Totally Ramification and Quadratic extension of p-adic field $ \mathbb{Q}_p$ $ \text{Totally Ramification}:$ 
Consider the quadratic extension $K=\mathbb{Q}_p(\sqrt 3)$ of the p-adic field $\mathbb{Q}_p$.
 Here,
$$  K=\mathbb{Q}_p(\sqrt 3)=\{a+b \sqrt 3: \ a,b \in \mathbb{Q}_p \}, $$ 
where $ \sqrt 3$ is root of $x^2-3$.
Consider $ O_K=\{ \alpha \in K: |\alpha|_p \leq 1 \}$ , $ \ m_K=\{ \alpha \in K: |\alpha|_p < 1 \}$ and $ \ O_K/m_K \ $, which is the residue field of $K$.
Assume that $ \ p=3 \ $ i.e., we are considering the 3-adic field $ \mathbb{Q}_3$ and $ \ K=\mathbb{Q}_3(\sqrt 3)$. 
We know that the value set of $|.|_p$ on $ \mathbb{Q}_p$ is the $ \{0 \} \cup \{p^m: \ m \in \mathbb{Z} \}$.
Now as $ \ |\sqrt 3|_3=\frac{1}{\sqrt 3}$, we conclude that $ \ |\sqrt 3|_3 \notin  \{0 \} \cup \{3^m : m \in \mathbb{Z} \}$, the value set of $ |.|_3$ on  $ \mathbb{Q}_3$.
Since it is a quadratic extension, we have $ \ n=[\mathbb{Q}_3(\sqrt 3): \mathbb{Q}_3]=2$.
Thus in genegral for any $ \alpha \in K=\mathbb{Q}_3(\sqrt 3)$, it has the form $ \ \alpha=a+b \sqrt 3 $, where $ \ a,b \in \mathbb{Q}_3$.
Now by the formula of extension of the p-norm $ \ |.|_3$ for the finite field extension from $ \mathbb{Q}_p$ to $ K=\mathbb{Q}_3$, we have 
$$ |a+b \sqrt 3|_3=\left(\left|\mathcal{N}_{\mathbb{Q}_3}^{K}(a+b \sqrt 3) \right|_3 \right)^{\frac{1}{n}}=\left(\left|\mathcal{N}_{\mathbb{Q}_3}^{K}(a+b \sqrt 3) \right|_3 \right)^{\frac{1}{2}} $$
$$ i.e., \ \  |a+b \sqrt 3|_3=\left|a^2-3b^2 \right|_3^{\frac{1}{2}} $$
$$ i.e., \ \  |a+b \sqrt 3|_3=\max  \left(|a^2|_3^{\frac{1}{2}}, \ |3b^2|_3^{\frac{1}{2}} \right) $$
$$ i.e., \ \  |a+b \sqrt 3|_3=\max  \left(|a|_3, \ \frac{1}{\sqrt 3}|b|_3 \right) $$ 
Therefore,
$ O_K=\{a+b \sqrt 3: \ a,b \in \mathbb{Z}_3 \}, \\ m_K=\sqrt 3 O_K=\{a+b \sqrt 3: \ a \in 3 \mathbb{Z}_3, \ b \in \mathbb{Z}_3 \}, \\O_K/m_K \simeq \mathbb{Z}_3/3 \mathbb{Z}_3 \simeq \mathbb{F}_3. $
Now there can be two options:
$ (i)$ $ \ \text{ $e_K$=ramification index=2 and  $f_K$=resdue degree=1}$
or,
$ (ii)$ $ \ \text{ $e_K$=ramification index=1 and  $f_K$=resdue degree=2}.$
In order to $K$ to be totally Ramified we must have 
$e_K=n=2, \ \ f_K=1$.
$ \text{So  the first condition should be true for totally Ramified. But I need help to understand this}$. $ \text{Also please explain me whether this holds for any $p \neq 3$ also.}$
Thanking you,
 A: You have $e=2$ and $f=1$. The set of nonzero valuations in $\Bbb Q_3$
is the multiplicative group $\{3^n:n\in\Bbb Z\}=G_1$.
The set of nonzero valuations in $K$
is the multiplicative group $\{3^{m/2}:m\in\Bbb Z\}=G_2$.
Then $$e=|G_2:G_1|=2.$$
The valuation ring of $K$ is $\mathcal{O}_K=\{a+b\sqrt3,a,b\in\Bbb Q_3,|a|_3,|b|_3\le1\}=\Bbb Z_3\oplus\sqrt3\Bbb Z_3$. Its maximal ideal is
$\sqrt 3\mathcal{O}_K=3\Bbb Z_3\oplus\sqrt3\Bbb Z_3$. Then
$$\frac{\mathcal{O}_K}{\sqrt3\mathcal{O}_K}=\frac{\Bbb Z_3\oplus\sqrt3\Bbb Z_3}{3\Bbb Z_3\oplus\sqrt3\Bbb Z_3}\cong\frac{\Bbb Z_3}{3\Bbb Z_3}.$$
Then
$$f=|\mathcal{O}_K/\sqrt3\mathcal{O}_K:\Bbb Z_3/3\Bbb Z_3|=1.$$
A: Everything follows straightforwardly from the definition of $K=\mathbf Q_3 (\sqrt 3)$. If $\pi:=\sqrt 3$ were in $\mathbf Q_3$, one would have $2 v_3(\pi)=v_3(3)=1$, impossible. So $K/\mathbf Q_3$ is quadratic, with ramification index $e=1$ or $2$ necessarily. If $e=1$ (non ramification), $3$ is a uniformizer, and again $2 v_K(\pi)=v_K(3)=1$, impossible. So $e=2$ (total ramification), $\pi$ is a uniformizer and $v_K(3)=2$.
