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There are at least two ways to construct cyclic cubic extensions of $\mathbb{Q}$ as explained below. (A third one is given in the answer to an earlier question).

  1. Given $A, B, C$ integers with $A\neq 0$ we put $$ \lambda=-\frac{C^2+27B^2}{4A^3} $$ Then, if $(a,b,c)=\lambda(A,B,C)$, we see that $c^2=-4a^3-27b^2$. Hence, the discriminant of $T^3+aT+b$ is a square in $\mathbb{Q}$. So the simple extension $\mathbb{Q}[z]$ obtained by adjoining a root $z$ of this polynomial is normal and cyclic. (Assume that $A$ and $B$ are chosen so that this polynomial is irreducible which is not difficult.)

  2. Given a positive integer $k$ such that $3$ divides the order of $(\mathbb{Z}/(k))^{*}$, there is a subgroup $H$ of this group that has index $3$. Choosing one such $H$, we see that if $E=\mathbb{Q}[\zeta_k]$, then $F=E^{H}$ is a cyclic cubic extension of $\mathbb{Q}$.

According to the Kronecker-Weber Theorem, every construction in (1) is associated with some construction in (2).

Question: Given $a$ and $b$, is there an algorithm to determine $k$? (The link referenced above can perhaps be used to determine an $a$ and $b$ given $H$.)

To begin with even a "brute force" algorithm based on giving a bound on $k$ in terms of $a$ and $b$ would be nice.

A specific case (which motivated this question) is when $(a,b)=(-7,-7)$. Then, serendipity led one to the solution $k=7$. This pair $(a,b)$ is different from the one produced by the referenced link. (Of course, multiple $(a,b)$ pairs could give the same extension.)

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  • $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$
    – GH from MO
    Mar 25, 2021 at 11:43
  • $\begingroup$ Some key ideas for an answer are in an answer to a related question on MSE. $\endgroup$
    – Kapil
    Mar 26, 2021 at 12:52

1 Answer 1

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This is very well known: see for instance Theorem 6.4.6 in my book GTM 138.

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    $\begingroup$ Thanks! This is a bit embarrasing since I even once taught some material from this book. Clearly, I didn't get far enough! $\endgroup$
    – Kapil
    Mar 25, 2021 at 12:10
  • $\begingroup$ The suggested text gives a canonical form for cyclic cubic fields. Hence, given $(a,b)$ one could compute the canonical form. This should lead to a bound on $k$ and then one can try various $H$'s. $\endgroup$
    – Kapil
    Mar 25, 2021 at 14:17
  • $\begingroup$ After reading GTM 138 and the proof of Lemma 6.4.5, one can see that given $(a,b)$ one can compute the canonical form for the cubic equation. So, further using Theorem 6.4.6 I can see how one can decide exactly which of the cases occur. However, I still could not see a way to bound $k$ given that the cubic field is written in this canonical form. $\endgroup$
    – Kapil
    Mar 26, 2021 at 12:42

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