Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Deveno

Math Help - Show this is Integral Domain

  1. #1
    Super Member
    Joined
    Feb 2008
    Posts
    535

    Show this is Integral Domain

    A is the set Z (integers), with the following "addition" + and "multiplication" * :

    a + b = a + b - 1 and a * b = ab - (a + b) + 2

    I've already proven that A is a commutative ring with unity. Now, I have to prove that A is an integral domain.

    To my understanding, in order to do this, I assume that a * b = 0 and show that either a = 0 or b = 0. Please correct me if I'm wrong...

    So, a * b = ab - (a + b) + 2 = ab - a - b + 2 = 0.

    I can't figure out how to proceed and show that either a or b equals zero. Can someone help here? Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    611
    Thanks
    248

    Re: Show this is Integral Domain

    I had to alter your definition slightly, the changed problem has solution:
    Show this is Integral Domain-anintegraldomain.png
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,310
    Thanks
    687

    Re: Show this is Integral Domain

    i don't think there is any need to alter the definition. if we use # for the "new" plus, we have:

    a#b = a + b - 1.

    to find an element z such that:

    a#z = a, for any a in Z, we solve:

    a + z - 1 = a
    a + z = a + 1
    z = 1, which shows that the "0" of our "modified" integers is 1, not -1.

    and indeed:

    a*1 = a - a - 1 + 2 = 1.

    the trouble you are having, jzellt, is that to prove A is an integral domain, you need to show that if a*b = 0A = 1, then either a = 0A = 1, or b = 0A = 1.

    so suppose a*b = 1. assume that a ≠ 1. we need to show that b = 1.

    so we have ab -a - b + 2 = 1 (this is an equation in our usual integers)

    ab - a - b = -1
    ab - b = a - 1
    b(a - 1) = a - 1

    since a - 1 ≠ 0 (since a ≠ 1), we can use the cancellation property of the integers to cancel the common factor of a - 1 from both sides:

    b = 1, QED.
    Thanks from jzellt
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. show why domain D is not simply connected.
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: August 14th 2011, 02:41 AM
  2. Integral Domain
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: May 22nd 2011, 11:23 AM
  3. [SOLVED] Show through C domain.
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 28th 2010, 08:18 AM
  4. Replies: 1
    Last Post: September 21st 2009, 10:02 AM
  5. show that a quotient ring is in integral domain
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: September 6th 2009, 08:01 AM

Search Tags


/mathhelpforum @mathhelpforum