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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?

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    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.2012-03-09
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    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."2012-03-09
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    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.2012-03-10
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    I should add that I am looking for the solution the author had in mind, the intended solution. One reason I chose this book is because "It is remarkably self-contained", according to a review on AMS MathSciNet.2012-03-11
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    @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.2012-03-12
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    @Marc van Leeuwen: Your proof below seems fine to me, and in fact I really like this proof. I did not know about Minkowski's theorem so I thank you for your response. I wrote to this forum because it seems to me that Exercise 6 is much much harder than the others in that exercise set. But I am not complaining. It may be that the book _is_ in fact "remarkably self-contained" and there _is_ a simpler, nonconstructive proof that none of us have found yet.2012-03-12
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    @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.2012-03-12

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