# “p-adic absolute value” in polynomial ring

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.

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).

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 $$|\alpha|_p = \sqrt[n]{|N_{L|\Bbb{Q}_p}(\alpha)|_p}$$

• Thanks for that, especially the answer on my second question was very useful. I fixed that $\alpha$ should be non-zero in my question. – Sebastian Bechtel Apr 27 '15 at 11:31
• @SebastianBechtel Depending on your knowledge of number theory you may wish to read through the first few sections of chapter 2 of Neukirch's book, where you can find a very nice (and concise) treatment of valuations from the algebraic point of view. – A.P. Apr 27 '15 at 18:05
• Unfortunately I have very few knowledge in number theory. However I have joined a seminar on quadratic forms (and my knowledge in algebra is sparse as well ;)). It's a great opportunity to explore stuff I would like to know better but which my studying doesn't allow. That's why I sadly don't have the time yet to dig that deep into details :-( – Sebastian Bechtel Apr 27 '15 at 20:50