# Show that $\mathbb{Q}_3(i)$ is an unramified extension

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 $$e_K=\text{ramification index}=1, \ f_K=\text{residue degree}=2 .$$

$$\mathbb{Q}_3(i)=\{a+bi : \ a,b \in \mathbb{Q}_3 \}$$, 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

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

• 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) – Diglett Oct 9 '18 at 22:09
• Do you already know that $v(\mathbb{Q}^\times_3) = \mathbb{Z}$ and $O_{\mathbb{Q}_3}/ m_{\mathbb{Q}_3} = \mathbb{F}_3$? – Diglett Oct 9 '18 at 22:11
• 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 ? – reuns Oct 17 '18 at 2:15

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 $$p$$.

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