# How to compute the minimal polynomial of the 8th root of unity over $\mathbb{Q}_3$?

We consider the extension $$\mathbb{Q}_3(\zeta_8)/\mathbb{Q}_3$$ where $$\zeta_8$$ is an $$8$$th root of unity.

Question What is the minimal polynomial of $$\zeta_8$$ over $$\mathbb{Q}_3$$?

In this post I thought the minimal polynomial of $$\zeta_8$$ is $$x^2+x+2$$. But after working with this relation, I noticed that I will not obtain $$\zeta_8^8=1$$, so something must be wrong.

I tried the following approach:

The cyclotomic extension $$\mathbb{Q}(\zeta_8)/\mathbb{Q}$$ is cyclic and has degree $$4$$, and it is $$\min_{\mathbb{Q}}(\zeta_8) = x^4+1$$. Therefore, I assume that $$x^4+1$$ as a polynomial over $$\mathbb{Q}_3$$ will factor into two (irreducible) polynomials both of degree $$2$$.

I also know that $$\zeta_8 \in \mathbb{Q}_3(i)$$ since $$\mathbb{Q}_3(i)/\mathbb{Q}_3$$ is ramified, and of course $$\min_{\mathbb{Q}_3}(i) = x^2+1$$.

But now I do not know how to make good use of these observations or if I eventually need more results. Could you please help me with this problem? Thank you in advance!

• $\zeta_8= \frac{-i\sqrt{-2}+\sqrt{-2}}{2}$ where $\sqrt{-2}=\sum_{k\ge 0} {1/2\choose k} (-3)^k\in \Bbb{Q}_3$ – reuns Feb 10 at 0:22
• To add a footnote to @reuns’s note, the Galois group is not cyclic, having the form $C_2\oplus C_2$, cyclic plus cyclic. – Lubin Feb 10 at 2:58

## 2 Answers

$$\zeta_8=a+bi$$ for some 3-adic rationals $$a,b$$. Raising to the 4th power, $$-1=a^4+4a^3bi-6a^2b^2-4ab^3i+b^4$$, so $$a^4-6a^2b^2+b^4=-1$$ and $$4a^3b-4ab^3=0$$. The second equation says $$a=0$$ or $$b=0$$ or $$a^2=b^2$$. The first two equations are impossible, so $$a^2=b^2$$, so $$-4a^4=-1$$, $$a^2=\pm1/2$$. Now $$2$$ is not a square in the 3-adics, but $$-2$$ is, so $$a^2=-1/2$$, and $$a=\sqrt{-1/2}$$. Now you should be able to get $$b$$, and then get the minimal polynomial for $$a+bi$$.

The way to use Hensel here is to make use of the (I hope) well-known factorization of $$x^4+4=(x^2+2x+2)(x^2-2x+2)$$. This tells you that in characteristic $$3$$, we have $$x^4+1=(x^2-x-1)(x^2+x-1)$$. The factors are relatively prime, so the factorization lifts to $$\Bbb Z_3$$- factorization.

EDIT, addition:
I’m sorry, I was being dumb-headedly thick when I gave the above answer, true as it is.

What I should have said was that the extension $$\Bbb Q(\zeta_8)\supset\Bbb Q$$ is unramified at $$3$$, so that the $$\Bbb Q_3(\zeta_8)$$ is unramified, and quadratic as we know. Thus the $$\Bbb Q_3$$-conjugate of $$\zeta_8$$ is $$\zeta_8^3$$, and the minimal polynomial for $$\zeta_8$$ over $$\Bbb Q_3$$ is: \begin{align} (X-\zeta_8)(X-\zeta_8^3)&=X^2-(\zeta_8+\zeta_8^3)X-1\\ &=X^2-\left(\frac{1+i}{\sqrt2}+\frac{-1+i}{\sqrt2}\right)X-1\\ &=X^2-\sqrt2iX-1\\ &=X^2-\sqrt{-2}X-1\,, \end{align} and since we know $$\sqrt{-2}\in\Bbb Q_3$$, there you are.

• And of course the primitive eighth roots of unity are not conjugate over $\Bbb Q_3$: there are two classes of them, the other two being the roots of $X^2+\sqrt{-2}X-1$. – Lubin Feb 15 at 23:36