Simple proof to show $\mathbb{Q}_2((-2)^{\frac{1}{2}})$ is contained in a cyclotomic extension

Question

Let $(-2)^{\frac{1}{2}}$ be any element in $\overline{\mathbb Q_2}$ satisfying $x^2+2=0$, then is there a simple way to show that $\mathbb{Q}_2((-2)^{\frac{1}{2}})$ is contained in a cyclotomic extension of $\mathbb{Q}_2$?

What I know is that the polynomial $x^2+2$ is Eisenstein over $\mathbb{Q}_2$, hence irreducible, but I don't see how to continue.

This is a step in the proof of local Kronecker-Weber theorem, so we can't use this theorem.

Thanks.

Answer 1

$$\zeta_8= \frac{1+i}{\sqrt2}$$ (or $\frac{\pm 1\pm i}{\pm \sqrt2}$ if you prefer)

$\sqrt {-2}= \zeta_8-\zeta_8^{-1}$

Answer 2

$\newcommand{\Qp}{\Bbb Q_p}$Let's first examine cyclotomic extensions of $\Qp$. For any $k$, let $\mu_k$ denote the set of $k$^th roots of unity.

If $k$ is coprime to $p$, then $\Qp\left(\mu_k\right)$ is an unramified extension, and so it is equal to $\Qp\left(\mu_{p^f-1}\right)$ where $f$ is the degree of the residue field extension. In fact, $f$ is the multiplicative order of $p$ mod $k$. In any case, it suffices to consider $p^f-1$ in the case of being coprime to $p$.

On the other end of the spectrum, $k$ is a power of $p$, say $k = p^d$. The minimal polynomial of $\mu_k$ over $\Bbb Q$ is $\Phi_{p^d} = \dfrac{X^{p^d}-1}{X^{p^{d-1}}-1} = \displaystyle \sum_{i=0}^{p-1} X^{ip^{d-1}}$. We don't know if it is irreducible over $\Qp$, so let $g(X) := \Phi_{p^d}(X+1)$, and we shall show that $g$ is Eisenstein at $p$:

• The leading coefficient of $g$ is $1$.
• The constant coefficient of $g$ is $p$.
• To show that the other coefficients are divisible by $p$, we reduce modulo $p$ and get $\Phi_{p^d} = \dfrac{(X-1)^{p^d}}{(X-1)^{p^{d-1}}} = (X-1)^{p^d - p^{d-1}}$, so $g = X^{p^d-p^{d-1}}$, so indeed the other coefficients are divisible by $p$.

Therefore, $\Phi_{p^d}$ is irreducible in $\Qp$ by Eisenstein, so $\Qp\left(\mu_{p^d}\right)$ is totally ramified over $\Qp$ with degree $p^d - p^{d-1}$.

In summary, here are our ingredients:

• $\Qp\left(\mu_{p^f-1}\right)$ is unramified with $f = f$.
• $\Qp\left(\mu_{p^d}\right)$ is totally ramified with $e = p^d - p^{d-1}$.

Since the extension we seek, namely $\Bbb Q_2\left(\sqrt{-2}\right)$, has $e=2$ and $f=1$, we see that we must try $\Bbb Q_2\left(\mu_{2^d}\right)$ first.

If $d=1$ then $e = 2 - 1 = 1$, which is not enough.

So let's examine $d=2$ where $e = 4 - 2 = 2$, i.e. we examine $\Bbb Q_2\left(\mu_4\right)$.

Let $\alpha$ be a primitive $4$^th root of unity, so $\alpha^2+1 = 0$. From the discussion above, we see that $\alpha-1$ is a uniformizer, so its square with differ from $-2$ by a unit. Indeed, $(\alpha-1)^2 = \alpha^2+1-2\alpha = -2\alpha$.

Now we see that if $-2$ is to be a square, then so must $\alpha$; but if $\alpha$ is a square, then its square root would be a primitive $8$^th root of unity, which cannot be found in $\Bbb Q_2\left(\mu_4\right)$ (since $\Bbb Q_2\left(\mu_8\right)$ has $e=4$).

So we know that $\Bbb Q_2\left(\mu_8\right)$ is the right place to find our square root of $-2$. Indeed, letting $\zeta_8$ be a primitive 8^th root of unity, we see that $\alpha := \zeta_8^2$ is a primitive 4^th root of unity, so $(\alpha-1)^2 = -2\alpha = -2\zeta_8^2$, so $-2 = \left[\zeta_8^{-1}(\alpha-1)\right]^2$.

Therefore, $\sqrt{-2} = \zeta_8^{-1}(\alpha-1) = \zeta_8 - \zeta_8^{-1}$