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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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.

Answer 5

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!).

Answer 6

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 $\bmod 9$.)

Answer 7

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.

Answer 8

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

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

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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.

2. 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.