dummit foote solutions2

dummit foote solutions2 - Solutions to Homework 9...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 2

dummit foote solutions2 - Solutions to Homework 9...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online