## 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 $R$.