# Extension of p-adic valuation to number field

I'm having some trouble understand the following thing.

Let $$p$$ : prime number and $$ord_p:\mathbb{C}_p\rightarrow \mathbb{Q}$$ be p-adic order, normalized so that $$ord_p(p)=1$$

Now let $$\beta_1,\beta_2$$ be the roots of the equation $$X^2-252X+3^{11}=0$$ ordered so that $$ord_3(\beta_1)\leq ord_3(\beta_2)$$

The author says that $$ord_3(\beta_1)=2$$. However I don't know how to get this value.

Since I have not enough knowledge for p-adic number theory, I will explain the concepts related to the question even though it may be basic stuffs. If there is something wrong, then please tell me. I will appreciate it :)

Given a prime $$p$$, the $$p$$-adic valuation(p-adic order) on $$\mathbb{Q}$$ is the map $$\nu_p:\mathbb{Q}^*\to\mathbb{Z}$$ given by $$\nu_p(p^ka/b)=k$$, where $$a,b$$ are prime to $$p$$.

So we have corresponding absolute value on $$\mathbb{Q}$$, namely $$|\cdot|_p$$

This absolute value gives a metric on $$\mathbb{Q}$$ and we have a completion $$\mathbb{Q}_p$$ w.r.s.t this metric.

We can extend $$|\cdot|_p$$ to $$\mathbb{Q}_p$$ and we can check that corresponding valuation $$\nu_p$$ on $$\mathbb{Q}_p$$ is extension of $$\nu_p:\mathbb{Q}^*\to\mathbb{Z}$$

Now we consider its algebraic closure $$\overline{\mathbb{Q}}_p$$. we have unique extension of $$\nu_p$$ in $$\overline{\mathbb{Q}}_p$$(For $$x\in\overline{\mathbb{Q}}_p, x\in L$$ for some finite extension $$L$$ of $$\mathbb{Q}_p$$. Then $$\nu_p(x):=\frac{1}{n}\nu_p(N_{L/\mathbb{Q}_p}(x))$$

Finally complete $$\overline{\mathbb{Q}}_p$$ w.r.s.t $$\nu_p$$, denoted by $$\mathbb{C}_p$$

We get $$ord_p:\mathbb{C}_p\rightarrow \mathbb{Q}$$(I think it may be $$\mathbb{Z}$$??)

It is known to be $$\mathbb{C}\cong \mathbb{C}_p$$. So we have an extension of p-adic valuation to number field.

Is it right process?

Now, come back to example, how to compute $$ord_3(\beta_1)$$?

Firstly, a few comments. $$\text{ord}_p:\mathbb{C}_p\to \mathbb{Q}$$ definitely cannot be restricted to codomain $$\mathbb{Z}$$ - note how you extend it to $$\overline{\mathbb{Q}_p}$$.

Also, there is a much easier way of extending the $$p$$-adic valuation to a number field, but it involves picking some prime ideal above $$p$$ in the number field. However, since you only care about a quadratic equation, the choice of prime in your quadratic extension doesn't matter much.

The way to talk about the $$3$$-valuation of $$x$$ is to use the basic properties of the valuation: $$\text{ord}_p(ab) = \text{ord}_p(a)+\text{ord}_p(b)$$ and $$\text{ord}_p(a+b)=\min\lbrace \text{ord}_p(a),\text{ord}_p(b)\rbrace$$ unless $$\text{ord}_p(a)=\text{ord}_p(b)$$. Since

$$\text{ord}_3(x^2-252x+3^{11}) = \infty ,$$

as $$\text{ord}_3(0)=\infty$$ by convention, we see that we must have $$\text{ord}_3(x^2-252x) = \text{ord}_3(3^{11}) = 11$$. Since $$\text{ord}_3(252) = 2$$, the only way to make this work is by picking $$\text{ord}_3(x)\geq 9$$, so that the minimum of the orders of $$x^2$$ and $$252x$$ is $$11$$, or $$\text{ord}_3(x)=2$$, so the orders agree. Can you then show that $$x$$ cannot have order $$11$$?

• Thank you very much!
– KS M
Oct 19, 2022 at 0:57

$$\newcommand{\ord}{\operatorname{ord}}$$You can also argue as follows: Note that $$X^2 -252X+3^{11} = (X-\beta_1)(X-\beta_2)$$. By expanding we find that $$252 = \beta_1 + \beta_2$$ and $$3^{11} = \beta_1\beta_2$$. Now, $$\ord_3(\beta_1) + \ord_3(\beta_2) = \ord_3(\beta_1\beta_2) = \ord_3(3^{11}) = 11$$ and $$2 = \ord_3(252) = \ord_3(\beta_1+\beta_2) \ge \min\{\ord_3(\beta_1), \ord_3(\beta_2)\} = \ord_3(\beta_1).$$ If we had $$\ord_3(\beta_1) = \ord_3(\beta_2)$$, we would get the contradiction $$2\ge \frac{11}{2}$$. Therefore, we must have $$\ord_3(\beta_1) < \ord_3(\beta_2)$$ in which case the inequality in the displayed formula is actually an equality, that is, $$\ord_3(\beta_1) = 2$$.

I like the other two answers. Note that both their methods generalize to the following result for arbitrary primes $$p$$ and monic polynomials $$x^2+bx+c$$:

Case I: If $$v_p(b) \ge \frac12 v_p(c)$$, then both solutions $$x_1, x_2$$ of $$x^2+bx+c=0$$ have the same value

$$\lvert x_i \rvert_p = \lvert c \lvert_p^{1/2} \Leftrightarrow v_p(x_i) = \frac12 v_p(c).$$ Case II: If $$v_p(b) < \frac12 v_p(c)$$, then one of the solutions has the same value as $$b$$, and the other the same as $$c/b$$: $$\lbrace \lvert x_1 \rvert_p, \lvert x_2 \rvert_p \rbrace = \lbrace \lvert b \rvert_p, \lvert \frac{c}{b} \rvert_p \rbrace \Leftrightarrow \{ v_p(x_1), v_p(x_2) \} = \{v_p(b), v_p(c)-v_p(b)\}.$$

as I wrote in this answer to a related question.

For fun, another way to show the special case you have is: Via the quadratic formula, the two solutions to your polynomial are

$$x_{1,2}= 3^2 \cdot (14 \pm \sqrt{-1991})$$

and since $$-1991 \equiv 1$$ modulo 3, one of its square roots (which exist in $$\mathbb Z_3$$) is $$\equiv 1$$ and the other $$\equiv 2$$ modulo $$3$$. Let's say by $$\sqrt{-1991}$$ in the above formula we mean the one which is $$\equiv 1$$. Then $$14 - \sqrt{-1991} \equiv 1$$ modulo $$3$$ i.e. $$v_p(14 + \sqrt{-1991}) = 0$$, but $$14 + \sqrt{-1991} \equiv 0$$ modulo $$3$$ i.e. $$v_p(14 + \sqrt{-1991}) \ge 1$$. So

$$v_p(3^2 \cdot (14 - \sqrt{-1991})) = 2+0 =2$$ but $$v_p(3^2 \cdot (14 + \sqrt{-1991})) \ge 2+1=3$$ (actually, it is $$=9$$ according to the finer statements above; indeed, $$14^2 = 196 \equiv -1991$$ mod $$3^7$$).