By J.L. Bueso

Integrates basic innovations from algebraic geometry, localization conception and ring concept, and demonstrates how every one subject is more desirable by means of interplay with others, offering new effects inside of a standard framework. Technical conclusions are offered and illustrated with concrete examples.

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**Extra resources for Compatibility, Stability, and Sheaves (Pure and Applied Mathematics)**

**Example text**

For all M E R—mod, one has aM C M and f(o'M) c aM' for any homomorphism f : M M' of left Rmodules. We will call2 the functor o idempotent resp. a preradical, if a(aM) = aM resp. cr(M/aM) = 0, for any M E R—mod. A left exact preradical is said to be a radical and is sometimes also referred to as an zdernpotent kernel functor, in the sense of Goldman (351. 2) Proposition. Any left exact subfunctor a of the identity in R—mod is idempotent. Proof. Let M be a left R-module. As aM C M, it clearly follows from the left exactness of a that a(aM) = Ker(aM —* a(M/aM)) =Ker(aM —* M/aM) = aM, which proves that a is idempotent, indeed.

Indeed, we claim that for any L e £(a) we have Qa(R)I = Actually, since R is a Dedekind domain, L is invertible, say with inverse = {q E K; Lq c R}, where K is the field of fractions of R. Since LL' c then R and since L E £(a), clearly L' c But L-'L = R, proving that 1 E QU(R)L, hence that = as claimed. 19) Note. If the radical a has property (T), then a automatically has finite type. Indeed, assume R to be a-torsionfree, for notational convenience and let L e £(a). Then so 1 = = and some E L. Let K = for some q.

For any left R-module M, define by Z(M) {x e M; is essential in R} the singular submodule of M. The functor Z: R-mod —* R-mod: M Z(M) is left exact (hence idempotent), but not a preradical, in general. 7) Let R be a commutative domain. An R-module M is said to be divisible if rM = M for any 0 r R. , aM is the largest divisible R-submodule of M. It is easy to see that this defines an idempotent radical a, which is not left exact in general, however. 2. 2) F = {N e R—mod; HomR(M, N) = 0 for all M E T}.