Let $K = \mathbb{Q}_3(i,\sqrt[4]{-3})$ and $L = K(\zeta_7)$ where $\zeta_7$ is a 7th root of unity.

Question What is $\min_K(\zeta_7)$?


Over $\mathbb{F}_9$, the residue field of $K$, we obtain the factorization of $x^7-1$ in irreducible factors: $$ x^7-1 = (x+2)(x^3+(a+1)x^2+ax+2)(x^3+2ax^2+(2a+2)x+2). $$ Here, $a \in \mathbb{F}_9$ with $a^2+a+2=0$, i.e. $a$ is a primitive 8th root of unity in $\mathbb{F}_9$.

Let $\bar{f} = x^3+(a+1)x^2+ax+2 \in \mathbb{F}_9[x]$. If we take $b \in \mathbb{F}_{9^3}$ with $\min_{\mathbb{F}_9}(b) = \bar{f}$, then $b$ is a $7$-th root of unity in $\mathbb{F}_{9^3} = \mathbb{F}_9(b)$.

Suppose $\beta \in L$ is a lift of $b$. By Hensel's Lemma, there must be a lift $f \in K[x]$ of $\bar{f}$ with $f(\zeta_7)=0$. I expected $\beta = \zeta_7$. I thought it might be $$ f = \min_K(\zeta_7) = x^3+(\zeta_8+1)x^2+\zeta_8 x + 2 $$ where $\zeta_8$ is the lift of $a \in \mathbb{F}_9$ with $\min_{\mathbb{Q}_3}(\zeta_8) = x^2+x+2$ which is a primitive 8th root of unity. But with this minimal polynomial, I obtain $\beta^8 \neq \beta$, which would have to be true if $\beta = \zeta_7$.

Could someone please point out my mistake in my line of thought and help me finding the relation for $\zeta_7$ (and if necessary $\zeta_8$?

Thank you in advance!

  • 2
    $\begingroup$ Well, as I’m sure you recognize, $\Bbb Q_3(i,\zeta_7)$ is already cubic and unramified over $\Bbb Q_3(i)$. So shouldn’t your minimal polynomial just be $(X-\zeta_7)(X-\zeta_7^9)(X-\zeta_7^{81})=(X-\zeta_7)(X-\zeta_7^2)(X-\zeta_7^4)$ ? $\endgroup$ – Lubin Feb 8 at 6:08

What went wrong is that by the quadratic formula the zeros of $x^2+x+2=0$ are $(-1\pm \sqrt{-7})/2$, and these are not roots of unity in $K$. Because $-7\equiv-1\pmod3$ they are elements of $\Bbb{Q}_3(i)$ though. Those roots of unity are only congruent to these numbers modulo the maximal ideal of the ring of integers of $K$ (is that not what being "a lift" means?).

I would approach a problem like this using Galois theory of cyclotomic extensions of $\Bbb{Q}$. We know that $\Bbb{Q}(\zeta_7)/\Bbb{Q}$ is cyclic of degree six, and that the only quadratic intermediate field is $\Bbb{Q}(\sqrt{-7})$. This suggests strongly to me that having $\sqrt{-7}$ around gives your answer.

With $\zeta_7=e^{2\pi i/7}$ we have $$ (x-\zeta_7)(x-\zeta_7^2)(x-\zeta_7^4)=x^3+\frac{1-i\sqrt{7}}2x^2+\frac{-1-i\sqrt{7}}2x-1. $$ Replacing $(-1\pm \sqrt{-7})/2$ with appropriate zeros of $x^2+x+2=0$ in your field gives the factorization of the seventh cyclotomic polynomial over $\Bbb{Q}_3(i)=\Bbb{Q}_3(\sqrt{-7})$. It depends on the choice of $\sqrt{-7}$ which half of the seventh roots of unity have this cubic as their minimal polynomial; the other half will need the conjugate.

  • $\begingroup$ Looks right to me. $\endgroup$ – Lubin Feb 8 at 6:16
  • $\begingroup$ Thanks @Lubin. Sorry about missing your comment. It must have come while I was typing. $\endgroup$ – Jyrki Lahtonen Feb 8 at 6:17
  • $\begingroup$ Apologies are not called for — I may have missed the alert that you had posted. $\endgroup$ – Lubin Feb 8 at 6:19
  • 2
    $\begingroup$ @Diglett I hate to blow my own trumpet, but just in case you have forgotten about the quadratic intermediate fields of cyclotomic fields of a prime conductor, you can take a look at this old answer of mine for the simple details of a calculation relating $\zeta_7$ and $\sqrt{-7}$. You will see that the only thing that matters is the fact that the quantities are seventh roots of unity. It is totally irrelevant that they happen to be complex numbers as opposed to $3$-adics. $\endgroup$ – Jyrki Lahtonen Feb 8 at 6:24

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.