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

Question

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!

Answer 1

$\zeta_8=a+bi$ 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 $a+bi$.

Answer 2

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.