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Unformatted text preview: Â· 3 = (1 + âˆš5)(1âˆš5), so 6 does not have a unique factorization. (d) The common divisors (besides units) of 6 and 2 + 2 âˆš5 are Â± 2 , Â± (1 + âˆš5). But they are not divisible by each other. So 6 and 2 + 2 âˆš5 have no weak gcd. GK 7.2: Let Ï† : Z [ i ] / (5 i4) â†’ Z / 41 be the given isomorphism. Then Ï† (5 i4) = Ï† (0) = 0 in Z / 41. 5 Ï† ( i ) â‰¡ 4 (mod 41). So Ï† ( i ) = 9, Ï† ( i + 2) = 11 in Z / 41. Solution from Shigenori Nakatsuka shows the correspondence between Z [ i ] / (5 i4) and Z / 41. 1 2 MAT 115B HOMEWORK 7 SOLUTION GK 7.3: (Solution from Greg Lauro) 7 becomes a unit since 7 Â· 1 7 = 1. So 7 is no longer prime. The non7 primes are still prime: if p 6 = 7 and p  a 7 m b 7 n , then p  ab . Since p is prime, p  a or p  b . Hence p  a 7 m or p  b 7 n . Nothing else is prime: Let a âˆˆ Z be composite. Then a âˆˆ Z [1 / 7] has the same factorization and is hence not prime....
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 Fall '08
 Shapiro,I
 Math, Number Theory, Factors, Prime number, Greatest common divisor, MAT 115B HOMEWORK, Z/41, nonunit shorter elements

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