Finding the Units of the Ring $R=\mathbb{Z}[w]$

[evinda]

Hello! :D
I am given the following exercise and I have some questions

Calculate the units of $R=\mathbb{Z}[w]=\{ a+bw, a,b \in \mathbb{Z}\} , w=\frac{-1+i \sqrt{3}}{2}$.

In my notes is the following solution:
We remark that $w$ is a root of $x^2+x+1$
So, $w^2+w+1=0 \Rightarrow w^2=-w-1$
$x^3-1=(x-1)(x^2+x+1)$
$w$ is a root of $x^2+x+1$,so $w^3-1=0 \Rightarrow w^3=1 \Rightarrow w^4=w$
$(w^2)^2+w^2+1=w^4+w^2+1=w-w-1+1=0$,so $w^2$ is a root of $x^2+x+1$.
So, $w^2, w$ are conjugate.
Let $x=a+bw \in \mathbb{Z}[w], a,b \in \mathbb{Z} $ an unit.
So,there is $(c+dw) \in \mathbb{Z}[w] $ such that $ (a+bw)(c+dw)=1 $
We take also the conjugate and we get $(a+bw^2)(c+dw^2)=1$
$(a+bw)(a+bw^2)(c+dw)(c+dw^2)=1 \Rightarrow ... \Rightarrow (a^2-ab+b^2)(c^2-cd+d^2)=1 \Rightarrow 4(a^2-ab+b^2)=4a^2-4ab+4b^2=(2a)^2-2 \cdot 2ab + b^2+3b^2=(2a-b)^2+3b^2 >0$
So,$a^2-ab+b^2>0$
So,$a^2-ab+b^2=1$
$4(a^2-ab+b^2)=4 \Rightarrow (2a-b)^2+3b^2=4$
We take cases:
i)$ (2a-b)^2=0$
$3b^2=4 \Rightarrow b^2=\frac{4}{3} \Rightarrow b= \pm \frac{2}{\sqrt{3}} \not\in \mathbb{Z} $,so we reject it!
ii) $(2a-b)^2=1$
$3b^2=3 \Rightarrow b=\pm 1$

• for $b=1$ : $(2a-1)^2=1 \Rightarrow 2a-1= \pm 1$
  $2a-1=1 \Rightarrow a=1$
  $2a-1=-1 \Rightarrow a=0$
• for $b=-1$ : $(2a+1)^2=1 \Rightarrow 2a+1= \pm 1$
  $2a+1=1 \Rightarrow a=0$
  $2a+1=-1 \Rightarrow a=-1$

iii) $(2a-b)^2=4 \Rightarrow 3b^2=0 \Rightarrow (a,b)=(1,0) , (a,b)=(-1,0)$

We have six couples:
$(a,b)=(1,1): x=1+1 \cdot w=-w^2$
$(a,b)=(0,1): x=w$
$(a,b)=(0,-1): x=-w$
$(a,b)=(-1,1): x=-1-w=w^2$
$(a,b)=(1,0): x=1$
$(a,b)=(-1,0): x=-1$

First of all, how do we know that $w$ is the root of $x^2+x+1$?
Why do we have to know that $w^2$ is also a root of $x^2+x+1$?
At the red part,why the relation is equal to $1$ ?
Why do we take these cases?
(Blush) (Blush)

 

[Deveno]

Hello! :D
I am given the following exercise and I have some questions

Calculate the units of $R=\mathbb{Z}[w]=\{ a+bw, a,b \in \mathbb{Z}\} , w=\frac{-1+i \sqrt{3}}{2}$.

In my notes is the following solution:
We remark that $w$ is a root of $x^2+x+1$
So, $w^2+w+1=0 \Rightarrow w^2=-w-1$
$x^3-1=(x-1)(x^2+x+1)$
$w$ is a root of $x^2+x+1$,so $w^3-1=0 \Rightarrow w^3=1 \Rightarrow w^4=w$
$(w^2)^2+w^2+1=w^4+w^2+1=w-w-1+1=0$,so $w^2$ is a root of $x^2+x+1$.
So, $w^2, w$ are conjugate.
Let $x=a+bw \in \mathbb{Z}[w], a,b \in \mathbb{Z} $ an unit.
So,there is $(c+dw) \in \mathbb{Z}[w] $ such that $ (a+bw)(c+dw)=1 $
We take also the conjugate and we get $(a+bw^2)(c+dw^2)=1$
$(a+bw)(a+bw^2)(c+dw)(c+dw^2)=1 \Rightarrow ... \Rightarrow (a^2-ab+b^2)(c^2-cd+d^2)=1 \Rightarrow 4(a^2-ab+b^2)=4a^2-4ab+4b^2=(2a)^2-2 \cdot 2ab + b^2+3b^2=(2a-b)^2+3b^2 >0$
So,$a^2-ab+b^2>0$
So,$a^2-ab+b^2=1$
$4(a^2-ab+b^2)=4 \Rightarrow (2a-b)^2+3b^2=4$
We take cases:
i)$ (2a-b)^2=0$
$3b^2=4 \Rightarrow b^2=\frac{4}{3} \Rightarrow b= \pm \frac{2}{\sqrt{3}} \not\in \mathbb{Z} $,so we reject it!
ii) $(2a-b)^2=1$
$3b^2=3 \Rightarrow b=\pm 1$

• for $b=1$ : $(2a-1)^2=1 \Rightarrow 2a-1= \pm 1$
  $2a-1=1 \Rightarrow a=1$
  $2a-1=-1 \Rightarrow a=0$
• for $b=-1$ : $(2a+1)^2=1 \Rightarrow 2a+1= \pm 1$
  $2a+1=1 \Rightarrow a=0$
  $2a+1=-1 \Rightarrow a=-1$

iii) $(2a-b)^2=4 \Rightarrow 3b^2=0 \Rightarrow (a,b)=(1,0) , (a,b)=(-1,0)$

We have six couples:
$(a,b)=(1,1): x=1+1 \cdot w=-w^2$
$(a,b)=(0,1): x=w$
$(a,b)=(0,-1): x=-w$
$(a,b)=(-1,1): x=-1-w=w^2$
$(a,b)=(1,0): x=1$
$(a,b)=(-1,0): x=-1$

First of all, how do we know that $w$ is the root of $x^2+x+1$?
Why do we have to know that $w^2$ is also a root of $x^2+x+1$?
At the red part,why the relation is equal to $1$ ?
Why do we take these cases?
(Blush) (Blush)

First of all note that:

$w = \cos\left(\dfrac{2\pi}{3}\right) + i\sin\left(\dfrac{2\pi}{3}\right)$

so that $w^3 = \cos\left(\dfrac{6\pi}{3}\right) + i\sin\left(\dfrac{6\pi}{3}\right)$

$= \cos(2\pi) + i\sin(2\pi) = 1 + i0 = 1$, by DeMoivre's Theorem

(this is easier to see writing $w = e^{i2\pi/3}$).

So $w$ is a root of $x^3 - 1$, and $w \neq 1$.

Therefore, $w$ must be a root of:

$\dfrac{x^3 - 1}{x - 1} = x^2 + x + 1$.

Now $(w^2)^2 + w^2 + 1 = w^4 + w^2 + 1 = w(w^3) + w^2 + 1$

and since $w^3 = 1$ this is equal to:

$w + w^2 + 1 = w^2 + w + 1 = 0$ ($w^4$ is 8/3 times around the unit circle, which is the same as 2/3 around).

Now we know that:

$a^2 - ab + b^2 > 0$ and also that it is an integer which is a FACTOR of 1. Hence it must be 1, for one has no other (positive integer) factors.

Finally we have two integers that sum to 4:

$k^2 + 3m^2 = 4$

where $k = 2a - b$ and $m = b$.

If $|b| > 1$, then $3b^2 \geq 12 > 4$, so $|b| \leq 1$.

If $|b| = 1$, then $3b^2 = 3$, so we must have $2a - b = \pm 1$.

This gives 4 of the 6 possible cases.

Otherwise, $b = 0$, which forces $(2a - b)^2 = 4a^2 = 4$, so that $a = \pm 1$.

The solution you give works from the premise that:

$(2a - b)^2 \leq 4 \implies (2a - b)^2 = 0,1,\text{ or } 4$ as these are the only non-negative square integers less than or equal to 4.

 

[I like Serena]

Hey hey! ;);)

First of all, how do we know that $w$ is the root of $x^2+x+1$?

What is the solution of $x^2+x+1=0$?
That's how we know.
The point is, $w$ has the typical form of the solution of a quadratic equation, so we can rewrite it in that form.

Why do we have to know that $w^2$ is also a root of $x^2+x+1$?

To be able to conclude that $w$ and $w^2$ are conjugate.
A quadratic equation with real coefficients is guaranteed to have conjugate solutions.
Truth be told, we could have skipped these steps completely.
If we draw $w$ in the complex plane, we can see that its argument (angle) happens to be $2\pi/3$ and it has unit length.
That means that is is one of the solutions of $z^3=1$, which is the desired intermediate result. From there it follows immediately that $w^2$ is the other imaginary solution and therefore the conjugate of $w$.

At the red part,why the relation is equal to $1$ ?

We're dealing with a positive integer.
When multiplying it with another integer, we get $1$.
Therefore both integers have to be $1$.

Why do we take these cases?

To figure out all possible solutions for the units! (Wasntme)

 

[Deveno]

Interestingly enough, the set of units actually forms a group, and as we have seen, this group has order 6. Since it is abelian (because complex multiplication is commutative), and since there are (up to isomorphism), only two groups of order 6: $S_3$ and $C_6$, we know this group must be cyclic ($S_3$ is non-abelian).

It turns out that $-w$ is a generator:

$(-w)^2 = w^2 = \overline{w} \neq 1$
$(-w)^3 = -w^3 = -1 \neq 1$
$(-w)^4 = w^4 = w \neq 1$
$(-w)^5 = -w^5 = -w^2 = -\overline{w} \neq 1$
$(-w)^6 = w^6 = (w^3)^2 = 1^2 = 1$

It follows that $-w$ and $-w^2$ are primitive 6th roots of unity, that is, solutions to:

$\dfrac{x^6 - 1}{(x^3 - 1)(x + 1)} = x^2 - x + 1$.

Convince yourself that $-w^2 = e^{\pi i/3}$.

 

[evinda]

Now we know that:

$a^2 - ab + b^2 > 0$ and also that it is an integer which is a FACTOR of 1. Hence it must be 1, for one has no other (positive integer) factors.

Finally we have two integers that sum to 4:

$k^2 + 3m^2 = 4$

where $k = 2a - b$ and $m = b$.

If $|b| > 1$, then $3b^2 \geq 12 > 4$, so $|b| \leq 1$.

If $|b| = 1$, then $3b^2 = 3$, so we must have $2a - b = \pm 1$.

This gives 4 of the 6 possible cases.

Otherwise, $b = 0$, which forces $(2a - b)^2 = 4a^2 = 4$, so that $a = \pm 1$.

The solution you give works from the premise that:

$(2a - b)^2 \leq 4 \implies (2a - b)^2 = 0,1,\text{ or } 4$ as these are the only non-negative square integers less than or equal to 4.

We're dealing with a positive integer.
When multiplying it with another integer, we get $1$.
Therefore both integers have to be $1$.

To figure out all possible solutions for the units!

I understood so far! (Blush)

Interestingly enough, the set of units actually forms a group, and as we have seen, this group has order 6. Since it is abelian (because complex multiplication is commutative), and since there are (up to isomorphism), only two groups of order 6: $S_3$ and $C_6$, we know this group must be cyclic ($S_3$ is non-abelian).

It turns out that $-w$ is a generator:

$(-w)^2 = w^2 = \overline{w} \neq 1$
$(-w)^3 = -w^3 = -1 \neq 1$
$(-w)^4 = w^4 = w \neq 1$
$(-w)^5 = -w^5 = -w^2 = -\overline{w} \neq 1$
$(-w)^6 = w^6 = (w^3)^2 = 1^2 = 1$

It follows that $-w$ and $-w^2$ are primitive 6th roots of unity, that is, solutions to:

$\dfrac{x^6 - 1}{(x^3 - 1)(x + 1)} = x^2 - x + 1$.

Convince yourself that $-w^2 = e^{\pi i/3}$.

I haven't got taught groups yet.

First of all note that:

$w = \cos\left(\dfrac{2\pi}{3}\right) + i\sin\left(\dfrac{2\pi}{3}\right)$

so that $w^3 = \cos\left(\dfrac{6\pi}{3}\right) + i\sin\left(\dfrac{6\pi}{3}\right)$

$= \cos(2\pi) + i\sin(2\pi) = 1 + i0 = 1$, by DeMoivre's Theorem

(this is easier to see writing $w = e^{i2\pi/3}$).

So $w$ is a root of $x^3 - 1$, and $w \neq 1$.

Therefore, $w$ must be a root of:

$\dfrac{x^3 - 1}{x - 1} = x^2 + x + 1$.

Now $(w^2)^2 + w^2 + 1 = w^4 + w^2 + 1 = w(w^3) + w^2 + 1$

and since $w^3 = 1$ this is equal to:

$w + w^2 + 1 = w^2 + w + 1 = 0$ ($w^4$ is 8/3 times around the unit circle, which is the same as 2/3 around).

What is the solution of $x^2+x+1=0$?
That's how we know.
The point is, $w$ has the typical form of the solution of a quadratic equation, so we can rewrite it in that form.

To be able to conclude that $w$ and $w^2$ are conjugate.
A quadratic equation with real coefficients is guaranteed to have conjugate solutions.

Truth be told, we could have skipped these steps completely.
If we draw $w$ in the complex plane, we can see that its argument (angle) happens to be $2\pi/3$ and it has unit length.
That means that is is one of the solutions of $z^3=1$, which is the desired intermediate result. From there it follows immediately that $w^2$ is the other imaginary solution and therefore the conjugate of $w$.

At the solution of the exercise where do we use the fact that $w$ and $w^2$ are conjugate? I haven't understood yet why we need that... (Blush)

 

[I like Serena]

I understood so far! (Blush)

Good! (Nod)

Let $x=a+bw \in \mathbb{Z}[w], a,b \in \mathbb{Z} $ an unit.
So,there is $(c+dw) \in \mathbb{Z}[w] $ such that $ (a+bw)(c+dw)=1 $
We take also the conjugate and we get $(a+bw^2)(c+dw^2)=1$
$(a+bw)(a+bw^2)(c+dw)(c+dw^2)=1$

At the solution of the exercise where do we use the fact that $w$ and $w^2$ are conjugate? I haven't understood yet why we need that... (Blush)

We want to take the conjugate of $a+bw$ and multiply with it.
Then we'll get a real number, which also happens to be an integer.

But rather than multiplying with $a+b\bar w$, we'd like an expression with only $w$ in it. So we use that $\bar w = w^2$.

 

[evinda]

Good! (Nod)
We want to take the conjugate of $a+bw$ and multiply with it.
Then we'll get a real number, which also happens to be an integer.

But rather than multiplying with $a+b\bar w$, we'd like an expression with only $w$ in it. So we use that $\bar w = w^2$.

I understand...Thank you very much! :p

 

[Deveno]

I find it odd that you are studying rings without any reference to a group, since any ring $R$ forms an abelian (commutative) group under addition. It's rather like studying fractions without any notion of what an integer is (this is possible, as $\Bbb Q$ is actually the field generated by 1 in the real numbers, but it's a bit odd).

In any case, groups are "similar" to rings, but with only "one operation" (usually called "multiplication", although this can be misleading, it's just a binary operation, many kinds of operations might be the operation of a group). This operation of a group satisfies 3 axioms:

$a\ast(b\ast c) = (a\ast b)\ast c$ (associativity) for all $a,b,c \in G$
there exists a "special element" $e \in G$ with $a\ast e = e\ast a = a$ for all $a \in G$ (existence of an identity)
for every $a \in G$ there exists a (unique) $b \in G$ with $a\ast b = b\ast a = e$ (existence of inverses). the element $b$ is typically denoted $a^{-1}$.

The prototypical example of a group is the set of all bijections $S \to S$, for a given set $S$, where the identity is the identity function:

$\text{id}_S: S \to S, \text{id}_S(s) = s$ for all $s \in S$,

and the inverse of a bijection:

$f: S \to S, f(s_k) = t_k$ (here, $k$ functions as an "index" for $S$) is the function:

$g: S \to S, g(t_k) = s_k$

This set of bijections is most interesting when $S$ is a FINITE set, in which case we call the bijections "permutations of $S$", or "the symmetric group on $S$".

For any ring $R$, the set of units of that ring, form a group, called the group of units of $R, U(R)$. For example, in the ring of $n \times n$ matrices with entries in a field $F$, the group of units is the set of invertible matrices, often called "the general linear group of degree $n$ over the field $F$". This is an IMPORTANT example, because the general linear group tells us which matrices correspond to "uniquely solvable" systems of linear equations (over the given field $F$) of $n$ equations in $n$ variables.