Archimedean absolute value on the set of real rational fractions, whose induced absolute value in $\mathbb{R}$ is the trivial one.
Can this be done?
So, my idea is to see it in $\mathbb{C}(t)$ so I can have my "irreducibles" easy to classify (i.e ($x-a$)).
So the absolute value must map:
$0 \to 0$
$k \neq 0, \,\, k \to 1$ (This is to make the absolute value trivial when restricted to $\mathbb{C}$)
$(x-a)\to x_a \in \mathbb{R_+}$.
$\frac{1}{(x-a)}\to \frac{1}{x_a}$ (same $x_a$ as above)
Now, we can easily force the property $\mid \, p\cdot q\mid = \mid p \mid \cdot \mid q \, \mid $ when defining the absolute value on the remaining elements of $\mathbb{C}(t)$ in a trivial manner.
So, your absolute value should map:
$ k\cdot \frac{(x-a_1)...(x-a_n)}{(x-b_1)...(x-b_m)} \to \frac{x_{a_1}....x_{a_n}}{x_{b_1}...x_{b_n}}$
And its easy to check that its well defined.
The problem arises when securing the triangle inequality (and also making the absolute value archimidean)
Because there is no way (at least to my knowledge) of "factorizing" a sum of rational functions. So there is no way to predict where will $\mid \frac{p_1}{q_1} + \frac{p_2}{q_2} \mid$ will land.
The other way around this is forcing the triangle inequality in the definition and then securing the multiplicativeness, but I see it as a dead-end also.
The hard part is making it so that it is the trivial absolute value when restricted to $\mathbb{C}$, because I have already come up with an example of an archimidean absolute value in $\mathbb{R(t)}$
My gut tells me this is impossible, but if anyone has any hints or ideas as to find an absolute value or to prove that it is impossible I would be very thankful.