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Math Help - Simple Abstract Algebra

  1. #1
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    Simple Abstract Algebra

    Looking for some help/explanations for these problems.

    1. R is a UFD => LCM's exist.

    2. R is Euclidean. If a, b are associates, then phi(a)=phi(b).

    3. R is a PID and S is an integral domain
    phi: R--->S is onto. show either phi is an isomorphism or S is a field.

    4. R is a commutative ring with a 1. Show R[x] is PID iff R is a field.

    5. Prove that if R is a UFD, then R[x_1,x_2,....,X_k] (polynomials in k indeterminates) is a UFD. Exhibit a nonprincipal ideal.
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  2. #2
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    Quote Originally Posted by mathisthebestpuzzle View Post
    1. R is a UFD => LCM's exist.
    Let S be a set of representatives for each irreducible element.
    For a\in R be non-zero and \pi \in S, define d_{\pi}(a)=n so that \pi^n |a but \pi^{n+1} \not | a.
    Then a = \alpha \prod_j \pi_j ^{d_{\pi_j}(a)} where \alpha is unit.
    Let b be non-zero so b= \beta \prod_j \pi_j^{d_{\pi_j}(b)}.
    Where the product is over \pi_j \in S.

    Notice that \prod_j \pi_j^{m_j} where m_j = \max\{ d_{\pi_j}(a), d_{\pi_j}(b)\} is a LCM.

    2. R is Euclidean. If a, b are associates, then phi(a)=phi(b)

    3. R is a PID and S is an integral domain
    phi: R--->S is onto. show either phi is an isomorphism or S is a field.
    What is \phi?

    4. R is a commutative ring with a 1. Show R[x] is PID iff R is a field.
    It should be a known result that R being a field implies that R[x] is a PID.
    Now if R[x] is a PID, let a\not = 0 be in R.
    For the other direction see this.

    5. Prove that if R is a UFD, then R[x_1,x_2,....,X_k] (polynomials in k indeterminates) is a UFD.
    If R is an UFD then R[x] is an UFD[/tex].
    Once you know this you can induct.

    Exhibit a nonprincipal ideal
    Consider \mathbb{Z}[x,y]. And consider the set of all polynomials that have constant term equal to zero. This set is an ideal but it is a non-pricipal ideal.
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