$ be commutative.

1. If $R$ is Noetherian, then $\text{gl.dim}(R[[X]]) = 1 + \text{gl.dim}(R)$.

2. If $R[[X]]$ is coherent, then $\text{w-gl.dim}(R[[X]]) = 1 + \text{w-gl.dim}(R)$.

Now let $R$ be Noetherian. Hence $R[[X]]$ is Noetherian and since global and weak-global dimension agree for Noetherian rings, we obtain:
$$\text{gl.dim}(R[[X]]) = 1 + \text{gl.dim}(R) = \text{w-gl.dim}(R[[X]]).$$

The first result is Theorem 1.12 of the paper

and the second is Lemma 1 of

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Thank you very much. –  Aimin Xu Feb 19 '12 at 1:54
Is point 1 true for not commutative and/or not noetherian ring R? Like in case of polynomial ring. –  jsfdlkdj Dec 5 '12 at 21:05
@jsfdlkdj: I don't know. At least the proof of Auslander, Buchsbaum doesn't work in general (since it use localization). –  Ralph Dec 5 '12 at 23:26