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Math Help - Show that Z_p is intgrally closed

  1. #1
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    Show that Z_p is intgrally closed

    Hello,

    Let Z_p ={ \frac{a}{b}:a,b\in Z, b\ne 0, gcd(a,b)=1, p does not divide  b }
    Z_p is a subring of Q
    1) Show that for each x\in Q either x\in Z_p or x^{-1} \in Z_p

    2) use 1) to show that Z_p is integrally closed.

    I answered the first part of the question but I need your help in solving part 2, how can we show that Z_p is integrally closed.

    All helps and guidance are highly appreciated
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  2. #2
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    Quote Originally Posted by student2011 View Post
    Hello,

    Let Z_p ={ \frac{a}{b}:a,b\in Z, b\ne 0, gcd(a,b)=1, p does not divide  b }
    Z_p is a subring of Q
    1) Show that for each x\in Q either x\in Z_p or x^{-1} \in Z_p

    2) use 1) to show that Z_p is integrally closed.

    I answered the first part of the question but I need your help in solving part 2, how can we show that Z_p is integrally closed.

    All helps and guidance are highly appreciated
    suppose that a \in \mathbb{Q} \setminus Z_p is integral over Z_p. then a^{-1} \in Z_p. let
    p(x) =x^n + c_1x^{n-1} + \ldots + c_{n-1}x + c_n \in Z_p[x]
    be a polynomial of minimal degree such that p(a)=0. if n = 1, then a \in Z_p, which is a contradiction. so n > 1. now let
    q(x)=x^{n-1} + c_1x^{n-2} + \ldots + c_{n-2}x + c_{n-1} + c_na^{-1}.
    then q(x) \in Z_p[x] because a^{-1} \in Z_p. see that q(a)=0, which contradicts the minimality of n.
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