An elementary way to extend the p-adic valuation to algebraic numbers?

Question

The $p$-adic valuation is first defined $v_p(n) = k$ for $n \in \mathbb{Z}$, where $n = p^kq$ for $q$ not divisible by $p$. Then it is extended to $\mathbb{Q}$ by $v_p(\frac{a}{b}) = v_p(a) - v_p(b)$. It satisfies $v_p(xy) = v_p(x) + v_p(y)$ and $v_p(x + y) \ge \min(v_p(x), v_p(y))$, with equality guaranteed when $v_p(x) \neq v_p(y)$. My question is about extending $v_p$ to algebraic numbers while continuing to satisfy these properties.

---

I've figured out the only plausible way to extend $v_p$ to algebraic numbers of degree 2. Let $x = a + b\sqrt{d}$, with $d$ square-free and $a, b \in \mathbb{Q}$. It has minimal polynomial $x^2 - 2ax = -a^2+b^2d$. Thus, $$\min(2v_p(x), v_p(2a) + v_p(x)) \le v_p(a^2 - b^2d)$$ with equality if $2v_p(x) \neq v_p(2a) + v_p(x)$. That means, either $2v_p(x) = v_p(2a) + v_p(x)$, giving $v_p(x) = v_p(2a)$, or $\min(2v_p(x), v_p(2a) + v_p(x)) = v_p(a^2 - b^2d)$, which can be solved with a little work to be $$v_p(x) = \begin{cases} \frac{v_p(a^2-b^2d)}{2} & \text{ if } \frac{v_p(a^2-b^2d)}{2} \le v_p(2a) \\ v(a^2 - b^2d) - v_p(2a) & \text{ if } \frac{v_p(a^2-b^2d)}{2} \ge v_p(2a)\end{cases}$$

Since the conjugate $y = a - b\sqrt{d}$ has the same minimal polynomial, $v_p(y)$ has the same possible solutions. Together, we also have $v_p(x) + v_p(y) = v_p(xy) = v_p(a^2 - b^2d)$. Putting this together, we get:

• If $\frac{v_p(a^2-b^2d)}{2} \le v_p(2a)$, then $v_p(x) = v_p(y) = \frac{v_p(a^2-b^2d)}{2}$.
• If $\frac{v_p(a^2-b^2d)}{2} > v_p(2a)$, then $v_p(x) = v_p(2a)$ and $v_p(y) = v(a^2 - b^2d) - v_p(2a)$, or vice versa.

---

I have two questions:

1. In the case $\frac{v_p(a^2-b^2d)}{2} > v_p(2a)$, how should we choose which of the two options to pick in a consistent way? I am aware that axiom of choice will be involved. I'm looking for an algorithm or similar method such that after $v_p(x_1), \dots, v_p(x_n)$ are picked, on input $x_{n+1}$, I can tell whether I have free choice for $v_p(x_{n+1})$ or if $v_p(x_{n+1})$ is forced by $v_p(x_1), \dots, v_p(x_n)$.
2. Is there a similar argument to explicitly calculate the possible values of $v_p(x)$ for algebraic $x$ of degree $n$? I'm having trouble because the ultrametric inequality has a more complicated equality condition when there are 3 or more terms.

(I'm aware of similar questions on this site about extending $p$-adic valuations to algebraic numbers. However, my question is more about explicit computations rather than algebraic structure.)

Answer 1

I have a feeling that you might not like this answer, because you state that you're looking for things like "algorithms" or "explicit computations rather than algebraic structures" and in some way, my answer does not address this. Nonetheless, I think that this answer is the correct one because the problem that you're asking lies at the heart of algebraic number theory which is all about algebraic structures. This does not mean that this question or algebraic number theory is not amenable to explicit computations, far from it! There's a whole subfield of computational algebraic number theory, and there are many things you can calculate explicitly, either by hand or by a computer algebra system such as Magma or Sagemath. But here the computations are not an alternative to employing and understanding algebraic structures, instead they rely on it.

(I don't think your approach of only using only the defining properties of a valuation leads very far.)

With that preamble out of the way, let me start by picking an explicit example that is a useful starting point for algebraic number theory. Consider the quadratic case with $d=-1$, i.e. we consider numbers of the form $a+bi$, where $a,b \in \Bbb Q$. Actually it suffices to consider the case $a,b \in \Bbb Z$, because just like from $\Bbb Z$ to $\Bbb Q$, every valuation on $\Bbb Z[i]$ extends to $\Bbb Q(i)$.

Now the basic result which can be proved by knowing a few basic concepts from ring theory (Euclidean domains, UFDs etc.) is that if the prime $p$ satisfies $p\equiv 3 \pmod{4}$, then there's a unique extension $v$ of $v_p$ to $\Bbb Z[i]$ that is given by $v_p(a+bi)=\min(v_p(a),v_p(b))$. If $p$ satisfies $p \equiv 1 \pmod{4}$, then there are always two distinct extensions. For example for $p=5$, we have $1=v(5)=v((1+2i)(1-2i))=v(1+2i)+v(1-2i)$. This leads to the two possibilities determined either by $v(1+2i)=1$ and $v(1-2i)=0$ or $v(1-2i)=1$ and $v(1+2i)=0$. For $p=2$, something special happens. There's a unique extension, but it does not take values in $\Bbb Z$, but rather in $\frac{1}{2}\Bbb Z$. To see that this is necessary, consider that $4v(i)=v(i^4)=v(1)=0$, so $v(i)=0$. But now $(1+i)^2=2i$, so $2v(1+i)=v((1+i)^2)=v(2i)=v(2)+v(i)=1$. Thus if such a $v$ exists, we need to have $v(1+i)=\frac{1}{2}$.

One can give a complete solution along similar lines in the degree two case by employing some algebraic number theory and quadratic reciprocity.

For a general degree $n$ number, there is no complete answer, but still a lot to be said. There exist algorithms, but general things for certain algebraic numbers can also be said, although the details can be very involved. For example, a complete description of extensions of $v_p$ for all $p$ is known for $\Bbb Q(\alpha)$, where $\alpha$ is a root of $X^6-6X^4+9X^2+23$, but understanding this involves a difficult theorem of Deligne and Serre featuring scary objects like modular forms and Galois representations! (This toy example application of their result is due to Gabor Wiese.) So the rabbit hole goes quite deep, depending on the algebraic number at hand.

My advice, if you are sufficiently motivated to understand more about this question, is to study some algebraic number theory. The book by Frazer Jarvis is particularly light on algebraic prerequisites and even features some algorithmic applications of the material (to integer factoring).