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Math Help - Prime Ideal Integral domain

  1. #1
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    Prime Ideal Integral domain

    Let R be a commutative ring with unity. Show that the ideal generated by <x>={xf(x)|f(x) is a polynomial in R[x]} is a prime ideal in R[x] if and only if R is an integral domain.

    For the forward direction I tried to use fact that an ideal N of R is prime if and only if R/N is an integral domain. But got stuck after that.

    And I am completely clueless about the backwards direction. Sorry. Any help would be appreciated.
    Last edited by mr fantastic; May 22nd 2009 at 02:48 PM. Reason: Restored original question deleted by OP
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  2. #2
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    Quote Originally Posted by curiousmuch View Post
    Let R be a commutative ring with unity. Show that the ideal generated by <x>={xf(x)|f(x) is a polynomial in R[x]} is a prime ideal in R[x] if and only if R is an integral domain.

    For the forward direction I tried to use fact that an ideal N of R is prime if and only if R/N is an integral domain. But got stuck after that.

    And I am completely clueless about the backwards direction. Sorry. Any help would be appreciated.
    define the map \varphi: R[x] \longrightarrow R by \varphi(f(x))=f(0). it's easy to see that \varphi is a surjective ring homomorphism and \ker \varphi = <x>. thus \frac{R[x]}{<x>} \cong R and the result follows.
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  3. #3
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    Oh wow this way both directions are taken care of at once.
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