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The answers given here seem very convoluted: The units of $\mathbb Z[\sqrt{2}]$.

Is it possible to provide a more explanatory proof?

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user5826
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  • As a start, mote that (by rationalizing the denominator) $\frac{1}{1+\sqrt{2}}=-1+\sqrt{2}$. The reciprocal of $1+\sqrt{2}$ is therefore in $\mathbb{Z}[\sqrt{2}]$, so $1+\sqrt{2}$ is a unit. – André Nicolas May 15 '15 at 00:13
  • $N(1+\sqrt 2) = -1$ which means $1+\sqrt 2$ is a unit. But how would I show it generates an infinite cyclic group? – user5826 May 15 '15 at 01:02
  • My calculation was more basic, since it did not use concept, properties of norm. But if you have done properties of norm, that;s good too. The integer powers of $(1+\sqrt{2}$ are, easily, all units, and a group. There is a natural isomorphism between this group and $\mathbb{Z}$, given by $(1+\sqrt{2})^n$ is mapped to $n$. – André Nicolas May 15 '15 at 01:08
  • How do we know that map will hold? – user5826 May 15 '15 at 01:15
  • Showing that map is the same problem as showing $\left< 1+ \sqrt 2 \right>$ is infinite, which seems like we're going in circles. – user5826 May 15 '15 at 01:16
  • It is infinite. For note that (as a real number) $\sqrt{2}+1\gt 1$. So all positive powers of $\sqrt{2}+1$ are different, since if $m\lt n$ we have $(1+\sqrt{2})^m\lt (1+\sqrt{2})^n$. – André Nicolas May 15 '15 at 01:19

1 Answers1

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You ask a quite different question from the one asked there. To show that $1 + \sqrt{2}$ generates an infinite group of units it suffices to show that:

  1. $1 + \sqrt{2}$ is a unit. (It then will generate a group of units that is of course cyclic.)

  2. $1 + \sqrt{2}$ has infinite order. (The the group is infinite.)

The first can be seen by observing $(1 + \sqrt{2})(-1 + \sqrt{2})=1$. The second by noting $|1 + \sqrt{2}|\neq 1$, so its order can never be finite.

quid
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  • What do you mean the order of $1 + \sqrt 2$? – user5826 May 15 '15 at 00:02
  • Don't we have to show that $\left|1 + \sqrt 2 \right| \neq 1$? – user5826 May 15 '15 at 00:04
  • The order of an element $g$ of a group is the smallest positive ineteger $n$ such that $g^n =1$ if such an $n$ exists, and infinity otherwise. On the second comment: "show" or "note" is about the same thing it merely depends on the effort. The absoluet value is just $1 + \sqrt{2}$ which is clealry not $1$. I could also have said it follows by noting $1 + \sqrt{2}>1$ so $(1 + \sqrt{2})^n >1$ for all $n>0$ – quid May 15 '15 at 00:12
  • In $\mathbb Z[i]$, $|i| \neq 1$, yet its order is finite. Doesn't that contradict what your saying? – user5826 May 15 '15 at 00:15
  • Yes it would contradict it. But it is false. So, there is no problem. What is $|i|$ in your opinion if not $1$? – quid May 15 '15 at 00:18
  • Is $|x|$ absolute value or order? How can we talk about absolute value in $\mathbb Z[\sqrt2]$? – user5826 May 15 '15 at 00:20
  • @quid The (simpler) question of generating an infinite group of units is not only answered but strengthened to the fact that all units are generated by this fundamental unit in the link provided by Al Jebr. – tc1729 May 15 '15 at 00:22
  • @SiddharthPrasad Yes. But they want a simpler argument. – quid May 15 '15 at 00:24
  • @AlJebr it is the absolute value. And, $\mathbb{Z}[\sqrt{2}] \subset \mathbb{R}$. If you do not consider $\sqrt{2}$ as a real number but merely as a solution of $X^2 - 2$ than that would be quite unusual. – quid May 15 '15 at 00:28
  • Got it. How do we know every element of $\left<1+\sqrt 2 \right>$ will be a unit in $\mathbb Z[\sqrt 2]$? – user5826 May 15 '15 at 00:31
  • Is it because units are associates? – user5826 May 15 '15 at 00:32
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    It's because the norm is multiplicative: $N((1+\sqrt{2})^n)=(N(1+\sqrt{2}))^n=1^n=1$. – tc1729 May 15 '15 at 00:33
  • @AlJebr being a unit is not an equicalence relation. You have that $1 + \sqrt{2}$ is an element of the group of units. The subgroup it generates is thus necessarily a subgroup of the group of units. Moreover the product of units is always a unit. As is the inverse of unit. – quid May 15 '15 at 00:35
  • I changed my comment, I meant to ask: units are associates? – user5826 May 15 '15 at 00:35
  • @AlJebr while all units are associates this is irrelevant. – quid May 15 '15 at 00:36
  • When you say absolute value, isn't that the same as norm? – user5826 May 15 '15 at 00:44
  • $N(1+ \sqrt 2)=-1$. I don't see how your argument for $1+\sqrt 2$ having infinite order is valid. – user5826 May 15 '15 at 00:46
  • $\mathbb Z[i] \subset \mathbb R$, but there is no ordering in gaussian integers. So, there is no absolute value. – user5826 May 15 '15 at 00:50
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    @AlJebr Actually $\Bbb Z[i]\not\subseteq\Bbb R$ because $i$ is not real. But even $\Bbb C$, despite not being ordered, has the absolute value defined on it - you should look it up and learn more about complex numbers. Let's recap the argument here. Since $(1+\sqrt{2})(-1+\sqrt{2})=1$ we know $1+\sqrt{2}$ is a unit. The power of any unit is a unit (because if $ab=1$ then $a^nb^n=1$ for all $n\in\Bbb Z$ assuming $a$ and $b$ commute). Hence $\langle1+\sqrt{2}\rangle$ is a cyclic group of units. – anon May 15 '15 at 04:04
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    Finally, if $(1+\sqrt{2})^n=1$ for some $n\ne0$, then $|1+\sqrt{2}|^n=1$ so $|1+\sqrt{2}|=1$ which is false, so $1+\sqrt{2}$ cannot have finite order. Which part are you having trouble understanding? (Please answer that after you learn what absolute value means. It's not the same as the norm from number theory. Absolute values are a concept someone should have taught you long before any number theory or abstract algebra.) If you want to forego absolute values instead of learning about them, you can read Andre's latest comment under your question. – anon May 15 '15 at 04:10