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Unformatted text preview: MATH 3360 Sketch of solutions for Assignment 10 17C.25 1 2 (3 x + x 2 ) 19A.10 There are many possible solutions, for instance 2, 2 = 59, etc. Additional Problem 1 Some students wrote U 17 , the group of units in Z 17 , is a field of 16 elements. However, this is NOT the case. One needs addition and multiplication operations to be closed on a field, yet addition on U 17 is not closed. To show F 2 [ ], for a root of x 4 + x + 1 (or other irreducible polynomials) is a field, many of you tried to find a primitive element in F 2 [ ]. Indeed one only needs some simple argument to show an inverse exists: For example, to find the inverse of a + b + c 2 + d 3 in F 2 [ ], one can consider solving ( a + bx + cx 2 + dx 3 ) g ( x ) 1(mod x 4 + x + 1) Since x 4 + x + 1 is irreducible, ( a + bx + cx 2 + dx 3 ,x 4 + x + 1) = 1 and g ( x ) always exists. Moreover, one can choose g ( x ) such that it is of degree less than 4, as in previous homeworks....
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 Spring '08
 BILLERA
 Algebra, Addition

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