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EDIT: Don't think about this. The problem statement is flawed. (See comments)

Let $K$ be a field, and let $K(T)$ be the quotient field of polynomials over $K$.

Then I define $v(f/g) = \deg(f) - \deg(g)$ if $f/g\neq 0$ and $v(0) = \infty$. I need to show that this is a discrete valuation.

The properties needed:

(i) $v(f/g) = \infty$ if and only if $f/g = 0$.

(ii) $v\Bigl((f/g)\cdot (p/q)\Bigr) = v(f/g) + f(p/q)$

(iii) $v\Bigl((f/g) + (p/q)\Bigr) \geq \min(v(f/g),v(p/q))$

The first two properties were easy, but I run into the following difficulty with the 3rd. Assume wlog that $v(f/g)\leq v(p/q)$.

I have $v\Bigl((f/g) + (p/q)\Bigr) = v\Bigl((fq + pg)/gq\Bigr) = \deg(fq + pg) - \deg(gq)$.

I somehow need to get $\deg(fq + pg) \geq \deg(fq)$, and then I can finish the chain to get what I want.

I feel like the following argument "almost" works.

Since $\deg(f/g)\leq \deg(p/q)$, then I think that $\deg(fq)\leq \deg(pg)$.

If the latter inequality were strict, then I would KNOW that (1) $\deg(fq)\leq \deg(fq + pg)$.

But if equality holds, I don't know how to deal with something like the following possibility.

$f(x)q(x) = x^2 - 1$, $p(x)g(x) = -x^2$, then this violates $(1)$.

  • 0
    What I described is the "valuation at $0$"; you can similarly define the "valuation at $a$" as the degree to which $f/g$ vanishes at $a$ (i.e., highest power of $x-a$ that divides $f$, minus highest power of $x-a$ that divides $g$. As Brandon Carter and Georges Elencwajg note, what you are doing is computing the "valuation at $\infty$".2012-03-17

1 Answers 1

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0) It is to your credit that you couldn't prove iii) because it is false! Indeed
$ v((x^2)+(x-x^2))\geq \operatorname {min}(v(x^2),v(x-x^2)) \;$ is equivalent to $1 \geq \operatorname {min}(2,2)$

1) So $v$ is not a valuation. The correct definition of the valuation is

$w(f/g)=\operatorname {deg} g-\operatorname {deg} f \; \text {for} \; f\neq 0 \quad (\text {and} \;w(0)=\infty)$

Of course it can't hurt to realize that $w$ just computes the order of vanishing of fractions at infinity . For example $w( \frac {T^3-1}{T^7+T^2+1})=4$

2) The proofs of i) and ii) are trivial and in iii) you may reduce (by taking a common denominator) to the case of non zero fractions $\frac {f}{g}, \frac {F}{g}$ .
You then have to show that:

$w( \frac {f+F}{g})= \operatorname {deg}(g)-\operatorname {deg} (f+F)\stackrel {?}{\geq} \operatorname {min}(w(\frac {f}{g}),(w(\frac {F}{g}))=\operatorname {min}(\operatorname {deg}g -\operatorname {deg}f,\operatorname {deg}g -\operatorname {deg}F)$

which boils down to the obvious $\operatorname {deg}(f+F)\leq \operatorname {max}(\operatorname {deg}(f),\operatorname {deg}(F)) $

  • 0
    thanks very much for this!2012-03-16