1-Dimensional Cohen-Macaulay Rings by Eben Matlis

By Eben Matlis

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Extra resources for 1-Dimensional Cohen-Macaulay Rings

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But this follows again from Fubini’s theorem. Now simplify the conditions for special cases. Perhaps the most annoying assumptions are the measurability and c-finiteness of 29. Let 7 be the family of ( p xp)-measurable parts of a-finite measurable sets (see Appendix B for the relevant definitions). Write R ( s , t ):= (s, s t ) (it is injective) and 29o(s,t) = cp(s)x(t) ; then we have to consider the sets d-’(B) = R(2901(B))for B open in C \ ( 0 ) . From the integrability of cp/e and x / e we infer that 2901(B) E 7 , so it is enough if R ( T ) E 7 for all T E 7 .

Ifn 35 Convolution and Regularization Set H = { z E G ; p ( B A x B ) = 0) . Since p is left invariant, H is asubgroup. If 3: 4 H then s B A s x B is not negligible and hence A A s x B is not negligible (for " A M s B " ) and therefore p ( A n s z B ) = 0 for almast every such x . Thus we obtain that the function vanishes almost everywhere outside H . On the other hand, H is exactly the set where h = 1 . Hence h2 - h = 0 almost everywhere but h is continuous and vanishing a t infinity because = p * @ with p = p ( B ) - ' I 2 l g E L 2 ( G , p ) .

I f p is a left Haar measure on G and p1, p2 are absolutely continuous finite measures (with respect to p ) with p, = fjp then p1 * p 2 = gp with (2-3) where we integrate with respect to p , and A is the modular function of G (the integrals may not exist for a negligible set of z ’s). PROOF:Replacing f1, f2 by Baire measurable functions does not affect the integrals here. Then for ‘p E C,(G) the function fi(y)f2(x)cp(yz) is integrable with respect to p x p and vanishes outside a countable union of compacts.

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