Vtyjkyuk

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 $K=mathbbQ(alpha)$, with $alpha$ a root of $f=X^3+4kX-k$. Here $kgeq 1$ 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 $mathrmGal(f)cong S_3$.

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 $K(sqrtDelta(f))=K(sqrt-256k-27)$, but $K(sqrt-256k-27)/K$ is ramified at the real place of $K$.

I next tried to consider $K(sqrtd)$ for any square-free integer $dneq 1$, but this also cannot happen: if $K(sqrtd)/K$ is unramified then $d>1$ because $K(sqrtd)$ has to be totally real. As $2$ doesn't divide $Delta(f)$ we see that $2$ is unramified in $K(sqrtd)/mathbbQ$, hence also in $mathbbQ(sqrtd)$ so $dequiv 1bmod 4$. Thus if $p$ is a rational prime dividing $d$ we have $pneq 2$ and $p$ ramifies in $mathbbQ(sqrtd)$. There are now three cases.

1. If $p$ is unramified in $K$ then clearly $K(sqrtd)/K$ cannot be unramified as then $K(sqrtd)/mathbbQ$ would be unramified while it contains $mathbbQ(sqrtd)$


2. If $p$ is ramified in $K$ as $pmathcalO_K=mathfrakp^2mathfrakq$, then we have tame ramification of $p$ in $K$, and Abhyankar's implies that $mathfrakp$ is unramified in $K(sqrtd)$.
   However $mathfrakq$ still ramifies in $K(sqrtd)$ because of $mathbbQ(sqrtd)$.


3. If $p$ is totally ramified in $K$ then $p$ is totally ramified in $K(sqrtd)$, because $e(mathfrakB|p)$ is divisible by both $2$ and $3$ in that case for a prime $mathfrakB$ of $K(sqrtd)$. 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 $K(sqrtalpha)/K$ should work as $mathcalO_K$ is not always equal to $mathbbZ[alpha]$: the index $[mathcalO_K:mathbbZ[alpha]]$ 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 $mathcalO_K[sqrtalpha]$.

Does anyone know the answer?

Edit: Okey, so after using Magma's command IsUnramified on $K(sqrtw)/K$ for various values of $k$ and $winmathbbZ[alpha]$, I found that for $k$ up to say $1000$ we have that $K(sqrtkalpha)/K$ 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?

This is not a complete answer but it is a proof of your observation that $Bbb Q(alpha) subset Bbb Q(alpha,sqrtkalpha)$ is unramified.

Let $v$ be the valuation of a non-archimedean place of $Bbb Q(alpha)$.

Since $alpha^3 + 4kalpha - k = 0$, the two terms with lowest valuation must have the same valuation.

We have $v(4kalpha) = v(k) + v(4alpha) ge v(k)$, so $v(k)$ has to be that lowest valuation.

If $v(alpha^3) = v(k)$ then $v(k) = 3v(alpha)$ and so $v(kalpha) = 4v(alpha)$, which is even.

If $v(4kalpha) = v(k)$ then $v(alpha)=0$ and so $v(k)=v(alpha)=0$, and then $v(kalpha)$ is also even.

In fact, this shows that $alpha^3/k = 1-4alpha$ is a unit $u$ of the ring of integers of $Bbb Q(alpha)$ (you can indeed compute its inverse, $1+64k+4alpha+16alpha^2$), and then the extension $Bbb Q(alpha) subset Bbb Q(alpha,sqrtkalpha) = Bbb Q(alpha,sqrt u)$ 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 $1-4alpha$ is congruent to $1$ mod $4$, this extension is also obtained by adjoining $(1pm sqrt u)/2$, 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.