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Math Help - UFD that is not a PID

  1. #1
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    UFD that is not a PID

    I have to prove that \mathbb{Z}[x] is a unique factorization domain that is not a principal ideal domain.

    Im supposed to notice that <2,x> cannot be generated by an ideal of one element.

    Can anyone explain me why is that?

    Thank you!
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  2. #2
    Senior Member vincisonfire's Avatar
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    Suppose  <2,x> can be generated by a single element. That means that there exist  \alpha such that \alpha|2 and  \alpha|x but since  (2,x) = 1 so it is impossible.
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  3. #3
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    Quote Originally Posted by vincisonfire View Post
    Suppose  <2,x> can be generated by a single element. That means that there exist  \alpha such that \alpha|2 and  \alpha|x but since  (2,x) = 1 so it is impossible.
    you also need to show that: <2,x> \neq \mathbb{Z}[x]: since the constant term of every polynomial in <2,x> is a multiple of 2, we have 1 \notin <2,x> and thus <2,x> is a proper ideal of \mathbb{Z}[x].

    similarly you can easily prove a general result which is quite useful: if R is a non-zero ring, then the polynomial ring R[x] is a PID if and only if R is a field.
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    NonCommAlg:
    Im trying to prove the last thing you said, but I cant get it.
    Can you give me any hints?
    Thank you!
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  5. #5
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    Quote Originally Posted by Inti View Post
    NonCommAlg:
    Im trying to prove the last thing you said, but I cant get it.
    Can you give me any hints?
    Thank you!
    if R is a field, we know that R[x] will be a PID. that's a well-known fact. conversely, suppose R[x] is a PID. then since R is a subring of R[x], it has to be an integral domain. we'll prove that

    every non-zero element of R is a unit, which means R is a field. so let 0 \neq r \in R[x]. let I=<r,x>. since R[x] is a PID, there exists f(x) \in R[x] such that I=<f(x)>. hence f(x) must

    be a unit element of R because f(x) \mid r, \ f(x) \mid x. so 1 \in I, i.e. there exists g(x), h(x) \in R[x] such that rg(x) + xh(x)=1. now put x=0 to get rg(0)=1, i.e. r is a unit element of R. \ \Box
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