# Ramification degree of field extension of the 5-adic field obtained by adjoining a primitive third root of unity

Let $$K = \mathbb{Q}_5$$ and $$K' = \mathbb{Q}_5(\xi_3)$$ where $$\xi_3$$ is a primitive third root of unity, i.e. $$\xi_3, \xi_3^2 \neq 1$$ but $$\xi_3^3=1$$. The minimal polynomial of $$\xi_3$$ over $$K$$ is $$x^2+x+1$$, so the degree of $$K'/K$$ is 2. This means that the ramification index of $$K'/K$$ is either $$1$$ or $$2$$, i.e. it is either unramified or totally ramified.

Question: Which case is true?

Ideas and Thoughts:

• My thought is that $$K'/K$$ is going to be unramified. I think this is true because the minimal polynomial of $$\xi_3$$, namely $$x^2+x+1$$, is not an Eisenstein polynomial over $$\mathbb{Z}_5$$. But maybe there might be a weird linear combination $$a + b \xi_3$$ with $$a,b \in \mathbb{Q}_5$$ whose minimal polynomial is indeed an Eisenstein polynomial. I don't know if that is possible thuogh.
• If $$K'/K$$ was totally ramified, then there must be an $$x=a + b \xi_3$$ with $$a,b \in \mathbb{Q}_5$$ such that the valuation is $$1/2$$, i.e. $$v_K'(x) = \frac{1}{2}$$. We have the general formula $$v_K'(x) = \frac{1}{2}v_5(N_{K'/K}(x)) = \frac{1}{2}v_5(a^2-ab+b^2)$$, so we have to check if $$a^2-ab +b^2 = 0$$ has a solution over $$\mathbb{F}_5$$. Is there a good way to check quickly (other than try out all possibilities) if it has a solution?
• I know that there is a unique unramified extension of $$K$$ of degree 2 which can be obtained by adjoining a primitive $$(5^2-1)$$-th root of unity, say $$\xi_{24}$$. So we could check if $$\xi_3 \in \mathbb{Q}_5(\xi_{24})$$. But this also seem to be really tedious.

Could you explain me what the best way is to approach this problem? Thank you!

• Note that in general it is true that if $L$ is a local field (i.e. a finite extension of $\mathbb{Q}_{p}$) and $\eta_{n}$ is a $n$'th root of unity and $(n,p)=1$, then the extension $L(\eta_{n})/L$ is unramified of degree $f$, where $f$ is the smallest natural number such that $p^{f}\equiv 1\mod{n}$. You can try and show this first for $\mathbb{Q}_{p}$ but the same proof goes through for the case of a local field. – YumekuiMath Sep 21 '18 at 15:45
• What YumekuiMath says. If you caught pisco's hint before it was deleted that gives another fine approach (I voted to undelete). Also, the discriminant of that quadratic is $-3$ meaning that $3$ is the only prime that might be ramified. But in $\Bbb{Q}_5$ it is not a prime at all. – Jyrki Lahtonen Sep 21 '18 at 17:18

I think that the answer-key to this question is the fundamental formula $$n=ef$$, where $$n$$ is the degree $$[K':K]$$ and $$e$$, $$f$$, are the ramification index and the residue-field extension degree, respectively.
To get all cube roots of unity into an extension of the residue field $$\Bbb F_5$$, you have to go to a field $$\Bbb F_{5^m}$$, whose multiplicative group is of order $$5^m-1$$, and cyclic, as I’m sure you know. Once $$3|(5^m-1)$$, you’ve got those cube roots of unity in your field. And of course $$m=2$$ serves very nicely, since $$3|24$$.
So by adjoining the cube roots of unity to $$\Bbb Q_5$$, you need a quadratic extension of the residue field. Thus here, $$f=2$$, forcing $$e=1$$. End of story.
• Very much agree that this is the best approach. I would add that due to this formula, over a local field, if one is given a minimal polynomial of prime degree (we have a quadratic one here), the field extension it generates is unramified iff the (residue of the) minimal polynomial remains irreducible over the residue field. In this concrete case one checks easily that $x^2+x+1$ remains irreducible over $\Bbb F_5$. – Torsten Schoeneberg Sep 22 '18 at 22:26