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3 Pages

PracticeExam3Solns

Course: MATH 3, Fall 2009
School: Los Angeles Southwest...
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__________________________ Name Math 547 Practice Exam #3 1. (a). Explicitly describe the elements of the field ! (! ) . Solution: The elements of ! (! ) are quotients of polynomials in ! . (b). Explicitly describe the elements of the field ! Solution: ! ( 2 ) = {a + b 3 3 2 + c 3 4 : a,b, c !! . } ( 2). 3 (c). Give a basis for the field ! Solution: 1,! 2,! 3,! 6,!i,!i 2,!i 3,!i { ( 3,! 2,!i over ! . )...

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__________________________ Name Math 547 Practice Exam #3 1. (a). Explicitly describe the elements of the field ! (! ) . Solution: The elements of ! (! ) are quotients of polynomials in ! . (b). Explicitly describe the elements of the field ! Solution: ! ( 2 ) = {a + b 3 3 2 + c 3 4 : a,b, c !! . } ( 2). 3 (c). Give a basis for the field ! Solution: 1,! 2,! 3,! 6,!i,!i 2,!i 3,!i { ( 3,! 2,!i over ! . ) 6} . (d). Define algebraic extension. Solution: An extension field E of a field F is an algebraic extension if every element of E is algebraic over F. 2. Prove: If F ! K ! E are fields and K is a finite extension of F and E is a finite extension of K, then [ E : F ] = [ E : K ][ K : F ] . Solution: See your notes. 3. Suppose that ! is a zero of p(x) = x 2 + 2x + 3 !Z 5 [x] in some extension field E. Note: p(x) = x 2 + 2x + 3 is irreducible in Z 5 [x] ; you need not verify this. (a). How many elements are there in Z 5 (! ) ? Explain. Solution: Z 5 (! ) = {a + b! : a,b "Z 5 } . Moreover, each element of Z 5 (! ) is uniquely expressible in the form a + b! . There are 5 ways to choose the values for each of a and b. Thus there are 25 elements in Z 5 (! ) . (b). Express the product (1 + 2! ) ( 3 + ! ) in the form a + b! , a,b !Z 5 . Solution: (1 + 2! ) ( 3 + ! ) = 2 + 3! . (c). Find an expression (in terms of ! ) for the other zero of p(x) = x 2 + 2x + 3 in E. Solution: dividing x 2 + 2x + 3 by x ! " gives a quotient of x + (! + 2 ) . Thus the other zero is !(" + 2) = 4" + 3 . 4. Let D be an integral domain with F ! D ! E where F and E are fields and E is a finite extension of F. Show that D is a field. Solution: See your notes this is a problem from your text. Name __________________________ 5. Show directly that ! = i + 3 is an algebraic number and determine its degree. Fully justify your answer. Hint: You may take as given that i + 3 !! i, 3 . ( ) Solution: Letting x = i + 3 , we get x 2 = i + 3 . So, x 4 = !1 + 2i 3 + 3 = 2 + 2i 3 " x 4 ! 2 = 2i 3 . Thus x 8 ! 4x 4 + 4 = !12 . This implies that ! = i + 3 satisfies the polynomial p(x) x = 8 ! 4x 4 + 16 . It might take some effort to show directly that this polynomial is irreducible and hence that the degree of ! is 8. ( 3 ) ! ",!i "! ( 3 ) . Thus i satisfies the irreducible polynomial x + 1 !! ( 3 )[ x ] and so !! ( i,! 3 ) : ! ( 3 ) # = 2 . " \$ Now it follows easily that !! ( i,! 3 ) : ! # = !! ( i,! 3 ) : ! ( 3 ) # !! ( 3 ) : ! # = 4 . " \$ " \$" \$ Also, !! ( i + 3 ) : ! ( i, 3 ) \$ = 2 , since ! = i + 3 !! ( i, 3 ) and ! # & " % satisfies p(x) = x ! ( i + 3 ) . However, since ! 2 2 Thus, !! # " ( i +! 3 : ! \$ = !! & # % " ) ( i + 3 : ! i, 3 \$ !! i, 3 : ! \$ = 8 . % & %" ) ( ) ( ) 6. Given that ! is transcendental, show that most 2. Solution: Suppose that So we have, 0 = p ! cannot be algebraic of degree at ! is algebraic of degree at most 2. Then there is a polynomial p(x) = ax 2 + bx + c !![x] such that p ( ! ) = a! + b ( ! )= 0. ! + c " a! + c = #b ! . Thus squaring both sides we get. Hence for the polynomial p(x) = a 2 x 2 + 2ac ! b 2 x + c 2 , p(! ) = 0. So, ! would be algebraic of degree at most 2. ! ( ) Note: A simple, but more wordy, extension of this idea shows that algebraic of any degree; i.e., ! is transcendental. ! is not Na...

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