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MA352s10midterm_1_solutions

MA352s10midterm_1_solutions - College of the Holy Cross...

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College of the Holy Cross, Spring 2010 Math 352, Midterm 1 Solutions Monday, March 22 All the numbered problems on this exam are weighted equally (some have two parts, but that counts as one problem). Choose 6 of the following 7 problems to do, and clearly mark which problem you do not want graded. You may use any results from class or from the text, except for homework exercises unless explicitly stated otherwise. To get full credit for your answers, you must fully explain your reasoning. 1. (a) Let R be a commutative ring, and let A be any subset of R . Show that the annihilator of A , Ann( A ) = { r R | ra = 0 for all a A } , is an ideal. (b) Give an example of a ring R and a non-zero subset A of R such that Ann( A ) has exactly three elements. (a) First, note that 0 Ann( A ) since 0 a = 0 for all a A . Thus Ann( A ) is not empty. Now, use the ideal test. Suppose that x, y Ann( A ), and show that x - y Ann( A ). Since x, y Ann( A ), we have xa = ya = 0 for all a A , and thus ( x - y ) a = xa - ya = 0 - 0 = 0 for all a A . Thus x - y Ann( A ). Suppose that x Ann( A ) and r R , and show that rx and xr are in Ann( A ). Note that ( rx ) a = r ( xa ) = r 0 = 0 for all a A , and thus rx Ann( A ). Since R is commutative, we have xr = rx Ann( A ). Thus Ann( A ) is an ideal. (b) Note that we had better not choose R to be an integral domain, since Ann( A ) = { 0 } for any A 6 = { 0 } in an integral domain. Let R = Z 6 , and take A = { 0 , 3 } . Then Ann( A ) = { 0 , 2 , 4 } , since any of these elements multiplied by either 0 or 3 yields 0.
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2. Find infinitely many monic polynomials f ( x ) Q [ x ] such that f (0) = 4 and Q [ x ] / h f ( x ) i is a field. Recall that a monic polynomial is a polynomial with leading coefficient 1. From Theorem 14.4, we know that Q [ x ] / h f ( x ) i is a field if and only if h f ( x ) i is max- imal. But from Theorem 17.5, we know that h f ( x ) i is maximal if and only if f ( x ) is irreducible over Q .
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