Proof of valuation property (function on $K(X_1,\,\dots,\, X_n)$)

Question

Let $K$ be a field and consider the field of rational multivariate polynomials $K(X_1,\,\dots,\, X_n)$ in $K$ for some $n \geq 1$.

Define $\mathbb{P}:= (0,\, \infty)$ as the set of positive reals and a map $$ v_0:K(X_1,\,\dots,\, X_n) \longrightarrow \{0\}^n \cup \mathbb{P}^n $$ as follows:

If $f=0$, then $v_0\left(\frac{f}{g}\right):=0$.

Otherwise, in $K[X_1,\,\dots,\, X_n] \hookrightarrow K(X_1,\,\dots,\, X_n)$, $$ f = X_i^{\text{ord}_i(f)}{f_i}'\,,\,\,X_i \nmid f_i,\quad i = 1,\,\dots,n $$ for suitable ${f_i}'$, where $\text{ord}_i$ is the lowest index of a $X_i$-monomial with nonzero coefficient of $f \neq 0$ when viewed as a polynomial in $X_i$ (e.g., $f$ can be uniquely written as $f = \sum_{i=\text{ord}_i(f)}^{\text{deg}_i(f)} \tilde f_i X^i$ for suitable $\tilde f_i \in K[X_1,\,\dots,\, X_{i-1},\,X_{i+1},\,\dots,\,X_n ]$).
so that we can put $$ v_0\left(\frac{f}{g}\right):= \left(2^{\text{ord}_i(g) - \text{ord}_i(f)}\right)_{i=1,\,\dots,\,n} $$

I want to show that this is a valuation on $K(X_1,\,\dots,\, X_n)$, so what needs to be shown is, that for $\frac{f}{g},\,\frac{\phi}{\gamma} \in K(X_1,\,\dots,\, X_n)$,

1. $v_0\left(\frac{f}{g}\right) = 0 \iff \frac{f}{g}=0$
2. $v_0\left(\frac{f}{g} \frac{\phi}{\gamma}\right) = v_0\left(\frac{f}{g}\right)v_0\left(\frac{\phi}{\gamma}\right)$
3. $v_0(\frac{f}{g} + \frac{\phi}{\gamma}) \leq \text{max}\left(v_0\left(\frac{f}{g}\right),\,v_0\left(\frac{\phi}{\gamma}\right)\right)$

Concerning 3.: We are considering $\mathbb{P}^n \cup \{0\}^n$ with the lexicographic order, so $$ \left(x_i\right)_{i=1,\,\dots,\,n} \lt \left(y_i\right)_{i=1,\,\dots,\,n} \iff \exists \, 1\leq \mu\leq n \left( \, \forall 1\leq j \lt \mu \left( \, x_j = y_j\right) \land x_\mu \lt y_\mu\right) $$

Well-definedness, 1. and 2. were easy, but I'm struggling with 3.. Admittedly, partially, because I at first used a wrong definition. And now I'm pretty exhausted and feel like I need to move on.

Please note, this isn't a homework. It's part of an exercise from Grillet's Abstract Algebra which I'm using to work my way towards Hensel's Lemma, which I need to use in my Bachelor's thesis...
The complete formal definition here is a part of that exercise I've already done, and I think it is correct. For reference, see the named book, 'Second Edition', p. 255, exercise 3 (exercises on the chapter for valuations).

Answer 1

What you've defined is usually called an absolute value $|x|$, whereas a valuation $v$ is a kind of logarithm of an absolute value. I'll be using the conventional notation instead.

Here are some properties you can verify individually:

• If $\pi$ is a prime in a UFD $D$ and $b>1$ then $|x|_{\pi}=b^{-\mathrm{ord}_{\large\pi}(x)}$ is an abs. val. on $\mathrm{Frac}(D)$.
• In fact, it is non-archimedean: it satisfies $|x+y|\le\max\{|x|,|y|\}$.
• Which also means that $|x|\le|y| \implies |x+y|\le|y|$ with equality if $|x|<|y|$. $~\color{Red}{(\star)}$

(See the Wikipedia proof for the last bullet point. Also note that $|x|$ is a non-archimedean absolute value if and only if $v(x)=-\log_b |x|$ is a valuation for a, or any, base $b>1$.)

If $G$ and $H$ are linearly ordered abelian groups, we can define the lexicographic ordering on the direct product $G\times H$ by the rule

$$ (x,y)\le (x',y') \quad \iff \quad x<x' ~\textrm{or}~(x=x' ~\textrm{and}~ y\le y') $$

Proposition. Suppose $|x|_1$ and $|x|_2$ are two non-archimedean absolute values on $F$ which take values in linearly ordered multiplicative groups $G$ and $H$. Then $(|x|_1,|x|_2)$ is an absolute value taking values in $G\times H$.

I assume you can verify properties (1) and (2), so it remains to see property (3):

$$ (|x+y|_1,|x+y|_2)\le\max\{(|x|_1,|x|_2),(|y|_1,|y|_2)\}. $$

Assume wlog that $(|x|_1,|x|_2)\le(|y|_1,|y|_2)$. That is, we have

$$ \color{Green}{|x|_1<|y|_1} ~\textrm{or}~(\color{Blue}{|x|_1=|y|_1 ~\textrm{and}~|x|_2\le|y|_2}) $$

and what we want to derive from this is

$$ |x+y|_1<|y|_1 ~\textrm{or}~(|x+y|_1=|y|_1 ~\textrm{and}~|x+y|_2\le |y|_2). $$

The proof makes use of $\color{Red}{(\star)}$.

Case I ($\color{Green}{|x|_1<|y|_1}$). This immediately implies $|x+y|_1<|y|_1$.

Case II ($\color{Blue}{|x|_1=|y|_1 ~\textrm{and}~|x|_2\le|y|_2}$). The fact $|x|_1\le |y|_1$ implies $|x+y|_1\le|y|_1$, so either we have $|x+y|_1<|y|_1$ and we're done, or else $|x+y|_1=|y|_1$ and we additionally want to derive the fact $|x+y|_2\le |y|_2$, but this follows from $|x|_2\le|y|_2$ and we're done.

So your problem reduces to verifying $(3)$ in the case of $K(X)$ (or really, $K[X]$ by clearing denominators as the user suggests in the comments). Can you do that?