Any finite field extension (in particular Galois extension) of $\mathbb{Q}$ is ramified. Is there an intuitive geometric explanation of this fact?

Suppose we have an number field $K$, is any Galois extension of $K$ ramified? I think the answer is no, but I do not have a clear picture, examples will be appreciated.

My main question is the following:

Can we reconstruct the absolute Galois group $\mathrm{Gal}_{\mathbb{Q}}(\overline{\mathbb{Q}})$ using only unramified Galois extensions?

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    $\begingroup$ The discriminant of any number field different from $\mathbb Q$ is not $\pm1$ by Minkowski's classical bound for discriminants--a quite geometric fact. And this is exactly what makes the field ramified. $\endgroup$ – Wadim Zudilin Nov 2 '15 at 12:37
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    $\begingroup$ Look up the Hilbert class field. $\endgroup$ – KConrad Nov 2 '15 at 13:34
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    $\begingroup$ @KConrad Class field theory isn't the full explanation, right? I thought I remembered that there were number fields $K$ with $h=1$ but unramified $A_5$ extensions $L/K$. $\endgroup$ – David E Speyer Nov 2 '15 at 16:18
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    $\begingroup$ @DavidSpeyer, you're right, class field theory doesn't explain all occurrences of unramified extensions, but if someone doesn't know the Hilbert class field then that is probably the first thing they should become aware of on this theme. I interpret the OP's question "is any Galois extension of $K$ ramified" as meaning "is every Galois extension of $K$ ramified" and the most concise way to show counterexamples (other than the trivial case of $K$ itself) is with Hilbert class fields. $\endgroup$ – KConrad Nov 2 '15 at 17:35
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    $\begingroup$ @David, there is a comment on a deleted answer to mathoverflow.net/questions/199705/… confirming your suspicion. See also (your own) answer to mathoverflow.net/questions/44801/… $\endgroup$ – Gerry Myerson Nov 2 '15 at 22:25

I am far from being an expert, but I can confirm that there exist number fields $K\neq\mathbb{Q}$ which have no nontrivial unramified extensions. For example, imaginary quadratic number fields of class number one have this property. See this paper by Yamamura for more examples and background information.

On the other hand, I don't understand your main question. What do you mean by "reconstructing" the Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$?

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  • $\begingroup$ It sounds like he wants to claim that $Gal(\bar{\mathbb{Q}}/\mathbb{Q}) = Gal(\mathbb{Q}^{ur}/\mathbb{Q})$, which is false. $\endgroup$ – Pig Dec 4 '15 at 3:31

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