Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A number field $K$ is said to be monogenic when $\mathcal{O}_K=\mathbb{Z}[\alpha]$ for some $\alpha\in\mathcal{O}_K$. What is currently known about which $K$ are monogenic? Which are not? From Marcus's Number Fields, I'm familiar with the proof that the cyclotomic fields are monogenic, and for example that $\mathbb{Q}(\sqrt{7},\sqrt{10})$ is not monogenic (it is exercise 30 of chapter 2), but because Marcus eschews anything local, I haven't seen any of the perhaps more natural proofs of these results.

If $K$ is monogenic, is there an effective method of determining those $\alpha\in\mathcal{O}_K$ for which $\mathcal{O}_K=\mathbb{Z}[\alpha]$?

More generally, what is known about the minimal number of generators of $\mathcal{O}_K$ as a $\mathbb{Z}$-algebra? That is, can we determine, or at least put non-trivial bounds on, the minimal $m$ such that $\mathcal{O}_K=\mathbb{Z}[\alpha_1,\ldots,\alpha_m]$ for some $\alpha_i\in\mathcal{O}_K$? We know that any $\mathcal{O}_K$ has an integral basis of $n=[K:\mathbb{Q}]$ elements, so certainly $m\leq n$ (I'm considering that trivial).

share|improve this question
2  
Marcus does indeed eschew everything local, but the reason Q(sqrt(7),sqrt(10)) is not monogenic is global. The analogous local extensions are all monogenic: the local extensions are all Galois so this follows from Prop. 1 on p33 of Cassels-Froehlich. –  Kevin Buzzard Apr 13 '10 at 22:08
18  
An analogy I learned from Schoof: minimal number of alg. generators over $\mathbf{Z}$ is like min dimension of projective space over $k = \mathbf{F}_ q$ into which a smooth projective (geom. conn'd) curve over $k$ embeds. A "dumb" obstruction to a low-dim proj. embedding over $k$ is having more $k$-points than the proj. space! (Can happen when $q$ is small compared to genus.) Analogy in char. 0 is a totally split small prime (small compared to degree). The first non-monogenic fields found were cubic with $2$ totally split; proof: $\mathbf{Z}[x]$ has < 3 maps to $\mathbf{F}_ 2$! –  BCnrd Apr 13 '10 at 22:18
6  
Is there any conceptual consequence of a ring of integers being monogenic? Off the top of my head I can think of one: it implies the different ideal is principal (generated by $f'(\alpha)$ where $f(x)$ is the minimal polynomial of the ring generator $\alpha$). This provides an amusing way to create examples of number fields whose ring of integers is not monogenic: any number field whose different ideal is not principal. Of course there is then the task of coming up with examples by that method... –  KConrad Apr 14 '10 at 2:01
2  
@Brian: it looks exactly the same as the old version inside. They just photocopied it! The only differences are a new cover, 8 pages of typos, and a blurb at the front mentioning you and Keith in the same sentence as Serre :-) –  Kevin Buzzard Apr 14 '10 at 8:37
1  
A cubic field with a nonprincipal different is ${\mathbf Q}(sqrt[3]{175})$, so its ring of integers has no power basis. (A ${\mathbf Z}$-basis is 1, $\sqrt[3]{175}$ and $\sqrt[3]{245}$; note $175 = 5^2 \cdot 7$ and $245 = 5 \cdot 7^2$.) The indices $[{\mathcal O}_K:{\mathbf Z}[\alpha]]$ as $\alpha$ varies don't all share a common prime factor, so you can't prove there's no power basis by showing all those indices are, say, even (and thus not 1). Kevin's example has a principal different since the class number is 2 and the different is a square in the class group upstairs (Hecke's theorem). –  KConrad Apr 14 '10 at 15:13

4 Answers 4

up vote 29 down vote accepted

Zev, when $[K:{\mathbf Q}] > 2$, finding all $\alpha$ which are ring generators for ${\mathcal O}_K$ is a hard problem in general: there are only finitely many choices modulo the obvious condition that if $\alpha$ works then so does $a + \alpha$ for any integer $a$. In other words, up to adding an integer there are only finitely many possible choices -- which could of course mean there are no choices.

Here is a nice example: what are the possible ring generators for the integers of ${\mathbf Q}(\sqrt[3]{2})$? We know a basis for the ring of integers is $1, \sqrt[3]{2}, \sqrt[3]{4}$, so a ring generator over $\mathbf Z$ would, up to addition by an integer, have the form $\alpha_{x,y} = x\sqrt[3]{2} + y\sqrt[3]{4}$ for some integers $x$ and $y$ which are not both 0. The index of the ring ${\mathbf Z}[\alpha_{x,y}]$ in the full ring of integers is the absolute value of the determinant of the matrix expressing $1, \alpha_{x,y}, \alpha_{x,y}^2$ in terms of $1, \sqrt[3]{2},\sqrt[3]{4}$, and after a computation that turns out to be $|x^3 - 2y^3|$. We want this to be 1 in order to have a ring generator, which means we have to find all the integral solutions to the equation $x^3 - 2y^3 = \pm 1$. Well, that's a pretty famous example of an equation with only finitely many integral solutions. Up to sign the only solutions are $(1,0)$ and $(1,1)$, so $\alpha_{x,y}$ is $\sqrt[3]{2}$ or $\sqrt[3]{2} + \sqrt[3]{4}$ up to sign (and then addition by an integer).

Here's a more general cubic exercise, just to put the previous example in some perspective (among concrete examples). Let ${\mathbf Q}(\alpha)$ be a cubic field where $\alpha^3 + b\alpha + c = 0$ for integers $b$ and $c$.

a) Show for $x, y \in {\mathbf Z}$ not both 0 that $[{\mathbf Z}[\alpha]:{\mathbf Z}[x\alpha + y\alpha^2]] = |x^3 + bxy^2 + cy^3|$. Therefore if $1,\alpha,\alpha^2$ is known to be a ${\mathbf Z}$-basis of the ring of integers, finding all other ring generators besides $\alpha$, up to addition by integers, amounts to solving $x^3 + bxy^2 + cy^3 = \pm 1$ in integers.

b) It is natural to guess from part a that if $\alpha^3 + a\alpha^2 + b\alpha + c = 0$ and $x, y \in {\mathbf Z}$ are not both 0 the index $[{\mathbf Z}[\alpha]:{\mathbf Z}[x\alpha + y\alpha^2]]$ should be $|x^3 + ax^2y + bxy^2 + cy^3|$. Decide if that natural guess is right!

In general, finding all possible ring generators (modulo addition by an integer) for the ring of integers in a number field amounts to solving some norm-form equation equal to $\pm 1$, and beyond the quadratic case that kind of equation will have just a finite number of integral solutions. A place to look for further discussion is Narkiewicz's massive tome on algebraic number theory: pp. 64--65 and especially p. 80. It turns out the question of finiteness of the number of possible ring generators up to addition by an integer goes back to Nagell. The general case was settled by Gyory in 1973; see MathSciNet MR0437489.

There's actually a whole book on this theme: Diophantine Equations and Power Integral Bases by István Gaál, Birkhauser, 2002.

share|improve this answer
    
Thanks for your excellent answer! I'll have to think about that cubic exercise for a bit. –  Zev Chonoles Apr 14 '10 at 5:05
    
Is there a simple argument why (a) has finitely many solutions? (Gyory does prove this, but using a general argument for all fields and is in French) –  Dror Speiser Apr 16 '10 at 9:35
1  
Dror: good question! Note $x^3 + bxy^2 + cy^3 = 1$ is an affine model of the proj. curve $x^3 + bxy^2 + cy^3 = z^3$. If it had a singular point then, after some algebra, one sees $4b^3 + 27c^2$ is 0. Up to sign that's the discriminant of the irreducible polynomial $x^3 + bx + c$ giving rise to our cubic field, so it must have distinct roots and thus nonzero discriminant. Hence our proj. curve is smooth and cubic, so it's an elliptic curve (it has the rational point [1,0,1]). Now use Siegel's finiteness theorem for integral points on affine models of smooth proj. curves with positive genus. –  KConrad Apr 17 '10 at 3:12

To add to Keith's answer, there are various classes of number fields which are known to be not monogenic. For instance, the following paper

Marie-Nicole Gras, Non monogénéité de l'anneau des entiers des extensions cycliques de $\mathbb{Q}$ de degré premier $l\ge 5$, J. Number Theory 23 (1986), 347-353

gives an elegant proof of the fact that no cyclic extension $K$ of the rationals of prime degree $l\ge 5$ is monogenic unless it happens to be the real part of a cyclotomic field.

share|improve this answer
    
+1 What an intriguing result! –  Zev Chonoles Apr 15 '10 at 17:40

In answer to the second part of your question, there is a paper by Pleasants [Zbl 0328.12008],``The number of generators of the integers a number field'' in which it is shown that the number of generators is at most $\lceil log_2(n) \rceil$, with equality when $2$ splits completely in the field.

share|improve this answer

I recently run across the following result in a paper of [Ash-Brakenhoff-Zarrabi][1] (Lemma 3.1), where it is called Dedekind's criterion:

Let $T(x)$ be a monic irreducible polynomial with root $\theta$ and let $K = \mathbb{Q}(\theta)$. Let $\mathcal{O}$ be the ring of integers in $K$ and let $p$ be a prime. We use overlines to denote reduction modulo $p$.

Factor $\bar{T}(x)$ in $\mathbb{F}_p[x]$ as $\prod \bar{t}_i(x)^{e_i}$, and choose lifts of the $\bar{t}_i$ to monic polynomials in $\mathbb{Z}[x]$. Define $g(x) = \prod t_i(x)$, $h(x)=\prod t_i(x)^{e_i-1}$ and $f(x) = (T(x) - g(x) h(x))/p$. Then $p$ divides the index of $\mathbb{Z}[\theta]$ in $\mathcal{O}$ if and only if $GCD(\bar{f}, \bar{g}, \bar{h})$ is not $1$.

share|improve this answer

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.