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Discrete Valuation

  1. Dec 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##p## and let ##\nu : \mathbb{Q}^\times \rightarrow \mathbb{Z}## be defined by ##\nu (\frac{a}{b}) = \alpha##, where ##\frac{a}{b} = p^\alpha \frac{c}{d}##, where ##p## divides neither ##c## nor ##d##. Prove that the corresponding valuation ring ##R := \{x \in \mathbb{Q}^\times ~|~ \nu(x) \ge 0 \} \cup \{0\}## is the ring of all rational numbers whose denominators are relatively prime to ##p##. Describe the units of this valuation ring.

    2. Relevant equations

    Let ##K## be some field. A function ##\nu : K^\times \rightarrow \mathbb{Z}## is a discrete valuation if

    ##\nu(ab) = \nu(a) + \nu(b)##

    ##\nu## is surjective

    ##\nu(x+y) \ge \min \{\nu(x), \nu(y)\}## if ##x+y \neq 0##.

    3. The attempt at a solution

    I find this problem rather annoyingly ill-posed. For one, it seems that the author of this problem presupposes that every rational number ##\frac{a}{b}## as ##p^\alpha \frac{c}{d}## for some integers ##\alpha##, ##b##, and ##c## satisfying the above conditions, without offering a proof of this or asking to prove this as a part of the problem. Moreover, we aren't even asked to show that this is a well-defined discrete valuation. So, I would like to prove some of these things and then discuss the difficulty I am having with this problem.

    First, let me prove every rational number ##\frac{a}{b}## can be written in the previously mentioned form. If neither ##a## nor ##b## have the prime ##p## appearing in their factorization, then ##\frac{a}{b} = p^0 \frac{a}{b}##. Now, if both ##a## and ##b## have ##p## appearing in their factorization in the form ##p^\alpha## and ##p^\beta##, respectively, then ##\frac{a}{b} = \frac{p^\alpha c}{p^\beta d} = p^{\alpha - \beta} \frac{c}{d}##.

    Now I will prove that ##\nu## is well-defined. Suppose that ##p^\alpha \frac{a}{b} = p^\beta \frac{c}{d}## but ##\alpha - \beta \neq 0##. Without loss of generality, suppose ##\alpha - \beta > 0##. Then ##p^{\alpha - \beta}ad = bc## implies ##p^{\alpha - \beta} |cd##, and because ##p|p^{\alpha-\beta}##, we have ##p|cd##. But this implies either ##p|c## or ##p|d##, both of which are contradictions. Hence, ##\alpha = \beta##. This gives us

    ##\nu(p^{\alpha} \frac{a}{b}) = \alpha = \beta = \nu (p^\beta \frac{c}{d})##,

    thereby establishing that ##\nu## is well-defined.

    How does this sound so far? The following is where difficulty begins to emerge.

    Now I am trying to show that it is a discrete valuation. Showing that ##\nu## satisfies ##\nu(ab) = \nu(a) + \nu(b)## is rather easy. For surjectivity, if ##n## is some integer, then ##\nu(p^n \frac{p+1}{p-1}) = n##, where ##p## clearly divides neither the numerator nor denominator. I am having a little trouble showing ##\nu## possesses the last property: Consider ##p^\alpha \frac{a}{b}## and ##p^\beta \frac{c}{d}##, and suppose that ##\alpha < \beta##. Then

    ##\nu(p^\alpha \frac{a}{b} + p^\beta \frac{c}{d}) = \nu( p^\beta \frac{p^{\alpha - \beta} ad + cd}{bd}) = \beta \ge \min \{\alpha, \beta \} = \alpha##

    However, it seems possible that ##p## could divide the numerator of ##\frac{p^{\alpha - \beta} ad + cd}{bd}##...
     
  2. jcsd
  3. Dec 21, 2016 #2

    andrewkirk

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    I don't follow this step. Why does that last statement follow?

    In any case, I don't think it's necessary since part of our supposition should also be that ##p## does not divide any of ##a,b,c,d## in this case. So by uniqueness of prime factorisations we get a contradiction by prime-factoring both sides of the equation ##p^{\alpha - \beta}ad = bc## and observing that the factorisation of the left-hand side has ##p## as a factor while that of the right-hand side does not.

    By the way, it would be better in this paragraph to use letters other than ##a,b## because they have already been used in the definition of ##\nu## for something different, where they can be divisible by ##p##, yet here we require them to not be divisible by ##p##. I would be inclined to instead write: suppose ##\frac ab=p^\beta\frac cd=p^\beta \frac{c'}{d'}## where ##p## does not divide any of ##c,d,c',d'## and ##\alpha,\beta## are non-negative integers with ##\alpha\geq \beta##. We can then use the prime factorisations of ##p^{\alpha-\beta}cd'=c'd## to prove that ##\alpha=\beta##, thereby pacifying the constructivist logicians that are uncomfortable with proof by contradiction (not that that matters, but I think it's good to be constructivist if one can do so without undue inconvenience).
     
  4. Dec 21, 2016 #3

    andrewkirk

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    There are two errors in here. Firstly the numerator of that fraction should be ##p^{\alpha - \beta} ad + cb##, not ##p^{\alpha - \beta} ad + cd##. Secondly, given your assumptions, ##p^{\alpha - \beta}## will not be an integer. It's best to be consistent with notation and assumptions or else one confuses oneself. Earlier you assumed ##\alpha>\beta## but now you have assumed ##\beta>\alpha##, yet you seem to still be using the earlier assumption some of the time.

    If you re-write this fraction carefully, paying attention to the assumptions, you should be able to write the sum of the two rationals as ##p## raised to the lower of the two indices, multiplied by a fraction whose denominator is not divisible by ##p##, and whose numerator may or may not be.
    That is not a problem. If it does it just makes the value of ##\nu## applied to the sum bigger, so the inequality will still be satisfied.
     
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