Unramified cubic extension of imaginary quadratic fields

Let $K=\mathbb Q(\sqrt{-m})$ be an imaginary quadratic field with class number $6.$ Then by Hilbert class field theory and Galois correspondence it is known that $K$ has a unramified cubic extension. Is there any explicit way to find out unramified cubic extension of quadratic fields.

• Any reason why you didn’t ask this question for class number $3$? – Lubin Mar 1 '15 at 23:02
• Mr. Lubin I thought if class number is $3$ then unramified cubic extension is same as Hilbert class field of $K$. Am I correct? – MKJ Mar 2 '15 at 3:26
• then in the case of class number $6$, can’t you take the requisite Hilbert class field and the unique subfield that’s a cubic extension of $K$? – Lubin Mar 2 '15 at 17:10
• Thank you Mr.Lubin still I am having some doubt how to write explicitly what is the cubic unrmified extension of $K.$ – MKJ Mar 9 '15 at 4:32
• it means that your cubic extension $K\supset k$ may not be of the form $K=k(\alpha^{1/3})$, that’s all I was saying. Indeed, any time that $k(\alpha^{1/3})$ is Galois over $k$, you must have $\omega\in K$, and I don’t believe that this is the case in most of the situations we’re talking about, since $3$ is not ramified in $K$. – Lubin Mar 11 '15 at 14:03
Let $L = K(\sqrt{-3})$; then your cubic extension $H/K$ becomes a Kummer extension $HL = L(\sqrt[3]{\alpha})$. The proofs of classical class field theory show that $\alpha$ can be chosen as a generator of an ideal ${\mathfrak a}^3$, where ${\mathfrak a}$ generates an ideal class of order $3$ in ${\mathbb Q}(\sqrt{3m})$. You have to pick the generator $\alpha$ in such a way that the cubic extension becomes unramified (the condition is something like $\alpha \equiv 1 \bmod 3\sqrt{-3}$ if memory serves). See Thm. 1.5.1 over here.