Math 115B Homework 7solution - 3 =(1 √-5(1-√-5 so 6...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
MAT 115B HOMEWORK 7 SOLUTION 14.2.16: (a) Observing that N (9 + i ) = 82 = 2(4 2 + 5 2 ), we try factors 1 + i , 5 + 4 i and 4 + 5 i . Have 9 + i = (1 + i )(4 + 5 i )( - i ). 14.2.28: (a) Using Euclidean algorithm to get 4 - i = (1 - i )(2+ i )+1. So x 3( - 1+ i ) 1 + 2 i (mod 4 - i ). 14.3.2: (d) 1001000 = 2 3 · 5 3 · 7 · 11 · 13. Since 7 3 (mod 4), 11 3 (mod 4) and their corresponding powers are odd, there is no way to write 1001000 as a sum of two squares. (cf. Theorem 13.6 from the textbook) GK 7.1: (b) The only numbers in Z [ - 5] that are smaller than 2 are 0 and ± 1. But none of 1 + - 5 , - 5 , 2 + - 5 is divisible by 2. Hence there does not exist a way to divide 1 + - 5 by 2 where the remainder is smaller than the divisor. (c) 2 is irreducible since the only shorter elements than 2 are all units. 3 is irreducible since the only non-unit shorter elements than 3 are ± 2 , ± - 5 , ± (1+ - 5). But none of them divides 3. 1 ± - 5 is irreducible since the only non-unit shorter elements than 1 ± - 5 are ± 2 , ± - 5. But none of them divides 1 ± - 5. So all four factors are irreducible. 6 = 2
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern