As in question, I'm trying to describe group of units of $\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$. I know how this ring looks like but I can't get any results. I don't know if it is helpful, but I see that $2+\sqrt{5}$ is a unit.
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There is a general result that this group is generated by $-1$ and the least unit greater than $1$. – Wojowu Aug 28 '17 at 20:55
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Do you know where I can find it? – Barabara Aug 28 '17 at 20:59
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2As explained by Wojowu basic facts from algebraic number theory tell us that the units are $\pm\phi^n$ where $\phi=(1+\sqrt5)/2$ and $n\in\Bbb{Z}$. Note that $\phi^{-1}=(1-\sqrt5)/2$. Proving that all those numbers are units is easy. Proving that no other units exist is harder. – Jyrki Lahtonen Aug 28 '17 at 21:25
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1Note that $2+\sqrt5=\phi^3$. – Jyrki Lahtonen Aug 28 '17 at 21:26
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@ JyrkiLahtonen i prove that if $\frac{\sqrt{5}-1}{2}$ is in this ring then we have that $\sqrt{5}= \frac{2a+b+1}{1-b}$ proving that $\sqrt{5}$ is an rational number which is not the case..Correct me if i am wrong of course – Marios Gretsas Aug 28 '17 at 21:30
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@MariosGretsas What are $a$ and $b$ in your comment? – anon Aug 29 '17 at 02:52
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@anon they are integers ....i assumed that if $\frac{\sqrt{5}-1}{2}=a+b \frac{\sqrt{5}+1}{2}$ then we have a contradiction..thats why i wanted to know if i'm wrong. – Marios Gretsas Aug 29 '17 at 06:26
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@MariosGretsas Since the solution is $a=-1,b=1$, you just divided by $0$. – anon Aug 29 '17 at 13:25
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ah ok then ..thanks – Marios Gretsas Aug 29 '17 at 13:29
2 Answers
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Here are some hints (for a general approach which works in any real quadratic ring):
Write a unit $\alpha\neq\pm 1$ as $a+b\sqrt{5}$, where $a,b$ are both integers or both nonintegers. Exactly one of $\alpha,-\alpha,\alpha^{-1},-\alpha^{-1}$ is greater than $1$. Also, exactly one of $\pm a\pm b\sqrt{5}$ is greater than $1$. Deduce that $\alpha>1$ iff $a,b>0$.
It follows that there exists a smallest unit $\beta$ greater than $1$. Consider any other unit $\gamma$, which we assume is greater than $1$. Then $\beta^k\leq\gamma<\beta^{k+1}$ for some integer $k$. Then $1\leq \beta^{-k}\gamma<\beta$. From minimality of $\beta$, $\beta^{-k}\gamma=1,\gamma=\beta^k$.
Hence every unit is of the form $\pm\beta^k$.
Wojowu
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Could you please elaborate on why it is not possible that a unit $a+b\sqrt{5}$ is greater than 1 if $a>0$ and $b<0$ for example ? – rae306 Nov 30 '20 at 11:48
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1@rae306 If $a>0,b<0$ then we know that $\alpha=a-b\sqrt{5}>1$. But then $a+b\sqrt{5}=\overline\alpha=\pm\alpha^{-1}$, which cannot be greater than $1$. – Wojowu Nov 30 '20 at 12:31
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In a real quadratic integer ring like this one, if a given number $u \neq \pm 1$ is a unit, then all $u^n$ and $(-u)^n$ are distinct numbers and also units, with $n \in \mathbb{Z}$. Using what you know, we see that $u = 2 + \sqrt{5}$, so $$1, 2 + \sqrt{5}, -682 + 305 \sqrt{5}, 161 - 72 \sqrt{5}, \ldots$$ are also units, as well as their reciprocals, additive inverses and additive inverses of reciprocals.
Although you now have found infinitely many units, you can't be sure you have identified all the units, and in fact you could be missing infinitely many units! Without knowing what the fundamental unit is, you are still in the dark.
What about that number $$\phi = \frac{1 + \sqrt{5}}{2}?$$ Could that be the fundamental unit? As it turns out, $$N(\phi) = \frac{1}{4} - \frac{5}{4} = -1,$$ so $\phi$ is in fact a unit, though it may or may not be the fundamental unit. Also notice that $1 < \phi < u$ (let's use $u$ to mean the same number as before). In fact, $\phi^3 = u$.
Then $\phi^n$ and $(-\phi)^n$ give us infinitely many units skipped over by $u^n$ and $(-u)^n$, as well as all those units we got before. That is to say $u^n = \phi^{3n}$.
I hope your textbook has a detailed explanation of how to find the fundamental unit. Since I'm not in the class and can avail myself to any calculator, I just asked Wolfram Alpha: NumberFieldFundamentalUnits[GoldenRatio]. It replied: $$\frac{1}{2}(1 + \sqrt{5}).$$
Mr. Brooks
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Could you perhaps sketch or give a reference for a proof that $(1 +\sqrt{5})/2$ is indeed a fundamental unit (i.e. we're not missing even more units)? Perhaps even better, do you have a reference for an algorithm for finding the fundamental unit? – Anakhand Oct 10 '23 at 12:45