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 $ord_3(\beta_1)$?