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Let $K$ be a number field containing $\mu_3$, the third roots of unity. Consider a monic irreducible cubic polynomial $f \in K[x]$ whose discriminant $\Delta$ is a square in $K$. Thus the splitting field $L$ of $f$ gives a $\mathbb{Z}/3\mathbb{Z}$-extension of $K$, and by Kummer Theory, we have

$$ L = K(\sqrt[3]{a})$$

for some non-cube $a \in K$. I would like to know how to find such an $a$.

Perhaps more concretely, the splitting field of $f = x^3 + Ax + B$ (say) is equal to the splitting field of an irreducible cubic of the form $x^3 - a$ (by Kummer Theory), and I guess I'm asking, how do I go from $x^3 + Ax + B$ to $x^3 - a$?

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Kummer theory says that since the norm of an n-th root of unity $\zeta$ is $1$, Hilbert 90 gives an element $\alpha \in L$ with $\sigma(\alpha) = \zeta \alpha$, where $\sigma$ generates the Galois group. Then $a = \alpha^3$.

Now the proof of Hilbert 90 is constructive, so there seem to be no problems.

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I see, so if $t_1, t_2, t_3$ are the roots of my given $f$, then a choice for $\alpha$ from Hilbert 90 is $t_1 + \zeta t_2 + \zeta^2 t_3$? Thank you Franz. – Giuseppe Apr 13 '12 at 14:52
Making this more explicit $(t_1 + \zeta t_2 + \zeta^2 t_3)^3 = m_3 - (3/2) m_{21} + 6 m_{111} + (3/2) \sqrt{-3} \Delta$ where the $m_{\lambda}$ are the monomial symmetric functions in the $t_i$. You should be able to rewrite this as an explicit formula in terms of $A$, $B$ and $\Delta$. – David Speyer Apr 15 '12 at 20:58

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