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Math Help - Ring Homomorphism

  1. #1
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    Ring Homomorphism

    Stuck on this problem.
    Let f: R to S be a ring homomorphism.
    -Let J-S be an ideal (the - is the triangular symbol). Prove that f^(-1) (J) (equal by definition to {r in R: f(r) in J}) is an ideal of R.
    -Prove that if f is surjective and I-R is an ideal then f(I) is an ideal (where f(I)={f(i): i in I}).
    -Show, by example, that if f is not surjective the assertion in the above need not hold.

    I appreciate any help.
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  2. #2
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    Quote Originally Posted by JoeDardeno23 View Post
    Stuck on this problem.
    Let f: R to S be a ring homomorphism.
    -Let J-S be an ideal (the - is the triangular symbol). Prove that f^(-1) (J) (equal by definition to {r in R: f(r) in J}) is an ideal of R.
    So we have a ring homomorphism,
    \phi: R\to S
    And J is an ideal in S.

    Thus, <J,+> is an additive subgroup of <S,+>. From group theory you should be know that the inverse image in a group homomorphism preserves subgroups. Thus, \phi^{-1}[J] is an additive subgroup of R.

    All we need to show is that,
    r\phi^{-1}[J],\phi^{-1}[J]r \subseteq \phi^{-1}[J], \forall r\in R

    I will show that,
    r\phi^{-1}[J]\subseteq \phi^{-1}[J] and leave the second anagolus part to thee to prove.
    Let x\in r\phi^{-1}[J] thus, x=rj where \phi(j)\in J
    Now, (by homomorphism)
    \phi(x)=\phi(rj)=\phi(r)\phi(j) and \phi(j)\in J and \phi(r)\in R but \phi(r)\phi(j)\in J because J is an ideal! Thus, r\phi^{-1}[J]\subseteq \phi^{-1}[J]
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  3. #3
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    Quote Originally Posted by JoeDardeno23 View Post
    -Prove that if f is surjective and I-R is an ideal then f(I) is an ideal (where f(I)={f(i): i in I}).
    -Show, by example, that if f is not surjective the assertion in the above need not hold.
    In general if,
    \phi: R\to S
    With I an ideal in R
    Then,
    \phi[ I ] is ideal of \phi[R].
    But since \phi is surjective \phi[R]=R. Thus, \phi[ I ] is an ideal of R.
    ---
    But it seems you never learned that theorem.
    In that case, we can still show it.

    First, I is an ideal of R that means that I is an additive subgroup of R. Thus, \phi[ I ] is an additive subgroup of S because any group homomorphism preserves the images of subgroups.

    Now, is it true that, s\phi[ I ],\phi[ I ]s\subseteq \phi[ I ] \, \forall s\in S. Again I will only show,
    s\phi[ I ]\subseteq \phi[ I ].
    The other half is anagolous.
    If, x\in s\phi[ I ] then,
    x=si where s \in S, i\in \phi[ I ]
    But since the map is surjective, for some,
    x=\phi(s')\phi(i') where s'\in R, i'\in I
    Since this is a ring homomorphism,
    x=\phi(s'i') \in \phi[ I ]
    Because s'i'\in I since it is an ideal!
    Thus,
    s\phi[ I ] \subseteq \phi[ I ]
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  4. #4
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    Thanks so much.

    Can you think of an example for part three that would work?
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  5. #5
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    Quote Originally Posted by JoeDardeno23 View Post
    Can you think of an example for part three that would work?
    Let,
    \phi: F\to F'
    Be the trivial homomorphism between fields.

    The trivial homomorphism is defined as,
    \phi(x)=0' \, \, \forall x\in F
    Let, \{0\} be the ideal of F.
    Then, \phi[\{0\}]=\{0'\}
    So the function is not surjective but yet we have an ideal preservation under the image of the function.

    --------------
    Maybe, you were asking where it would not work?
    Consider,
    \phi: \mathbb{Z}\to \mathbb{R}.
    Defined as, \phi(x)=x
    This is not a surjective map.

    The set \mathbb{Z} is an ideal (improper ideal).
    Yet,
    \phi [\mathbb{Z}]=\mathbb{Z}
    But this image is not an ideal in \mathbb{R} because,
    \sqrt{2} \mathbb{Z}\not \subseteq \mathbb{Z}
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