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 $ord_3(\beta_1)$?


3 Answers 3


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 $11$?

  • $\begingroup$ Thank you very much! $\endgroup$
    – KS M
    Commented 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 $3^7$).


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