# valuation ring questions

• Dec 8th 2008, 07:02 PM
xianghu21
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 $k \geq 0$.
• Dec 9th 2008, 08:26 AM
NonCommAlg
Quote:

Originally Posted by xianghu21
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 $ \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): $x = up^n=up^{n-k}p^k \in .$ this proves that $A_k \subseteq ,$ 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 $A_k== \subseteq I. \ \ \Box$