dummit foote solutions2

# dummit foote solutions2 - Solutions to Homework 9...

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Solutions to Homework 9 46. (Dummit-Foote 10.3 #2) Suppose R n ' R m . Let I be a maximal ideal of R , then R n /IR n ' R m /IR m . By the exercise 12 of section 2, this implies that ( R/I ) n ' ( R/I ) m , and as these are vector spaces over the ﬁeld R/I , we have n = m . 47. (a) Suppose that Q were a free Z -module. Let a/b,c/d Q . Then bc ( a/b ) + ( - ad )( c/d ) = 0, which says that there is a non-trivial linear relation between any two basis elements of Q . Hence Q is a free Z -module of rank at most 1. But a/ 2 b is not in the Z -span of a/b for any a/b Q , which says that Q cannot have rank 1 either. Thus Q is not a free Z -module. (b) Suppose that I were a free R -module. If f and g are basis elements, then the non-trivial relation - gf + fg = 0 forces I to have rank at most 1. Thus I = Rf for some f , but this implies that I is a principal ideal. To see that I cannot be principal, suppose we had ( X,Y ) = I = ( f ). Then there exist g,h R such that gf = X,hf = Y . The ﬁrst equation says that the Y -degree of f (the degree of f when viewed as a polynomial in Y

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## This note was uploaded on 11/08/2011 for the course MATH 224 taught by Professor Gurel during the Spring '05 term at Boğaziçi University.

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dummit foote solutions2 - Solutions to Homework 9...

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