92IIalgmidsol - Topics in algebra Solutions to midterm All...

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Topics in algebra Solutions to midterm All rings are commutative with identity. Total credits: 110. (1) (6pts) Prove that n 1 2 n Z = 0. Proof . Let I = n 1 2 n Z . Since Z is a P.I.D., I = m Z for some integer m . However 2 n | m for every n . Therefore m = 0. (2) (6pts) If I and J are co-maximal in a ring R , then IJ = I J . Proof . I J = ( I J )( I + J ) = ( I J ) I + ( I J ) J JI + IJ = IJ . (3) (6pts) Prove that Z p n = Z / ( p n ) n 2 is a local ring. Proof . Since the maximal ideals of Z / ( p n ) are of the form m / ( p n ), where m is a maximal ideal of Z that contains ( p n ), m = ( p ). Therefore Z / ( p n ) is a local ring with the unique maximal ideal ( p ) / ( p n ). (4) (6pts) Give an example of a ring R with only two maximal ideals m 1 and m 2 such that m 1 m 2 = (0). Solution: Let R = Z 6 , m 1 = 2 R and m 2 = 3 R ; then m i is a maximal ideal of R for every i and m 1 m 2 = (0). (5) (6pts) Let a be a nilpotent element of a ring R . Prove that 1 - a is a unit. Proof . Since a is nilpotent, there is an integer n such that a n = 0. Moreover, as (1 - a )(1 + a + · · · + a n - 1 ) = 1 - a n = 1, we see that 1 - a is a unit. (6) (8pts) Find all solutions of the equations: x 1 ( mod 3), x 2 ( mod 7) and x 4 ( mod 11). Solution: Since 154 1 ( mod 3) 154 0 ( mod 7) 154 0 ( mod 11) , - 33 0 ( mod 3) - 33 2 ( mod 7) - 33 0 ( mod 11) and 147 0 ( mod 3) 147 0 ( mod 7) 147 4 ( mod 11) , { 37 + 231 k | k Z } are solutions of the equations.
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