Describe the units of Z[sqrt 2]

[Math Amateur]

In John Stillwell's book: Elements of Number Theory, Exercise 6.1.2 reads as follows:

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Describe the units of \(\displaystyle \mathbb{Z} [ \sqrt{2}]\)

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Now ... \(\displaystyle \mathbb{Z} [ \sqrt{2}] = \{ a + b \sqrt{2} \ | \ a,b \in \mathbb{Z} \}\)We have that \(\displaystyle (a_1 + b_1 \sqrt{2} )\) is a unit of \(\displaystyle \mathbb{Z} [ \sqrt{2}]\) if there exists an element \(\displaystyle (a_2 + b_2 \sqrt{2} )\) in \(\displaystyle \mathbb{Z} [ \sqrt{2}]\) such that:

\(\displaystyle (a_1 + b_1 \sqrt{2} ) (a_2 + b_2 \sqrt{2} ) = (a_2 + b_2 \sqrt{2} ) (a_1 + b_1 \sqrt{2} ) = 1 = 1 + 0 \sqrt{2}\)So then, to determine all the units of \(\displaystyle \mathbb{Z} [ \sqrt{2}]\) we need to determine all the solutions of the following equation:

\(\displaystyle (a_1 + b_1 \sqrt{2} ) (a_2 + b_2 \sqrt{2} ) = 1 = 1 + 0 \sqrt{2}\)

So proceeding, we have:

\(\displaystyle (a_1 + b_1 \sqrt{2} ) (a_2 + b_2 \sqrt{2} ) = 1 = 1 + 0 \sqrt{2}\)\(\displaystyle \Longrightarrow ( a_1a_2 + 2b_1b_2) + ( a_1b_2 + a_2b_1) \sqrt{2} = 1 + 0 \sqrt{2}\)\(\displaystyle \Longrightarrow a_1a_2 + 2b_1b_2 = 1\)

and \(\displaystyle a_2 + b_2 \sqrt{2} = 0\)But ... where to from here ... ?... ... unlike the case for \(\displaystyle \mathbb{Z} \) a straightforward application on the two square identity does not seem to work ...Can someone please help?

Peter

 

[mathbalarka]

A norm over the integral domain $\Bbb Z[\sqrt{2}]$ is defined as $N : \Bbb Z[\sqrt{2}] \to \Bbb Z$ by $N(a + b\sqrt{2}) = a^2 - 2b^2$. Note that a norm is multiplicative : $N(\alpha \beta) = N(\alpha)N(\beta)$ (exercise for you).

Thus, if $\alpha = a + b\sqrt{2}$ is a unit of $\Bbb Z[\sqrt{2}]$, i.e., $\alpha \beta = 1$ for some other $\beta$ in the domain, $N(\alpha\beta) = N(1) = 1$ by definition. But then norm is multiplicative, so $N(\alpha)N(\beta) = 1$ and $N(\alpha)$ is a unit of $\Bbb Z$.

The only units of $\Bbb Z$ are $1$ and $-1$, so $N(\alpha) = \pm 1$, hence $a^2 - 2b^2 = \pm 1$. This is actually known as a Pell equation. A plausible way to do this would be to note that $(a, b) = (1, 0)$ is a solution, so a unit is $1 + \sqrt{2}$. Thus, you automatically have infinitely many units $(1 + \sqrt{2})^k$ for positive integers $k$. What else can you think of?

 

[Math Amateur]

A norm over the integral domain $\Bbb Z[\sqrt{2}]$ is defined as $N : \Bbb Z[\sqrt{2}] \to \Bbb Z$ by $N(a + b\sqrt{2}) = a^2 - 2b^2$. Note that a norm is multiplicative : $N(\alpha \beta) = N(\alpha)N(\beta)$ (exercise for you).

Thus, if $\alpha = a + b\sqrt{2}$ is a unit of $\Bbb Z[\sqrt{2}]$, i.e., $\alpha \beta = 1$ for some other $\beta$ in the domain, $N(\alpha\beta) = N(1) = 1$ by definition. But then norm is multiplicative, so $N(\alpha)N(\beta) = 1$ and $N(\alpha)$ is a unit of $\Bbb Z$.

The only units of $\Bbb Z$ are $1$ and $-1$, so $N(\alpha) = \pm 1$, hence $a^2 - 2b^2 = \pm 1$. This is actually known as a Pell equation. A plausible way to do this would be to note that $(a, b) = (1, 0)$ is a solution, so a unit is $1 + \sqrt{2}$. Thus, you automatically have infinitely many units $(1 + \sqrt{2})^k$ for positive integers $k$. What else can you think of?

Thanks Mathbalarka ... it is great to have your help ...

It is late now here in Tasmania ... but I will be working through your post in the morning and also dealing with some new number theory ideas ...

Thanks again ... and goodnight from Tasmania ...

Peter

 

[mathbalarka]

Good night, sleep tight, don't let the Tasmanian monster bi... whoops did I violate Rule #12? (Lipssealed)

 

[Math Amateur]

Good night, sleep tight, don't let the Tasmanian monster bi... whoops did I violate Rule #12? (Lipssealed)

Just going in going to sleep Mathbalarka ... Did you mean the Tasmanian Devil? ... Yes it is a real wild creature in the bush in Tasmania ...

Good night,

Peter

Hope I am not violating Rule 12!

 

[Math Amateur]

A norm over the integral domain $\Bbb Z[\sqrt{2}]$ is defined as $N : \Bbb Z[\sqrt{2}] \to \Bbb Z$ by $N(a + b\sqrt{2}) = a^2 - 2b^2$. Note that a norm is multiplicative : $N(\alpha \beta) = N(\alpha)N(\beta)$ (exercise for you).

Thus, if $\alpha = a + b\sqrt{2}$ is a unit of $\Bbb Z[\sqrt{2}]$, i.e., $\alpha \beta = 1$ for some other $\beta$ in the domain, $N(\alpha\beta) = N(1) = 1$ by definition. But then norm is multiplicative, so $N(\alpha)N(\beta) = 1$ and $N(\alpha)$ is a unit of $\Bbb Z$.

The only units of $\Bbb Z$ are $1$ and $-1$, so $N(\alpha) = \pm 1$, hence $a^2 - 2b^2 = \pm 1$. This is actually known as a Pell equation. A plausible way to do this would be to note that $(a, b) = (1, 0)$ is a solution, so a unit is $1 + \sqrt{2}$. Thus, you automatically have infinitely many units $(1 + \sqrt{2})^k$ for positive integers $k$. What else can you think of?

Can someone please tell me if a norm in a Euclidean Domain is the same as the Euclidean Degree Function ...

... and a further question ... how is a particular norm determined ... I think Euclidean Domains can have several norms ... so it it anything that fits with the degree function and suits our purposes ...

I would really appreciate some clarification ...

Peter

 

[Euge]

Can someone please tell me if a norm in a Euclidean Domain is the same as the Euclidean Degree Function ...

... and a further question ... how is a particular norm determined ... I think Euclidean Domains can have several norms ... so it it anything that fits with the degree function and suits our purposes ...

I would really appreciate some clarification ...

Peter

By the degree function, you mean the map that sends every polynomial to its degree, right? If that's the case, then you can see that the answer to the first question is no by considering the absolute value norm on $\Bbb Z$. In fact, given an arbitrary euclidean domain $D$ and any norm $N$ defined on it, we can create infinitely many norms from $N$ by setting $N_m(x) = mN(x)$ for all $m\in \Bbb Z^+$ and $x\in D\setminus\{0\}$.

 

[Math Amateur]

By the degree function, you mean the map that sends every polynomial to its degree, right? If that's the case, then you can see that the answer to the first question is no by considering the absolute value norm on $\Bbb Z$. In fact, given an arbitrary euclidean domain $D$ and any norm $N$ defined on it, we can create infinitely many norms from $N$ by setting $N_m(x) = mN(x)$ for all $m\in \Bbb Z^+$ and $x\in D\setminus\{0\}$.

Hi Euge,

Thanks for your post ... ... and in particular thanks for the interesting point that " ... ... given an arbitrary euclidean domain $D$ and any norm $N$ defined on it, we can create infinitely many norms from $N$ by setting $N_m(x) = mN(x)$ for all $m\in \Bbb Z^+$ and \(\displaystyle x\in D\setminus\{0\ \)To answer you question though ... ... No I did not mean that ... I was having a look at Joseph Rotman's book: Advanced Modern Algebra - specifically, the Section on Euclidean Rings and Principal Ideal Domains, and found the following:

https://www.physicsforums.com/attachments/3399

Mind you, I should have read past a few examples and onto the next page and then I would have found his answer to my question - as follows:

View attachment 3400

Your post caused me to go back to Rotman and read more carefully ... thanks again ...

By the way, when helped by you and Mathbalarka on number theory problems lately I have begun to wonder how in particular examples, such as the Eisenstein Domain, \(\displaystyle \mathbb{Z} + \mathbb{Z} \omega\), you choose a particular norm in order to solve a problem, such as determining the units. I suppose it is intuition and insight based on the particular problem structure ... is that right?

Peter

 

[Euge]

Hi Euge,

Thanks for your post ... ... and in particular thanks for the interesting point that " ... ... given an arbitrary euclidean domain $D$ and any norm $N$ defined on it, we can create infinitely many norms from $N$ by setting $N_m(x) = mN(x)$ for all $m\in \Bbb Z^+$ and \(\displaystyle x\in D\setminus\{0\ \)To answer you question though ... ... No I did not mean that ... I was having a look at Joseph Rotman's book: Advanced Modern Algebra - specifically, the Section on Euclidean Rings and Principal Ideal Domains, and found the following:

https://www.physicsforums.com/attachments/3399

Mind you, I should have read past a few examples and onto the next page and then I would have found his answer to my question - as follows:

View attachment 3400

Your post caused me to go back to Rotman and read more carefully ... thanks again ...

By the way, when helped by you and Mathbalarka on number theory problems lately I have begun to wonder how in particular examples, such as the Eisenstein Domain, \(\displaystyle \mathbb{Z} + \mathbb{Z} \omega\), you choose a particular norm in order to solve a problem, such as determining the units. I suppose it is intuition and insight based on the particular problem structure ... is that right?

Peter

There are few things to say about all this. When I used the term "norm" in this thread I removed the multiplicative condition, so it would be the same as a valuation, or degree function in your case. Having said that, to have $N_r$ to be a norm in Rotman's sense, $r$ needs to be idempotent, i.e., $r^2 = r$. So there may only be finitely many (multiplicative) norms of the form $N_r$ on a given integral domain $D$.

In the problems you've posted recently, finding a norm of an element $r$ can be obtained by computing the product of all the Galois conjugates of $r$ (by that I mean the product of the roots of the minimal polynomial of $r$) or by computing the determinant of the mapping $x \mapsto rx$. I used the latter to get $N(a + b\omega) = a^2 - ab + b^2$ in $\Bbb Z + \Bbb Z \omega$, but used the former to get $N(a + b\sqrt{n}) = a^2 - nb^2$ in $\Bbb Z[\sqrt{n}]$.

 

[Math Amateur]

There are few things to say about all this. When I used the term "norm" in this thread I removed the multiplicative condition, so it would be the same as a valuation, or degree function in your case. Having said that, to have $N_r$ to be a norm in Rotman's sense, $r$ needs to be idempotent, i.e., $r^2 = r$. So there may only be finitely many (multiplicative) norms of the form $N_r$ on a given integral domain $D$.

In the problems you've posted recently, finding a norm of an element $r$ can be obtained by computing the product of all the Galois conjugates of $r$ (by that I mean the product of the roots of the minimal polynomial of $r$) or by computing the determinant of the mapping $x \mapsto rx$. I used the latter to get $N(a + b\omega) = a^2 - ab + b^2$ in $\Bbb Z + \Bbb Z \omega$, but used the former to get $N(a + b\sqrt{n}) = a^2 - nb^2$ in $\Bbb Z[\sqrt{n}]$.

Well ... ... thanks Euge ... indeed ... that gives me a lot to think about ... obviously I need to back up a bit and cover some theory ...

Thanks so much for your help ...

Peter

 

[Euge]

Well ... ... thanks Euge ... indeed ... that gives me a lot to think about ... obviously I need to back up a bit and cover some theory ...

Thanks so much for your help ...

Peter

Let me show you in more detail how I found a norm on $\Bbb Z + \Bbb Z\omega$. Fix an element $a + b\omega\in \Bbb Z + \Bbb Z\omega$. To find $N(a + b\omega)$, I computed the determinant of the transformation $L : c + d\omega \mapsto (a + b\omega)(c + d\omega)$. Since

$(a + b\omega)(c + d\omega) = (ac - bd) + (ad + bc - bd)\omega$

for all $c + d\omega$ in $\Bbb Z + \Bbb Z\omega$, the matrix of $L$ with respect to $\{1, \omega\}$ is

\(\displaystyle \begin{pmatrix}a & -b\\ b & a - b\end{pmatrix}\)

Its determinant is

\(\displaystyle a(a - b) - b(-b) = a^2 - ab + b^2\)

Therefore, $N(a + b\omega) = a^2 - ab + b^2$.

 

[Math Amateur]

Let me show you in more detail how I found a norm on $\Bbb Z + \Bbb Z\omega$. Fix an element $a + b\omega\in \Bbb Z + \Bbb Z\omega$. To find $N(a + b\omega)$, I computed the determinant of the transformation $L : c + d\omega \mapsto (a + b\omega)(c + d\omega)$. Since

$(a + b\omega)(c + d\omega) = (ac - bd) + (ad + bc - bd)\omega$

for all $c + d\omega$ in $\Bbb Z + \Bbb Z\omega$, the matrix of $L$ with respect to $\{1, \omega\}$ is

\(\displaystyle \begin{pmatrix}a & -b\\ b & a - b\end{pmatrix}\)

Its determinant is

\(\displaystyle a(a - b) - b(-b) = a^2 - ab + b^2\)

Therefore, $N(a + b\omega) = a^2 - ab + b^2$.

Thanks for the insight into your working Euge ... Still reflecting over the implications of your post ...

Peter

 

[Euge]

Let me just add that the multiplicative property of $N$ follows from the homomorphism property of determinants. Given $\alpha \in \Bbb Z + \Bbb Z\omega$, let $L_\alpha : x \mapsto \alpha x$. Then

\(\displaystyle N(\alpha\beta) = \text{det}(L_{\alpha\beta}) = \text{det}(L_\alpha \circ L_\beta) = \text{det}(L_\alpha) \text{det}(L_\beta) = N(\alpha)N(\beta).\)

 

[MupptMath]

A norm over the integral domain $\Bbb Z[\sqrt{2}]$ is defined as $N : \Bbb Z[\sqrt{2}] \to \Bbb Z$ by $N(a + b\sqrt{2}) = a^2 - 2b^2$. Note that a norm is multiplicative : $N(\alpha \beta) = N(\alpha)N(\beta)$ (exercise for you).

Thus, if $\alpha = a + b\sqrt{2}$ is a unit of $\Bbb Z[\sqrt{2}]$, i.e., $\alpha \beta = 1$ for some other $\beta$ in the domain, $N(\alpha\beta) = N(1) = 1$ by definition. But then norm is multiplicative, so $N(\alpha)N(\beta) = 1$ and $N(\alpha)$ is a unit of $\Bbb Z$.

The only units of $\Bbb Z$ are $1$ and $-1$, so $N(\alpha) = \pm 1$, hence $a^2 - 2b^2 = \pm 1$. This is actually known as a Pell equation. A plausible way to do this would be to note that $(a, b) = (1, 0)$ is a solution, so a unit is $1 + \sqrt{2}$. Thus, you automatically have infinitely many units $(1 + \sqrt{2})^k$ for positive integers $k$. What else can you think of?

Hi, I'm new here - just to clarify my understanding:

How does (a,b)=(1,0) get to 1+sqrt{2}? Do you mean (a,b)=(1,1) is a solution so that 1+1*sqrt{2} is a unit?
In order to generalize solutions to (1+sqrt{2})^k, k in Z, do we think about is as such.. if a is a unit, there exists b, such that ab=1. Therefore, aa will be a unit and have bb as an inverse since aabb=abab=1*1=1?

 

[Deveno]

Hi, I'm new here - just to clarify my understanding:

How does (a,b)=(1,0) get to 1+sqrt{2}? Do you mean (a,b)=(1,1) is a solution so that 1+1*sqrt{2} is a unit?
In order to generalize solutions to (1+sqrt{2})^k, k in Z, do we think about is as such.. if a is a unit, there exists b, such that ab=1. Therefore, aa will be a unit and have bb as an inverse since aabb=abab=1*1=1?

It appears to be a typo on mathbalarka's part. It should be $(a,b) = (1,1)$ which has a norm of -1. It is not hard to see that its inverse is $-1+\sqrt{2}$, since:

$(1 + \sqrt{2})(-1 + \sqrt{2}) = -1 + \sqrt{2} - \sqrt{2} + 2 = 2 - 1 = 1$.

Your second conjecture is correct, since $\Bbb Z[\sqrt{2}]$ is a sub-ring of $\Bbb R$ it is necessarily commutative (this can also be shown directly from the commutativity of $\Bbb Z$, and assuming $\sqrt{2}$ commutes with all of $\Bbb Z$ -a subtle assumption that derives from the assumption that $x$ commutes with $\Bbb Z$ in the polynomial ring $\Bbb Z[x]$), and for such a ring we have:

$(xy)^k = x^ky^k$, for any positive integer $k$, so if $x$ is a unit (with inverse $y$) then so is $x^k$ (with inverse $y^k$).

(Note: even if our ring $R$ is non-commutative, as long as it is associative, we still have $x \in U(R) \implies x^k \in U(R)$, by induction:

$x \in U(R) \implies \exists y\in R: xy = yx = 1_R$

Assume $x^{k-1}y^{k-1} = y^{k-1}x^{k-1} = 1_R$.

Then $x^ky^k = x(x^{k-1})(y^{k-1})y = x(x^{k-1}y^{k-1})y = x(1_R)y = xy = 1_R$, and:

$y^kx^k = y(y^{k-1})(x^{k-1})x = y(y^{k-1}x^{k-1})x = y(1_R)x = yx = 1_R$).

Even more generally, we have $x \in U(R) \implies x^k \in U(R)$ by the fact the $U(R)$ is a group, which is closed under multiplication.

 

[didigames]

It appears to be a typo on mathbalarka's part. It should be $(a,b) = (1,1)$ which has a norm of -1. It is not hard to see that its inverse is $-1+\sqrt{2}$, since:

$(1 + \sqrt{2})(-1 + \sqrt{2}) = -1 + \sqrt{2} - \sqrt{2} + 2 = 2 - 1 = 1$.

Your second conjecture is correct, since $\Bbb Z[\sqrt{2}]$ is a sub-ring of $\Bbb R$ it is necessarily commutative (this can also be shown directly from the commutativity of $\Bbb Z$, and assuming $\sqrt{2}$ commutes with all of $\Bbb Z$ -a subtle assumption that derives from the assumption that $x$ commutes with $\Bbb Z$ in the polynomial ring $\Bbb Z[x]$), and for such a ring we have:

$(xy)^k = x^ky^k$, for any positive integer $k$, so if $x$ is a unit (with inverse $y$) then so is $x^k$ (with inverse $y^k$).

(Note: even if our ring $R$ is non-commutative, as long as it is associative, we still have $x \in U(R) \implies x^k \in U(R)$, by induction:

$x \in U(R) \implies \exists y\in R: xy = yx = 1_R$

Assume $x^{k-1}y^{k-1} = y^{k-1}x^{k-1} = 1_R$.

Then $x^ky^k = x(x^{k-1})(y^{k-1})y = x(x^{k-1}y^{k-1})y = x(1_R)y = xy = 1_R$, and:

$y^kx^k = y(y^{k-1})(x^{k-1})x = y(y^{k-1}x^{k-1})x = y(1_R)x = yx = 1_R$).

Even more generally, we have $x \in U(R) \implies x^k \in U(R)$ by the fact the $U(R)$ is a group, which is closed under multiplication.

Thank you. I like it