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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

$[v(K^{\times}): v(\mathbb{Q}_3^{\times})]=e_K=?$

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    $\begingroup$ I think we have $|a^2-b^2|$ instead of $|a^2+b^2|$, right? (not sure if that's going to be crucial though) $\endgroup$ – Diglett Oct 9 '18 at 22:09
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    $\begingroup$ Do you already know that $v(\mathbb{Q}^\times_3) = \mathbb{Z}$ and $O_{\mathbb{Q}_3}/ m_{\mathbb{Q}_3} = \mathbb{F}_3$? $\endgroup$ – Diglett Oct 9 '18 at 22:11
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    $\begingroup$ The ramification index $e$ of $K/F$ is such that $|\varpi_K|_K^e = |\varpi_{F}|_K$, the degree of extension of the residue fields is $f = [O_K/(\varpi_K):O_{F}/(\varpi_F)]$, and when $K/F$ is Galois then $ef =[K:F]$, right ? $\endgroup$ – reuns Oct 17 '18 at 2:15
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The last equality in (1) deserves more justification, but I do think it's true, because $a$ and $bi$ are linearly independent over $\mathbb{Q}_3$, so there can be no cancellation between $a$ and $bi$ to realize a strict ultrametric inequality (UI).

More explicitly, since $i^2=-1$ has valuation $0$, $i$ must have valuation $0$ too, so $|a+bi| \leq \max \{ |a|, |bi| = |b| \}$ immediately from UI. All I am saying previously is that this inequality is in fact an equality.

If you are already convinced that $\left| a+bi \right| = \max \{ |a|, |b| \}$, then $\mathcal{O}_K = \{ x \in K : |x| \leq 1\}$ by definition (from the theory of complete discrete valuation fields). Since $|a+bi| \leq 1$ iff $\max \{ |a|, |b| \} \leq 1$ iff $|a| \leq 1$ and $|b| \leq 1$, this realizes your description of $\mathcal{O}_K$. (In fact when you write $a,b \in \mathbb{Z}_3$, you already have $|a|,|b| \leq 1$, they are the same thing). Another description of $\mathcal{O}_K$ that might be more hands on is $\mathbb{Z}_3[T]/(T^2+1)$.

Similarly $\mathfrak{m}_K = \{ x \in K : |x| <1\}$, and you can check (similar argument as above) this is exactly what you have written down: it's $3 \mathcal{O}_K$.

Now by definition of unramifiedness, $K/\mathbb{Q}_3$ is unramified iff $f= [ K:\mathbb{Q}_3]=2$ (and iff $e=1$). To see this, observe that $$ \mathcal{O}_K/\mathfrak{m}_K = \mathbb{Z}_3[T]/(3,T^2+1) = \mathbb{F}_3[T]/(T^2+1)$$ which is a degree $2$ extension of $\mathbb{F}_3$ since $T^2+1$ admits no roots in $\mathbb{F}_3$. This proves the unramified claim.

Some additional remarks: it seems you are confused as to why $f=2$ is equivalent to $e=1$: this is simply because for an extension of local fields, $ef$ is the same as the degree of extension. There is a more generalized version of this you can prove for global fields.

Finally (and most importantly), I personally like to think about unramified extensions of complete discrete valuation fields in the following way: finite unramified extensions correspond to finite extensions on residue fields. Since for finite fields, and a fixed $n$, there is a unique extension of degree $n$, this says that for each $K/\mathbb{Q}_p$ finite, there is exactly one $L$ with $L/K$ unramified of degree $n$, and you can describe $L$ in the following explicit way:

I'll use your question as an example. You find extensions of $\mathbb{F}_3$ by adjoining prime-to-$3$-th roots of unity (in this case $i$), and this immediately means all unramified extensions of $\mathbb{Q}_3$ are obtained by adjoining prime-to-$3$-th roots of unity. So in particular $\mathbb{Q}_3(i)$ is unramified over $\mathbb{Q}_3$, and is the unique such of degree $2$. This works when you replace $3$ by your favourite prime $p$.

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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 $f=[\mathbf F_3(i):\mathbf F_3]=2$.

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