Let's consider the definition of (algebraic) absolute value given by Wikipedia (https://en.wikipedia.org/wiki/Absolute_value_%28algebra%29), and focus the attention under the voice "Fields and integral domains".
There is written that: "If $D$ is an integral domain with absolute value $|x|$, then we may extend the definition of the absolute value to the field of fractions of $D$ by setting $|x/y| = |x|/|y|$. My question is: does this extension work for any non-trivial localization, or only on the field of fractions?
Just to be more precise, the question is:
Let $A$ be an integral domain with absolute value $|\cdot|$, and consider a multiplicative set $S \subseteq A$ such that $0 \notin S$. Is it always possible to extend the absolute value function from $A$ to $S^{-1}A$ defining $|a/s| = |a|/|s|$?
I tried to prove it by direct computation, and it seems to be definitely true, although it sound a bit "strange" considering some statement I wrote quickly on my exercise book during a lecture. I should be particularly grateful if someone could clarify this point, although it surely is something not so particularly interesting or challenging.
Thank you in advance for any suggestions. Cheers
That's actually fairly easy to prove. To set the notation, let's define $$ \left| \frac{a}{s} \right|_S := \frac{|a|}{|s|} $$ for any $a/s \in S^{-1}A$. Note that this is well defined, because $|s| \neq 0$ since by hypothesis $|\cdot|$ is an absolute value on $A$ and $0 \notin S$. Furthermore, it is indeed an extension of $|\cdot|$ because $$ \left| \frac{a}{1} \right|_S := |a| $$ since $|1| = |1|^n$ for every $n \in \Bbb{Z}_{>0}$ implies that $|1| = 1$.
Now to the absolute value properties:
$|a/s|_S \geq 0$ if and only if $|a| \geq 0$ and $|s| \geq 0$.
$|a/s|_S = 0 \iff |a| = 0 \iff a = 0$.
$ \left| \dfrac{a}{s} \dfrac{b}{r} \right|_S = \dfrac{|ab|}{|sr|} = \dfrac{|a|}{|s|} \dfrac{|b|}{|r|} $
$ \left|\dfrac{a}{s} + \dfrac{b}{r} \right|_S = \left|\dfrac{ra + sb}{sr} \right|_S = \dfrac{|ra + sb|}{|sr|} \leq \dfrac{|ra| + |sb|}{|sr|} = \dfrac{|a|}{|s|} + \dfrac{|b|}{|r|} $
Note: If $A$ is a domain, then for any multiplicatively closed set $S$ without $0$ there is a natural injection of $S^{-1}A$ into the field of fractions $K$ of $A$, and $|\cdot|_S$ is nothing more than the restriction of $|\cdot|$ (on $K$) to $S^{-1}A$.