"p-adic absolute value" in polynomial ring

Question

I'm working with Gouvea's book on p-adic numbers. In problem 34 I'm asked to give a "p-adic" (I put it in quotes as in my understanding its just a p-adic-like) valuation and absolute value for an irreducible polynomial $p(t)\in F[t]$.

Valuation is clear, for $q(t)\in F[t]$ it's the maximal $k$ s.t. $q(t)=p(t)^kq'(t)$ where $q(t)$ is the remainder polynomial in $F[t]$.

I know that for $c>1$ I get a non-archimedean absolute value by $|\cdot|=c^{-v(\cdot)}$ where $v(\cdot)$ is a valuation.

Now I could just use my valuation from above and fix some basis greater $1$, e.g. $e$, and get an absolute value but I'm not sure whether that as really p-adic-like. I've read that the choice of $p$ as basis for the valuation $v_p(\cdot)$ is sensible (why is this so?) and hence I would guess that in this case a smarter choice of the basis would be sensible as well.

If $F=\mathbb{C}$, then $p(t)$ would be of the form $(t-\alpha)$ for some $\alpha\in\mathbb{C}^*$ so in this case I would choose $|\alpha|$ as basis but this strategy is of course not applicable for other fields.

Answer 1

To answer your main question, so far I've only seen (exponential) valuations used on algebraic function fields. To continue your example for $F = \Bbb{C}$, if $F \subseteq \Bbb{C}$ you may use $|\alpha|$ where $\alpha$ is any root of $p(t)$ (which is non-zero because $p$ is irreducible).

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For your other question, recall the following

Fact: Two absolute values $|\cdot|_1$ and $|\cdot|_2$ on a field $F$ are equivalent if and only if there is a real number $s > 0$ such that $$ |x|_1 = |x|_2^s $$ for every $x \in F$.

Proof: See Neukirch's Algebraic Number Theory, proposition 3.3, chapter II.

Now, you probably defined the valuation $v_p(x)$ for $x \in \Bbb{Q}$ and $p \in \Bbb{Z}$ prime as the unique integer such that $$ x = \frac{f}{g} \, p^{v_p(x)} \quad \text{with} \quad \gcd(fg,p) = 1 $$ By the above fact we could use any real number $c > 1$ to define $|x|_p = c^{-v_p(x)}$, but $c = p$ gives us the ever so useful product formula for free:

For every non-zero rational number $x$ $$ \prod_{p} |x|_p = 1 $$ where ranges in $\{p \in \Bbb{Z}: p \text{ is prime}\} \cup \{\infty\}$ and $|x|_{\infty}$ is the Archimedean absolute value on $\Bbb{Q}$.

Finally, note that the choice of $c$ we make on $\Bbb{Q}$ immediately extends to every finite algebraic extension $L$ of $\Bbb{Q}_p$ of degree $n$, because in this case $|\cdot|_p$ extends uniquely to $L$ as $$ |\alpha|_p = \sqrt[n]{|N_{L|\Bbb{Q}_p}(\alpha)|_p} $$