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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.

Thank You in advance.

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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.

Essentially this is the only "trivial" construction of Hilbert class fields in the pure cubic case; all the others require actual computation of unit groups, class groups or other tricks.

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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 $\mathbb{Q}(\sqrt[3]{11})(\sqrt{9-4\sqrt[3]{11}})$.
The standard example for computing the Hilbert class field with Magma also explains some of the "theory behind the computation".

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  • $\begingroup$ Thank you for your response will you explain please how to get that polynomial. $\endgroup$ – MKJ Sep 10 '14 at 3:04
  • $\begingroup$ Did you have a look at Lemmermayer's article ? $\endgroup$ – Dietrich Burde Sep 10 '14 at 20:37
  • $\begingroup$ 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? $\endgroup$ – MKJ Sep 12 '14 at 2:48

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