# Hilbert class field of cubic field

Let $K=\mathbb Q(\sqrt[3]7)$ be a pure cubic field with class number 3. I want know how to compute its Hilbert Class Field. I know that its degree of extension is 3.

• One can compute it with, say, Magma: magma.maths.usyd.edu.au/magma/handbook/text/397#4146. The class group is isomorphic to $\mathbb{Z}/3$. – Dietrich Burde Sep 9 '14 at 9:48
• Actually I want to know the theory behind the computation of Hilbert class field of pure cubic fields – MKJ Sep 9 '14 at 16:46
• – Dietrich Burde Sep 9 '14 at 18:16

In this case the answer is easy: fields $K = {\mathbb Q}(\sqrt[3]{m})$ for which $m$ is divisible by some prime number $p = 3n+1$ have an unramified cubic extension guaranteed by Abhyankar's lemma, namely the extension $KF/K$, where $F$ is the cubic subfield of the cyclotomic field of $p$-th roots of unity. The extension is of course abelian, and unramified by Abhyankar's Lemma. This also works for general degrees, except that when the degree is even you have to take care of possible ramification at infinity.
There is a way of computing the Hilbert class field of $\mathbb{Q}(\sqrt[3]{m})$ via elliptic curves in certain cases. For details see section $7$ and $8$ of Franz Lemmermayer's article. He does it explicitly for the example of $\mathbb{Q}(\sqrt[3]{11})$. The polynomial then is (after a further simplification) $$x^6-3x^5+9x^4-1,$$ and $\mathbb{Q}(\sqrt[3]{11})$ has the unramified quadratic extension $\mathbb{Q}(\sqrt[3]{11})(\sqrt{9-4\sqrt[3]{11}})$.
• Dietrich Burde thank you I am reading that article itself in that he is taking only $m=8b^3+3$forms what about in general pure cubic fields? – MKJ Sep 12 '14 at 2:48