valuation ring, discrete valuation

**Erdos32212** · November 18th 2008, 07:47 PM

Background: Let $K$ be a field. A discrete valuation on $K$ is a function $\nu$ on $K$ is a function $\nu$ $: K^* \rightarrow \mathbb{Z}$ such that
(i) $\nu(ab)=\nu(a)+\nu(b)$ for all $a, b \in K^*.$
(ii) $\nu$ is surjective
(iii) For all $x, y$ $\in K^*$ such that $y \neq -x$ , $\nu(x+y)\geq$ min $\{ \nu(x), \nu(y) \}$ . The set $R=$ $\{ x \in K^* | \nu(x) \geq 0 \} \cup \{ 0 \}$ is called the valuation ring of $\nu$ .

Prove:
(a) Prove that $R$ is a subring of $K$ .
(b) Prove that for each nonzero element $x \in K$ , $x \in R$ , or $x^{-1} \in R$ .
(c) Prove that $x \in R$ is a unit if and only if $\nu(x)=0$ .
(d) Now let $K = \mathbb{Q}$ and let $p$ be a prime. Given a nonzero element $x \in \mathbb{Q}$ , write $x=p^e \cdot \frac{c}{d}$ where $e$ is an integer and $p$ divides neither $c$ nor $d$ ; then define $\nu_p=e$ . Prove $e$ is uniquely defined and that $\nu_p$ is a discrete valuation on $\mathbb{Q}.$

**NonCommAlg** · November 18th 2008, 10:47 PM

Originally Posted by Erdos32212

Background: Let $K$ be a field. A discrete valuation on $K$ is a function $\nu$ on $K$ is a function $\nu$ $: K^* \rightarrow \mathbb{Z}$ such that
(i) $\nu(ab)=\nu(a)+\nu(b)$ for all $a, b \in K^*.$
(ii) $\nu$ is surjective
(iii) For all $x, y$ $\in K^*$ such that $y \neq -x$ , $\nu(x+y)\geq$ min $\{ \nu(x), \nu(y) \}$ . The set $R=$ $\{ x \in K^* | \nu(x) \geq 0 \} \cup \{ 0 \}$ is called the valuation ring of $\nu$ .

Prove:
(a) Prove that $R$ is a subring of $K$ .
(b) Prove that for each nonzero element $x \in K$ , $x \in R$ , or $x^{-1} \in R$ .
(c) Prove that $x \in R$ is a unit if and only if $\nu(x)=0$ .
(d) Now let $K = \mathbb{Q}$ and let $p$ be a prime. Given a nonzero element $x \in \mathbb{Q}$ , write $x=p^e \cdot \frac{c}{d}$ where $e$ is an integer and $p$ divides neither $c$ nor $d$ ; then define $\nu_p=e$ . Prove $e$ is uniquely defined and that $\nu_p$ is a discrete valuation on $\mathbb{Q}.$

(a) you only need to show that R is closed under addition and multiplication: let $x,y \in R,$ and put $u=x+y.$ if u = 0, then u obviously is in R. if $u \neq 0,$ then by (iii): $\nu(u) \geq \min \{\nu(x), \nu(y) \} \geq 0,$

because $\nu(x) \geq 0, \ \nu(y) \geq 0.$ hence $u \in R.$ to prove that $xy \in R,$ by (i): $\nu(xy) = \nu(x) + \nu(y) \geq 0.$ so $xy \in R.$

(b) first note that by (i): $\nu(1)=\nu(1) + \nu(1).$ thus: $\nu(1)=0.$ now suppose $x \in K.$ if x = 0, then x clearly is in R. so let $x \neq 0.$ suppose $x \notin R, \ x^{-1} \notin R.$ then $\nu(x) < 0, \ \nu(x^{-1}) < 0.$

thus: $0=\nu(1)=\nu(x) + \nu(x^{-1}) < 0.$ contradiction!

(c) if $x \in R$ is a unit, then $x^{-1} \in R.$ we showed in (b) that $\nu(1)=0.$ hence: $\nu(x) + \nu(x^{-1})=0,$ which is possible only if $\nu(x)=\nu(x^{-1})=0,$ because $\nu(x) \geq 0, \ \nu(x^{-1}) \geq 0.$ now suppose that

$x \in R$ and $\nu(x)=0.$ we want to show that $x^{-1} \in R,$ that is $\nu(x^{-1}) \geq 0.$ well, as we've seen: $\nu(x^{-1}) = -\nu(x)=0.$

(d) it's obvious that $e$ is uniquely defined because of the unique factorization in the ring of integers. it's also trivial that $v_p$ satisfies the properties (i), (ii). the property (iii) is less obvious: suppose

that $x=p^r \frac{a}{b}, \ y=p^s \frac{c}{d},$ with $r \leq s$ and $a,b,c,d$ not divisible by p. so $\nu_p(x)=r, \ \nu_p(y)=s.$ then $x+y=p^r \frac{ad + p^{s-r}bc}{bd}.$ so since $p$ does not divide $bd,$ we have: $\nu_p(x+y) \geq r = \min \{r,s \}.$

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