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Show that an element $ m + n \sqrt{2} $ of $ \mathbb{Z}[\sqrt{2}] $ is a unit if and only if $ m^{2} - 2 n^{2} \in \{ 1,-1 \} $.

Okay, I have a pretty big hint as to how to do this problem, but I'm having a problem connecting the dots. Here's the hint:

For the $\Rightarrow$ direction, suppose $m+n\sqrt2$ is a unit, so there exists $x+y\sqrt2\in\Bbb{Z}[\sqrt2]$ such that $(m+n\sqrt2)(x+y\sqrt2)=1$. Show that this implies $(m-n\sqrt2)(x-y\sqrt2)=1$ also, then multiply these equations.
For the $\Leftarrow$ direction, you need to suppose that $m^2-2n^2\in \{1,-1\}$, then use this assumption to define a multiplicative inverse for $m+n\sqrt2$.

I understand that there exists an element $x+y\sqrt2$ when multiplied by $m+n\sqrt2$ equals $1$. I don't understand how this implies the next statement and then I don't understand how to go from there to the fact that $(m+n\sqrt2)(m-n\sqrt2)=m^2+2n^2$.

Also, I know that $\Bbb{Z}[\sqrt2]$ is a subring of $\Bbb{R}$, so I can use those properties.

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    Last line $\mathbb Z[\sqrt 2]$ is a subring of $\mathbb R$ – 2017-02-24

4 Answers 4

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In general :To determine invertible elements in $\mathbb Z[\sqrt n]$ where $n$ free of square::

Let be $u=a+b\sqrt n $ an invertible element in $\mathbb Z[\sqrt n]$, then exists $v=c+d\sqrt n \in \mathbb Z[\sqrt n] $ where $uv=1$ , then $N(uv)=N(u)N(v)=1\Rightarrow N(u)=1$. we have three cases:

$1)$ $n=-1 \Rightarrow N(u)=a^2+b^2=1\Rightarrow a=\pm 1, b=0$ or $a=0, b=\pm 1$, then $u=\pm 1$ or $u=\pm i$

$2)$ $n<-1 \Rightarrow N(u)=a^2-nb^2=1 \Rightarrow a=\pm 1, b=0$ ,then $u=\pm 1$

$3)$ $n>0 \Rightarrow N(u)=a^2-nb^2=1$ has infinitely many of solutions , then invertible element is $\{-1,+1 \}$ and infinite many of elements.

for example for case $(3)$ in $\mathbb Z[\sqrt3]$ we have $u=2-\sqrt3$ is invertible element because $(2-\sqrt3)(2+\sqrt3)=1$

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    I haven't learned norm. – 2017-02-24
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    norm element is $N(a+b\sqrt n)=(a+b\sqrt n)(a-b\sqrt n)=a^2-nb^2$ – 2017-02-24
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    Okay. i understand that. However, since I haven't learned it...I don't think I'm supposed to use it for the problem. – 2017-02-24
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    The norm is easy things to prove unit element in a ring $\mathbb Z [\sqrt n ]$ – 2017-02-24
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If there exists $x,y\in\mathbf Z$ such that $(m+n\sqrt 2)(x+y\sqrt2)=1$ means the linear system $$\begin{cases}mx+2ny=1\\nx+my=0\end{cases}$$ has a unique solution $(x,y)\in\mathbf Z^2$. It has a unique solution if and only if the matrix $\begin{pmatrix}m&2n\\n&m\end{pmatrix}$ is invertible, which is the case if and only if its determinant $m^2-2n^2$ is a unit in $\mathbf Z$.

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    How do we know that $mx+2ny=1$ and $nx+my=0$? Are there no other possiblities? – 2017-02-24
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    $a+b\sqrt2=c+d\sqrt 2$ $\;(a,b,c,d\in\mathbf Z)$ if and only if $a=c,\;b=d$. – 2017-02-24
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I'm going to follow your hint. Given $a+b\sqrt{2}$, we define its "conjugate" as $\overline{a+b\sqrt{2}}=a-b\sqrt{2}$. This conjugate behaves like the usual conjugate for complex numbers, e.g., given $a+b\sqrt2, c+d\sqrt2\in \Bbb{Z}[\sqrt{2}]$, it holds $$\overline{a+b\sqrt{2}}\cdot \overline{c+d\sqrt{2}}=\overline{(a+b\sqrt{2})(c+d\sqrt{2})}\;.$$

Try to prove by yourself that this is indeed true. Now, if you have $(m+n\sqrt2)(x+y\sqrt2)=1 ...(\alpha)$, applying the conjugate to both sides of $(\alpha)$ we get $$\overline{(m+n\sqrt2)(x+y\sqrt2)}=\overline{1}=1$$ $$\implies \overline{(m+n\sqrt2)}\cdot \overline{(x+y\sqrt2)}=1$$ $$\implies (m-n\sqrt2)\cdot (x-y\sqrt2)=1 ... (\beta)$$

Then if we multiply $(\alpha)$ and $(\beta)$ we deduce that $$(m+n\sqrt2)(m-n\sqrt2)(x+y\sqrt2)(x-y\sqrt2)=1,$$ which leads to $(m^2-2n^2)(x^2-2y^2)=1$, so $m^2-2n^2\mid 1$, i.e., $m^2-2n^2\in \{\pm1\}$.

For the other implication, if $m^2-2n^2\in \{\pm 1\}$, then show that $\frac{m-n\sqrt{2}}{m^2-2n^2}$ is the inverse of $m+n\sqrt2$ in $\Bbb{Z}[\sqrt2] $.

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    Btw, the above solution is somehow artificial. – 2017-02-24
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    I'm a little confused how you go from $(m^2-2n^2)(x^2-2y^2)=1$ to $m^2-2n^2|1$? Is that because we know that $x^2-2y^2$ is an integer? Then we know that since both $m^2-2n^2$ and $x^2-2y^2$ are integers, then they both $\in$ {-1,1}? Am i going through the right reasoning? – 2017-02-24
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    Also, for the other implication...I get that multipying $(m+n^2)(\frac{m-n\sqrt2}{m^2-2n^2}) = m^2-2n^2$, which is equal to -1 or 1. Is that correct reasoning as well? If so, wouldn't it have to be equal to 1 for it to be the multiplicative identity? And how did you come up with $\frac{m-n\sqrt2}{m^2-2n^2}$? – 2017-02-24
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    Actually, I think I got it. Multiplying those just gives you 1. I messed something up. And I understand how you came up with that number. Thanks so much. I still have a couple more parts on the question that I'm struggling with, so I'll get to those. Thanks a lot. – 2017-02-24
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    @johnie4usc yes, your reasoning is right. I think you should study norms because that will give you a better insight about how to find the inverse (when exists) of an arbitrary element $a+b\sqrt{d}\in \Bbb{Z}[\sqrt{d}]$. – 2017-02-24
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    Yeah, I'm in a modern algebra class so I'm sure we'll get to norms pretty soon. Is there any way you could help me show that $3+2\sqrt2$ has infinite order in ($Z[2]^x$)? I also need to show that and that -1 and 1 are the only units with finite order in ($Z[2]^x$). He gave us a hint, but the hint doesn't really help me. The hint states If $a^n$ = 1, then |$a^n$| = 1. If 0 < |a| < 1, how does |$a^n$| relate to |a| for a positive integer n? If |a| > 1, how does |$a^n$| relate to |a| for a positive integer n? No worries if you can't help. – 2017-02-24
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    I understand $a^n$ < $a$ for 0 < |$a$| < 1 and $a^n$ > $a$ for $a$ > 1. I'm just not seeing how that helps prove either of those statements. – 2017-02-24
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    @johnie4usc I think it'd better if you post your question and show your thoughts and what you've tried. – 2017-02-24
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    I understand. I've been posting lots of questions from my classes so I didn't want to post again. I don't know the limits. Thanks though! – 2017-02-24
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First of all, notice that the irrationnality of $\sqrt{2}$ entails the uniqueness of the $(a,b) \in \mathbb{Z}^{2}$ in the $a+b\sqrt{2}$ notation. This allow you to define $N(a+b\sqrt{2})=a^{2}-2b^{2} \in \mathbb{Z}$. Show that $N(uv)=N(u)N(v)$ for all $x,y \in \mathbb{Z}[\sqrt{2}]$. So, if $u$ is invertible and $v$ is it's inverse, you get $N(u)N(v)=1$ and $N(u),N(v) \in \mathbb{Z}$, so $N(u) \in \{-1,1\}$.

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    I haven't learned norm yet. – 2017-02-24
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    I think you want 'Uniqueness'. not 'unicity', in the first line. – 2017-02-24
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    You have now (LOL). – 2017-02-24
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    @johnie4usc you nevertheless know how it's called :) there's nothing much to know about it than what I suggest you to do in my answer. StevenStadnicki thx, i'm not a native english speaker – 2017-02-24
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    I've been searching on stack exchange and actually found the same question [here](http://math.stackexchange.com/questions/571393/showing-mn-sqrt-2-is-a-unit-in-set-r?rq=1). I'm confused how the first answer notices that $(nx+my)=0$ and $(mx+2ny)=1$. Why is that true? – 2017-02-24
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    @johnie4usc It's the uniqueness of the form $a+b\sqrt{2}$: if this number is equal to $a'+b'\sqrt{2}$ and $b' \neq b$, then $\sqrt{2}=(a-a')/(b'-b) \in \mathbb{Q}$, a contradiction, thus $b'=b$, then $a'=a$. – 2017-02-24
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    Hm. Okay. That doesn't really help me honestly. It seems above my level. – 2017-02-24
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    If $nx + my \ne 0$ then $\sqrt 2 = \frac {1 - mx + 2ny}{nx+ym}$ is rational. – 2017-02-24
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    If $(m + n\sqrt{2})(x + y \sqrt{2}) = 1$ then $mx + 2ny + (nx+ym)\sqrt{2} = 1$. If $nx +ym \ne 0$ then $\sqrt 2 = \frac {1 - mx + 2ny}{nx+ym}$ is rational. So $nx + ym = 0$ and $mx + 2ny + (nx+ym)\sqrt{2} = mx + 2ny = 1$. – 2017-02-24
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    Thanks for the explanation. That makes sense. But, do we know that $\frac{1-mx+2ny}{nx+ym}$ isn't rational? Sorry if that's a dumb question. – 2017-02-24
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    @johnie4usc We know that $\sqrt{2} \notin \mathbb{Q}$, subch as $\sqrt{d}$ for every integer $d$ which is not a square. You need this fact to study your ring. Sketch of proof: if $\sqrt{2}=\frac{a}{b}$ with $a$ an $b$ coprime integers, you get $a^{2}=2b^{2}$, and $\gcd(a,b)=1$ gives, by Gauss lemma, $b=1$, hence $\sqrt{2} \in \mathbb{Z}$, a contradiction. You can show by this way that, if $P$ is an integer-coefficients polynomial with leading coefficient $1$, all it's rationnal roots are integers. – 2017-02-24
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    Yeah I understand that $\sqrt2$ isn't rational. I didn't notice the equality sign. my bad. – 2017-02-24