So if we have a Galois extension $K/\mathbb{Q}$, then the Hilbert Class Field $H$ of $K$ is certainly Galois over $\mathbb{Q}$. But is the converse true? I know many examples of nongalois $K/\mathbb{Q}$ with $H$ nongalois over $\mathbb{Q}$, for example any such $K$ with class number one, but I don't know if this happens uniformly. If it's not so simple, are there known necessary/sufficient conditions for $H/\mathbb{Q}$ to be Galois?
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2$\begingroup$ Take $K=\mathbf{Q}(\alpha)$, where $\alpha$ is a root of $T^3-T-1$. Then $K|\mathbf{Q}$ is not galoisian, but $H|\mathbf{Q}$ is an $\mathfrak{S}_3$-extension. $\endgroup$– Chandan Singh DalawatApr 12, 2014 at 4:03
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$\begingroup$ Chandan, your $K$ has class number $1$ and is its own class field, $\endgroup$– Franz LemmermeyerSep 6, 2021 at 5:10
2 Answers
The answer is no. Take $K = {\mathbb Q}(\sqrt[4]{-5})$. Then $Cl(K)$ is cyclic of order $4$, and its Hilbert class field is given by $H = L(i,\sqrt{1+2i})$. This field is the compositum $H = KF$ of the two dihedral extensions $K$ and $F = {\mathbb Q}(i,\sqrt{5},\sqrt{1+2i})$, hence is normal over the rationals.
Edit (2015). You can construct a lot more examples along the following lines: take a number field $K$ with finite nonabelian class field tower $L/K$, and let $F$ be a subextension with $L/F$ cyclic of prime degree such that $F/K$ is not normal. A typical example: let $K$ be a cyclic cubic field with class group of type $(2,2)$ and Hilbert class field with class number $1$; then any of the three sextic subextensions is not normal over the rationals, but its Hilbert class field has Galois group $A_4$. For explicit examples see J. Théor. Nombres Bordx. 11, No.2, 387-406 (1999).
I would first look at the corresponding local question. So $\mathbf{Q}$ gets replaced by $\mathbf{Q}_p$ (where $p$ is a prime), $K$ becomes a finite extension of $\mathbf{Q}_p$, and $H$ is the maximal abelian unramified extension of $K$ (here the adjective "abelian" is superfluous).
Is it possible for $H$ to be galoisian over $\mathbf{Q}_p$ while $K$ itself fails to be so ? If such is the case, the galoisian closure $\tilde K$ of $K$ over $\mathbf{Q}_p$ will be unramified over $K$, so let's look for a $K$ for which $\tilde K$ over $K$ is ramified (of degree $>1$).
To begin somewhere, let's look for $K$ which have degree $p$ over $\mathbf{Q}_p$. If $p=2$, there are no examples because every quadratic extension is galoisian. But if $p\neq2$, by looking at the parametrisation of degree-$p$ extensions of $\mathbf{Q}_p$ as in arXiv:1005.2016, it is easy to see that such $K$ do indeed exist.