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Math Help - valuation ring, discrete valuation

  1. #1
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    valuation ring, discrete valuation

    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}.
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  2. #2
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    Quote Originally Posted by Erdos32212 View Post
    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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