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 $ord_3(\beta_1)$?
Best Answer
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 $11$?