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Math Help - Prove it is not a PID.

  1. #1
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    Prove it is not a PID.

    Fix a field k and consider the subring k[x^2,x^3] \subseteq k[x] (the former ring is all polynomials with zero coefficient in front of x). Prove that k[x^2,x^3] is not a principal ideal domain.



    I just need to find an ideal I that is not principal, right?

    I=(x^2,x^3)= \{ polynomials in x s.t. a_0=a_1=0 \}

    Is this actually true? (x^2,x^3)= \{ polynomials in x s.t. a_0=a_1=0 \}

    Is I principal?
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  2. #2
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    Suppose it is. Then \exists r \in k[x^2,x^3] such that I = (r).

    r = a_{0} + a_{1}x + ... + a_{n}x^n

    If a_{n} \neq 0 for any n > 2 then clearly (r) won't generate I since it will only contain polynomials of degree > 2. So r = a_{0} + a_{1}x + a_{2}x^2.

    If a_{2} = 0 then show I is a strict subset of (r). Then a_{2} \neq 0 and show (r) contains no polynomials of degree 3 and so it can't be I.

    I think that's fine anyways.
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  3. #3
    Senior Member
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    Sorry, correction..

    Let I:=(x^2,x^3)=\{\sum_n a_ix^i:a_0=a_1=0,n\in\mathbb{N}\}

    Suppose towards a contradiction that I=(p(x)) for some nonzero p\in(x^2,x^3). Then x^2=p(x)q_1(x) and x^3=p(x)q_2(x) for some q\in k[x^2,x^3]. So \deg p+\deg q_1=2. We know that \deg p\geq 2 since p\neq 0, so 0=\deg q_1. This gives us \deg p=2 and therefore 2+\deg q_2=\deg p+\deg q_2=3. Then \deg q_2=1, a contradiction.
    Last edited by hatsoff; March 29th 2011 at 03:15 PM.
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