I am working through Neukirch's Algebraic Number Theory on my own. Exercise 6 in Section 1 (page 5) is to show that the ring $\mathbb{Z}[\sqrt{d}]$, for any squarefree rational integer $d>1$, has infinitely many units.

I know that in $\mathbb{Z}[\sqrt 2]$ there are infinitely many units, because $(\sqrt{2} + 1)(\sqrt{2} - 1) = 1$ and then taking $n$th powers shows that $(\sqrt{2} + 1)^n$ is a unit for any $n\ge 1$.

Similarly in $\mathbb{Z}[\sqrt{3}]$, we have $(2+\sqrt{3})(2-\sqrt{3}) = 1$, and then $(2+\sqrt{3})^n$ for $n\ge 1$ is an infinite family of units.

I can find other "fundamental units" for other specific values of $d$.

But it seems I have to show that the (Pell) equation $a^2 - db^2 = \pm 1$, for any $d>1$, has an integer solution $(a, b) \ne (\pm 1, 0)$, because I know if I can find one solution, then I can get infinitely many. But from my limited knowledge of Pell's equation this is a difficult problem (using techniques such as continued fractions.)

Maybe there is a simpler nonconstructive proof that I'm missing. Any hints or suggestions?

  • 3
    $\begingroup$ You can write down a proof by specializing one of the standard proofs of Dirichlet's unit theorem (certain aspects of the proof simplify). I don't know if this is the intended solution though. $\endgroup$ – Qiaochu Yuan Mar 9 '12 at 19:21
  • $\begingroup$ Actually finding a solution to the Pell equation is a computational pain, but as for existence, to quote Wikipedia's page on it: "Lagrange proved that for any natural number n that is not a perfect square there are x and y > 0 that satisfy Pell's equation." $\endgroup$ – user14972 Mar 9 '12 at 19:24
  • $\begingroup$ I'm deleting my answer because I realized that it didn't really answer your question. Assuming there is a fundamental unit (which is assured by Dirichlet's unit theorem), the method I gave will work to find the unit. However, I can't think of a easier way of showing existence than either Qiaochu's suggestion above or using continued fractions. $\endgroup$ – Cardboard Box Mar 10 '12 at 5:43
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    $\begingroup$ @rgb: So you first ask for a proof, and now it turns out that what you really want is to read Neukirch's mind. Or to complain that the book is not as self-contained as advertised. From the book, all I can tell is that (1) the chapter "answers to all exercises" is missing, and (2) it is too late to ask Jürgen Neukirch (from the Foreword). For all I can tell your exercise 6 quite a bit harder than its place would suggest. However the proof I gave is easier than a first glance suggests; it's just that a few technical details need to be settled for a complete and explicit proof. $\endgroup$ – Marc van Leeuwen Mar 12 '12 at 15:19
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    $\begingroup$ @rgb: My proof is non-constructive on a double count, if that is any merit: it is by contradiction form the non-existence of a fundamental unit, and Minkowski's theorem is also existence throught contradiction. I rather doubt there is a really simple proof; I've done all points except existence of a fundamental unit with my students without much pain, but upon consulting collegues who've also done this we could not come up with better than below for existence. Also note that the known bounds for where the unit can be found are extremely bad. $\endgroup$ – Marc van Leeuwen Mar 12 '12 at 17:47

I don't actually know any of the standard proofs of Dirichlet's unit theorem that Qiaochu Yuan refers to, but I think they might use Minkowski's theorem about the existence of nonzero lattice points in sufficiently large centrally symmetric convex subsets of $\mathbf R^n$. I'll give a proof for the specific question asked here, which uses Minkowski's theorem, but only in the very simple case of a parallelogram. I'll write the proof top-down, so that one sees how the theorem is used before I'll state (and prove) it.

First I recall some generalities about the ring $R=\mathbf Z[\sqrt d]$ for $d$ a positive squarefree number, to get started. The additive group of $R$ is free Abelian of rank $2$ with generators $1,\sqrt d$. The norm map $N:R\to\mathbf Z$ given by $a+b\sqrt d\mapsto a^2-db^2$ is multiplicative, and the units of $R$ are precisely the elements with norm $\pm1$. One has the non-trivial unit $-1$, but it is of finite order; the point to prove is therefore that the subgroup of positive units is non-trivial (once a positive unit${}\neq1$ is found, its powers form an infinite set of units). I will reason by contradiction, so assume that $1$ is the unique positive unit of $R$. The first step will be to show that this would imply that for any $n\in\mathbf N$, the number of positive $r\in R$ with $|N(r)|=n$ is finite, in fact at most $n^2$.

Lemma. For any $n\in\mathbf N_{>0}$, the number of principal ideals of $R$ that contain $n$ is at most $n^2$.

Proof. Since these ideals all contain $nR$, they all map to principal ideals of $R/nR$, and the mapping is injective. The number of principal ideals of $R/nR$ cannot exceed the number $n^2$ of its elements. QED

The bound given here is far from sharp, but finiteness is all we need. Under the hypothesis that $1$ is the unique positive unit of $R$, two positive elements of $R$ generate distinct principal ideals, and if $|N(r)|=n$, the ideal generated by $r$ contains $n$, so the lemma justifies our claim that the number of of such$~r$ is finite.

This means that for every $M>0$ there is some $\varepsilon_M>0$ such that for all $a,b\in\mathbf Z$ with $0\neq|a+b\sqrt d|<\varepsilon_M$ one has $|N(a+b\sqrt d)|\geq M$. We shall show that for sufficiently large $M$ this contradicts Minkowski's theorem. Note that $N(a+b\sqrt d)=(a+b\sqrt d)(a-b\sqrt d)$, so we can bound $N(a+b\sqrt d)$ above if in addition to the value $a+b\sqrt d$ we also bound its conjugate $a-b\sqrt d$. Now the linear endomorphism of $\mathbf R^2$ sending $\binom ab\mapsto\binom{a+b\sqrt d}{a-b\sqrt d}$ has determinant $-2\sqrt d$, so the conditions $|a+b\sqrt d|<x$ and $|a-b\sqrt d|<y$ define the interior of a parallelogram of area $4\frac{xy}{2\sqrt d}$, for any $x,y>0$. Minkowski's theorem says such a parallelogram will contain a nonzero lattice point whenever its area is greater than $2^2=4$.

So here is how to obtain a contradiction: take any $M>2\sqrt d$ and put $$ x_0=\min\{ z\in R \mid z>0 \land |N(z)|\leq M\}\qquad\text{and}\qquad y_0=\frac M{x_0}. $$ Then $\frac{x_0y_0}{2\sqrt d}>1$, so Minkowski's theorem ensures the existence of $a,b\in\mathbf Z$, not both $0$, with $|a+b\sqrt d|<x_0$ and $|a-b\sqrt d|<y_0$. But then on one hand $|N(a+b\sqrt d)|>M$ by the choice of $x_0$ (and the fact $N(-z)=N(z)$), but on the other hand $|N(a+b\sqrt d)|=|a+b\sqrt d||a-b\sqrt d|<x_0y_0=M$, a contradiction.

Minkowski's theorem. Any centrally symmetric convex subset $S$ of $\mathbf R^d$ of volume greater than $2^d$ contains a nonzero element of $\mathbf Z^d$.

Proof. The map $f:\mathbf R^d\to(2\mathbf Z)^d$ is locally area-preserving, so its restriction to $S$ cannot be injective since the total area at arrival is $2^d$. If $s,s'\in S$ have $s\neq s'$ and $f(s)=f(s')$ then by central symmetry $-s'\in S$, and by convexity $\frac{s-s'}2\in S$; therefore, since $f(s-s')\in(2\mathbf Z)^d$, one has $\frac{s-s'}2\in S\cap(\mathbf Z^d\setminus\{0\})$. QED


Here is a link to incredible notes from a class given by Keith Conrad on ANT:


On page 18 and 19 (of the actual notes, rather than the pdf count) is a method for the case of $\mathbb{Z}[\sqrt{2}]$.

Maybe you can apply it to other cases.

  • $\begingroup$ That link is corrupted. $\endgroup$ – nilo de roock Nov 2 '14 at 10:24
  • $\begingroup$ @ndroock1 Can't find it. Maybe these pieces by Keith Conrad will be of use to you: math.uconn.edu/~kconrad/blurbs $\endgroup$ – user12802 Nov 2 '14 at 11:50

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