I asked Wolfram Alpha to tell me the fundamental unit of $\mathbb{Z}[\root 3 \of 2]$, it replied $1 - \root 3 \of 2$. Then I tried asking it for $(1 - \root 3 \of 2)^n$ for $-5 \leq n \leq 5$. If I run $n$ over all the integers, does that give me all the units of this domain? Or does this give me some of the units and some non-units?

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    $\begingroup$ Powers of units are units, so as long as this is a unit, you'll never get non-units. $\endgroup$ Commented May 17, 2016 at 23:07
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    $\begingroup$ Don't forget the roots of unity. $\endgroup$
    – Mathmo123
    Commented May 17, 2016 at 23:47
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    $\begingroup$ Indeed, if that number has a norm of 1 or $-1$, then all its powers will have a norm of 1 or $-1$ as well. As for whether your formula gives all units, I suspect that it does not, but I'm not going to pretend that I know more about this than I actually do. $\endgroup$ Commented May 18, 2016 at 1:54
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    $\begingroup$ It seems that $u=1-\sqrt[3]2$ is indeed a fundamental unit of $\Bbb Q(\sqrt[3]2\,)$. Since the rank $r$ of the unit group of this field is $1$, this means that every unit is of form $\pm u^m$ for some $m\in\Bbb Z$. $\endgroup$
    – Lubin
    Commented May 20, 2016 at 3:54

1 Answer 1


As I'm filling in the gaps in my answer I see that it has become very, very similar to sans answer. This is in no way an attempt to top that wonderful and very comprehensive answer: I've decided to still post this because I felt it would be a shame to let the time I've put into this answer go to waste. If it would be better to remove this answer because it is a duplicate (or for whatever other reason), don't hesitate to point this out.

The ring $\Bbb{Z}[\sqrt[3]{2}]$ is contained in the ring of integers $\mathcal{O}_K$ of $K:=\Bbb{Q}(\sqrt[3]{2})$. This is a cubic field with one real embedding and one pair of non-real embeddings into $\Bbb{C}$, so by Dirichlet's unit theorem $$\operatorname{rank}\mathcal{O}_K^{\times}=1.$$ This means that $\mathcal{O}_K^{\times}=\langle-1,\varepsilon\rangle$ for some fundamental unit $\varepsilon\in\mathcal{O}_K^{\times}$, and the units of $\Bbb{Z}[\sqrt[3]{2}]$ are a subgroup of $\mathcal{O}_K^{\times}$. If $\varepsilon$ is a fundamental unit then also $-\varepsilon$, $\varepsilon^{-1}$ and $-\varepsilon^{-1}$ are fundamental units, and there are no others. Clearly there is precisely one fundamental unit $\eta$ such that $\rho(\eta)>1$, where $\rho$ is the unique real embedding of $K$.

It is a nice exercise to show that a cubic number field $L$ with a unique real embedding $\rho$ has a unique fundamental unit $\eta$ with $\rho(\eta)>1$, and satisfies $$|\Delta_L|\leq4\rho(\eta)^3+24.$$ See exercise 5.21 on page 62 of these lecture notes for a hint, or see Lemma 5.13 on page 92 of these course notes for all the details.

In this particular case $\Delta_K=-108$, from which it follows that $$\rho(\eta)\geq\sqrt[3]{21}.$$ Because $1<\rho(1-\sqrt[3]{2})<0$ we know that $1-\sqrt[3]{2}=-\eta^{-k}$ for some $k>0$, and hence that $$\sqrt[3]{2}-1=\eta^{-k}\leq\sqrt[3]{21}^{-k},$$ which implies $k=1$; if we suppose $k\geq2$ then we get $\sqrt[3]{2}-1\leq\sqrt[3]{21}^{-2}<2^{-2}$ which implies that $$\sqrt[3]{2}<1+2^{-2}=\tfrac{5}{4}\qquad\text{ and so }\qquad2<\tfrac{125}{64},$$ a contradiction. Therefore $k=1$ and $1-\sqrt[3]{2}=-\eta^{-1}$ is a fundamental unit.

This shows that $\mathcal{O}_K^{\times}=\langle-1,1-\sqrt[3]{2}\rangle$, and because both generators are in $\Bbb{Z}[\sqrt[3]{2}]$ it follows that $$\Bbb{Z}[\sqrt[3]{2}]^{\times}=\langle-1,1-\sqrt[3]{2}\rangle.$$ So you don't get all units in $\Bbb{Z}[\sqrt[3]{2}]$ by letting $n$ run over the integers for $(1-\sqrt[3]{2})^n$, but you do get them all by letting $n$ run over the integers for $(1-\sqrt[3]{2})^n$ and for $-(1-\sqrt[3]{2})^n$. This doesn't give you any non-units, because the product of two units is always a unit; the units form a group!


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