If $\alpha^3+4k\alpha-k=0$ with $k\geq 1$ is odd, then $\mathbb{Q}(\alpha)$ has even class number. This is an exercise found in course notes I have for an introductory course in class field theory and it has been bugging me a for long time.
So $K=\mathbb{Q}(\alpha)$, with $\alpha$ a root of $f=X^3+4kX-k$. Here $k\geq 1$ is an odd integer. The polynomial $f$ is irreducible by Perron's criterion, although that uses maybe a bit much. We have $\Delta(f)=-k^2(256k+27)$ hence $\mathrm{Gal}(f)\cong S_3$.
I tried to find an unramified quadratic extension of $K$, which would imply the result as such an extension is clearly Abelian and hence contained in the Hilbert class field of $K$. The first candidate would be its normal closure $K(\sqrt{\Delta(f)})=K(\sqrt{-256k-27})$, but $K(\sqrt{-256k-27})/K$ is ramified at the real place of $K$.
I next tried to consider $K(\sqrt{d})$ for any square-free integer $d\neq 1$, but this also cannot happen: if $K(\sqrt{d})/K$ is unramified then $d>1$ because $K(\sqrt{d})$ has to be totally real. As $2$ doesn't divide $\Delta(f)$ we see that $2$ is unramified in $K(\sqrt{d})/\mathbb{Q}$, hence also in $\mathbb{Q}(\sqrt{d})$ so $d\equiv 1\bmod 4$. Thus if $p$ is a rational prime dividing $d$ we have $p\neq 2$ and $p$ ramifies in $\mathbb{Q}(\sqrt{d})$. There are now three cases.


*

*If $p$ is unramified in $K$ then clearly $K(\sqrt{d})/K$ cannot be unramified as then $K(\sqrt{d})/\mathbb{Q}$ would be unramified while it contains $\mathbb{Q}(\sqrt{d})$

*If $p$ is ramified in $K$ as $p\mathcal{O}_K=\mathfrak{p}^2\mathfrak{q}$, then we have tame ramification of $p$ in $K$, and Abhyankar's implies that $\mathfrak{p}$ is unramified in $K(\sqrt{d})$. 
However $\mathfrak{q}$ still ramifies in $K(\sqrt{d})$ because of $\mathbb{Q}(\sqrt{d})$.

*If $p$ is totally ramified in $K$ then $p$ is totally ramified in $K(\sqrt{d})$, because $e(\mathfrak{B}|p)$ is divisible by both $2$ and $3$ in that case for a prime $\mathfrak{B}$ of $K(\sqrt{d})$. Thus also in this last case we have a contradiction.


So this also doesn't work. I also don't see how working with something like $K(\sqrt{\alpha})/K$ should work as $\mathcal{O}_K$ is not always equal to $\mathbb{Z}[\alpha]$: the index $[\mathcal{O}_K:\mathbb{Z}[\alpha]]$ can be anything as far as I can tell. Thus I don't see how I can do calculations on explicit primes of a ring like $\mathcal{O}_K[\sqrt{\alpha}]$.
Does anyone know the answer?
Edit: Okey, so after using Magma's command IsUnramified on $K(\sqrt{w})/K$ for various values of $k$ and $w\in\mathbb{Z}[\alpha]$, I found that for $k$ up to say $1000$ we have that $K(\sqrt{k\alpha})/K$ is unramified. However I'm still a bit annoyed with this because a) I have no idea how to prove this (for the finite primes), and b) I have no idea how this seems like a logical thing to try. Surely I must be missing something?
 A: This is not a complete answer but it is a proof of your observation that $\Bbb Q(\alpha) \subset \Bbb Q(\alpha,\sqrt{k\alpha})$ is unramified.
Let $v$ be the valuation of a non-archimedean place of $\Bbb Q(\alpha)$.
Since $\alpha^3 + 4k\alpha - k = 0$, the two terms with lowest valuation must have the same valuation.  
We have $v(4k\alpha) = v(k) + v(4\alpha) \ge v(k)$, so $v(k)$ has to be that lowest valuation.
If $v(\alpha^3) = v(k)$ then $v(k) = 3v(\alpha)$ and so $v(k\alpha) = 4v(\alpha)$, which is even.
If $v(4k\alpha) = v(k)$ then $v(\alpha)=0$ and so $v(k)=v(\alpha)=0$, and then $v(k\alpha)$ is also even.
In fact, this shows that $\alpha^3/k = 1-4\alpha$ is a unit $u$ of the ring of integers of $\Bbb Q(\alpha)$ (you can indeed compute its inverse, $1+64k+4\alpha+16\alpha^2$), and then the extension $\Bbb Q(\alpha) \subset \Bbb Q(\alpha,\sqrt{k\alpha}) = \Bbb Q(\alpha,\sqrt u)$ can only be ramified at $2$ and infinity. In the real embedding of $\Bbb Q(\alpha)$, $\alpha$ is positive, thus so is $u$, which only leaves $2$.
Since $1-4\alpha$ is congruent to $1$ mod $4$, this extension is also obtained by adjoining $(1\pm \sqrt u)/2$, the roots of $X^2-X+\alpha$. So the extension is unramified. 
We have left to show that the extension is nontrivial, by showing that $u$ is actually not a square of another unit. Also, we haven't used yet the assumption that $k$ was odd.
