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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
up vote 4 down vote accepted

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

Auslander, Buchsbaum: Homological dimension in Noetherian rings, Trans. Amer. Math. Soc. 88(1958),194-206

and the second is Lemma 1 of

Jondrup, Small: Power Series over coherent rings, Math. Scand. 35(1974), 21-24.

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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

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