1
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ $\zeta_8= \frac{-i\sqrt{-2}+\sqrt{-2}}{2}$ where $\sqrt{-2}=\sum_{k\ge 0} {1/2\choose k} (-3)^k\in \Bbb{Q}_3$ $\endgroup$ – reuns Feb 10 at 0:22
  • 2
    $\begingroup$ 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. $\endgroup$ – Lubin Feb 10 at 2:58
4
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ 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 $X^2+\sqrt{-2}X-1$. $\endgroup$ – Lubin Feb 15 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.