Let $l$ be a prime number, $\mathbb{Z}_l$ be the ring of $l$-adic integers, then what is the projective dimension of the ring $A:=\mathbb{Z}_l[T,T^{-1}]$?
Is it two?
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$\begingroup$ What is the difference between the projective dimension of A and the projective dimension of the category of A-modules? $\endgroup$– Steven LandsburgCommented May 1, 2011 at 1:42
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$\begingroup$ Don't forget the words 'continuous' and 'open' should be inserted appropriately. As a crude ring, $\mathbb{Z}_l$ is infinitely-generated and almost certainly has infinite projective dimension, but there should be no earthly reason you want to forget the topology. $\endgroup$– Greg MullerCommented May 1, 2011 at 4:04
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2$\begingroup$ @Greg: $\mathbb Z_l$ is a principal ideal domain, so its global dimension is $1$. $\endgroup$– Mariano Suárez-ÁlvarezCommented May 1, 2011 at 4:11
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$\begingroup$ @Steven: Thanks for reminding me the ambiguity about proj. dim. $\endgroup$– HeerCommented May 1, 2011 at 14:57
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$\begingroup$ Heer: I wasn't trying to remind you of anything; I was asking a question. $\endgroup$– Steven LandsburgCommented May 1, 2011 at 15:11
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1 Answer
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$\mathbb Z_p$ is a principal ideal domain, so it is Noetherian and its global dimension is $1$. Now, there is a general theorem that tells you that for all right Noetherian rings $R$ one has $$\operatorname{gldim}R[X,X^{-1}]=\operatorname{gldim}R+1.$$ So your answer is indeed $2$.
You'll find that theorem proved pretty much anywhere where global dimension is discussed. For example, J. C. McConnell and J. C. Robson's bible Noncommutative Noetherian Rings (which should be called, as usual, "non-necesarily commutative noetherian rings"...)
answered May 1, 2011 at 3:00