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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 $(1)$.

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As you note: if $\frac{f}{g} = \frac{1+t}{t}$ and $\frac{p}{q} = \frac{1-t}{t}$, then $v(f/g)=v(p/q) = 0$, and $\frac{f}{g}+\frac{p}{q} = \frac{1+t}{t}+\frac{1-t}{t} = \frac{2}{t}$, so $v((f/g)+(p/q))=-1$, which is not greater than or equal to $0$. So something is wrong. I think the standard valuation on this ring is $v(f)-v(g)$, where $v(f)$ is the largest power of $t$ that divides $f$ (i.e., multiplicity of the root $0$); in that situation, both $(1+t)/t$ and $(1-t)/t$ would have valuation $-1$. –  Arturo Magidin Mar 16 '12 at 21:44
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So I guess the question is: are you sure you got the right definition of $v$? –  Arturo Magidin Mar 16 '12 at 21:53
    
Yes this is how $v$ was defined in the problem. But as you pointed it out it doesn't work. Thanks for saving me an awful lot of time! I'll ask my professor about it. –  Kyle Schlitt Mar 16 '12 at 21:58
    
Note: I should have mentioned that the notation for the discrete valuation is $v_{\infty}$. That may help determine if the valuation is the one you mentioned or not. This is important since it is heavily used in a later problem, which is probably also incorrect unless the mapping is changed accordingly. –  Kyle Schlitt Mar 16 '12 at 22:16
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The infinite valuation on $K(T)$ is given by $v_\infty(\frac{p}{q}) = \deg q - \deg p$. –  Brandon Carter Mar 16 '12 at 22:29

1 Answer 1

up vote 3 down vote accepted

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)) $

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thanks very much for this! –  Kyle Schlitt Mar 16 '12 at 22:33

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