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Math Help - Ring Isomorphism

  1. #1
    Senior Member vincisonfire's Avatar
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    Ring Isomorphism

    Let R and S be rings and let I ▹R, J ▹S be ideals. Prove that
     (R\times S)/(I\times J) \simeq (R/I) \times (S/J)
    I think
     (R\times S)/(I\times J)  is of the form  (a,b) + (I,J) : a \in R, b \in S   that is also of the form  (a + I, b + J) : a \in R, b \in S   because of definition of addition.
    This is the "cross set"  (R/I) \times (S/J) .
    Something tells me it can't be that simple. Can you tell me why?
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  2. #2
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    Quote Originally Posted by vincisonfire View Post
    Let R and S be rings and let I ▹R, J ▹S be ideals. Prove that
     (R\times S)/(I\times J) \simeq (R/I) \times (S/J)
    I think
     (R\times S)/(I\times J)  is of the form  (a,b) + (I,J) : a \in R, b \in S   that is also of the form  (a + I, b + J) : a \in R, b \in S   because of definition of addition.
    This is the "cross set"  (R/I) \times (S/J) .
    Something tells me it can't be that simple. Can you tell me why?
    1)First prove I\times J\triangleleft R\times S
    2)Define \phi : (R\times S)/(I\times J) \to (R/I)\times (S/J) by \phi ((a,b)(I\times J)) = aI\times bJ
    3)Prove \phi is well-defined.
    4)Prove \phi is one-to-one.
    5)Prove \phi is a ring homorphism.
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  3. #3
    Senior Member vincisonfire's Avatar
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    Aren't aI and  bJ always equal to zero in <br />
R/I and  S/J .
    The function maps everything to (0 , 0) = 0?
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