# Easy way to show that $\mathbb{Z}[\sqrt[3]{2}]$ is the ring of integers of $\mathbb{Q}[\sqrt[3]{2}]$

This seems to be one of those tricky examples. I only know one proof which is quite complicated and follows by localizing $\mathbb{Z}[\sqrt[3]{2}]$ at different primes and then showing it's a DVR. Does anyone know any simple quick proof?

• Thanks for some very enlightening answers!
– pki
Commented Jan 18, 2012 at 21:12
• The ring of integers of a pure cubic field is entirely known, see e.g. math.stackexchange.com/a/3508522/300700 Commented Feb 5, 2020 at 17:28
• You can read kconrad.math.uconn.edu/blurbs/gradnumthy/integersradical.pdf. It is not true that the ring of integers of $\Bbb Q(\sqrt[n]{2})$ is equal to $\Bbb Z[\sqrt[n]{2}]$ in general! This is true for $n=2, 3, 4$ and in fact for all $n \leq 1000$, but not for $n=1093$... ! Commented Jun 4, 2021 at 9:47

The short answer is no, in that you almost certainly have to perform separate checks "one prime at a time." For that matter, there's no really slick of doing the quadratic case, either. You either have to do some grunt work with mod-4 conditions on coefficients of minimal polynomials, etc., or build up the theory of the different, etc., and start hitting problems with bigger hammers. When you get past the quadratic case, the grunt work becomes increasingly tedious (/impossible), and you're only left with hammers. So you need technical lemmas on how to conclude that a subring of a ring of integers is really the whole thing, and I don't think it's possible to do that without considering the various primes which could possibly divide the index. Keith Conrad's notes that Álvaro mentions give one solution (his Lemma 1) -- here's another slightly different approach. At the very least, it avoids working explicitly with local rings, even if it doesn't avoid the fact that philosophically we're working locally anyway.

Let $\mathcal{O}$ be the ring of integers of $\mathbb{Q}(\sqrt[3]{2})$. We have $\mathbb{Z}[\sqrt[3]{2}]\subset\mathcal{O}$, and we wish to show equality. It suffices to show that for each prime $\mathfrak{p}$ of $\mathbb{Z}[\sqrt[3]{2}]$, we have $\mathcal{O}=\mathbb{Z}[\sqrt[3]{2}]+\mathfrak{p}$ (this is basically using Nakayama's Lemma to disguise a collection of local things to check with a collection of global things to check). Since for $\alpha:=\sqrt[3]{2}$, the minimal polynomial of $\alpha$ is $f_\alpha(x)=x^3-2$, we also know that $$\mathcal{O}\subset \tfrac{1}{f'(\alpha)}\mathbb{Z}[\sqrt[3]{2}]=\frac{1}{3\sqrt[3]{4}}\mathbb{Z}[\sqrt[3]{2}],$$ making it trivial to check the desired equality for everything but $p=2$ and $p=3$. Now (this part is basically the same as in Keith Conrad's notes) we observe that it suffices to demonstrate $p$-Eisenstein polynomials $h_p(x)$ for a generator $x_p\in\mathbb{Z}[\sqrt[3]{2}]$ for $p=2$ and $p=3$. But these are easy to come by: For $p=2$, take $x_2=\sqrt[3]{2}$ and $h_2(x)=f_\alpha(x)$ and for $p=3$, take $x_3=\sqrt[3]{2}+1$ and $h_3(x)=f_\alpha(x-1)$. Ta-da.

• I was hoping to ask for a little clarification of why $$\mathcal{O}\subset \frac{1}{f'(\alpha)} \mathbb{Z}[\sqrt[3]{2}]=\frac{1}{3\sqrt[3]{4}}\mathbb{Z}[\sqrt[3]{2}]$$ based on the fact that $f_\alpha(x)=x^3-2$. And then why exactly in the case of $p=2,3$ is it enough to find $p$-Eisenstein polynomials for generators $x_p\in\mathbb{Z}[\sqrt[3]{2}]$? Commented Jan 20, 2012 at 19:53
• Neither are particularly obvious, hence my comment about front-loading the computation with theoretical conclusions. The first of your questions is through the theory of the different: Well-known is that if you have a generator $\alpha$ of $\mathcal{O}$, then the different is precisely the principal ideal generated by $(f'(\alpha))$. The generalization is that if $\alpha$ generates only a subring, then you at least get $f'(\alpha)\mathcal{O}$ contained in the different ideal. The second is a slightly technical piece of local commutative algebra. I will try to find a good reference. Commented Jan 20, 2012 at 20:16

It is inescapable that one has to do some work here. The methods sketched by Cam McLeman, and surely what is in KConrad's notes, and also in Lang's Alg No Th, are probably the minimum, because it is not always the case that the ring of integers in $\mathbb Q({\root 3 \of a})$ is $\mathbb Z({\root 3 \of a})$ for square-free $a$. Just as ${1+\sqrt{D}\over 2}$ is an algebraic integer for $D=1\mod 2^2$, ${1+{\root 3\of a}+{\root 3\of a^2}\over 3}$ is an algebraic integer for $a=1\mod 3^2$. Similarly with $3$ replaced by $p$ prime, and so on.

• As a reference for this, and several special cases of this form, I strongly recommend A Course in Computational Algebraic Number Theory by Henri Cohen. Commented Aug 15, 2012 at 14:17

Notations. Let $p$ be a prime number. Let $n$ be an integer. If $n$ is divisible by $p$, but not divisible by $p^2$, we write $p\mid\mid n$.

Let $A$ be a Dedekind domain. Let $P$ be a non-zero prime ideal of $A$. Let $\alpha \in A$. If $\alpha$ is divisible by $P$, but not divisible by $P^2$, we write $P\mid\mid\alpha$.

Let $A$ be an integral domain containing $\mathbb{Z}$. Let $p$ be a prime number. Let $S = \mathbb{Z} - p\mathbb{Z}$. $S$ is a multiplicative subset of $\mathbb{Z}$. We denote by $A_p$ the localization of $A$ with respect to $S$.

Lemma 1. Let $A$ be a discrete valuation ring, $K$ its field of fractions. Let $P$ be the maximal ideal of $A$. Let $L$ be a finite separable extension of $K$. Let $B$ be the integral closure of $A$ in $L$. Suppose $P$ is totally ramified in $L$. Let $Q$ be the unique prime ideal of $B$ lying over $P$. Let $\pi$ be an element of $B$ such that $Q\mid\mid\pi$. Then $B = A[\pi]$.

Proof. Let $n = [L : K]$. Since $PB = Q^n$, $[B/Q : A/P] = 1$. Hence for every $\alpha \in B$, there exists $a_0 \in A$ such that $\alpha \equiv a_0$ (mod $Q$). Consider the congruence equation $\pi x \equiv \alpha - a_0$ (mod $Q^2$). Since $(\pi, Q^2) = Q$ and $\alpha - a_0 \in Q$, there exists a solution $x = a_1 \in A$. Hence $\alpha \equiv a_0 + a_1\pi$ (mod $Q^2$). Similarly there exist $a_0, a_1,\dots, a_{n-1} \in A$ such that $\alpha \equiv a_0 + a_1\pi +\cdots+ a_n\pi^{n-1}$ (mod $Q^n$). Since $PB = Q^n$, $B = A[\pi] + PB$. Let $M = B/A[\pi]$. $M$ is a finitely generated $A$-module. Since $PM = (A[\pi] + PB)/A[\pi] = M$, $M = 0$ by Nakayama's lemma. Hence $B = A[\pi]$. QED

Lemma 2. Let $K$ be an algebraic number field. Let $A$ be the ring of algebraic integers in $K$. Let $p$ be a prime number. Suppose $p$ is totally ramified in $K$. Let $P$ be a prime ideal of $A$ lying over $p$. Let $\pi$ be an element of $A$ such that $P\mid\mid\pi$. Let $S = \mathbb{Z} - p\mathbb{Z}$. Let $\mathbb{Z}_p$ be the localization of $\mathbb{Z}$ with respect to $S$. Let $A_p$ be the localization of $A$ with respect to $S$. Then $A_p = \mathbb{Z}_p[\pi]$.

Proof. Since $A_p$ is integrally closed and integral over $\mathbb{Z}_p$, the assertion follows from Lemma 1. QED

Lemma 3. Let $p$ be a prime number. Let $f(X) = X^n + a_{n-1}X^{n-1} +\cdots+ a_1X + a_0 \in \mathbb{Z}[X]$ be an Eisenstein polynomial at $p$. That is, $p\mid a_i, i = 0,\dots,a_{n-1}$ and $p\mid\mid a_0$. Let $\theta$ be a root of $f(X)$. Let $K = \mathbb{Q}(\theta)$. Let $A$ be the ring of algebraic integers in $K$. Let $P$ be a prime ideal of $A$ lying over $p$.

Then $p$ is totally ramified in $K$ and $P\mid\mid\theta$.

Proof. Since $f(X)$ is irreducible in $\mathbb{Q}[X]$, $n = [K : \mathbb{Q}]$. Let $v_P$ be the discrete valuation associated with $P$. Let $e = v_P(p)$. Since $f(\theta) = 0$ and $p\mid a_i, i = 0,\dots,a_{n-1}$, $\theta^n \equiv 0$ (mod $P$). Hence $\theta\equiv 0$ (mod $P$). Since $p\mid a_i, a_i\theta^i \equiv 0$ (mod $P^{e+1}$) for $i = 1,\dots,a_{n-1}$. Hence $\theta^n + a_0 \equiv 0$ (mod $P^{e+1}$). Since $p\mid\mid a_0$, $v_P(a_0) = e$. Hence $v_P(\theta^n) = e$. On the other hand, $v_P(\theta^n) \geq n$. Hence $e = n$. Hence $v_P(\theta) = 1$ and $p$ is is totally ramified in $K$. QED

Lemma 4. Let $n > 1$ be an integer. Let $m$ be an an integer. Let $p$ be a prime number such that $p\mid\mid m$. Let $\theta$ be a root of $X^n - m$. Let $K = \mathbb{Q}(\theta)$. Let $A$ be the ring of algebraic integers in $K$. Then the following assertions hold.

(1) $X^n - m$ is irreducible in $\mathbb{Q}[X]$.

(2) $p$ is totally ramified in $K$.

(3) Let $P$ be a prime ideal of $A$ lying over $p$. Then $P\mid\mid\theta$.

Proof. $X^n - m$ is an Eisenstein polynomial at $p$. Hence the assertions immediately follows from Lemma 3. QED

Lemma 5. Let $p$ be a prime number. Let $m$ be an integer. Let $\theta$ be a root of $X^p - m$. Let $K = \mathbb{Q}(\theta)$. Let $A$ be the ring of algebraic integers in $K$. Suppose there exists $a \in \mathbb{Z}$ such that $p\mid\mid (m - a^p)$. Then the following assertions hold.

(1) $X^p - m$ is irreducible in $\mathbb{Q}[X]$.

(2) $p$ is totally ramified in $K$.

(3) Let $P$ be a prime ideal of $A$ lying over $p$. Then $P\mid\mid (\theta - a)$.

Proof. $(X + a)^p - m$ is an Eisenstein polynomial at $p$. $\theta - a$ is a root of this polynomial. Hence $\mathbb{Q}(\theta) = \mathbb{Q}(\theta - a)$ has degree $p$ over $\mathbb{Q}$. This proves (1). (2) and (3) follows from Lemma 3. QED

Lemma 6. Let $K$ be an algebraic number field. Let $A$ be an order of $K$. Suppose $A_p$ is integrally closed for all prime numbers $p$. Then $A$ is the ring of algebraic integers in $K$.

Proof. Let $B$ be the ring of algebraic integers in $K$. Let $p$ be a prime number. Since $B$ is integral over $A$, $B_p$ is integral over $A_p$. Since $A_p$ is integrally closed and $K$ is the field of fractions of $A_p$, $B_p = A_p$.
Let $I$ = {$a \in \mathbb{Z}; aB \subset A$}. $I$ is an ideal of $\mathbb{Z}$. Suppose $I \neq \mathbb{Z}$. There exists a prime number $p$ such that $I \subset p\mathbb{Z}$. Since $B \subset A_p$ and $B$ is a finite $\mathbb{Z}$-module, there exists $s \in \mathbb{Z} - p\mathbb{Z}$ such that $sB \subset A$. Hence $s \in I$. This is a contradiction. Hence $I = \mathbb{Z}$. Hence $B = A$. QED

Proposition. Let $p, q$ be distinct prime numbers. Let $\theta$ be a root of $X^p - q$. Let $K = \mathbb{Q}(\theta)$. Suppose there exists $a \in \mathbb{Z}$ such that $p\mid\mid(q - a^p)$.

Then $\mathbb{Z}[\theta]$ is the ring of algebraic integers in $K$.

Proof. Let $B$ be the ring of algebraic integers in $K$. Let $A = \mathbb{Z}[\theta]$. Let $f(X) = X^p - q$. Since $f(X)$ is Eisenstein at $q$, it is irreducible in $\mathbb{Q}[X]$. Let $d$ be the discriminant of $f(X)$. $|N_{K/\mathbb{Q}}(\theta)| = |q|$. Hence $|d| = |N_{K/\mathbb{Q}}(f'(\theta))| = |N_{K/\mathbb{Q}}(p\theta^{p-1})| = p^p q^{p-1}$.

Let $r$ be a prime number other than $p$, $q$. Since $\mid d\mid = p^p q^{p-1}$, $r$ does not divide $d$. Let $R$ be a prime ideal of $A$ lying over $r$. By this, $A_R$ is a discrete valuation ring. Hence $A_r$ is integrally closed. Hence $A_r = B_r$.

On the other hand, by Lemma 4 and Lemma 2, $B_q = \mathbb{Z}_q[\theta] = A_q$. By Lemma 5 and Lemma 2, $B_p = \mathbb{Z}_p[\theta - a] = \mathbb{Z}_p[\theta] = A_p$.

Hence we are done by Lemma 6. QED

Corollary. Let $\theta$ be a root of $X^3 - 2$. Let $K = \mathbb{Q}(\theta)$. Then $\mathbb{Z}[\theta]$ is the ring of algebraic integers in $K$.

Proof: $3\mid\mid(2 - 2^3)$. QED

• "Does anyone know any simple quick proof?" I gather that your answer to this question is no...? Commented Aug 15, 2012 at 14:23
• @MTurgeon According to the answers of Cam McLeman and Paul Garrett, it seems that there's no simple quick proof. Mine is longer than others', but it's simply because I didn't omit the details of the proof. Commented Aug 15, 2012 at 19:14
• “Suppose there exists $a \in \mathbb{Z}$ such that $p || a^p - q$.” This begs the question, does such an integer always exist? Does $a=q$ work always? Commented Apr 2 at 20:37

The following is simple, but perhaps not as quick as you'd like. Hopefully I've written it such that generalizing to other examples isn't difficult. Let $\alpha = \sqrt[3]3$ and let $\mathcal O$ be the ring of integers in $\mathbf Q[\alpha]$. Recall that $$\DeclareMathOperator\disc{disc} \newcommand{\bZ}{\mathbf{Z}}\disc(\bZ[\alpha]) = (\mathcal O : \bZ[\alpha])^2\disc\mathcal(O).$$ The discriminant of $\bZ[\alpha]$ is $-2^23^3$. So certainly $6\mathcal O \subset \bZ[\alpha]$ and hence I can write an $x \in \mathcal O$ as $$x = \frac16(x_0 + x_1\alpha + x_2\alpha^2)$$ for some $x_0, x_1, x_2 \in \bZ$. If $x$ is not in $\bZ[\alpha]$ then one of these, call it $x_i$, is not divisible by $6$, hence is not divisible by $p$, where $p$ is $2$ or $3$. If we multiply by the integer $6/p$, then the coefficient of $\alpha^i$ is the reduced fraction $x_i/p$.

By some other simple manipulations, we can obtain an element of $\mathcal O$ not in $\bZ[\alpha]$, $$\frac1p(y_0 + y_1\alpha + y_2\alpha^2)$$ in which $y_i = 1$ and all $y_j$ satisfy $0 \leq y_j < p$. Since $p$ is small there are not so many combinations to check, and if I've added correctly then the trace and norm suffice to prove that none of these can actually be in $\mathcal O$.

Added. I was going to add some more remarks in response to Prof Emerton's comments, but I stumbled upon these nice notes of by Matt Baker's that explain the local computations as simply as is possible. See Proposition 2.9 there.

• Dear Dylan, You could add that a local argument at $2$ (the polynomial $X^3 - 2$ is Eisenstein at $2$) shows that $2$ can't be in the denominator, so you only have to consider a possible denominator of $3$. Regards, Commented Jan 24, 2012 at 18:21
• @Matt That's a good point. I wanted to avoid local methods because of how the question was posed, but maybe I will add an appendix. Commented Jan 24, 2012 at 19:19
• Dear Dylan, Ah yes, it had been a while since I read the question, and I forgot about the "non-local" stipulation. (Of course, I wouldn't describe those methods as "quite complicated", but to each their own.) Best wishes, Commented Jan 24, 2012 at 22:05
• why "certainly $6\mathcal{O}\subset \mathbb{Z}[\alpha]$"? Commented Feb 2, 2017 at 19:42
• And why 'not divisible by 6' implies 'not divisible by 2 or 3'? 2 is not divisible by 6 but divisible by 2. Commented Aug 25, 2017 at 5:55

Most introductory textbooks find the integers in quadratic fields and maybe cyclotomic fields, and then leave it at that. One that pays a lot of attention to cubic fields is Alaca and Williams, Introductory Algebraic Number Theory. ${\bf Q}(\root3\of2)$ is done as Example 7.1.6, starting on page 153 (and filling three pages!). Many other examples are done in detail, and Dedekind's general formula for pure cubic fields is given as Theorem 7.3.2 on page 176 (with the proof left to the reader!).

• Dear Gerry, I have seen and admired your remarks on m.se. In that you provide the only comment on "Alaca and Williams," I would very much appreciate your opinion. I am looking for an ANT book for self study. One that would go after "Dummit & Foote" and a bit less than "Marcus" or "Samuel." I was also considering "Mollin." Any preferences regarding these or others would be great. Thanks, Andrew
– user12802
Commented Apr 2, 2013 at 15:07
• @Andrew, I like Stewart & Tall, also Pollard & Diamond. But if you post a new question, asking for advice, you may get some very useful answers. In fact, you might check to see whether such a question has already been asked on this site. Commented Apr 3, 2013 at 12:45

Here's an elementary proof. It is similar to Dylan Moreland's but we will use every coefficient of the characteristic polynomial instead of computing the discriminant.

Let $$D$$ be a squarefree integer not divisible by $$3$$, let $$\theta = \sqrt[3]{D}$$, let $$K = \mathbb{Q}(\theta)$$, and let $$x = \frac{a + b \theta + c \theta^2}{3} \in \mathcal{O}_K$$. We want to determine the possible $$x$$ which do not lie in $$\mathbb{Z}[\theta]$$. Taking the traces of $$x, \theta x, \theta^2 x$$ we conclude that $$a, Db, Dc \in \mathbb{Z}$$. The coefficients of the characteristic polynomial of $$x$$ are $$e_1 = a, e_2 = \frac{a^2 - Dbc}{3}, e_3 = \frac{a^3 + Db^3 + D^2 c^3 - 3Dabc}{27}$$

and these must all be integers. In particular, $$27 D^2 e_3 = D^2 a^3 + (Db)^3 + D(Dc)^3 - 3Da Db Dc$$

is an integer divisible by $$D$$, so it follows that $$D | (Db)^3$$, hence $$D | Db$$ (since $$D$$ is squarefree), from which we conclude that $$b \in \mathbb{Z}$$. The above expression is also divisible by $$D^2$$, and so we conclude that $$D | (Dc)^3$$, hence (again since $$D$$ is squarefree) $$D | Dc$$, so $$c \in \mathbb{Z}$$.

By adding integer multiples of $$1, \theta, \theta^2$$ to $$x$$ we may assume WLOG that $$a, b, c \in \{ 0, 1, -1 \}$$. If $$a = 0$$, then $$e_2 \in \mathbb{Z}$$ implies $$bc = 0$$ and $$e_3 \in \mathbb{Z}$$ implies $$b = c = 0$$. If $$bc = 0$$, then $$e_2 \in \mathbb{Z}$$ gives $$a = 0$$, and again $$b = c = 0$$. So WLOG $$a, b, c \in \{ 1, -1 \}$$.

Now suppose that $$D \equiv -1 \bmod 3$$. Then by multiplying by $$\theta$$ we conclude that $$\frac{a + b \theta + c \theta^2}{3} \in \mathcal{O}_K \Rightarrow \frac{-c + a \theta + b \theta^2}{3} \in \mathcal{O}_K$$

so we can work up to a signed cyclic permutation. Now $$e_2 \in \mathbb{Z}$$ if and only if $$bc = -1$$, and cyclically permuting this condition gives $$ab = -1, ca = 1$$. WLOG $$a = 1$$ since we can also work up to negation; then $$b = -1, c = 1$$. Now $$e_3 \in \mathbb{Z}$$ if and only if $$(D + 1)^2 \equiv 0 \bmod 27$$, which is equivalent to $$D \equiv -1 \bmod 9$$.

If $$D \equiv -1 \bmod 3$$ but $$D \not\equiv -1 \bmod 9$$, then we conclude that there are no elements of $$\mathcal{O}_K$$ not in $$\mathbb{Z}[\theta]$$. (Small modifications to the last part of argument tell you what happens for $$D$$ any residue class $$\bmod 9$$.)

• Maybe this is the "grunt work" Cam McLeman's answer mentions, but I don't think it's so bad if you take advantage of symmetries to simplify the casework as above. Commented Aug 16, 2012 at 5:42
• Really nice argument. But: by replacing $D$ by $-D$ if necessary, you’ve allowed yourself to assume that $D\equiv-1\pmod3$, and concluded that only $D\equiv-1\pmod9$ gives the exceptional situation. But your last paragraph, standing alone, is not right, since $\frac13(1+\sqrt[3]{10}+\sqrt[3]{100})$ is an algebraic integer. Commented Oct 25, 2017 at 2:22
• Oops, sorry. I think the condition I meant to write, and what I actually get out of the proof, is $D \equiv -1 \bmod 3$ but $D \not \equiv -1 \bmod 9$. We get your example with $D = -10, a = 1, b = -1, c = 1$ in accordance with the second-to-last paragraph. Commented Oct 25, 2017 at 2:33
• Sure. I recognized perfectly well that $D=-10$ fits into your framework. But since I have your attention, I think I have a much simpler argument than any of those here; I’ve asked Michael Rosen to check it out, and if it’s sound, I’ll post it here. Commented Oct 25, 2017 at 15:43

Here is an elementary method that doesn't require too many calculations. This answer is long only because every detail is given. Let $$\alpha = \sqrt[3]{2}$$ and $$K = \mathbb Q(\alpha)$$.

We begin by computing the norm and trace. Consider an element $$x \in K$$ where $$x = a + b\alpha +c\alpha^2$$. Multiplication by $$x$$ gives a $$\mathbb Q$$-linear map $$K \to K$$ represented by the matrix $$\begin{bmatrix} a&b&c\\2c&a&b\\2b&2c&a\end{bmatrix}$$ Thus $$N(x) = a^3+2b^3+4c^3-6abc$$ and $$\mathrm{Tr}(x) = 3a$$.

We compute the discriminant of $$\mathbb Z[\alpha]$$ with the basis $$1, \alpha, \alpha^2$$. As this is a power basis, the discriminant of the basis is the discriminant of the minimal polynomial of $$\alpha$$, which is $$x^3-2$$. By the formula $$-4p^3-27q^2$$ for the discriminant of cubic polynomials, the discriminant is $$-4\cdot27$$.

Alternatively, the discriminant can be calculated directly as the determinant of the matrix of traces. \begin{align}\begin{vmatrix} Tr(1\cdot 1)&Tr(1 \cdot\alpha)&Tr(1 \cdot \alpha^2)\\Tr(\alpha\cdot 1)&Tr(\alpha \cdot\alpha)&Tr(\alpha \cdot \alpha^2)\\Tr(\alpha^2\cdot 1)&Tr(\alpha^2 \cdot\alpha)&Tr(\alpha^2 \cdot \alpha^2)\end{vmatrix} &= \begin{vmatrix} Tr(1)&Tr(\alpha)&Tr(\alpha^2)\\Tr(\alpha)&Tr(\alpha^2)&Tr(2)\\Tr(\alpha^2)&Tr(2)&Tr(2\alpha)\end{vmatrix}\\ &=\begin{vmatrix} 3&0&0\\0&0&6\\0&6&0 \end{vmatrix}\\ &= -4 \cdot 27\end{align}

Now, recall that $$\mathrm{disc}(\mathbb Z[\alpha]) = (\mathcal O_K : \mathbb Z[\alpha])^2 (\mathrm{disc}(\mathcal O_K)),$$ as clearly $$\mathbb Z[\alpha] \subseteq \mathcal O_K$$. Thus we know that $$(\mathcal O_K : \mathbb Z[\alpha])^2 \mid (-4 \cdot 27)$$, equivalently $$(\mathcal O_K : \mathbb Z[\alpha]) \mid 6$$.

Suppose that $$2 \mid (\mathcal O_K : \mathbb Z[\alpha])$$, or equivalently $$O_K / \mathbb Z[\alpha] \cong \mathbb Z/2\mathbb Z$$ or $$\mathbb Z/6\mathbb Z$$. In either case, let $$x = a + b\alpha +c\alpha^2 \in \mathcal O_K$$ be an element of order 2 in the quotient. Thus $$2x \in \mathbb Z[\alpha]$$, so $$a, b, c \in \frac{1}{2}\mathbb Z$$. We may assume that $$a,b,c \in \{ 0, 1 \}$$ because we can subtract integer multiples of $$1, \alpha, \alpha^2$$ and stay in $$\mathcal O_K$$. Notice that $$N(x) = \frac{1}{8}(a^3+2b^3+4c^3-6abc)$$. As the norm of $$x \in O_K$$ must be an integer, we have that $$8 \mid a^3 + 2b^3 + 4c^3 - 6abc$$. If one of $$a,b,c$$ is zero, the last term is zero, and $$|a^3+2b^3+4c^3| \le 7$$ . Otherwise $$a=b=c=1$$, which gives $$a^3+2b^3+4c^3-6abc = 1$$. Thus $$a^3+2b^3+4c^3 = 0$$ implying $$N(x) = 0$$, so $$x = 0$$, contradicting our hypothesis there was a element of order 2 in $$\mathcal O_K/\mathbb Z[\alpha]$$.

The case where $$3 \mid (\mathcal O_K : \mathbb Z[\alpha])$$ is similar. By definition, $$O_K / \mathbb Z[\alpha] \cong \mathbb Z/3\mathbb Z$$, so let $$x = a + b\alpha +c\alpha^2 \in \mathcal O_K$$ be an element of order 3 in the quotient. Thus $$3x \in \mathbb Z[\alpha]$$, so $$a, b, c \in \frac{1}{3}\mathbb Z$$. Similarly, we may assume that $$a,b,c \in \{-1, 0, 1 \}$$. Notice that $$N(x) = \frac{1}{27}(a^3+2b^3+4c^3-6abc)$$. As the norm of $$x \in O_K$$ must be an integer, we have that $$27 \mid a^3 + 2b^3 + 4c^3 - 6abc$$. By the triangle inequality $$-13 \leq a^3 + 2b^3 + 4c^3 - 6abc \leq 13$$. It follows that $$a^3 + 2b^3 + 4c^3 - 6abc = 0$$, implying that $$N(x) = 0$$, so $$x = 0$$, contradicting our hypothesis there was a element of order 3 in $$\mathcal O_K/\mathbb Z[\alpha]$$.

Now, $$(\mathcal O_K : \mathbb Z[\alpha]) = 1$$, as it divides $$6$$ and cannot be $$2$$ nor $$3$$ nor $$6$$. So $$\mathcal O_K \cong \mathbb Z[\alpha]$$ as required.

• It should be "contradicting our hypothesis that there was an element of order 2 in $\mathcal{O}_K/\mathbb{Z}[\alpha]$" (in the quotient). The meaning of $a,b,c$ seems also to be confused with $2a,2b,2c$ (what is in $\{0,1\}$?) Commented Oct 8, 2018 at 9:04
• Suggestion: you can use the fact that $x^3-2$ is Eisenstein in $2$ to conclude directly that $(\mathcal{O}_K:\mathbb{Z}[\alpha])=1$ or $3$. That wouldn't be so elementary, but could save some time Commented May 25, 2019 at 22:26
• @rmdmc89 Could you tell me why we can conclude $(O_K:\Bbb{Z}[\alpha])=1$or $3$ from the fact that $x^3-2$ is 2-Eisenstein ? I guess that is something related to $\Bbb{Q}_2(2^{1/3})/\Bbb{Q}_2$ is totally ramified, but I want to know the detail. Commented May 16 at 10:39
• @Pont, I imagine some kind of ramification argument should work, but I don't know how to elaborate it. Anyway, here is a great material on this Eisenstein argument (Theorem 2.3): kconrad.math.uconn.edu/blurbs/gradnumthy/totram.pdf. Commented May 16 at 19:07

(Note: Corrected after @user26857: flagged some clear errors)

In this particular case we can combine divisibility and the fact that $$D=2$$ is small.

Consider an algebraic integer $$x = a+ b \sqrt[3]{2}+c\sqrt[3]{4}$$. Two steps in the argument:

1. Its trace $$3 a$$ is an integer, and

$$3x - 3 a = 3b \sqrt[3]{2}+3c\sqrt[3]{4}$$

is also an algebraic integer.

Now from $$3b$$, $$3c$$ subtract some integers so that we are left with the algebraic integer

$$u \sqrt[3]{2} + v\sqrt[3]{4}$$ with $$|u|,|v|\le \frac{1}{2}$$. Now the norm

$$N(u \sqrt[3]{2} + v\sqrt[3]{4}) = 2 u^3 + 4 v^3$$ is in absolute value $$< 1$$ so, being an integer, it must be $$0$$.

Conclusion: $$3a$$, $$3b$$, $$3c$$ are integers.

1. Subtract from $$x$$ integer multiples of $$1$$, $$\sqrt[3]{2}$$, $$\sqrt[3]{4}$$ and get an algebraic integer

$$x' =a'+ b' \sqrt[3]{2}+c'\sqrt[3]{4}$$ with $$a'$$, $$b'$$, $$c' \in \{-\frac{1}{3}, 0, \frac{1}{3}\}$$. The norm of $$x'$$ is an integer. However:

$$|N(x')| = |a'^3+2 b'^3 + 4 c'^3- 3 \cdot 2 a' b' c'| \le \frac{1+2+4+6}{27}< 1$$ and we conclude $$x'=0$$.

Therefore $$a$$, $$b$$, $$c$$ are integers.