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Math Help - Homomorphisms

  1. #1
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    Homomorphisms

    This is way over my head.

    Let R and S be commutative runs and phi:R---->S be a ring homomorphism. a) Given an ideal J < S (that < means subset), define
    ={a (element of) R: phi(a) (element of) J] < R. (usually this is denoted by phi^-1(J)). PRove this is an ideal
    b) Given an ideal I < R define
    ={phi(aO:a (element of) I] < S (usually denoted by phi(I)). Prove this is an ideal provided phi maps onto S.
    c) In a case with b, then prove that if phi maps onto S then phi(<a>)=<phi(a)>.
    Last edited by ThePerfectHacker; April 18th 2007 at 12:10 PM.
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  2. #2
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    Quote Originally Posted by chadlyter View Post
    This is way over my head.

    Let R and S be commutative runs and phi:R---->S be a ring homomorphism. a) Given an ideal J < S (that < means subset), define
    ={a (element of) R: phi(a) (element of) J] < R. (usually this is denoted by phi^-1(J)). PRove this is an ideal
    It is just a question whether you know your definitions or note. The definitions in Field Theory are more harder to understrand and use than in most areas in mathematics.
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  3. #3
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    Quote Originally Posted by chadlyter View Post
    b) Given an ideal I < R define
    ={phi(aO:a (element of) I] < S (usually denoted by phi(I)).
    Here.
    Attached Thumbnails Attached Thumbnails Homomorphisms-picture17.gif  
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  4. #4
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    Quote Originally Posted by chadlyter View Post
    c) In a case with b, then prove that if phi maps onto S then phi(<a>)=<phi(a)>.
    Let us assume these are rings with unity.

    To remind, a "principal ideal" <a> is the ideal {ra|r in R}.

    We want to show,

    phi[<a>] = <phi(a)>

    The standard trick is to show each is a subset of eachother.

    First note that,
    phi[<a>] = phi [{ra| r in R}] = {phi(r)phi(a)| r in R}.

    Then since phi(a) in <phi(a)> and phi(r) in R we have that,
    phi(r)phi(a) in <phi(a)>

    Second note that,
    <phi(a)> = {s phi(a) | s in S}

    Then since phi(a) in phi[<a>] and s = phi(r) for some r in R (because it is onto) we have that,
    s phi(a) in <phi(a)>

    Thus, they are equal.
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  5. #5
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    Ideal being principal

    If I wanted to continue this problem how could I prove that every ideal in Z_m is principal.
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  6. #6
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    Quote Originally Posted by chadlyter View Post
    If I wanted to continue this problem how could I prove that every ideal in Z_m is principal.
    That is easy.

    Theorem: A subgroup of a cyclic group is a cylic group.

    Now, if "I" is an ideal of Z_m it must be an additive subgroup of Z_m. But as an additive subgroup it must be cyclic. That is basically the meaning of what a principla ideal means.

    See if you can do it from there.


    Is this an undergraduate or a graduate course?
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  7. #7
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    Graduate

    Graduate course. I am just really confused by homomosphisms and isomorphisms.
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