251-EX03-solns - MATH 251: ABSTRACT ALGEBRA I EXAM #3...

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Unformatted text preview: MATH 251: ABSTRACT ALGEBRA I EXAM #3 Problem 1 . Let F be a field. For a polynomial f ( x ) = ∑ n i =0 a i x i ∈ F [ x ], we denote by f ( x ) = n X i =1 ia i x i- 1 = a 1 + 2 a 2 x + ··· + na n x n- 1 the formal derivative of f . (a) Let I = { f ( x ) ∈ F [ x ] : f (1) = f (1) = 0 } . Show that I is an ideal of F [ x ]. (b) Let J = { f ( x ) ∈ F [ x ] : f (1) = f (0) = 0 } . Is J an ideal of F [ x ]? Prove or disprove. Solution. A subset I ⊂ R = F [ x ] is an ideal if I is a subgroup under + and is closed under multiplication by R . For (a), we clearly have 0 ∈ I so that I 6 = ∅ ; if f, g ∈ I then ( f + g )(1) = f (1) + g (1) = 0 and ( f + g ) (1) = f (1)+ g (1) = 0 so f + g ∈ I , and if f ∈ I then (- f )(1) =- f (1) = 0 and (- f ) (1) =- f (1) = 0 so- f ∈ I , hence I is a subgroup under +. Next, if f ∈ I and p ∈ R , then ( pf )(1) = p (1) f (1) = 0 and ( pf ) (1) = ( p f + pf )(1) = p (1) f (1) + p (1) f (1) = 0 + 0 = 0 . Thus I is an ideal. Indeed, I is the principal ideal generated by ( x- 1) 2 ....
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This note was uploaded on 11/08/2011 for the course MATH 321 taught by Professor Ergenc during the Spring '11 term at Boğaziçi University.

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251-EX03-solns - MATH 251: ABSTRACT ALGEBRA I EXAM #3...

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