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.

Thank You in advance.

  • $\begingroup$ Any reason why you didn’t ask this question for class number $3$? $\endgroup$ – Lubin Mar 1 '15 at 23:02
  • $\begingroup$ Mr. Lubin I thought if class number is $ 3$ then unramified cubic extension is same as Hilbert class field of $ K$. Am I correct? $\endgroup$ – MKJ Mar 2 '15 at 3:26
  • $\begingroup$ 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 $K$? $\endgroup$ – Lubin Mar 2 '15 at 17:10
  • $\begingroup$ Thank you Mr.Lubin still I am having some doubt how to write explicitly what is the cubic unrmified extension of $ K.$ $\endgroup$ – MKJ Mar 9 '15 at 4:32
  • 2
    $\begingroup$ 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 $K$. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.