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From time to time I ask about units in Cubic fields. I noticed for $\mathbb{Z}[\sqrt[3]{2}]$ I get an analogue of the Pell equation:

$$ \det \left[ \begin{array}{ccc} a & 2c & 2b \\ b & a & 2c \\ c & b & a \end{array} \right] = a^3 + 2b^3 + 4c^3 - 6abc = 1 $$

without citing the Dirichlet unit theorem.

Clearly the answers are $a + b\sqrt[3]{2} + c\sqrt[3]{4} = (1 + \sqrt[3]{2} + \sqrt[3]{4})^n$ with $n \in \mathbb{Z}$ (since we this number is unit we can have negative exponent.

Except, I have no way of ruling out other families of solutions. And is there any constructive way of solving this, i.e. without taking a wild guess?

In the $\text{deg}=2$ case (such as $x^2 - 2y^2 = 1$) there is an answer using the pell eq and the continued fraction $\sqrt{2} = [1;\overline{2}]$. Can we do something analogous here?

There are continued fractions you can do on triples of numbers. I think the first step here is:

$$ (1, \sqrt[3]{2}, \sqrt[3]{4}) \to (1, \sqrt[3]{2}-1, \sqrt[3]{4}-1) \to \dots $$

not sure what the smallest number is here. I wonder if the familiar story from Pell works here? Does this euclidean algorithm repeat? Maybe this does not lead to finding units.

Since I am basically asking for a walk-through explication of the Dirichlet unit theorem, there are numerous resources that discuss this in generality.

Avoiding Minkowski's theorem in algebraic number theory.

What is your favorite use of the pigeonhole principle?

Another one that comes to mind is Hasse's Lectures on Number Theory if you have an English copy of the book.

The equation $z^3 - 2 = (z - \sqrt[3]{2})(z - \omega\sqrt[3]{2})(z - \omega\sqrt[3]{2}) = 0$ has a real root and a complex conjugate pair of roots. So the field $\mathbb{Q}(\sqrt[3]{2})$ has a real embedding and a complex embedding.

Then define an embedding $K=\mathbb{Q}(\sqrt[3]{2}) \to \mathbb{R}^3$

$$ (a+b\sqrt[3]{2}+c\sqrt[3]{4})(a+ b\omega\sqrt[3]{2}+c\omega^2\sqrt[3]{4}) (a+b\omega^2\sqrt[3]{2}+c\omega\sqrt[3]{4}) = a^3 + 2b^3 + 4c^3 - 6abc$$

This is not a familiar norm since it's cubic. And then define a region:

\begin{eqnarray} |a+b\sqrt[3]{2}+c\sqrt[3]{4}| &=& c_1 \\ (a+ b\omega\sqrt[3]{2}+c\omega^2\sqrt[3]{4}) (a+b\omega^2\sqrt[3]{2}+c\omega\sqrt[3]{4}) &\leq & c_2^2 \end{eqnarray}

The second equation defines the interior an conic section.

$$ a^2 + \sqrt[3]{2}b^2 + 2\sqrt[3]{4}c^2 + ab + ac + 4bc \leq c_2^2 $$

I am following arguments here and here trying to get the case $K= \mathbb{Q}(\sqrt[3]{2})$.

If $c_1 c_2^2 < A$ can we show there is a lattice point in $\mathcal{O}_K$ inside this 3D region? This should work in a similar way for $\mathbb{Q}(\sqrt[3]{m})$ maybe the constant $A$ changes.

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  • $\begingroup$ You could possibly go through the proof of the unit theorem in this special case. $\endgroup$
    – Fan Zheng
    Commented Nov 11, 2016 at 17:11
  • $\begingroup$ we may try to prove that for a unit $x$ the unit $x(\sqrt[3]{2}-1)^{\pm 1}$ has smaller `something', and so make a descent $\endgroup$
    – Fedor Petrov
    Commented Nov 11, 2016 at 19:56
  • $\begingroup$ Please clarify your question: do you want a proof without citing Dirichlet's unit theorem and without going through the usual proof of Dirichlet's unit theorem in this instance? $\endgroup$
    – GH from MO
    Commented Nov 12, 2016 at 3:09
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    $\begingroup$ [Sorry to see that you deleted your answer; the references you gave are indeed useful, and for the record I did not downvote your answer. Since the references are no longer generally visible, I copy them here: Example 1 in Section V.3 of Fröhlich-Taylor: Algebraic number theory (Cambridge University Press, 1991) and SAGE using the commands K.<a> = NumberField(x^3 - 2) // K.maximal_order().basis() // UnitGroup(K).gens_values() BTW GP will also do this quickly, e.g. bnfinit(x^3-2)[8][5] reports that $x-1$ (i.e. $\root3\of2-1$) is a generator of the units mod $\pm1$.] $\endgroup$
    – Noam D. Elkies
    Commented Nov 12, 2016 at 4:05
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    $\begingroup$ @NoamD.Elkies Thanks for your support. I understand those who downvoted. Amusingly, I also copied back here my textbook reference, and I hit "enter" 10 seconds after you. $\endgroup$
    – GH from MO
    Commented Nov 12, 2016 at 4:07

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