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This is problem taken from Problem 4.

I couldn't find the solution anywhere and I am curious to see a solution for this problem, as i can atleast comprehend the question and it seems, that the mechanism for the solution involved will be a somewhat understandable.

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Using the norm function (standard for any algebraic extension) you can reduce the problem to finding polynomials with integer coefficients p(t) and q(t) such that p^2(t) - q^2(t)(t^2-1) = 1 or -1. Have you tried that? –  Arturo Magidin Aug 31 '10 at 20:04
    
@IvoTerek, I don't agree with that edit. A very minor edit to an old question. Also the edit did not improve the question. The Miklos Schweitzer competition is a famous contest, and giving that bit visible would help readers. Don't go on an editing-spree of old questions. There are several meta threads about this bad habit. –  Jyrki Lahtonen Jul 12 at 19:31
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4 Answers 4

up vote 6 down vote accepted

For any non-negative integer $n$ the element $\pm (t \pm \sqrt{t^2 - 1})^n$ has norm $1$, and there are no elements $a(t) + b(t) \sqrt{t^2 - 1}$ of norm $-1$ because $a(1)^2 - b(1)^2 \sqrt{1^2 - 1} \ge 0$.

Suppose $a(t) + b(t) \sqrt{t^2 - 1}$ is a unit which is not one of the above units such that $\deg b$ is minimal. Then it has norm $1$, so $a(t)^2 - b(t)^2 (t^2 - 1) = 1$. It is not hard to see that the units $\pm 1$ are the only units for which $b = 0$, so WLOG $b$ is nonzero. This implies that $\deg a = d+1, \deg b = d$ for some non-negative integer $d$, and moreover the leading terms of $a$ and $b$ must agree up to sign. If the leading terms agree, then

$$( a(t) + b(t) \sqrt{t^2 - 1})(t - \sqrt{t^2 - 1}) = (ta(t) - b(t) (t^2 - 1)) + (tb(t) - a(t)) \sqrt{t^2 - 1}$$

is a unit with the property that the coefficient of $\sqrt{t^2 - 1}$ has degree strictly less than that of $b$ which is not on the above list, which contradicts the assumption of minimality. Similarly, if the leading terms of $a$ and $b$ are opposite in sign, then

$$( a(t) + b(t) \sqrt{t^2 - 1})(t + \sqrt{t^2 - 1}) = (ta(t) + b(t) (t^2 - 1)) + (tb(t) + a(t)) \sqrt{t^2 - 1}$$

is a unit with the property that the coefficient of $\sqrt{t^2 - 1}$ has degree strictly less than that of $b$ which is not on the above list, again contradicting the assumption of minimality. So no such units exist.

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Out of curiosity, I suppose one could have deduced this abstractly from some sufficiently general version of Dirichlet's unit theorem. Does anyone know a precise statement of such a theorem? –  Qiaochu Yuan Aug 31 '10 at 20:43
    
Excellent! –  anonymous Sep 1 '10 at 2:31
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Qiaochu: I doubt there is such a theorem. Consider the ring $\mathbb{Q}[t][\sqrt{t^4+at^3+bt^2+ct+d}]$, for some $a$, $b$, $c$ and $d$ in $\mathbb{Q}$. The unit group of this is either $\mathbb{Q}^* \times \mathbb{Z}$ or $\mathbb{Q}^*$. Which one is it? Let $E$ be the elliptic curve which is the completion of $u^2=t^4+at^3+bt^2+ct+d$, and let $p$ and $q$ be the two points at infinity. Then the unit group has a $\mathbb{Z}$ factor if and only if $p-q$ is torsion in the group law of $E$. Any generalization you are thinking of has to be capable of seeing this distinction. –  David Speyer Sep 1 '10 at 18:04
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Here is another argument:

Write $x = t + \sqrt{t^2 - 1}$ and $y = t - \sqrt{t^2 - 1}$. Then $x y = 1$, i.e. $y = x^{-1}$, and $t = (x + y)/2$ , $\sqrt{t^2 - 1} = (x - y)/2$.
Thus we see that $\mathbb Z[t][\sqrt{t^2 - 1}] \subset \mathbb Z[1/2][x,x^{-1}].$

Now it is easy to check that if $A$ is an integral domain, then the only units in $A[x,x^{-1}]$ are of the form $a x^n,$ where $a \in A$ is a unit and $n$ is an integer. Since the units in $\mathbb Z[1/2]$ are precisely the elements $\pm 2^m$ (for some $m$) we see that the units in $\mathbb Z[1/2][x,x^{-1}]$ are of the form $\pm 2^m x^n = \pm 2^m (t + \sqrt{t^2 - 1})^n.$

It is easy to check that if such an element and its inverse both actually lie in $\mathbb Z[t][\sqrt{t^2 - 1}],$ then necessarily $m = 0$ (e.g. for norm reasons, or just looking explicitly at their denominators), and so we get the answer that the units are precisely the elements of the form $\pm (t + \sqrt{t^2 -1 })^n$. (Here $n$ runs over $\mathbb Z$; this is the same set as $\pm (t \pm \sqrt{t^2 - 1})^n$, where $n$ is now non-negative.)

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Very nice! (and extra characters to satisfy the silly limit) –  Mariano Suárez-Alvarez Sep 1 '10 at 3:22
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Nice. I was beginning to think along these lines, but then I remembered the descent argument. –  Qiaochu Yuan Sep 1 '10 at 18:47
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PROOF $\rm\;$ It's simply a typical unit group descent. Let $\rm\: U \:$ be the group generated by the units $\rm\: (\pm t,\: 1)\:$.
If unit $\rm\; (a,\;b) \not\in U \;$ then $\rm\; (a,\;b)(\pm t,\; 1) = (\cdots,\;a\pm bt) \;$ is a "smaller" unit $\rm\:\not\in U\;$ since one of $\rm\: a\pm bt \;$ has smaller degree than $\rm\: b\:,\:$ as is easily verified using $\rm\; (a-bt)(a+bt) = 1-b^2 \:. \:\;$ QED

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Just a "geometric translation" of Matt's "algebraic" proof:

It is clear that the ring $\mathbb{Z}[t, \sqrt{t^2 - 1}]$ is equal to $A = \mathbb{Z}[x,y] / (x^2 - y^2 + 1)$. Consider the ring $B = A \otimes_\mathbb{Z} \mathbb{C} = \mathbb{C}[x,y] / (x^2 - y^2 + 1)$.

$B$ is the ring of regular functions of the hyperbola $X \subseteq \mathbb{A}^2_\mathbb{C}$ defined by the equation $x^2 - y^2 + 1 = 0$. The projection of $X$ into one of its asymptotes gives an isomorphism $$ (x,y) \mapsto x-y $$ which maps $X$ onto $\mathbb{A}^1_\mathbb{C} \setminus \{ 0 \}$. Therefore $B$ is isomorphic, as a $\mathbb{C}$-algebra, to $\mathbb{C}[u,u^{-1}]$, where $u = x-y$ is trascendental over $\mathbb{C}$. So every unit in $B$ is of the form $\lambda u^n$, for some $\lambda \in \mathbb{C}^*$ and $n \in \mathbb{Z}$. Since $A$ is flat over $\mathbb{Z}$, $A \subseteq B$. Therefore every unit in $A$ is of the form $\pm (x-y)^n$, for some $n \in \mathbb{Z}$.

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Could you explain what you mean exactly by "Since $\mathbb{C}$ is flat over $\mathbb{Z}$, $A\subseteq B$"? I'm starting to get out of practice with my commutative algebra, perhaps this is something obvious about flatness I'm forgetting. –  Zev Chonoles Aug 18 '11 at 22:22
    
Sorry, I need that $A$ is flat over $\mathbb{Z}$. But it is true because $A$ is torsion-free over $\mathbb{Z}$. –  Andrea Aug 18 '11 at 22:34
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