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Given that $v$ is a non-archimedean valuation on $K$, we can extend it to $|\cdot|:K[x]\to\mathbb{R}$ by $|a_0 + a_1x + \cdots +a_nx^n|=\max\left\{|a_1|,\ldots,|a_n|\right\}$. My question is how can we extend $|\cdot|$ to a valuations on $K(x)$?

My guess is that we can extend it to $\|\cdot\|$ given by $$ \left\| \frac{a_0 + a_1x + \cdots +a_nx^n}{b_0 + b_1x + \cdots +b_mx^m} \right\| = \max\left\{|a_1|,\ldots,|a_n|\right\}-\max\left\{|b_1|,\ldots,|b_m|\right\}$$ but I'm not sure if this is true, nor do I know how to show that $\|\cdot\|$ extends $|\cdot|$. Any help would be appreciated.

user26857
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Galois
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1 Answers1

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In general, suppose you have a valuation $v$ on an integral domain $A$. Then you can define a valuation $v'$ on the fraction field $K$ of $A$ by setting $v'(a/b)=v(a)-v(b)$. It's clear that $v'$ is well defined because $v(xy)=v(x)+v(y)$ for all $x,y\in A$. Moreover, $v'$ extends $v$ since $v(1)=0$. Now you just have to check that $v'$ satisfies the axioms for a valuation. The facts that $v'(x)=\infty$ iff $x=0$ and that $v'(xy)=v'(x)+v'(y)$ for $x,y\in K$ are clear. It remains to prove that $v'(a/b+c/d)\geq \min\{v'(a/b),v'(c/d)\}$. But $v'(a/b+c/d)=v(ad+bc)-v(b)-v(d)\geq \min\{v(a)+v(d),v(b)+v(c)\}-v(b)-v(d)$. Now it's very easy to see that $v'(a/b)\leq v'(c/d)$ iff $v(a)+v(d)\leq v(b)+v(c)$, which proves the claim.

Ferra
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