hw7 - Math 113 Homework 7 selected solutions Fraleigh 18.12...

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

View Full Document Right Arrow Icon
Math 113 Homework # 7, selected solutions Fraleigh 18.12. Let R = { a + b 2 | a, b Q } . Then R is closed under addition because ( a 1 + b 1 2) + ( a 2 + b 2 2) = ( a 1 + a 2 ) + ( b 1 + b 2 ) 2 and Q is closed under addition. Indeed ( R, +) is a subgroup of ( R , +), since 0 = 0 + 0 2 R and - ( a + b 2) = ( - a ) + ( - b ) 2. R is closed under multiplication because ( a 1 + b 1 2)( a 2 + b 2 2) = ( a 1 a 2 +2 b 1 b 2 )+ ( a 1 b 2 + b 1 a 2 ) 2. Thus R is a subring of R . Since R is commutative, so is R . R has a multiplicative identity, namely 1 + 0 2. Finally, R is a field, because the multiplicative inverse of a + b 2 6 = 0 is ( a - b 2) / ( a 2 - 2 b 2 ). Note that the denominator is nonzero because since 2 is irrational and a and b are both nonzero, a - b 2 6 = 0, so the product a 2 - 2 b 2 = ( a + b 2)( a - b 2) 6 = 0, since R is an integral domain. Fraleigh 18.13. This is not a ring since it is not closed under multiplication: i · i is not in the set. Fraleigh 18.18. An element of a direct product is a unit if and only if each component is a unit. So the units in Z × Q × Z are ( ±
Image of page 1

Info iconThis 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.

{[ 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