Characterization of the ideals in a discrete valuation ring

Let K be a field, let v be a discrete valuation on K, and let R be the valuation ring of v. For each integer k \geq 0, define A_k = \{ r \in R \ |\ v(r) \geq k \} \cup \{0\}.

  1. Prove that A_k is a principal ideal and that A_{k+1} \subseteq A_k for all k \in \mathbb{N}.
  2. Prove that if I is a nonzero ideal of R, then I = A_k for some k \geq 0. Deduce that R is a local ring with maximal ideal A_1.

  1. It is clear that A_{k+1} \subseteq A_k for all k.

    Now we show that A_k is an ideal. We have 0 \in A_k, so A_k is nonempty. Now let a,b \in A_k; if one or both of a and b is 0, then a+b \in A_k. If a+b = 0, then a+b \in A_k. If a+b \neq 0, we have v(a+b) \geq \min(v(a),v(b)) \geq k, so that a+b \in A_k. Moreover, v(-a) = v(a), so that -a \in A_k. Finally, if r \in R, we have v(r) \geq 0. Then v(ra) = v(r) + v(a) \geq k, so that ra \in A_k. Since K is commutative, we have that A_k is an ideal of R.

    We now show that A_k is principal. Choose \alpha \in A_k such that v(\alpha) = k; such an element exists because v is surjective. Note that \alpha^{-1} exists in K, and that v(\alpha^{-1}) = -v(\alpha). Let r \in A_k; then v(r) \geq k. Note that v(\alpha^{-1}r) = v(\alpha^{-1}) + v(r) = v(r) - v(\alpha) \geq 0, so that \alpha^{-1}r \in R. Moreover, r = \alpha\alpha^{-1}r. Thus r \in (\alpha), and we have A_k \subseteq (\alpha), and thus A_k = (\alpha). In particular, A_k is generated by any element of valuation k.

  2. Let I \subseteq A_k be a nonzero ideal of R. Let k be minimal among v(r) for r \in I, and let a \in I such that v(a) = k. In particular, we have I \subseteq A_k. Moreover, since A_k = (a), we have A_k \subseteq I. Thus I = A_k.

    From part (1), every proper ideal of R is contained in A_1; thus A_1 is maximal, and moreover, is the unique maximal ideal of R.

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