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Math Help - valuation ring questions

  1. #1
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    valuation ring questions

    Let K be a field, \nu : K^* \rightarrow \mathbb{Z} a discrete valuation on K, and R=\{x \in K^* : \nu(x) \geq 0 \} \cup \{0\} the valuation ring of \nu. For each integer k \geq 0, define A_k=\{r \in R : \nu(r) \geq k \} \cup \{0\}.

    (a) Prove that for any k, A_k is a principal ideal, and that A_0 \supseteq A_1 \supseteq A_2 \supseteq\ldots
    (b) Prove that if I is any nonzero ideal of R, then I=A_k for some k \geq 0.
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  2. #2
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    Quote Originally Posted by xianghu21 View Post
    Let K be a field, \nu : K^* \rightarrow \mathbb{Z} a discrete valuation on K, and R=\{x \in K^* : \nu(x) \geq 0 \} \cup \{0\} the valuation ring of \nu. For each integer k \geq 0, define A_k=\{r \in R : \nu(r) \geq k \} \cup \{0\}.

    (a) Prove that for any k, A_k is a principal ideal, and that A_0 \supseteq A_1 \supseteq A_2 \supseteq\ldots
    (b) Prove that if I is any nonzero ideal of R, then I=A_k for some k \geq 0.
    three things that you need to recall first:

    1) since \nu is surjective, \nu(p)=1, for some p, which is clearly in R because \nu(p)=1 > 0.

    2) \nu(x)=0 if and only if x \in R is a unit: this is a quick result of this fact that \forall x \in K^*: \ \nu(x^{-1})=-\nu(x).

    3) \nu(x)=\nu(y) for some 0 \neq x,y \in R, if and only if y=xu, for sime unit u \in R: this is an immediate result of 2).

    now we want to show that A_k is a principal ideal. first from 1) we have: \nu(p^k)=k\nu(p)=k. thus p^k \in A_k. hence <p^k> \subseteq A_k. suppose now that x \in A_k and \nu(x)=n \geq k. then \nu(x)=\nu(p^n).

    hence by 3): x = up^n=up^{n-k}p^k \in <p^k>. this proves that A_k \subseteq <p^k>, and part (a) of your problem is solved.

    for part (b), let k=\min \{\nu(a): \ a \in I \}. obviously I \subseteq A_k. choose a \in I with \nu(a)=k. then \nu(a)=\nu(p^k), and so a=up^k, for some unit u \in R. therefore A_k=<p^k>=<a> \subseteq I. \ \ \Box
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