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Math Help - Commutative UFD but not a PID

  1. #1
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    Commutative UFD but not a PID

    Hi guys,

    I'm trying to understand an example of a commutative UFD that is not a PID.

    Let k be a field and consider the ring of polynomials over k, k[x,y]. Assuming that this is a UFD, I'm trying to show that this is not a PID.

    It was shown to me a while ago before I had any 'further' knowledge and looking back at my notes now, I have

    "xR+yR is not a principal ideal and not free..."
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  2. #2
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     Q[x,y] is a commutative UFD that is not a PID. A polynomial ring over a field is always a UFD. But in Q[x,y], the ideal (x,y) is not principal. Thus, Q[x,y] is not a principal domain...
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