Does $\mathbb{Q}_2(\sqrt{-1},\sqrt{10})$ contain $\mathbb{Q}_2(\sqrt{5})$?

Question

My suspicion is yes, because I think $\mathbb{Q}_2(\sqrt{-1},\sqrt{10})$ is not totally ramified over $\mathbb{Q}_2$, so it should contain the unique unramified extension, $\mathbb{Q}_2(\sqrt{5})$. If it did contain it, it would also contain $\sqrt{2}$, so somehow we would have $\mathbb{Q}_2(\sqrt{-1},\sqrt{10},\sqrt{5},\sqrt{2})=\mathbb{Q}_2(\sqrt{-1},\sqrt{10})$?

Answer 1

Let

$$\alpha = \sqrt{-1} + \sqrt{10} \left(\frac{1 - \sqrt{-1}}{2}\right).$$

Then $\alpha$ is a root of the polynomial

$$x^4 + 2 x^2 - 20 x - 26.$$

Since this is Eisenstein at $2$, it follows that $\mathbf{Q}_2(\alpha) = \mathbf{Q}_2(\sqrt{10},\sqrt{-1})$ is a totally ramified degree $4$ extension of $\mathbf{Q}_2$. In particular, it cannot contain the non-trivial unramified extension $\mathbf{Q}_2(\sqrt{5})$.

Answer 2

This can be explained without any references to ramification. It's "just" Galois theory: for a field $K$ of characteristic not $2$ and nonsquares $a$ and $b$ in $K$ such that $ab$ is not a square, the extension $K(\sqrt{a},\sqrt{b})/K$ is Galois of degree $4$ with intermediate fields $K(\sqrt{a})$, $K(\sqrt{b})$, and $K(\sqrt{ab})$. By algebra, $K(\sqrt{c}) = K(\sqrt{d})$ for nonsquares $c$ and $d$ in $K$ iff $cd$ is a square in $K^\times$.

Let's take $K = \mathbf Q_2$, $a = -1$, and $b = 10$: $-1$ and $10$ are not squares in $\mathbf Q_2$ since $x^2 + 1$ and $x^2 - 10$ are irreducible over $\mathbf Q_2$, as a root would be in $\mathbf Z_2$ and $-1$ and $10$ are not squares mod $8$ (or even mod $4$). For the same reason, $x^2 - 5$ is irreducible over $\mathbf Q_2$.

The quadratic extensions of $\mathbf Q_2$ inside $\mathbf Q_2(\sqrt{-1},\sqrt{10})$ are $\mathbf Q_2(\sqrt{-1})$, $\mathbf Q_2(\sqrt{10})$, and $\mathbf Q_2(\sqrt{-10})$. If any of these were equal to $\mathbf Q_2(\sqrt{5})$ then $-5$, $50$, or $-50$ would be a square in $\mathbf Q_2$, and thus in fact in $\mathbf Z_2$, but none of them is even a square mod $8$.