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Math Help - Polynomial Rings

  1. #1
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    Polynomial Rings

    Let R be a commutative ring with identity.

    Let p(x)=a_nx^n+ a_{n-1}x^{n-1} . . . +a_2x^2+a_1x^1+a_0 be a polynomial in R[x].

    p(x) is a unit if and only if a_0 unit and a_i for i>1 are nilpotent.

    So I can easily prove the first term, a_0 is a unit.

    However I feel that I cannot guarantee that a_n is nilpotent.

    Since p is a unit, then there exists a q in R[x] (let's say the coefficients of q are of the form b_i and that it has degree m) where p*q = 1.

    So taking the product of the 2 polynomials:

    a_nb_mx^{n+m}+ . . . . . +a_0b_0=1

    and it must equal zero because a_0 is a unit, which implies that the last term = 1, and the rest equals 0.

    So I can show that a_n is a zero divisor but I don't feel I can guarantee that it's nilpotent.

    The hint says to look at the x^(n-1) term and multiply by a_n.

    a_nb_{n-1}+a_{n-1}b_n=0
    a_n*a_nb_{n-1}+a_n*a_{n-1}b_n=0

    then the term on the right dissapears because a_nb_m =0.

    a_n^2b_{n-1}=0
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  2. #2
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    Quote Originally Posted by seld View Post
    Let R be a commutative ring with identity.

    Let p(x)=a_nx^n+ a_{n-1}x^{n-1} . . . +a_2x^2+a_1x^1+a_0 be a polynomial in R[x].

    p(x) is a unit if and only if a_0 unit and a_i for i>1 are nilpotent.
    suppose first that a_n, \cdots , a_1 are nilpotent and a_0 is a unit. then p(x) - a_0 is nilpotent and thus p(x)=p(x) - a_0 + a_0 is a unit becasue, in any commutative ring, nilpotent + unit is

    a unit. for the converse, we'll use induction on n, the degree of p(x): it's clear if n = 0. so suppose the claim is true for any polynomial which is a unit and has degree < n.

    let p(x) = \sum_{j=0}^n a_jx^j, \ n \geq 1, be a unit. so there exists q(x)=\sum_{j=0}^m b_jx^j such that p(x)q(x)=1. if m=0, then a_n = \cdots = a_1=0 and a_0 is a unit. so we may assume

    that m > 0. clearly b_0 must be a unit and we also have:

    a_nb_m = 0, \ a_nb_{m-1}+ a_{n-1}b_m = 0, \ \cdots , a_nb_0 +a_{n-1}b_1 + \cdots = 0. write these relations in terms of matrices: AX=0, where

    A=\begin{pmatrix}a_n & 0 & 0 & . & . & . & 0 \\ a_{n-1} & a_n & 0 & . & . & . & 0 \\ . & . & . & & . & . & . \\ . & . & . & & . & . & . \\ . & . & . & & . & . & . \\ * & * & * & . & . & . & a_n \end{pmatrix}, \ \ X=\begin{pmatrix}b_m \\ b_{m-1} \\ . \\ . \\ . \\ b_0 \end{pmatrix}, where * means that we don't care what those entries are. the important point is that A is a (lower) triangular matrix with a_n

    on its main diagonal. now multiply AX=0, from the left, by the adjoint matrix of A, to get a_n^m X =(\det A)X = \text{adj}(A)A X = 0. thus a_n^mb_0=0, and hence a_n^m = 0 because b_0

    is a unit. so a_n and therefore -a_nx^n is nilpotent. hence p_1(x)=p(x) -a_nx^n is a unit, because, as i already mentioned, nilpotent + unit is a unit. now since \deg p_1(x) < n, we can

    apply the induction hypothesis to finish the proof.
    Last edited by NonCommAlg; September 2nd 2010 at 07:04 AM.
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