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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$\begingroup$ is this homework? $\endgroup$– Fernando MuroFeb 18, 2012 at 14:49
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$\begingroup$ 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. $\endgroup$– RalphFeb 18, 2012 at 15:06
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$\begingroup$ @FernandoMuro It's still an open problem as of end of 2023. $\endgroup$– Denis TNov 15, 2023 at 1:10
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1 Answer
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Let $R$ be commutative.
If $R$ is Noetherian, then $\text{gl.dim}(R[[X]]) = 1 + \text{gl.dim}(R)$.
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
Jondrup, Small: Power Series over coherent rings, Math. Scand. 35(1974), 21-24.
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$\begingroup$ Is point 1 true for not commutative and/or not noetherian ring R? Like in case of polynomial ring. $\endgroup$– jsfdlkdjDec 5, 2012 at 21:05
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$\begingroup$ @jsfdlkdj: I don't know. At least the proof of Auslander, Buchsbaum doesn't work in general (since it use localization). $\endgroup$– RalphDec 5, 2012 at 23:26