Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
6  
Have you had a look at these notes by Keith Conrad? math.uconn.edu/~kconrad/blurbs/gradnumthy/Qw2.pdf –  Álvaro Lozano-Robledo Jan 18 '12 at 3:03
    
Thanks for some very enlightening answers! –  pki Jan 18 '12 at 21:12
add comment

6 Answers 6

up vote 16 down vote accepted

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.

share|improve this answer
    
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 $x_p\in\mathbb{Z}[\sqrt[3]{2}]$? –  Ashleigh Jan 20 '12 at 19:53
5  
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. –  Cam McLeman Jan 20 '12 at 20:16
add comment

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.

share|improve this answer
1  
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. –  M Turgeon Aug 15 '12 at 14:17
add comment

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.

share|improve this answer
1  
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, –  Matt E Jan 24 '12 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. –  Dylan Moreland Jan 24 '12 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, –  Matt E Jan 24 '12 at 22:05
add comment

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 $p||n$.

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 $P||\alpha$.

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 $Q||\pi$. 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 $P||\pi$. Let $S = \mathbb{Z} - p\mathbb{Z}$. Let $\mathbb{Z}_p$ be the localizations of $\mathbb{Z}$ with respect to $S$. Let $A_p$ be the localizations 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, $p|a_i, i = 0,\dots,a_{n-1}$ and $p||a_0$. 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 raimified in $K$ and $P||\theta$.

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 $p|a_i, i = 0,\dots,a_{n-1}$, $\theta^n \equiv 0$ (mod $P$). Hence $\theta \equiv 0$ (mod $P$). Since $p|a_i, a_i\theta^i \equiv 0$ (mod $P^{e+1}$) for $i = 1,\dots,a_{n-1}$. Hence $\theta^n + a_0 \equiv 0$ (mod $P^{e+1}$). Since $p||a_0$, $v_P(a_0) = e$. 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 raimified 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 $p||m$. 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 raimified in $K$.

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

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 $p||(m - a^p)$. Then the following assertions hold.

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

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

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

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 number $p$. 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 $p||(q - a^p)$.

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 $f(X) = X^p - q$. Snce $f(X)$ 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 $|d| = p^p q^{p-1}$, $r$ 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 colosed. 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||(2 - 2^3) QED

share|improve this answer
2  
"Does anyone know any simple quick proof?" I gather that your answer to this question is no...? –  M Turgeon Aug 15 '12 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. –  Makoto Kato Aug 15 '12 at 19:14
    
@MTurgeon In my experiences, a detailed proof does not usually get upvotes, while a rough sketch answer gets many. Actually I didn't understand well Cam McLeman's answer. This is one of the reasons why I posted my answer. –  Makoto Kato Aug 15 '12 at 20:44
1  
I think the main reason why a long post will not get upvoted is because most people will find it too long to read. @Makoto –  M Turgeon Aug 15 '12 at 20:47
    
@MTurgeon I think what I wrote are basics(perhaps "must" if you want to learn algebraic number theory). It would be strange if people who upvoted Cam McLeman find my answer too long to read. –  Makoto Kato Aug 15 '12 at 21:00
show 5 more comments

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!).

share|improve this answer
    
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 –  Andrew Apr 2 '13 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. –  Gerry Myerson Apr 3 '13 at 12:45
    
Thanks for your advice. –  Andrew Apr 3 '13 at 13:49
add comment

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 $a, Db, Dz \in \mathbb{Z}$. 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 $D \not\equiv -1 \bmod 9$, 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 $\bmod 9$.)

share|improve this answer
    
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. –  Qiaochu Yuan Aug 16 '12 at 5:42
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.