Consider the p-adic field $ \mathbb{Q}_3$ and its finite extension $ \ K=\mathbb{Q}_3(i)$, where $ \ i$ are roots of $ x^2+1$ .
Show that $ \ \mathbb{Q}_3(i)$ is unramified. In fact $ e_K=\text{ramification index}=1, \ f_K=\text{residue degree}=2 .$
Answer:
$ \mathbb{Q}_3(i)=\{a+bi : \ a,b \in \mathbb{Q}_3 \}$, where $ \ i $ is a root of $x^2+1$. The polynomial $ x^2+1$ does not have roots modulo $3$, so it is irreducible in $ \mathbb{Q}_3[x]$.
Now for $ a,b \in \mathbb{Q}_3$, we have
$$|a+bi|_3=|a^2+b^2|_3^{\frac{1}{\large \left[\mathbb{Q}_3(i): \mathbb{Q}_3\right]}}=|a^2+b^2|_3^{1/2}=\max (|a|_3,|b|_3), ……(1)$$
How can we get the following result from the result $(1)$?
$ O_K=\{a+bi: \ a,b \in \mathbb{Z}_3, \ \ |a|_3 \leq 1, \ |b|_3 \leq 1 \} \\ m_K=\{a+bi: \ a,b \in 3 \mathbb{Z}_3 \ , \ |a|_3 < 1, \ |b|_3 < 1 \}=3O_K, \\ O_K/m_K=\{a+bi: \ a,b \in \mathbb{Z}_3/3 \mathbb{Z}_3=\mathbb{F}_3 \}=\mathbb{F}_3(i). $
Here $O_K=\text{ring of integers} , \ m_K=\text{maximal ideal} \ $
I am confused right here how we get these from $(1)$.
Please someone check my work and explain also.
Now,
$[O_K/m_K: \mathbb{Z}_3/3 \mathbb{Z}_3]=[O_K/3O_K: \mathbb{Z}_3/3 \mathbb{Z}_3]=f_K=? $
and
$[v(K^{\times}): v(\mathbb{Q}_3^{\times})]=e_K=?$
Best Answer
You are searching too far. Consider the extension of residual fields, which is $\mathbf F_3(i)/\mathbf F_3$. As $i^2=-1$, the multiplicative order of $i$ is $4$, so $i\notin \mathbf F^*_3$ and $f=[\mathbf F_3(i):\mathbf F_3]=2$.