TITLE: Easy way to show that $\mathbb{Z}[\sqrt[3]{2}]$ is the ring of integers of $\mathbb{Q}[\sqrt[3]{2}]$
QUESTION [44 upvotes]: 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?

REPLY [7 votes]: Here is an elementary method that doesn't require too many calculations. This answer is only long 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 $Tr(x) = 3a$.
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 or 3 or 6. So $\mathcal O_K \cong \mathbb Z[\alpha]$ as required.