Applications of categorical algebra;: [proceedings by American Mathematical Society (ed.)

By American Mathematical Society (ed.)

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Extra info for Applications of categorical algebra;: [proceedings

Sample text

Thus R is h-local if and only if modulo any non-zero prime ideal it is a quasi-local ring, and modulo any non-zero ideal it is a quasi-semilocal ring. The notion of an h-local ring has proved to be extremely useful, and it crops up under an amazing variety of conditions. We will explore a number of equivalent definitions of this kind of ring. It is this richness of description which gives the notion its tremendous power. But first we need some lemmas. THEOREM 19. Let Mand N be maximal ideals of R.

B' - 0 . Therefore, rank B' =rank A' - rank I'= n-1 =rank B. Hence B and B' are torsionless R-modules of rank n-1 by Theorem 27. Since B' is a submodule of a free R-module by the same theorem, we have Ext~(B•, R) rows: = 0. I is an isomorphism, since I is reflexive; isomorphism by induction on n = rank A. B is an Therefore A. A is an iso- is a reflexive R-module. Thus R is a reflexive ring. THEOREM 30. Let R be a reflexive ring .. Then R is an h-local ring. Proof. Let maximal ideals of R. {M } , y e y r , be the collection of all of the Jy = R/M y Let by Theorem 29 we can assume that E y be the injective envelope of ,i , and let = ~~ ye r C K and that K = E( ~ in K.

Thus ~ is 54 onto and we have r ~ 1 - 1 /r S. But J/I c J- 1 - 1 /r, and therefore J = J-1-1_ Remarks. There are examples to slx>w that every ideal of R can be reflexive without R being a reflexive ring. There are also examples to show that K can be injective without being a universal injective. However, the next theorem shows that for Noetherian rings neither of Compare this theorem with Theorem 29 ! these things can happen. THEOREM 40. Let R be a Noetherian integral domain. Then the following statements are equivalent: (1) R is a reflexive ring.