practice_midterm_1_problems_solns

practice_midterm_1_problems_solns - College of the Holy...

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College of the Holy Cross, Fall 2009 Math 352, Practice Midterm 1 solutions Prof. Jones 1. Let I be an ideal in a ring R , and set N ( I ) = { x R : xi = 0 for all i I } . a) Show that N ( I ) is an ideal in R . Using the ideal test, we must show first that N ( I ) is not empty, and then that that if x, y N ( I ) and r R , then x - y , rx , and xr are all in N ( I ). Note that 0 N ( I ), proving that N ( I ) isn’t empty. Now x, y N ( I ) means that xi = yi = 0 for all i I , and so ( x - y ) i = xi - yi = 0 - 0 = 0 for all i I . Hence x - y N ( I ). Also, ( rx ) i = rxi = r 0 = 0 for all i I , and thus rx N ( I ). Consider ( xr ) i , which is the same as x ( ri ). Since I is an ideal, ri I . But x times any element of I must be 0 (by the definition of N ( I )), and thus x ( ri ) = 0. b) Show that if R is an integral domain, then either I = { 0 } or N ( I ) = { 0 } . Suppose that I 6 = { 0 } , and suppose that x N ( I ). Then xi = 0 for all i I . Since I 6 = { 0 } , there is some i I with i = 0. Hence xi = 0 for some i 6 = 0. Since R is an integral domain, this implies that x = 0 (otherwise both x and i would be zero-divisors). We’ve thus taken an arbitrary x N ( I ) and shown that x = 0. Thus N ( I ) = { 0 } . c) Give an example of a ring R and an ideal I 6 = { 0 } in R such that N ( I ) = R . Consider the subring R = { 0 , 2 } of Z 4 . Let I = R . Then for any x R , we have xi = 0 for all i I . If x = 0 then this is obvious, and if x = 2 it is still true that 2 i = 0 for all i R . Thus N ( I ) = R . 2. Let φ : R R 0 be a ring homomorphism, and let I be an ideal of R .
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