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Math Help - Prove k[x]/(f(x)g(x)) \cong k[x]/(f(x)) \bigoplus k[x]/(g(x)).

  1. #1
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    Prove k[x]/(f(x)g(x)) \cong k[x]/(f(x)) \bigoplus k[x]/(g(x)).

    Suppose k is a field and f(x), g(x) \in k[x] such that (f(x),g(x)) = k[x].
    Prove k[x]/(f(x)g(x)) \cong k[x]/(f(x)) \bigoplus k[x]/(g(x)).
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  2. #2
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    Quote Originally Posted by joestevens View Post
    ...................... (f(x),g(x)) = k[x].
    what this means?
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  3. #3
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    Quote Originally Posted by zoek View Post
    what this means?
    f(x) and g(x) generate an ideal that is equal to the entire ring.
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    Quote Originally Posted by joestevens View Post
    f(x) and g(x) generate an ideal that is equal to the entire ring.
    so, it means that every element of k[x] can be written as follows: s(x)f(x)+t(x)g(x) where s(x) and t(x) \in k[x]? or not?
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  5. #5
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    Quote Originally Posted by zoek View Post
    so, it means that every element of k[x] can be written as follows: s(x)f(x)+t(x)g(x) where s(x) and t(x) \in k[x]? or not?
    Correct.
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  6. #6
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    Quote Originally Posted by joestevens View Post
    Suppose k is a field and f(x), g(x) \in k[x] such that (f(x),g(x)) = k[x].
    Prove k[x]/(f(x)g(x)) \cong k[x]/(f(x)) \bigoplus k[x]/(g(x)).
    if you define the following map:

    p(x)--> (p(x)+I, p(x)+J) where I=(f(x)), J=(g(x)) and p(x) \in k[x]

    you can prove that it is a ring epimorphism with kernel = I-intersection-J which is equal to IJ =(f(x)g(x)) since I+J = (f(x), g(x)) = k[x] and then you can apply the 1st isomorphism theorem for rings.
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  7. #7
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    Quote Originally Posted by zoek View Post
    if you define the following map:

    p(x)--> (p(x)+I, p(x)+J) where I=(f(x)), J=(g(x)) and p(x) \in k[x]

    you can prove that it is a ring epimorphism with kernel = I-intersection-J which is equal to IJ =(f(x)g(x)) since I+J = (f(x), g(x)) = k[x] and then you can apply the 1st isomorphism theorem for rings.
    Thanks. We never covered ring epimorphisms in my class. Is a ring epimorphism just a surjective ring homomorphism?
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  8. #8
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    Quote Originally Posted by joestevens View Post
    Thanks. We never covered ring epimorphisms in my class. Is a ring epimorphism just a surjective ring homomorphism?
    exactly!

    I think that this is the most difficult part of the proof.

    you can work as follows:

    there exist s(x) and t(x) such that 1 = s(x)f(x)+t(x)g(x), since 1 \in k --> 1\in k[x] --> 1\in (f(x), g(x)).

    if p(x) = a(x)t(x)g(x)+b(x)s(x)f(x) then you can verify that p(x)+I = a(x) +I and p(x)+J = b(x) +J

    so (p(x)+I, p(x)+J) =(a(x)+I, b(x)+J)

    with (a(x)+I, b(x)+J) an arbitrary element \in k[x]/(f(x)) \bigoplus k[x]/(g(x)) .
    Last edited by zoek; April 16th 2011 at 01:35 PM.
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