What are the sixth roots of three in the 2-adic numbers?

Question

Question

What are the sixth roots of three in the 2-adic numbers?

My Attempt

I guess I want solutions to $x^6-3=0$ modulo increasing powers of $2^n\Bbb Z$.

I get $x\equiv1\pmod2$ but then I think neither $1$ nor $3$ is a solution $\pmod4$ because they both require $2=0$, a contradiction, so any solution must lie within an extension, if it exists.

I know there are finitely many quadratic extensions to $\Bbb Q_p$ but I guess sixth roots are a step more complicated than quadratics. I know $\sqrt3$ is one of the valid square roots so I assume I want a cube root of that but that's about as far as I can go.

I got from here that for an unramified extension of degree $6$ I would want to append a $2^6-1$ root of unity but "reading a textbook" (which I am not apt to do) told me that:

an extension $L/K$ of local fields is unramified if $[L:K] = [l:k]$ with $l=O_L/\pi_L$ and $K=O_K/\pi_K$ where $\pi_L,\pi_K$ are uniformizers of $L,K$ which is equivalent to saying that $\pi_K$ is inert in $L$ i.e. that ramification index $e=\mu_K(\pi_L)$ is $1$

, and more words of that were meaningless to me than meaningful so at this point I was stuck working out whether it mattered to me whether I had an unramified extension or not.

Answer 1

We can try to take a sixth root by first taking a cube root $\sqrt[3]{3}$, then trying to take a square root of this. The cube root is actually not a problem: $\sqrt[3]{3} \equiv 3 \bmod 8$, so we can apply Hensel's lemma to find a cube root in $\mathbb{Q}_2$ congruent to $3 \bmod 8$. $\mathbb{Q}_2$ has no nontrivial cube roots of unity so this cube root is unique, and we can compute its $2$-adic expansion to arbitrary precision; it ends

$$\sqrt[3]{3} = \dots 10111011_2.$$

This means $\frac{\sqrt[3]{3}}{3} \equiv 1 \bmod 8$, and every element of $\mathbb{Q}_2$ congruent to $1 \bmod 8$ has a square root (again by Hensel's lemma), whose $2$-adic expansion we can also compute to arbitrary precision; so actually there is a sixth root of $3$ already living in the quadratic extension

$$\mathbb{Q}_2(\sqrt{3})$$

and the problem can be reduced to computing the square root of $3$. (This extension contains one of the three cube roots of $3$; to get the others we need to multiply by cube roots of unity as usual.)

This is a problem since squares are congruent to $0, 1 \bmod 4$, so no square root exists $\bmod 4$. This means we have to pass to an extension $\mathbb{Q}_2(\sqrt{3})$, and this extension is ramified, and that does matter.

By contrast, $\mathbb{Q}_2(\sqrt{5})$ is unramified, which means it is easy to understand in the following sense: unramified extensions of $\mathbb{Q}_p$ are in one-to-one correspondence with extensions of $\mathbb{F}_p$, and this correspondence preserves degrees. So there is a unique unramified quadratic extension of $\mathbb{Q}_2$ corresponding to the unique quadratic extension $\mathbb{F}_4$ of $\mathbb{F}_2$. $\mathbb{F}_4$ can be thought of as being obtained by adjoining a cube root of unity, and it turns out that the same can be done with the corresponding extension of $\mathbb{Q}_2$, meaning it is

$$\mathbb{Q}_2(\omega) \cong \mathbb{Q}_2(\sqrt{-3}).$$

This is not obviously the same as $\mathbb{Q}_2(\sqrt{5})$, but we have $\frac{\sqrt{5}}{\sqrt{-3}} = \sqrt{-\frac{5}{3}}$ and we can check that this square root exists in $\mathbb{Q}_2$ already (it suffices for $-\frac{5}{3} \equiv 1 \bmod 8$ and then we can apply Hensel's lemma again), so $\sqrt{5}$ is a rational multiple of $\sqrt{-3}$ and $\sqrt{-3}$ can be expressed in terms of $\omega$.

One upshot of all this is that if you want a more explicit description of $\omega$, $\sqrt{5}$, or $\sqrt{-3}$, they all turn out to have what one might call "$4$-adic expansions," where by $4$-adic here I don't mean having coefficients $\bmod 4$ but having coefficients living in $\mathbb{F}_4$. Actually to extend the analogy further it would not be too unreasonable to call the unique unramified quadratic extension $\mathbb{Q}_4$ although I think it is usually just called $\mathbb{Q}_2(\zeta_3)$.

Unfortunately $\mathbb{Q}_2(\sqrt{3})$ is ramified so it does not admit this nice description in terms of $\mathbb{F}_4$. This is where my number theory knowledge runs out; I don't understand in what sense it's possible to describe this extension more explicitly, which complicates the question of what it means to "find" $\sqrt{3}$ over $\mathbb{Q}_2$. Possibly this is already the simplest description.

In any case, we at least get that we can write down two of the possible values of $\sqrt[6]{3}$ explicitly as a rational multiple of $\sqrt{3}$. We can get the other four by adjoining $\omega$, which will give us $-\omega$, a primitive sixth root of unity. The conclusion is that the splitting field of $x^6 - 3$ is the compositum of two quadratic extensions $\mathbb{Q}_2(\sqrt{3}, \omega)$, one ramified and one unramified.

Answer 2

Your proposed extension is, in fact, ramified.

There are actually seven quadratic extensions of the $2$-adics, correlated with possible residues of the squarefree radicands $\bmod16$; to wit

$\mathbb{Q}_2[\sqrt2]$

$\mathbb{Q}_2[\sqrt3]$

$\mathbb{Q}_2[\frac12(1+\sqrt5)]$

$\mathbb{Q}_2[\sqrt6]$

$\mathbb{Q}_2[\sqrt7]$

$\mathbb{Q}_2[\sqrt{10}]$

$\mathbb{Q}_2[\sqrt{14}]$

where the third one listed involves a radicand that's one greater than a multiple of $4$ and thus we admit integers $(a+b\sqrt5)/2$ with $a$ and $b$ both odd. Residues $1,9\bmod16$ are not listed because these correspond to $\mathbb{Q}_2$ without an extension. Residues $11,13,15$ give the same extensions as residues $3,5,7$ respectively.

The minimal quadratic polynomials of the appended square roots all have even discriminants. Thus the prime $2$ is ramified, and with that goes the $2$-adic extension – except for the extension given by

$\mathbb{Q}_2[\frac12(1+\sqrt5)].$

That, alone, is your unramified quadratic extension. You may also use $\sqrt{-3}$ in place of $\sqrt5$ because $-3\equiv5\bmod8$; that amounts to directly appending complex cube roots of unity.

So, the extension you propose is $\mathbb{Q}[\sqrt3]$, which is ramified, because the cube root of $3\bmod 8$ is $3$, not $5$.

If you were seeking sixth roots of $-3$ instead of $+3$, then you would have the unramified extension identified above because $-3\equiv5^3\bmod8$.