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Thread: valuation ring, discrete valuation

  1. #1
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    valuation ring, discrete valuation

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

    Prove:
    (a) Prove that $\displaystyle R$ is a subring of $\displaystyle K$.
    (b) Prove that for each nonzero element $\displaystyle x \in K$, $\displaystyle x \in R$, or $\displaystyle x^{-1} \in R$.
    (c) Prove that $\displaystyle x \in R$ is a unit if and only if $\displaystyle \nu(x)=0$.
    (d) Now let $\displaystyle K = \mathbb{Q}$ and let $\displaystyle p$ be a prime. Given a nonzero element $\displaystyle x \in \mathbb{Q}$, write $\displaystyle x=p^e \cdot \frac{c}{d}$ where $\displaystyle e$ is an integer and $\displaystyle p$ divides neither $\displaystyle c$ nor $\displaystyle d$; then define $\displaystyle \nu_p=e$. Prove $\displaystyle e$ is uniquely defined and that $\displaystyle \nu_p$ is a discrete valuation on $\displaystyle \mathbb{Q}.$
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  2. #2
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    Quote Originally Posted by Erdos32212 View Post
    Background: Let $\displaystyle K$ be a field. A discrete valuation on $\displaystyle K$ is a function $\displaystyle \nu$ on $\displaystyle K$ is a function $\displaystyle \nu$ $\displaystyle : K^* \rightarrow \mathbb{Z}$ such that
    (i) $\displaystyle \nu(ab)=\nu(a)+\nu(b)$ for all $\displaystyle a, b \in K^*.$
    (ii) $\displaystyle \nu$ is surjective
    (iii) For all $\displaystyle x, y$ $\displaystyle \in K^*$ such that $\displaystyle y \neq -x$, $\displaystyle \nu(x+y)\geq$ min$\displaystyle \{ \nu(x), \nu(y) \}$. The set $\displaystyle R=$$\displaystyle \{ x \in K^* | \nu(x) \geq 0 \} \cup \{ 0 \}$ is called the valuation ring of $\displaystyle \nu$.

    Prove:
    (a) Prove that $\displaystyle R$ is a subring of $\displaystyle K$.
    (b) Prove that for each nonzero element $\displaystyle x \in K$, $\displaystyle x \in R$, or $\displaystyle x^{-1} \in R$.
    (c) Prove that $\displaystyle x \in R$ is a unit if and only if $\displaystyle \nu(x)=0$.
    (d) Now let $\displaystyle K = \mathbb{Q}$ and let $\displaystyle p$ be a prime. Given a nonzero element $\displaystyle x \in \mathbb{Q}$, write $\displaystyle x=p^e \cdot \frac{c}{d}$ where $\displaystyle e$ is an integer and $\displaystyle p$ divides neither $\displaystyle c$ nor $\displaystyle d$; then define $\displaystyle \nu_p=e$. Prove $\displaystyle e$ is uniquely defined and that $\displaystyle \nu_p$ is a discrete valuation on $\displaystyle \mathbb{Q}.$
    (a) you only need to show that R is closed under addition and multiplication: let $\displaystyle x,y \in R,$ and put $\displaystyle u=x+y.$ if u = 0, then u obviously is in R. if $\displaystyle u \neq 0,$ then by (iii): $\displaystyle \nu(u) \geq \min \{\nu(x), \nu(y) \} \geq 0,$

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

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

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

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

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

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

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