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Math Help - [SOLVED] Show that G forms a ring.

  1. #1
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    Exclamation [SOLVED] Show that G forms a ring.

    Can someone look over what I have done and help me finish my proof? I have no idea how to prove distribution.
    Thanks.

    Exercise: Suppose G is an abelian group wrt addition and with identity 0. Define a multiplication in G by ab=0 for all a,b in G. Show that G forms a ring wrt these operations.

    Since G is abelian wrt multiplication, I would like to use the last two conditions of the following definition.

    Definition 5.1b. Alternative Definition of a Ring
    Suppose R is a set in which a relation of equality and operation of addition and multiplication are defined. Then R is a ring wrt these operations if these conditions hold:
    R forms an abelian group wrt addition.
    R is closed wrt and associative multiplication.
    Two distributive laws hold in R: x(y + z) = xy + xz, and (x + y)z = xz + yz for all x, y, and z in R.

    Proof. Since G is an abelian group with respect to addition, then we only need to prove the last two conditions of the Alternative Definition of a Ring.
    Let ab=0 and bc=0 for all a, b, c in G. This means that a(bc)=a(0)=0, and (ab)c=(0)a=0. This shows that G is closed under associative multiplication.
    Now we need to show that the distributive laws hold.
    ...?
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    You were correct up to this point.

    Quote Originally Posted by yvonnehr View Post
    Let ab=0 and bc=0 for all a, b, c in G. This means that a(bc)=a(0)=0, and (ab)c=(0)a=0. This shows that G is closed under associative multiplication.
    I think you mean: Let a,b,c\in G. Then ab=0\in G proving closure and (ab)c=0c=0=a0=a(bc) proving associativity.

    Distributivity is true because 0=0+0. Think about it.
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    Quote Originally Posted by TheAbstractionist View Post
    You were correct up to this point.


    I think you mean: Let a,b,c\in G. Then ab=0\in G proving closure and (ab)c=0c=0=a0=a(bc) proving associativity.

    Distributivity is true because 0=0+0. Think about it.
    How did you get from 0 = a0 = a(bc)? I'm not sure if I missed an implication somewhere.

    Thanks.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by yvonnehr View Post
    How did you get from 0 = a0 = a(bc)? I'm not sure if I missed an implication somewhere.

    Thanks.
    Remember that you let ab=0=bc...
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