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The classical definition of a field $K$ with an absolute value $|\cdot|:K\to \mathbb{R}_{\geq 0}$ is that $\forall x,y\in K$

  1. $x=0\Leftrightarrow|x|=0$
  2. $|xy|=|x|\cdot|y|$
  3. $|x+y|\leq |x|+|y|$

If the last one can be replaced by a stronger condition

  1. $|x+y|\leq \max\{|x|,|y|\}$

then we say it is a non-Archimedean absolute value, otherwise it is Archimedean.

But when we generalize this concept, we have only the non-Archimedean requirement:

Here $\Gamma$ is a totally ordered abelian group with group operation written as multiplication. enter image description here

My problem is, what is the motivation of abandaoning the Archimedean version? Did that path exist in the history? Is the structure theorem of that generalization too trivial to be written?

Update: I found a literature that mentioned this set up: normed ring/field, in A.G. Kurosh, "Lectures in general algebra" , Chelsea (1963) (Translated from Russian), and likewise we have Archimedean/non-Archimedean normed ring/field. But there is no any deep result with this definition. enter image description here

Z Wu
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    $\mathbb{R}_{\geq 0}$ has both addition and multiplication defined. The ordered group only has multiplication and a linear order. What would you use for the $+$ in the third property? – Chad K May 26 '23 at 17:38
  • Not sure if this is what you're asking, but property 4 implies 3, so you're not "abandoning" the Archimedan property 3. – Ted May 26 '23 at 18:12
  • @John We could use a totally ordered commutative ring. – Z Wu May 26 '23 at 18:45
  • @Ted Archimedean would be (3) holds while (4) does not hold, not just having (3) hold. – Z Wu May 26 '23 at 18:48
  • @ZWu What totally ordered commutative rings would you suggest? – Torsten Schoeneberg May 26 '23 at 23:48
  • @TorstenSchoeneberg e.g. any ordered field with $|g|=\max{g,-g}$ would satisfy the triangle inequality and not the ultrametric one. But for the purpose of defining the concept, any totally ordered commutative ring could work. – Z Wu May 27 '23 at 04:09
  • As for history: Van der Waerden in his book Algebra II introduces valuations of fields with values in an arbitrary ordered field. He then quickly narrows the treatment down to the case of an archimedian ordered field, mentioning that such a field can be embedded into the reals preserving the order relation. – Hagen Knaf Jul 31 '23 at 08:10

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