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Math Help - Prove that if R is a PID then every ideal in S is also principal.

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    Prove that if R is a PID then every ideal in S is also principal.

    (a) Suppose that  \phi : R \rightarrow S is a surjective ring homomorphism. Prove that if R is a principal ideal domain then every ideal in S is also principal.

    (b) Give an example to show that S need not be an integral domain.
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    For part (a), let J be an ideal in S. What can you say about \phi^{-1}(J)?
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    Quote Originally Posted by ojones View Post
    For part (a), let J be an ideal in S. What can you say about \phi^{-1}(J)?
    \phi^{-1}(J)=I\subseteq R
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    You can say a lot more than that.
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    \phi^{-1}(J)=I\subseteq R is principal
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    Correct. Therefore?
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    \phi^{-1}(J)=I\subseteq R is principal; therefore, all ideals J \subseteq S are principal.
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    Nice try. I principal means I=aR for some a\in R. Can you guess the form of J?
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    Quote Originally Posted by ojones View Post
    Nice try. I principal means I=aR for some a\in R. Can you guess the form of J?
    J=\phi(I)=\phi(aR)=\phi(a)*\phi(R)=\phi(a)*S
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    There's no *. Conclusion?

    For part (b), if S fails to be an integral domain, then what?
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    Quote Originally Posted by ojones View Post
    There's no *. Conclusion?
    You mean J=\phi(I)=\phi(aR)=\phi(a)\phi(R)=\phi(a)S?
    Conclusion: J is principal.

    Quote Originally Posted by ojones View Post
    For part (b), if S fails to be an integral domain, then what?
    Then S has at least one nonzero zero-divisor. So if b,c\in S then if bc=0, b and c can both be nonzero.
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    Quote Originally Posted by JJMC89 View Post
    You mean J=\phi(I)=\phi(aR)=\phi(a)\phi(R)=\phi(a)S?

    Well yes, but this isn't the best way to argue. You want to show that J=\phi(a)S or equivalently that if b\in J then b=\phi(a)s for some s\in S.

    Conclusion: J is principal..
    Right.


    Then S has at least one nonzero zero-divisor. So if b,c\in S then if bc=0, b and c can both be nonzero.
    OK, but what's an obvious example of a ring which is a PID, one that's not, and an obvious homorphism between the two? Hint: \mathbb{Z} is involved.
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  13. #13
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    \mathbb{Z} is a PID
    n\mathbb{Z}, (n>1) is not an integral domain.
    \phi(z)=nz
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  14. #14
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    You need to be careful here. Does your definition of an integral domain require the ring to have an identity element? If so, your example is fine. I was thinking of S=\mathbb{Z}/n\mathbb{Z} for n not prime.
    Last edited by ojones; March 29th 2011 at 10:54 PM.
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    Yes, my definition requires unity.


    Thanks for your help! This thread is one of the most helpful responses I've gotten (for any problem).
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