HW3_selected_solns

# HW3_selected_solns - College of the Holy Cross Fall 2009...

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College of the Holy Cross, Fall 2009 Math 351 – Homework 3 selected solutions Chapter 14, #28 First one must verify that A = { (3 x,y ) | x,y Z } , the set of all elements whose ﬁrst entry is a multiple of 3 and whose second entry is allowed to be anything, is in fact an ideal of Z Z . This is straightforward to do using the ideal test, and everyone basically got it. To show that A is maximal, let B be an ideal of Z Z , and assume that B properly contains A (that is, contains A but is not equal to A ). Then B contains at least one element not in A . Say this element is ( r,s ). Since ( r,s ) 6∈ A , we know that r cannot be a multiple of 3. Say that r = 3 n + 1 for some n Z (the other case is r = 3 n + 2 for some n Z ). Then (3 n,s - 1) A , and so (3 n,s - 1) B . Thus since B is closed under subtraction, we get (3 n + 1 ,s ) - (3 n,s - 1) = (1 , 1) B . However, (1 , 1) is the multiplicative identity in Z Z , and it follows from remarks in class that B
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## This document was uploaded on 11/03/2010.

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