781hw7sol

781hw7sol - 18.781, Fall 2007 Problem Set 7 Solutions to...

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18.781, Fall 2007 Problem Set 7 Solutions to Selected Problems Problem 1 First, observe that N (13) = 13 2 - 13 · 0+0 2 = 169 = 13 2 . If ab = 13 for some nonunit a, b R , then N ( a ) N ( b ) = 13 2 and N ( a ) = N ( b ) = 13 since 13 is a prime number in Z and N ( a ) , N ( b ) > 1. For any a = r + , we have N ( a ) = ( r - s ) 2 + rs . By some computations, we can find the set of ( r, s ) which gives N ( a ) = 13. (Actually, they are (4 , 1) , ( - 4 , - 1) , (4 , 3) , ( - 4 , - 3)). And we can easily find that 13 = (3 + 4 ω )( - 1 - 4 ω ) . Each element of left hand side is clearly non-unit in R since the value by N is not 1. 2 Problem 2 For any element a = s + ti in Z [ i ], define the function λ by λ ( a ) = s 2 + t 2 . Since s + ti = 0 if and only if both s and t are 0, we see that λ ( s + ti ) 1 when s + ti 0. It is easy to find that λ ( ab ) = λ ( a ) λ ( b ) for a, b Z [ i ]. Then when b 6 = 0 we have λ ( a ) = λ ( a ) · 1 λ ( a ) λ ( b ) = λ ( ab ) . If
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781hw7sol - 18.781, Fall 2007 Problem Set 7 Solutions to...

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