HW6Sol - Homework 6: Math 3360 Applicable Algebra Chapter...

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Homework 6: Math 3360 Applicable Algebra Chapter 7: Rings and Fields 44. Let F be a field of characteristic 2. Show that for any a,b F (i) a = - a (ii) ( a + b ) 2 = a 2 + b 2 Solution: (i) Since F has characteristic 2, then we know 1+1 = 2 = 0. So by the distributive property 0 = a (0) = a (1 + 1) = a + a , and then a = - a . (ii) We use the distributive property again to get ( a + b )( a + b ) = a 2 + ab + ba + b 2 . Since F is a field, then it is commutative under multiplication, so we have ab = ba . Then ( a + b ) 2 = a 2 + ab + ab + b 2 = a 2 + 2 ab + b 2 = a 2 + 0( ab ) + b 2 = a 2 + b 2 . Chapter 9: Fermat’s and Euler’s Theorems 20. Find the order of [3] in Z / 23 Z . Solution: Now 23 is prime so 3 is a unit and then we have 3 22 1 mod 23. Thus the order of [3] must divide 22, i.e. it is either 2 , 11, or 22. Now 3 2 9 mod 23 6≡ 1 mod 23 and 3 11 (3 3 ) 3 (3 2 ) mod 23 (27) 3 (9) mod 23 (4) 3 (9) mod 23 64(9) mod 23 ( - 5)(9) mod 23 ≡ - 45 mod 23 1 mod 23 33. Show that for prime p , then [( p - 1)!] p = [ - 1] p . Solution: Note that since p is prime, then every element in Z /p Z is a unit, so every every ele- ment has an inverse. First we will show that the only elements that are their own inverses are [1] and [ p - 1]. Suppose that [ a ] is its own inverse modulo p . Then we have that
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HW6Sol - Homework 6: Math 3360 Applicable Algebra Chapter...

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