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I know that there are $7$ field extensions of $\mathbb{Q}_2$ of degree $2$ (this follows from Hensel's lemma) and I think these are $$\mathbb{Q}_2(\sqrt{2}), \mathbb{Q}_2(\sqrt{3}), \mathbb{Q}_2(\sqrt{5}), \mathbb{Q}_2(\sqrt{6}), \mathbb{Q}_2(\sqrt{7}), \mathbb{Q}_2(\sqrt{10}), \mathbb{Q}_2(\sqrt{14}).$$ Is it easy to see which of these is the unramified extension?

My intuition is $\mathbb{Q}_2(\sqrt{5})$ but I am not sure how to prove it.

Jyrki Lahtonen
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user134871
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1 Answers1

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The unramified extension contains primitive roots of unity of order $2^2-1=3$. Therefore it has an element that looks like $\omega=(-1+\sqrt{-3})/2$. This tells us that $\Bbb{Q}_2(\sqrt{-3})$ is the unramified one.

But which one on the list is equal to $\Bbb{Q}_2(\sqrt{-3})$?

Hint: A known result tells us that an integer $m$ has a square root in $\Bbb{Q}_2$ if and only if $m=4^\ell\cdot k$ with $k\equiv1\pmod{8}$.

Another route to your conclusion would be to observe that the third root of unity is a zero of the polynomial $x^2+x+1$. As this polynomial is separable modulo two, and $$ x^2+x+1\equiv x^2-x-1\pmod2 $$ Hensel then tells us that the zeros of $x^2-x-1$ belong to the field $\Bbb{Q}_2(\omega)$. But you probably know the zeros of this polynomial from its Fibonacci context.

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Jyrki Lahtonen
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    Great, I didn't think of that. Then it must be $\sqrt{5}$ because $5 \equiv -3$ mod $8$. I had guessed that just because $5 \equiv 1$ mod $4$ sets it apart from the others, but this is much better – user134871 Mar 12 '14 at 05:51
  • Correct. Well done! I added another route to the same destination. – Jyrki Lahtonen Mar 12 '14 at 05:54