Show that ##\mathbb{Z}[\sqrt 3]## is a Unique factorisation domain

[chwala]

Homework Statement

see attached

Relevant Equations

Ring Theory

I am studying this now...i just want to check that my reasoning is fine,...

$[1+\sqrt{-3}]$ can be factorised to

$[2⋅-1]$ and $-1$ is a unit and $2$ is irreducible, therefore $\mathbb{z}[\sqrt 3]$ is a UFD.

 

[fresh_42]

Homework Statement:: see attached
Relevant Equations:: Ring Theory

View attachment 323812

I am studying this now...i just want to check that my reasoning is fine,...

$[1+\sqrt{-3}]$ can be factorised to

$[2⋅-1]$ and $-1$ is a unit and $2$ is irreducible.

There is no element $\sqrt{-3}$ in $\mathbb{Z}[\sqrt{3}]$.
A proof needs three parts: integral domain, representation of non-units, and uniqueness.

 

[chwala]

Thanks @fresh_42 i made a typo mistake there...let me amend...this is quite an interesting area to read on...really enjoying it...for integral domain is just to note that there are no zero divisors which i quite understand...

 

[chwala]

Homework Statement:: see attached
Relevant Equations:: Ring Theory

View attachment 323812

I am studying this now...i just want to check that my reasoning is fine,...

$[1+\sqrt{-3}]$ can be factorised to

$[2⋅-1]$ and $-1$ is a unit and $2$ is irreducible, therefore $\mathbb{z}[\sqrt 3]$ is a UFD.

I am studying this now...i just want to check that my reasoning is fine,...

$[1+\sqrt{3}]$ can be factorised to

$[2⋅-1]$ and $-1$ is a unit and $2$ is irreducible, therefore $\mathbb{z}[\sqrt 3]$ is a UFD.

 

[fresh_42]

$1+\sqrt{3}$ is probably irreducible.

a) integral domain
\begin{align*}
0&=(a+b\sqrt{3})\cdot (c+d\sqrt{3})=(ac+3bd) + (ad+bc)\sqrt{3} \\
&\Rightarrow ac+3bd=0 \wedge ad+bc=0 \\
&\Rightarrow 0=acd + 3bd^2= -bc^2+3bd^2\\
&\Rightarrow 0=b(3d^2-c^2)
\end{align*}

If $3d^2=c^2$ then $3$ occurs unevenly many times on the left and on the right. This case is therefore impossible. If $b=0$ then $0=ac+ad\sqrt{3},$ so either $a=0$ or $c=d=0.$

I would first determine the units of $\mathbb{Z}[\sqrt{3}].$
Say $(a+b\sqrt{3})\cdot (c+d\sqrt{3})=1.$
Then $ac+3bd=1 \wedge ad+bc=0$ etc.

 

[martinbn]

I am studying this now...i just want to check that my reasoning is fine,...

$[1+\sqrt{3}]$ can be factorised to

$[2⋅-1]$ and $-1$ is a unit and $2$ is irreducible, therefore $\mathbb{z}[\sqrt 3]$ is a UFD.

Why would you think that this is enough! To prove that it is an UFD, every element needs to factorize in a unique way. Not just one element!

 

[chwala]

Just to share on my further analysis of this...; with your indulgence of course, i noted for UFD's, the non zero unit has to be written as a product of irreducibles. I noted that $pε\mathbb{R}$ is irreducible if;

1. $p≠0$

2. $p=a⋅b$ where $a$ or $b$ is a unit.

Now going to the attached pdf on page $1$ we can say that,

$\mathbb{z}[\i]$ is a UFD because;

we can have for example,

$\mathbb{z}[\i]=(1+i)(1-i)=2⋅1$

$\mathbb{z}[\i]=(10+i)(10-i)=101⋅1$

Now coming to,

$\mathbb{z}[\sqrt 2]=(1+\sqrt 2)⋅(1-\sqrt 2)=1-2=1⋅-1$

$\mathbb{z}[\sqrt 2]=(14+\sqrt 2)⋅(14-\sqrt 2)=196-2=194=97⋅2$

$\mathbb{z}[\sqrt 2]=(17+\sqrt 2)⋅(17-\sqrt 2)=287⋅1$

Now coming to,

$\mathbb{z}[\sqrt 3]=(11+[\sqrt 3)⋅(11-[\sqrt 3)=121-3=59⋅2$

The integer $2$ above can be written as a Norm in $\mathbb{z}$ i.e in the form;

$a^2+b^2$ where $n=(a+bi)(a-bi)$ in essence,

$2=(1+i)(1-i)$

...before i continue further i think i need to wait for input...

 

[fresh_42]

Just to share on my further analysis of this...; with your indulgence of course, i noted for UFD's, the non zero unit has to be written as a product of irreducibles. I noted that $pε\mathbb{R}$ is irreducible if;

1. $p≠0$

2. $p=a⋅b$ where $a$ or $b$ is a unit.

Now going to the attached pdf on page $1$ we can say that,

$\mathbb{z}[\i]$ is a UFD because;

we can have for example,

$\mathbb{z}[\i]=(1+i)(1-i)=2⋅1$

$\mathbb{z}[\i]=(10+i)(10-i)=101⋅1$

Now coming to,

$\mathbb{z}[\sqrt 2]=(1+\sqrt 2)⋅(1-\sqrt 2)=1-2=1⋅-1$

$\mathbb{z}[\sqrt 2]=(14+\sqrt 2)⋅(14-\sqrt 2)=196-2=194=97⋅2$

$\mathbb{z}[\sqrt 2]=(17+\sqrt 2)⋅(17-\sqrt 2)=287⋅1$

Now coming to,

$\mathbb{z}[\sqrt 3]=(11+[\sqrt 3)⋅(11-[\sqrt 3)=121-3=59⋅2$

The integer $2$ above can be written as a Norm in $\mathbb{z}$ i.e in the form;

$a^2+b^2$ where $n=(a+bi)(a-bi)$ in essence,

$2=(1+i)(1-i)$

...before i continue further i think i need to wait for input...

You seem to be a bit confused. It would be better to concentrate on one specific ring at a time. Let us set $R=\mathbb{Z}_3$ for convenience and because it is the one you underlined in red.

I already showed you why it is an integral domain (a). Did you understand the proof?

How would you determine units?
Are prime elements and irreducible elements the same?
What are they?

 

[WWGD]

This is still just an isolated case, as @martinbn pointed out.

 

[chwala]

You seem to be a bit confused. It would be better to concentrate on one specific ring at a time. Let us set $R=\mathbb{Z}_3$ for convenience and because it is the one you underlined in red.

I already showed you why it is an integral domain (a). Did you understand the proof?

How would you determine units?
Are prime elements and irreducible elements the same?
What are they?

Yes, i was able to follow your proof; i understand the integral domain part; the part i am interested on is the unit part where you have indicated that the product has to satisfy
$(a+b\sqrt3) ⋅ (c+d\sqrt3)=1$ this is the part i want to understand. I seem to get it now, what they probably mean is that if $a,b ε \mathbb{z}$ then we have to check and see that $a$ and $b$ satisfy $a±b\sqrt3=1$ and not necessarily the way i was doing it (considered conjugates as products). For $\mathbb{z}[\sqrt -5]$, we cannot determine the $a$ and $b$ values satisfying
$(a+b\sqrt-5) ⋅ (c+d\sqrt-5)=1$

 

[fresh_42]

Two remarks before I answer.

a) There is probably a more elegant answer than mine, e.g. by using the fact that $\mathbb{Z}[\sqrt{3}]$ is Euclidean, and therefore a principal ideal domain, and therefore a unique factorization domain.

b) Instead, I try a brute force method.

Yes, i was able to follow your proof; i understand the integral domain part; the part i am interested on is the unit part where you have indicated that the product has to satisfy
$(a+b\sqrt3) ⋅ (c+d\sqrt3)=1$ this is the part i want to understand.

Multiplication gets us
$$
1=\underbrace{(a+b\sqrt{3})}_{\neq 0}\cdot \underbrace{(c+d\sqrt{3})}_{\neq 0}=(ac+3bd)+(ad+bc)\sqrt{3} \Longrightarrow ad+bc = 0 \wedge ac+3bd=1
$$
If $a=0$ then $bc=0.$ And since $b\neq 0,$ we have $c=0$ and $3bd=1$ which is impossible, so $a\neq 0.$ By the same argument, we get $c\neq 0.$ Now

\begin{align*}
\dfrac{1}{ac}=1+3\dfrac{b}{a}\cdot \dfrac{d}{c}=1-3 \dfrac{b^2}{a^2} &\Longrightarrow 0=a^2-\dfrac{a}{c}-3 b^2\\
&\Longrightarrow a=\dfrac{1}{2c}\pm \dfrac{1}{2c}\sqrt{1+12b^2c^2} \\
&\Longrightarrow (2ac-1)^2=4a^2c^2-4ac+1= 1+12b^c^2\\
&\Longrightarrow ac(ac-1)=3b^2c^2\\
&\Longrightarrow a^2-3b^2=\dfrac{1}{c} \in \mathbb{Z}\\
&\Longrightarrow c=\pm 1 =a^2-3b^2
\end{align*}

All square number modulo $3$ are either $0$ or $1.$ Since $\pm 1 \equiv a^2 \pmod{3}$ we have $a^2\equiv 1\pmod{3}$ and so $3\,|\,(a-1)$ or $3\,|\,(a+1)$ and $a=3k\pm 1$ for some $k\in \mathbb{Z}$ and only one sign is possible because $3$ cannot divide both. Now, we get $c=\pm 1=(3k\pm 1)^2-3b^2=9k^2\pm 6k +1-3b^2$ which modulo $3$ yields $c=1.$ The same calculations with switched factors gets $a=1$ and
\begin{align*}
1=(1+b\sqrt{3})(1+d\sqrt{3})=1+3bd+(b+d)\sqrt{3} &\Longrightarrow b=-d \wedge 3bd=-3b^2=0\Longrightarrow b=d=0
\end{align*}

This makes the units very easy, namely $\{\pm 1\}.$

I would use the Euclidean algorithm with the quadratic norm to show the existence of the factorization.

 

[chwala]

Two remarks before I answer.

a) There is probably a more elegant answer than mine, e.g. by using the fact that $\mathbb{Z}[\sqrt{3}]$ is Euclidean, and therefore a principal ideal domain, and therefore a unique factorization domain.

b) Instead, I try a brute force method.Multiplication gets us
$$
1=\underbrace{(a+b\sqrt{3})}_{\neq 0}\cdot \underbrace{(c+d\sqrt{3})}_{\neq 0}=(ac+3bd)+(ad+bc)\sqrt{3} \Longrightarrow ad+bc = 0 \wedge ac+3bd=1
$$
If $a=0$ then $bc=0.$ And since $b\neq 0,$ we have $c=0$ and $3bd=1$ which is impossible, so $a\neq 0.$ By the same argument, we get $c\neq 0.$ Now

\begin{align*}
\dfrac{1}{ac}=1+3\dfrac{b}{a}\cdot \dfrac{d}{c}=1-3 \dfrac{b^2}{a^2} &\Longrightarrow 0=a^2-\dfrac{a}{c}-3 b^2\\
&\Longrightarrow a=\dfrac{1}{2c}\pm \dfrac{1}{2c}\sqrt{1+12b^2c^2} \\
&\Longrightarrow (2ac-1)^2=4a^2c^2-4ac+1= 1+12b^c^2\\
&\Longrightarrow ac(ac-1)=3b^2c^2\\
&\Longrightarrow a^2-3b^2=\dfrac{1}{c} \in \mathbb{Z}\\
&\Longrightarrow c=\pm 1 =a^2-3b^2
\end{align*}

All square number modulo $3$ are either $0$ or $1.$ Since $\pm 1 \equiv a^2 \pmod{3}$ we have $a^2\equiv 1\pmod{3}$ and so $3\,|\,(a-1)$ or $3\,|\,(a+1)$ and $a=3k\pm 1$ for some $k\in \mathbb{Z}$ and only one sign is possible because $3$ cannot divide both. Now, we get $c=\pm 1=(3k\pm 1)^2-3b^2=9k^2\pm 6k +1-3b^2$ which modulo $3$ yields $c=1.$ The same calculations with switched factors gets $a=1$ and
\begin{align*}
1=(1+b\sqrt{3})(1+d\sqrt{3})=1+3bd+(b+d)\sqrt{3} &\Longrightarrow b=-d \wedge 3bd=-3b^2=0\Longrightarrow b=d=0
\end{align*}

This makes the units very easy, namely $\{\pm 1\}.$

I would use the Euclidean algorithm with the quadratic norm to show the existence of the factorization.

Thanks so much need to read more on this...

 

[SammyS]

Yes, i was able to follow your proof; i understand the integral domain part; the part i am interested on is the unit part where you have indicated that the product has to satisfy
$(a+b\sqrt3) ⋅ (c+d\sqrt3)=1$ this is the part i want to understand. I seem to get it now, what they probably mean is that if $a,b ε \mathbb{z}$ then we have to check and see that $a$ and $b$ satisfy $a±b\sqrt3=1$ and . . .

Do you know what a unit is ? I don't think so. When you wrote about finding $\displaystyle a,\,b \in \mathbb{Z}$ , satisfying $\displaystyle a\pm b\sqrt3=1$, that's referring to unity, i.e. the multiplicative identity. In a ring, a unit is any element having a multiplicative inverse.

In the ring of rational numbers, every non-zero element is a unit.

 

[chwala]

Do you know what a unit is ? I don't think so. When you wrote about finding $\displaystyle a,/,b \in \mathbb{Z}$ , satisfying $\displaystyle a\pm b\sqrt3=1$, that's referring to unity, i.e. the multiplicative identity. In a ring, a unit is any element having a multiplicative inverse.

In the ring of rational numbers, every non-zero element is a unit.

Correct, @SammyS I didn't get that right! You have given me insight...yes, I am aware of the Ring properties...
I guess, the challenge i had here was on the language used In reference to unity or unit in that case- didn't see the whole picture on Multiplicative inverse...

My understanding of unit as used in my context was totally misplaced!

 

[SammyS]

@chwala ,

You may find it more direct to find a unit (other than $1,\,-1$) by finding the multiplicative inverse of the general element, $\displaystyle a+b\sqrt 3$, and tailoring the values of $a \text{ and } b$ as needed.

Again, working in the ring $\displaystyle \mathbb{Z}[\sqrt{3}] \,$ .

$\displaystyle \dfrac{1}{a+b\sqrt 3 } = \dfrac{(a-b\sqrt 3 \,)}{(a+b\sqrt 3 \,)(a-b\sqrt 3 \,)} = \dfrac{(a-b\sqrt 3 \,)}{a^2-3b^2}$

It seems to me that the denominator needs to be $\pm 1$. It certainly would be convenient if we could find values for $a$ and $b$ which make it $1$ . Finding such values would mean that the inverse of the element is the element's conjugate .

We want $\displaystyle a^2-3b^2 = 1$ .

Borrowing the idea from @fresh_42 , let's look at this modulo $3$.

$\displaystyle a^2\equiv 1\pmod{3}$.

$a^2 = 1$ doesn't appear to work well. It just gives $b=0$ .

Trying $a^2 = 4$ gives $\displaystyle 4-3b^2 = 1$ , so $b=\pm1$.

So we have that $\displaystyle (2+1\sqrt 3 \,)(2-1\sqrt 3 \,) = 1$

Both $(2+1\sqrt 3) \text { and } (2-1\sqrt 3)$ are units in $\displaystyle \mathbb{Z}[\sqrt{3}\,] \,$.

 

[chwala]

@chwala ,

You may find it more direct to find a unit (other than $1,\,-1$) by finding the multiplicative inverse of the general element, $\displaystyle a+b\sqrt 3$, and tailoring the values of $a \text{ and } b$ as needed.

Again, working in the ring $\displaystyle \mathbb{Z}[\sqrt{3}] \,$ .

$\displaystyle \dfrac{1}{a+b\sqrt 3 } = \dfrac{(a-b\sqrt 3 \,)}{(a+b\sqrt 3 \,)(a-b\sqrt 3 \,)} = \dfrac{(a-b\sqrt 3 \,)}{a^2-3b^2}$

It seems to me that the denominator needs to be $\pm 1$. It certainly would be convenient if we could find values for $a$ and $b$ which make it $1$ . Finding such values would mean that the inverse of the element is the element's conjugate .

We want $\displaystyle a^2-3b^2 = 1$ .

Borrowing the idea from @fresh_42 , let's look at this modulo $3$.

$\displaystyle a^2\equiv 1\pmod{3}$.

$a^2 = 1$ doesn't appear to work well. It just gives $b=0$ .

Trying $a^2 = 4$ gives $\displaystyle 4-3b^2 = 1$ , so $b=\pm1$.

So we have that $\displaystyle (2+1\sqrt 3 \,)(2-1\sqrt 3 \,) = 1$

Both $(2+1\sqrt 3) \text { and } (2-1\sqrt 3)$ are units in $\displaystyle \mathbb{Z}[\sqrt{3}\,] \,$.

I see ...

Using the same approach we can find the multiplicative inverse of $a+b\sqrt 2$ in a similar manner i.e

Trying $a=3$ gives us $a^2 = 9$ thus

$\displaystyle 9-3b^2 = 1$ , so $b=2$.

...

thus $\displaystyle (3+2\sqrt 2 \,)(3-2\sqrt 2 \,) = 1$

Both $\displaystyle (3+2\sqrt 2 \,)$ and $(3-2\sqrt 2 \,)$ are units in $\mathbb{z}[\sqrt 2]$