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Thread: Ideals

  1. #1
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    Ideals

    I need some help on this problem as well:
    Let R and S be rings and let I be an ideal of R and J be an ideal of S.
    Prove that:
    (R x S)/(I x J) is isomorphic with (R/I) x (S/J).

    Thank you.
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  2. #2
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    Quote Originally Posted by DaisukeMatsuzaka View Post
    I need some help on this problem as well:
    Let R and S be rings and let I be an ideal of R and J be an ideal of S.
    Prove that:
    (R x S)/(I x J) is isomorphic with (R/I) x (S/J).

    Thank you.
    Did you consider the map,
    $\displaystyle \phi: (R\times S)/(I\times J)\to R/I \times S/J$
    Well defined as,
    $\displaystyle \phi((r,s)(I\times J))=rI\times sJ$
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  3. #3
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    Thanks for the response. I really don't know what the second half means. Could you elaborate?
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  4. #4
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    Quote Originally Posted by DaisukeMatsuzaka View Post
    Thanks for the response. I really don't know what the second half means. Could you elaborate?
    Yes.

    How do we know that this map is well-defined?
    Independent from the choices we can choose for $\displaystyle r,s$?

    Well, let us assume there are two ways to expess an element in $\displaystyle (R\times S)/(I\times J)$
    $\displaystyle (r_1,s_1)(I\times J)$ and $\displaystyle (r_2,s_2)(I\times J)$
    Then the function maps them into,
    $\displaystyle r_1I\times s_1J$, $\displaystyle r_2I\times s_2J$ respectively.
    But,
    $\displaystyle r_1I=r_2I$ and $\displaystyle s_1J=s_2J$ because the conditon we had said that,
    $\displaystyle (r_1,s_1)(I\times J)=(r_2,s_2)(I\times J)$
    Thus, the map is well-defined.

    The function is a group homomorphism,
    $\displaystyle \phi((r_1,s_1)(I\times J)\cdot (r_2,s_2)(I\times J))$=$\displaystyle \phi((r_1r_2,s_1s_2)(I\times J))=r_1r_2I\times s_1s_2J=(r_1I\times s_1J)(r_2I\times s_2J)$$\displaystyle =\phi((r_1,s_1)(I\times J))\cdot \phi((r_2,s_2)(I\times J))$

    The function is one-to-one,
    $\displaystyle \phi((r_1,s_1)(I\times J))=\phi((r_2,s_2)(I\times J))$
    $\displaystyle r_1I\times s_1J=r_2I\times s_2J$
    Thus,
    $\displaystyle r_1,r_2$ are both in same coset of $\displaystyle I$ and,
    $\displaystyle s_1,s_2$ are both in same coset of $\displaystyle J$.

    The function is onto,
    That result is trivial.
    Any, $\displaystyle rI\times sJ$ can be obtained from mapping the element,
    $\displaystyle (r,s)(I\times J)$
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