Let $K$ be a number field. Global class field theory tell's us that if the class number of $K$ is 1, then there's no unramified abelian extension (including archimedean places) of $K$. But I'm curious if this is still true for unramified extensions.
The most simplest such example is $\mathbb{Q}$. Minkowski bound tells us that $|d_{K}|>1$ for any number field $K\neq \mathbb{Q}$, so $K/\mathbb{Q}$ cannot be unramified.
Here is one more example (from past qual problems): this is true for $K = \mathbb{Q}(\sqrt{3})$.
To show this, I computed the discriminant $d_{L/\mathbb{Q}}$ and show that if $L/\mathbb{Q}(\sqrt{3})$ is unramified, then Minkowski's bound proves that $[L:\mathbb{Q}(\sqrt{3})]\leq 2$, which gives a contradiction (since every degree 2 extension is abelian and we can use CFT again).
But if field is big, then discriminant will be also big and the Minkowski bound may not imply that the extension should be abelian. So I want to know if there's any proofs or counterexamples. Thanks in advance.
c.f. $\mathbb{Q}(\sqrt{3}, \sqrt{-1})/\mathbb{Q}(\sqrt{3})$ is unramified at every finite places, but not at the infinite place.
Since discriminant $d_{\mathbb{Q}(\sqrt{-1})/\mathbb{Q}} = -4$, any $p\neq 2$ is unramified in $\mathbb{Q}(\sqrt{-1})$, and this impliies that the primes in $\mathbb{Q}(\sqrt{3})$ lying above $p\neq 2$ are unramified in $\mathbb{Q}(\sqrt{3}, \sqrt{-1})$. For primes lying over $p=2$, assume that it ramifies.
Then the ramification degree of $2$ in $\mathbb{Q}(\sqrt{3}, \sqrt{-1})$ will be 4. However, since $2$ unramifies in the subfield $\mathbb{Q}(\sqrt{-3})$, this can't happen.
Best Answer
To complement mathmos's nice answer (Ariel is that you?):
It turns out several examples of such fields have been constructed:
In Yamamura, Ken. “On Unramified Galois Extensions of Real Quadratic Number Fields.” Osaka Journal of Mathematics 23, no. 2 (1986): 471–478 the author constructs infinitely many examples of real quadratic fields with (strictly) unramified extensions with Galois group $A_5$. He also finds examples where the real quadratic has class number one.
Yamamura gives the example $\mathbf Q(\sqrt{36497})$ which we see has class number one in http://www.lmfdb.org/NumberField/2.2.36497.1, moreover $f(x) = x^{5} \mathstrut -\mathstrut 2 x^{4} \mathstrut -\mathstrut 3 x^{3} \mathstrut +\mathstrut 5 x^{2} \mathstrut +\mathstrut x \mathstrut -\mathstrut 1 $ has Galois group $S_5$, all roots real and discriminant $36497$. So if $K$ is the splitting field of $f$ we have
$$K/\mathbf Q(\sqrt{36497})$$
and the author shows that this extension is unramified at all finite primes (but we have real roots so this is everywhere unramified) so this is the sort of extension you were looking for, with non-abelian Galois group $A_5$.
In Brink, David. “Remark on Infinite Unramified Extensions of Number Fields with Class Number One.” Journal of Number Theory 130, no. 2 (February 1, 2010): 304–6. https://doi.org/10.1016/j.jnt.2009.08.013. You can find a $\operatorname{PSL}_2(\mathbf F_7)$ example over a real biquadratic field.
He also expands on the work of Yamamura and shows $\mathbf{Q}(\sqrt{36497}, \sqrt{2819 \cdot 103})$ is class number one and has an infinite unramified extension!