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Showing a subset is a subring?

  1. Apr 9, 2012 #1
    Showing a subset is a subring???

    1. The problem statement, all variables and given/known data

    Let R be a ring and a a fixed element in R. Let Ia={x in R l ax=0}

    2. Relevant equations



    3. The attempt at a solution

    I saw these conditions in my book, but I'm not sure are these conditions sufficient in showing Ia is a subring?

    (1) 0 is in Ia:

    Let 0 be in R, then
    a(0)=0

    (2) (a-b)is in Ia, for a, b in Ia:

    I'm not sure how I should start this.

    (3) Ia is closed under multipication.
     
  2. jcsd
  3. Apr 9, 2012 #2
    Re: Showing a subset is a subring???

    For 2) Let m, n be any two elements from Ia. Can you show that m+n is still in Ia? What do m and n look like? What does their sum look like?

    It is similar for showing that m*n is in Ia, as well.
     
  4. Apr 9, 2012 #3
    Re: Showing a subset is a subring???

    Okay, so for 2 this is what I got:
    Let m,n be in Ia. Then this means that am=0 and an=0. So,
    am=an
    and am-an=0
    so, a(m-n)=0 therefore (m-n) is in Ia. Is this correct?
    Should I follow the same flow for 3?

    Thanks
     
  5. Apr 9, 2012 #4
    Re: Showing a subset is a subring???

    Looks good.
     
  6. Apr 9, 2012 #5
    Re: Showing a subset is a subring???

    Note, you also need to show the distributive property of multiplication over addition holds, and associativity of multiplication and addition.
     
  7. Apr 10, 2012 #6
    Re: Showing a subset is a subring???

    Thank you
     
  8. Apr 10, 2012 #7

    micromass

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    Re: Showing a subset is a subring???

    No, you do not have to show those. Just showing the three conditions in the OP is good enough.
     
  9. Apr 10, 2012 #8
    Re: Showing a subset is a subring???

    Haha. Not according to my old abstract professor! We know they are inherited by R, but failure to show is -2 points..

    sigh..
     
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