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This is the problem from Schikhof's Ultrametric Calculus:

Let $D$ be an integral domain and $\|\cdot\|:D\to\mathbb{R}$ be a norm. Show that $\|\cdot\|$ may be uniquely extended to a norm on the quotient field of $D$ and if $\|\cdot\|$ satisfies the strong triangle inequality then so does the extension.

My attempt at a solution was that by analogy with $\mathbb{Z}$ and its quotient field $\mathbb{Q}$ I expect that the extension may be $\|\cdot\|^\star:\tfrac{x}{y}\mapsto\tfrac{\|x\|}{\|y\|}$ wherever $\tfrac{x}{y}$ is defined. Non-negativity is obvious, the norm is zero if and only if $x=0$ is too, and so is $\|ab\|^\star=\|a\|^\star\|b\|^\star$ for $a,b$ in the quotient field, since $\|\cdot\|^\star$ is a norm on $D$. If we take elements $\tfrac{x_1}{y_1},\tfrac{x_2}{y_2}$ in the quotient field, then $$\begin{align*} \left\|\frac{x_1}{y_1}+\frac{x_2}{y_2}\right\|^\star=\frac{\|x_1y_2+x_2y_1\|^\star}{\|y_1\|^\star\|y_2\|^\star}&\leq\tfrac{1}{\|y_1\|^\star\|y_2\|^\star}\max\left\lbrace\|x_1\|^\star\|y_2\|^\star,\|x_2\|^\star\|y_1\|^\star\right\rbrace \\&=\max\left\lbrace\left\|\frac{x_1}{y_1}\right\|^\star,\left\|\frac{x_2}{y_2}\right\|^\star\right\rbrace \end{align*}$$ so we have a good extension. But now it remains to show that it's unique, and I'm not sure how to continue. How can I show that this is the unique extension? This is a problem right at the beginning of the book, so I get the feeling there should be some simple solution, but I'm just not really seeing it. Help appreciated.

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Maybe I am missing something, but since your norm is multiplicative, can't you just argue that if $ \|\|^*$ extends $\|\|$, you have for nonzero $a\in D$ $$\|aa^{-1}\|^* = 1$$ giving $$\|a^{-1}\|^* = \frac{1}{\|a\|^*} = \frac{1}{\|a\|}$$ because the starred norm extends $\|\|$? Thus you know for all invertible $a\in D$, your norm on inverses is the reciprocal of the norm. Now multiplicativity again finishes the claim for all fractions.

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