By Forsberg, Johansson

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**Example text**

Then g ( c i z 0 1’ axi) = x', proving exactness. D. - 1- a * . , - We are ready for the first inequality. dim R [ I ; a ] Igl. dim R pd, o M 1 for any RIA; o]-module M. + + 1. In fact, pdRIA;al M I Proof: Let n = gl. dim R. For any R [ I ;u]-module M we have P~R[,;,]M[n; D l I pd, M 5 n, and pd,,,,,,(aM)[I;o] Ipd,oM I n. 18'thus shows pd,[,;,,M I n + 1. D. Our next goal is to see, in fact, that equality holds. 20: pd(@,,, Mi) = sup{pd(MJ: i E I}. , 9 is formed by taking the direct sum of the respective terms.

1 1) is a category equiualence 9r-#mj(R0) -+ Yt-&j(R). Proof of Claim 1: Take the functor G = R, OR-: R-Yt-Aud -+ R,-YaMod. Since GF is naturally equivalent to 1 it suffices to show P z FGP (graded) for every P in Yr-fiwj(R). 6. Hence there is a graded monic f: Q 4 P. Now FGP = R ORoQ so we can define a graded map cp: FGP + P such that cp(r 0 x) = r f x for r in R and x in Q. 34’. But then cp is epic and thus split, so G(ker cp) = 0, likewise implying ker cp = 0. Thus cp is an isomorphism, proving claim 1.

Dim, which also is 0 if gl. dim R = 0 since “semisimple Artinian” is leftright symmetric. However, for gl. dim 2 1. A thorough discussion of global dimension and its peculiar connection to the continuum hypothesis can be found in Osofsky [73B]. g. free resolution of length n. ) Our interest in FFR is derived from the next result. 28: If P i s projective with FFR then P is stably free. g. free resolution 0 -+ F, -+F,... F , fi Fo 5 P 0. g. free resolution 0 -, F, + F,- -+ F, -+ P’ + O of length n - 1.