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Math Help - Cancelation law and existence of identity

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    Cancelation law and existence of identity

    Hello

    If I have, in a ring R, ab=a and the law of cancelation is valid, can I say that b=1_R? Why?
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    Re: Cancelation law and existence of identity

    Quote Originally Posted by ModusPonens View Post
    Hello

    If I have, in a ring R, ab=a and the law of cancelation is valid, can I say that b=1_R? Why?
    \displaystyle \begin{align*} ab &= a \\ abb^{-1} &= ab^{-1} \\ ae &= ab^{-1} \\ a &= ab^{-1} \end{align*}.

    So this means

    \displaystyle \begin{align*} ab &= ab^{-1} \\ b &= b^{-1} \end{align*}

    What value is its own multiplicative inverse?
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    Re: Cancelation law and existence of identity

    Quote Originally Posted by ModusPonens View Post
    Hello

    If I have, in a ring R, ab=a and the law of cancelation is valid, can I say that b=1_R? Why?
    What do you mean by "the law of cancellation"? Do you mean that every non-zero element has an inverse?

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} ab &= a \\ abb^{-1} &= ab^{-1} \\ ae &= ab^{-1} \\ a &= ab^{-1} \end{align*}.

    So this means

    \displaystyle \begin{align*} ab &= ab^{-1} \\ b &= b^{-1} \end{align*}

    What value is its own multiplicative inverse?
    If every nonzero element has an inverse, then couldn't you just do

    ab=a

    a^{-1}ab=a^{-1}a

    b=1_{R}?
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    Re: Cancelation law and existence of identity

    Quote Originally Posted by Ackbeet View Post
    What do you mean by "the law of cancellation"? Do you mean that every non-zero element has an inverse?



    If every nonzero element has an inverse, then couldn't you just do

    ab=a

    a^{-1}ab=a^{-1}a

    b=1_{R}?
    Yes, yes you could
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    Re: Cancelation law and existence of identity

    A cancellation ring is one in which if ab= 0 then either a=0 or b= 0. It is not necessarily true that a cancellation ring has inverses or even has a multiplicative identity. In particular, the set of all even integers, with usual addition and multiplication, is a cancellation ring that has no multiplicative identity. Obviously, if the ring does not have a multiplicative identity, then this cannot be true.

    You cannot use the argument "ab- a= a(b- 1)" because, if the ring has no multiplicative identity, ab- a= a(b- 1) is not true.

    If it is given that the ring has a multiplicative identity (some texts, but not all, require a "ring" to have a multiplicative identity), you still cannot use "ab=a so a^{-1}ab= a^{-1}a because you do not know that a has an inverse. You can, however argue that if ab= a, then ab- a= a(b- 1)= 0. Since this is a cancellation ring, either a= 0 or b-1= 0. Since it is given that a is not 0, b- 1= 0 so b= 1.
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    Re: Cancelation law and existence of identity

    I searched wikipedia and it's called the cancellation property, not law. Sorry.

    Anyway, I'm using Hungerford to study for my algebra course and a number of exercises depend on this dubious conclusion to work. For example, how do I solve the following exercise without recurring to the conclusion ab=a => b=1?

    Let f:R-->S be a non zero homomorphism such that R has identity and S has no zero divisors. Prove that f(1_R) is the identity of S.
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    Re: Cancelation law and existence of identity

    Ok, I got it.

    If ab=a and the cancellation property is valid, Then abc=ac and therefore, bc=c. The analogous for right identity; and b is the identity. So simple and yet I missed it.
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