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**Unformatted text preview: **· 3 = (1 + √-5)(1-√-5), so 6 does not have a unique factorization. (d) The common divisors (besides units) of 6 and 2 + 2 √-5 are ± 2 , ± (1 + √-5). But they are not divisible by each other. So 6 and 2 + 2 √-5 have no weak gcd. GK 7.2: Let φ : Z [ i ] / (5 i-4) → Z / 41 be the given isomorphism. Then φ (5 i-4) = φ (0) = 0 in Z / 41. 5 φ ( i ) ≡ 4 (mod 41). So φ ( i ) = 9, φ ( i + 2) = 11 in Z / 41. Solution from Shigenori Nakatsuka shows the correspondence between Z [ i ] / (5 i-4) and Z / 41. 1 2 MAT 115B HOMEWORK 7 SOLUTION GK 7.3: (Solution from Greg Lauro) 7 becomes a unit since 7 · 1 7 = 1. So 7 is no longer prime. The non-7 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 ∈ Z be composite. Then a ∈ 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, non-unit shorter elements