By Prabhat Choudhary
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Extra info for Abstract Algebra
Then I is an ideal, necessarily principal by hypothesis. If I = (b) then b belongs to some an, so I ~ (an). Thus if i 2: n we have (a, ) ~ J ~ (an) ~ (a, ). Therefore (a = (an) j ) for all i 2: n, so that R satisfies the acc on principal ideals. Now suppose that a is irreducible. Then (a) is a proper ideal, for if (a) = R then I E a, so that a is a unit. By the acc on principal ideals, (a) is contained in a maximal ideal 1. (b) then b divides the irreducible element (a), and (b) is not a unit since J is proper.
Since PI'" Pn = (P2 '" Pn)PI E I, we have P2 ... Pn E J. Use the induction hypothesis to conclude that I is principal. 7. Le~ P and Q q be prime elements in the integral domain R, and let P = (p) and = (q) be the corresponding prime ideals. Show that it is not possible for P to be a proper subset of Q. 8. If R is a UFD and P is a nonzero prime ideal of R, show that P contains a nonzero principal prime ideal. Principal Ideal Domains and Euclidean Domains A principal ideal domain is a unique factorization domain, and this exhibits a class of rings in which unique factorization occurs.
Let R be an integral domain with quotient field F,and let h be a ring monomorphism from R to a field L. Show that h has a unique extension to a monomorphism from F to L. 6. Let h be the ring homomorphism from Z to Z p' P prime, given by hex) = x mod p. Why can't the analysis of Problem 5 be used to show that h extends to a monomorphism of the rationals to Z p? ) p 7. Let S be a multiplicative subset of the commutative ring R,with/: R - t ~lR defined by j(a) = all. , such that the diagram below is commutative.
Abstract Algebra by Prabhat Choudhary