Definition of an irreducible element in an integral domain

[fresh_42]

Multiply $s\cdot t = (f + g \sqrt{5} i)\cdot (h + k \sqrt{5} i)$ because it is the number you know it's divisible by $p$.
If you then calculate $\overline{p} \cdot s \cdot t = (- \sqrt{5} i)\cdot s \cdot t$ you get a number that is divisible by $5$ and $5$ is a prime. This should be good for something.

 

[Math Amateur]

Thanks fresh_42 ... will definitely try that ...

Peter

 

[fresh_42]

Thanks fresh_42 ... will definitely try that ...

Peter

I haven't done it, but it's all we have.

 

[Math Amateur]

Just a clarifying question ... why did you multiply by the conjugate of p ... and not p itself ..

Peter

 

[fresh_42]

I tried to get to $\frac{st}{p}$ which is similar to your $w$. The assumption is, that this is in $R$.
And this is the same as $\frac{\overline{p}st}{\overline{p}p}=\frac{\overline{p}st}{5}$. Then I have a prime integer which divides the nominator.
I don't know the rest, maybe taking it modulo $5$ can help, or the consideration of real and imaginary part.
But as I said, only an idea, since I haven't gone further.

 

[fresh_42]

You can also calculate in $R/(p)$ and prove it has no zero divisors.

 

[lavinia]

This can also be solved using the norm squared.

 

[Math Amateur]

Thanks Lavinia ...

 

[Math Amateur]

You can also calculate in $R/(p)$ and prove it has no zero divisors.

Thanks for the suggestion ...

Peter

 

[Math Amateur]

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We wish to prove that $p = \sqrt{5} i$ is prime in $R = \{ a + b \sqrt{5} i \ | \ a, b \in \mathbb{Z} \}$

If $p$ is prime then it is a non-unit and would have the property that whenever $p$ divides $s \cdot t$ we have that $p$ divides $s$ or $t$ ...To show that $p$ is not a unit ... ...

If $p = \sqrt{5} i$ is a unit then $\exists \ x \in R$ such that $px = 1$ ...

Suppose $x = y + z \sqrt{5} i$

... then we have ...

$px = 1$

$\Longrightarrow ( \sqrt{5} i ) ( y + z \sqrt{5} i ) = 1$

$\Longrightarrow N( \sqrt{5} i ) N( y + z \sqrt{5} i ) = N(1) = 1$

$\Longrightarrow 5 ( y^2 + 5 z^2 ) = 1$ ... ... ... ... (1)

Now ... no integers $y, z$ satisfy equation (1) and hence there is no element $x \in R$ such that $px = 1$ ...

So ... $p$ is not a unit ...
To show that $p = \sqrt{5} i$ is prime in $R$

Let $p = \sqrt{5} i$ and $s = f + g \sqrt{5} i$ and $t = h + k \sqrt{5} i$

Suppose $p|st$ ... ...We have ... ...

$p|st$

$\Longrightarrow \exists \ v = c + d \sqrt{5} i$ such that $st = pv$

$\Longrightarrow \exists \ v = c + d \sqrt{5} i$ such that $( f + g \sqrt{5} i ) ( h + k \sqrt{5} i ) = ( \sqrt{5} i ) ( c + d \sqrt{5} i )$
Now ... have to show $p|s$ or $p|t$ ...

Focus on $p|s$ ... ...$p|s$

$\Longrightarrow \exists \ w = m + n \sqrt{5} i$ such that $s = pw$ ... ...

$\Longrightarrow \exists \ w = m + n \sqrt{5} i$ such that $( f + g \sqrt{5} i ) = ( \sqrt{5} i ) ( m + n \sqrt{5} i )$ ... ... ... ... (2)We have to show that there exists $( m + n \sqrt{5} i )$ such that (2) is satisfied ... ...BUT ... how do we proceed ... where do we go from here ...

Can you help ...

Peter

Thanks for the suggestion ...

Peter

=========================================================================================

Try using Lavinia's suggestion to use the norm ...We wish to prove that $p = \sqrt{5} i$ is prime in $R = \{ a + b \sqrt{5} i \ | \ a, b \in \mathbb{Z} \}$

If $p$ is prime then it is a non-unit and would have the property that whenever $p$ divides $s \cdot t$ we have that $p$ divides $s$ or $t$ ...

Have shown that $p$ is not a unit ...

Have to show that whenever $p \mid st$ then $p \mid s$ or $p \mid t$ ... ...

... so ...

Let $p = \sqrt{5} i$ and $s = f + g \sqrt{5} i$ and $t = h + k \sqrt{5} i$

Suppose $p|st$ ... ...We have ... ...

$p|st$

$\Longrightarrow \exists \ v = c + d \sqrt{5} i$ such that $st = pv$

$\Longrightarrow \exists \ c, d \in \mathbb{Z}$ such that $( f + g \sqrt{5} i ) ( h + k \sqrt{5} i ) = ( \sqrt{5} i ) ( c + d \sqrt{5} i )$

$\Longrightarrow \exists \ c, d \in \mathbb{Z}$ such that $N( f + g \sqrt{5} i ) N( h + k \sqrt{5} i ) = N( \sqrt{5} i ) N( c + d \sqrt{5} i )$

$\Longrightarrow \exists \ c, d \in \mathbb{Z}$ such that $(f^2 + 5g^2) (h^2 + 5K^2) = 5(d^2 + 5d^2)$ ... ... ... ... (1)Now ... if (1) holds then we need $p \mid s$ or #p \mid t# ...Suppose #p \mid s# ... ... then we have ... $p|s$

$\Longrightarrow \exists \ w = m + n \sqrt{5} i$ such that $s = pw$ ... ...

$\Longrightarrow \exists \ m, n \in \mathbb{Z} i$ such that $( f + g \sqrt{5} i ) = ( \sqrt{5} i ) ( m + n \sqrt{5} i )$

$\Longrightarrow \exists \ m, n \in \mathbb{Z} i$ such that $N( f + g \sqrt{5} i ) = N( \sqrt{5} i ) N( m + n \sqrt{5} i )$

$\Longrightarrow \exists \ m, n \in \mathbb{Z} i$ such that $(f^2 + 5g^2) = 5(m^2 + 5n^2)$ ... ... ... (2)BUT ... cannot see a way forward from here ...

Maybe assume $p \nmid s$ and $p \nmid t$ ... and then try for contradiction ...?Can you help?

Peter

 

[fresh_42]

To be honest, this is a bit of a mess. (Means, too late here ...)
From $pv=st$ and with the notation $N(a+b\sqrt{5}i)=a^2+5b^2$ (norm squared), you get $N(p)N(v)=N(s)N(t)$.
If you multiply the right hand side and take both sides modulo $5$, you get $5 \,\vert \, f^2h^2$ and therefore $5 \,\vert \, f$ or $5\,\vert \,h$.
Let's say $5\,\vert \, f$. Can you divide $s\, : \,p$ with this information?

 

[lavinia]

To be honest, this is a bit of a mess. (Means, too late here ...)
From $pv=st$ and with the notation $N(a+b\sqrt{5}i)=a^2+5b^2$ (norm squared), you get $N(p)N(v)=N(s)N(t)$.
If you multiply the right hand side and take both sides modulo $5$, you get $5 \,\vert \, f^2h^2$ and therefore $5 \,\vert \, f$ or $5\,\vert \,h$.
Let's say $5\,\vert \, f$. Can you divide $s\, : \,p$ with this information?

Right you get $5$ divides $||f||^2||h||^2$ so since 5 is prime in the integers it divides one of the factors $||f||^2$ or $||h||^2$. Say it divides $||f||^2$.

 

[Math Amateur]

Hi Lavinia, fresh_42

Still puzzling over this exercise ... but just a few thoughts ...

If $5 \mid f^2$ ... then since $5$ is prime, $5 \mid f$ ...Now ... ...

Showing that $p \mid s$ means showing that $\sqrt{5}i \mid ( f+ g \sqrt{5}i )$ ...

But if $\sqrt{5} i \mid f$ and $\sqrt{5}i \mid g \sqrt{5}i$ then $\sqrt{5}i \mid ( f+ g \sqrt{5}i )$ ... ...

BUT ... we certainly have $\sqrt{5}i \mid g \sqrt{5}i$ ... ...

We need to show $\sqrt{5}i \mid f$ ... but how ...

... ... we have that $5 \mid f$ ... but how do we use this, exactly ...Can you help further ...

Peter*** EDIT ***

Just some further thoughts on showing that $\sqrt{5} i \mid f$ ... ...

We have $5 \mid f$ where $f \in \mathbb{Z}$ ...

so ... $f = 5y$ where $y \in \mathbb{Z}$

Thus ...

$f = \sqrt{5} \sqrt{5} y$

$= -i^2 \sqrt{5} \sqrt{5} y$

$= ( \sqrt{5} i ) ( \sqrt{5} i ) (-y)$

... ... hmmm ... not at all sure of this ... but it seems to indicate (demonstrate?) that $\sqrt{5} i \mid f$ ...Is that correct ... or am I sadly way off (it's late here in southern Tasmania ... ... )

Peter

 

[fresh_42]

Right track. Use the distributive law to factor out $p$ from $s$.

Edit: Btw, @lavinia , I like this exercise, as it combines several otherwise abstract concepts.

 

[lavinia]

Hi Lavinia, fresh_42

If $5 \mid f^2$ ... then since $5$ is prime, $5 \mid f$ ...

 

[lavinia]

$5$ is prime in the integers but it is not prime in $R$. In $R$, $5=(\sqrt -5)(-\sqrt -5)$

So you cannot conclude that if $5$ divides $||f||^2$ that it divides $f$.

For instance, $||5+\sqrt -5||^2 = 30$ but $5$ does not divide $5+\sqrt -5$

Note that $f^2$ is not the same as $||f||^2$ unless $f$ is an integer. For instance $|| 6+ \sqrt -5||^2 = 41$ but $(6 + \sqrt -5)^2 = 31 +12\sqrt -5$

 

[fresh_42]

So you cannot conclude that if $5$ divides $||f||^2$ that it divides $f$.

Why not? $5 = ||p||^2, f^2=||f||^2 \in \mathbb{Z} \subseteq \mathbb{R}$ . We left the ring $R$ the moment we applied the norm. And in $\mathbb{Z} \subseteq \mathbb{R}$ we have a prime factor decomposition. As the second factor $y$ is an integer there, it is as well an element of $R$. And because the multiplication rules are identical to those in $\mathbb{R}$, we can transport the result back to $R$.
I don't see a flaw here. The rest in done by the distributive law of $R$ again.

The example doesn't count, because we conclude $5\,\vert \,30$ and $30=5\cdot 6$. We already know that $f \in \mathbb{Z}$.

But even without the deviation to the reals, the prime factor decomposition still holds in the first summand of $R$, which is simply $\mathbb{Z}$. Applying a projection here would help, but is a bit of an overkill.

 

[Math Amateur]

Hi Lavinia...

It would help me if you could comment on fresh_42's post ...

I am still reflecting on the issue that fresh_42 raises but his post looks compelling to me ...

Therefore I would like to hear your analysis...

... by the way ... thanks to you and fresh_42 for all your interesting comments and help ...

Peter

 

[lavinia]

My mistake. I reread Fresh_42's post and saw that $f$ is already an integer.

His point which is exactly right is that if $5$ divides $N(s)$ or $N(t)$ then it must divides $f^2$ or $h^2$ because the other terms are multiples of $5$.
The since $5$ is a prime in the integers it divides $f$ or $h$.

Apologies to both of you for creating confusion. What I was writing was getting to the same point that Fresh_42 is making.

Your argument was correct but as Fresh_42 said you need to clinch it by showing that $\sqrt 5$i divides $s$.

 

[Math Amateur]

Hi Lavinia, fresh_42

Still puzzling over this exercise ... but just a few thoughts ...

If $5 \mid f^2$ ... then since $5$ is prime, $5 \mid f$ ...Now ... ...

Showing that $p \mid s$ means showing that $\sqrt{5}i \mid ( f+ g \sqrt{5}i )$ ...

But if $\sqrt{5} i \mid f$ and $\sqrt{5}i \mid g \sqrt{5}i$ then $\sqrt{5}i \mid ( f+ g \sqrt{5}i )$ ... ...

BUT ... we certainly have $\sqrt{5}i \mid g \sqrt{5}i$ ... ...

We need to show $\sqrt{5}i \mid f$ ... but how ...

... ... we have that $5 \mid f$ ... but how do we use this, exactly ...Can you help further ...

Peter*** EDIT ***

Just some further thoughts on showing that $\sqrt{5} i \mid f$ ... ...

We have $5 \mid f$ where $f \in \mathbb{Z}$ ...

so ... $f = 5y$ where $y \in \mathbb{Z}$

Thus ...

$f = \sqrt{5} \sqrt{5} y$

$= -i^2 \sqrt{5} \sqrt{5} y$

$= ( \sqrt{5} i ) ( \sqrt{5} i ) (-y)$

... ... hmmm ... not at all sure of this ... but it seems to indicate (demonstrate?) that $\sqrt{5} i \mid f$ ...Is that correct ... or am I sadly way off (it's late here in southern Tasmania ... ... )

Peter

============================================================================
============================================================================

Thanks Lavinia ...

But ... see above ... did have some thoughts about proving $p \mid s$ ...Basically ... to repeat my thoughts ...
Showing that $p \mid s$ means showing that $\sqrt{5}i \mid ( f+ g \sqrt{5}i )$ ...

But if $\sqrt{5} i \mid f$ and $\sqrt{5}i \mid g \sqrt{5}i$ then $\sqrt{5}i \mid ( f+ g \sqrt{5}i )$ ... ...

BUT ... we certainly have $\sqrt{5}i \mid g \sqrt{5}i$ ... ...

We need to show $\sqrt{5}i \mid f$ ... but how ...

... ... we have that $5 \mid f$ ... but how do we use this, exactly ...Well ... ...

$5 \mid f \Longrightarrow \exists \ y$ such that $f = 5y$ where $f$ is an integer ... and hence $y$ is an integer ...

So we have ...

$f = \sqrt{5} \sqrt{5} y$

$= -i^2 \sqrt{5} \sqrt{5} y$

$= ( \sqrt{5} i ) ( \sqrt{5} i ) (-y)$

... ... hmmm ... not at all sure of this ... but it seems to indicate (demonstrate?) that $\sqrt{5} i \mid f$ ...Is that correct ... ?

Peter

 

[lavinia]

Correct.

 

[Math Amateur]

Thanks Lavinia ...

... and many thanks to you and fresh_42 for all your help and guidance ... I really appreciate it ...

Peter

 

[lavinia]

Thanks Lavinia ...

... and many thanks to you and fresh_42 for all your help and guidance ... I really appreciate it ...

Peter

I hope working through this exercise clarified the idea of integral domain and the definitions of unit, irreducible, and prime in a ring.
This example can be taken much further if you are interested.

 

[Math Amateur]

Hi Lavinia ... yes, the exercise/example has advanced my understanding ...

I would certainly appreciate the opportunity to take it further ...

Thanks once again for your considerable help ...

Peter

 

[fresh_42]

Hi Peter,

how about this?

A commutative ring $R$ with $1$ is called Boolean, if every element is idempotent, that is $x^2=x$.
1.) Show that $2x=0$ for all $x \in R$.
2.) Every prime ideal $P$ is maximal and $R/P$ is a field with two elements.

 

[Math Amateur]

Hi Peter,

how about this?

A commutative ring $R$ with $1$ is called Boolean, if every element is idempotent, that is $x^2=x$.
1.) Show that $2x=0$ for all $x \in R$.
2.) Every prime ideal $P$ is maximal and $R/P$ is a field with two elements.

Hi fresh_42, Lavinia

We have a Boolean ring $R$ (with $1$) in which, by definition every element is idempotent, that is $x^2=x$.

We wish to show that $2x=0$ for all $x \in R$ ...Proof

Let $x \in R$ ...

$(x + x)^2 = x + x$ because $a^2 = a$ for every element $a \in R$ ... ... including $a = (x + x)$ ...

But ... we also have that $(x + x)^2 = x^2 + 2x + x^2 = x + 2x + x$ ... since $x^2 = x$

Thus $x + x = x + 2x + x$

$\Longrightarrow 2x = 0$Is that correct?

Peter

*** NOTE *** Solution to 2) will follow soon ... (I hope ... ... )

 

[fresh_42]

Yes, that's correct. For the other part, a general hint, which often applies in various situations:
You can start to think from the end. If you know, where you expect to end up, then you can think about what this means and develop ideas what to consider. Sometimes proofs work even in both directions, and it doesn't matter where to start. But even if not, it can help to approach a problem. In the given case, $R/P$ is expected to have two elements. But there are already two elements, which have to be at least in a field.

 

[Math Amateur]

Yes, that's correct. For the other part, a general hint, which often applies in various situations:
You can start to think from the end. If you know, where you expect to end up, then you can think about what this means and develop ideas what to consider. Sometimes proofs work even in both directions, and it doesn't matter where to start. But even if not, it can help to approach a problem. In the given case, $R/P$ is expected to have two elements. But there are already two elements, which have to be at least in a field.

=======================================================================================================================

Hi fresh_42, Lavinia

Now to prove:

Every prime ideal $P$ is maximal and $R/P$ is a field with two elements.

========================================================================================

Establish some Lemmas first ...Lemma 1 Let $R$ be a Boolean ring ... ($R$ is a ring with $1$ in which every element is idempotent ... i.e. $x^2 = x \ \forall \ x \in R$)

If $P$ is a prime ideal then $R/P$ is a Boolean ring ...

Proof

Let $x \in R$ and then consider the coset $(x +P) \in R/P$

We have $(x+ P)(x+P) = (x^2 + P) = (x+P)$ ...

Also note that we have $(1 + P) \in R/P$

Therefore $R/P$ is Boolean ...

================================

Will do separate Lemmas in separate posts for clarity and ease of comments on them ...

Is above Lemma correct?

Peter

 

[fresh_42]

Yes it is. And you know something else about quotients with a prime ideal.

 

[Math Amateur]

Hi fresh_42, Lavinia

Here is Lemma 2 ...

===================================

Lemma 2 Let $R$ be a Boolean ring ... ($R$ is a ring with $1$ in which every element is idempotent ... i.e. $x^2 = x \ \forall \ x \in R$)

If $P$ is a prime ideal then $R/P$ is an integral domain ... ...

Proof

Consider $ab \in P$ ... we have $ab + P \in R/P$

Because $P$ is prime, $ab \in P \ \Longrightarrow \ a \in P$ or $b \in P \ \Longrightarrow a + P = 0$ or $b + P = 0$

Thus in $R/P$ we have that:

$ab + P = 0 \ \Longrightarrow \ a + P = 0 \$ or $\ b + P = 0$

That is ... there are no zero divisors in $R/P$ ... in other words $R/P$ is an integral domain ...

==================================================================

Is above Lemma correct?

Peter