By H. Matsumura, Miles Reid
As well as being a fascinating and profound topic in its personal correct, commutative ring idea is necessary as a starting place for algebraic geometry and intricate analytical geometry. Matsumura covers the fundamental fabric, together with size thought, intensity, Cohen-Macaulay jewelry, Gorenstein jewelry, Krull earrings and valuation earrings. extra complex themes similar to Ratliff's theorems on chains of best beliefs also are explored. The paintings is basically self-contained, the single prerequisite being a valid wisdom of contemporary algebra, but the reader is taken to the frontiers of the topic. routines are supplied on the finish of every part and suggestions or tricks to a couple of them are given on the finish of the booklet.
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Extra resources for Commutative Ring Theory
10. We say that a non-empty closed set Ii in a topological space is reducible if it can be expressed as a union V = V, u V, of two strictly smaller closed sets V, and Vz, and irreducible if it does not have any such expression. If peSpec A then V(p) is an irreducible closed set, and conversely every irreducible closed set of Spec A can be written as V(p) for some pESpec A. 11. Any closed subset of a Noetherian union of finitely many irreducible topological space can be written closed sets. 12. Use the results of the previous two exercises to prove the following: for I a proper ideal of a Noetherian ring, the set of prime ideals containing I has only finitely many minimal elements.
First of all, A has only finitely many maximal ideals. Indeed,ifp,,p,,... is an infinite set of distinct maximal ideals then it is easy to see that p1 3 plpz =) p1p2p3”. is an infinite descending chain of ideals, which contradicts the assumption. Thus, we let pl, pz,. . , pr be all the maximal ideals of A and set I = p1p2.. p, = rad (A). The descending chain III2 2 ‘.. stops after finitely many steps, so that there is an s such that I” = Z’+i. If we set (0:I”) = J then (J:Z) = ((0:P):I) = (O:r+l) = J; let’s prove that J = A.
IIEX Proof. For XEK the set I= (aEAlaxEA} is an ideal of A. Now XEA, is equivalent to I $ p, so that if XEA,,, for every maximal ideal m then 1~1, that is XEA. w Remark. The above I is the ideal consisting of all possible denominators of x when written as a fraction of elements of A, together with 0, and this can be called the ideal of denominators of x; similarly Ix can be called the ideal of numerators of x. 8. Let A be a ring and M a finite A-module. If M OAIc(m) = 0 for every maximal ideal m then M = 0.