# valuation ring questions

• Dec 8th 2008, 07:02 PM
xianghu21
valuation ring questions
Let $\displaystyle K$ be a field, $\displaystyle \nu : K^* \rightarrow \mathbb{Z}$ a discrete valuation on $\displaystyle K$, and $\displaystyle R=\{x \in K^* : \nu(x) \geq 0 \} \cup \{0\}$ the valuation ring of $\displaystyle \nu$. For each integer $\displaystyle k \geq 0$, define $\displaystyle A_k=\{r \in R : \nu(r) \geq k \} \cup \{0\}$.

(a) Prove that for any $\displaystyle k$, $\displaystyle A_k$ is a principal ideal, and that $\displaystyle A_0 \supseteq A_1 \supseteq A_2 \supseteq\ldots$
(b) Prove that if $\displaystyle I$ is any nonzero ideal of $\displaystyle R$, then $\displaystyle I=A_k$ for some $\displaystyle k \geq 0$.
• Dec 9th 2008, 08:26 AM
NonCommAlg
Quote:

Originally Posted by xianghu21
Let $\displaystyle K$ be a field, $\displaystyle \nu : K^* \rightarrow \mathbb{Z}$ a discrete valuation on $\displaystyle K$, and $\displaystyle R=\{x \in K^* : \nu(x) \geq 0 \} \cup \{0\}$ the valuation ring of $\displaystyle \nu$. For each integer $\displaystyle k \geq 0$, define $\displaystyle A_k=\{r \in R : \nu(r) \geq k \} \cup \{0\}$.

(a) Prove that for any $\displaystyle k$, $\displaystyle A_k$ is a principal ideal, and that $\displaystyle A_0 \supseteq A_1 \supseteq A_2 \supseteq\ldots$
(b) Prove that if $\displaystyle I$ is any nonzero ideal of $\displaystyle R$, then $\displaystyle I=A_k$ for some $\displaystyle k \geq 0$.

three things that you need to recall first:

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

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

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

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

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

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