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Commutative Rings [bad OCR] by Irving Kaplansky

By Irving Kaplansky

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Suppose, in addition to the hypotheses of Theorem 110, that a n~ultiplicativelyclosed set in R is given: let it be denoted by S. Then R s is a locallyjnite intersection of the Vi’s that contain Rs. Proof. Let us use the subscript j for a typical V, containing Rs. To prove RS = A V , we take x E AV, and have to prove x E Rs. Let W,, . , W , be the finite number of V,’s not containing x. Thus sh is a nonunit in wk. By Theorem 108, sknlxE wk for some nr. Then with s = we have sx E R and so x E Rs.

Then some power x lies in ( I , a), say x = il + + ya (il e I , y e R). Similarly xn = i2 zb (i2e I, z e R). By multiplying these two equations we get x+" e I, x e I , as required. Remark. Since any prime ideal above a given ideal contains a minimal one (Theorem lo), the intersection in Theorem 87 might as well be 59 SEC. 2-2/ZERO-DIVlSOl7S confined to the prime ideals minimal over the radical ideal. In that case one readily sees that the expression is unique, and that the prime ideals that occur are exactly all the minimal primes over the radical ideal.

This is done by a semi-direct sum T of R and A . T = R A is an additite direct sum, and the rule of multiplication is + (11 T) Equation (24) gives us u = ah(t - atl) E ahT. So u E ahT n R h > n we have I h = Ihfl. This means that we can write (25) SEC. 2-2/ZERO-DIVISORS + al)(r~+ ad = rlrz + (rlaz + rzad where r, E R , a, e A . With the same Z as in Ex. 6, prove that J C <(T), but that no non-zero element of T annihilates J , where J = IT, 8. Let I be a finitely generated ideal contained in a minimal prime ideal of a ring R.

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