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Math Help - Units

  1. #1
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    Units

    Show that  \left( \mathbb{Z}_{[x]} \right)^* = \{-1,1\}
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Suppose  a_n x^n + \cdots + a_1 x + a_0 \in \mathbb{Z} [x] is a unit.

    Then there exists  p(x) s.t.  (a_n x^n + \cdots + a_1 x + a_0)p(x) = 1 . It is evident  p(x) \neq 0 .

     (a_n p(x)) x^n + \cdots + (a_1 p(x)) x + a_0 p(x) = 1

    Therefore  a_i p(x) = 0 \; 1 \leq i \leq n and  a_0 p(x) = 1 .

    Since  \mathbb{Z} [x] is an integral domain and  p(x) \neq 0 ,  a_i = 0 \; 1 \leq i \leq n .

    The only way for  a_0 p(x) = 1 is if  p(x) = c , where  c \in \mathbb{Z} . So  a_0 c = 1 \Longleftrightarrow a_0 = 1 , \; c = 1 or  a_0 = -1 , \; c = -1

    Hence  a_n x^n + \cdots + a_1 x + a_0 = -1 or  1 . Which means  \left( \mathbb{Z} [x] \right)^* = \{-1,1\} .
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  3. #3
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    Quote Originally Posted by chiph588@ View Post

    Therefore  a_i p(x) = 0 \; 1 \leq i \leq n and  a_0 p(x) = 1 .
    i don't think you can make this conclusion!

    well, if (a_nx^n + \cdots + a_0)(b_mx^m + \cdots + b_0)=1, \ a_n \neq 0, \ b_m \neq 0, and n > 0 or m > 0, then we must have a_nb_m=0, which is not possible in an integral domain. so n = m = 0, and the rest is

    obvious. this also shows that, in general, for any integral domain R, the set of units of R[x] is exactly equal to the set of units of R.

    Remark: there's a well-known result that in any commutative ring with unity R, a polynomial \sum_{k=0}^n a_kx^k \in R[x] is a unit iff a_0 is a unit of R and all a_k, \ k \neq 0, are nilpotent.
    Last edited by NonCommAlg; April 22nd 2009 at 01:44 AM. Reason: added the remark!
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