I Definition of an irreducible element in an integral domain

[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

 

[fresh_42]

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

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

 

[Math Amateur]

Hi fresh_42, Lavinia

Here is Lemma 3

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

Lemma 3 Let $R$ be a Boolean ring ... ($R$ is a ring with $1$ in which every element is idempotent ... i.e. $x^2 = x \ \forall \ x \in R$)

Then $R/P \cong \mathbb{Z}/(2)$ ... that is $R/P = \{ 0 + P, 1 + P \}$Proof

Let $x \in R$ ... then $x^2 = x$

$x^2 = x$

$\Longrightarrow x^3 = x^2 \cdot x = x \cdot x = x^2 = x$

$\Longrightarrow x^3 = 1.x$

$\Longrightarrow x^2 \cdot x = 1.x$ in $R$ ...

$\Longrightarrow (x^2 + P)(x + P) = (1 + P)(x + P)$ in $R/P$

... ... or we can write $\overline{x}^2 \cdot \overline{x} = \overline{1} \cdot \overline{x}$ in $R/P$ ...But ... $R/P$ is an integral domain (Lemma 2) ... so the cancellation law applies ...

Thus for arbitrary $\overline{x} \in R/P$ we have that $\overline{x}^2 = \overline{1}$ ...

Thus $\overline{x} = \overline{1}$ or $\overline{-1}$ ...

But $\overline{-1} = \overline{-1}^2 = \overline{1}$ ...

Thus $R/P$ has two elements, namely $(0 + P)$ and $(1 + P)$ ...


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

Is above Lemma correct?

Peter*** EDIT *** Have to confess ... not sure whether i should write $\overline{x}^2$ in the above ... or should I be writing $\overline{x^2}$ ... can you help?

 

[Math Amateur]

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

Sorry fresh_42 ... typo only ...

Peter

 

[fresh_42]

... ... or we can write $\overline{x}^2 \cdot \overline{x} = \overline{1} \cdot \overline{x} \in R/P$ ...But ... $R/P$ is an integral domain (Lemma 2) ... so the cancellation law applies ...

Fine until here. The cancellation rule doesn't apply if one factor is zero. Therefore you have to distinguish the two cases here which gives you the result. Do you know why $P$ is maximal then?

Thus for arbitrary $\overline{x} \in R/P$ we have that $\overline{x}^2 = \overline{x} .$..

Thus $R/P$ is a Boolean ring ... ...

You already had this by Lemma 1.

 

[Math Amateur]

Hi fresh_42, Lavinia

Now to show that ...

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

Lemmas 1 and 2 show that $R/P$ is an integral domain and a Boolean ring containing two elements, namely $(0 + P)$ and $(1 + P)$

But then $R/P$ is a field since its only non-zero element has an inverse, namely itself ...

But if $R/P$ is a field then $P$ is a maximal ideal of $R$ ... (see Proposition 12, Section 7. 4 Dummit and Foote ... ... Proposition 12 follows from the Lattice Isomorphism Theorem for Rings ... sometimes called the Correspondence Theorem for Rings)Is this correct?

Peter

 

[fresh_42]

Yes, that's correct.

 

[lavinia]

Here is another problem in the ring $R=Z[\sqrt -5]$

Show that if the normed squared $N(s)$ is a prime in the integers then $s$ is a prime in $R$.

Here is a sequence of steps for one proof.

- Let $p = N(s)$ be a prime in the integers. Show that the ring $R/R(p)$ the quotient of $R$ by the principal ideal generated by $p$ is a finite ring of order $p^2$. (For instance, in the case of $\sqrt -5$ the quotient ring has order 25.)

- Show that the principal ideal generated by the coset of $s$ in $R/R(p)$ is a non-zero proper subset of $R/R(p)$. Use LaGrange's Theorem for subgroups of a finite group to conclude that this principal ideal has order $p$.

- Conclude that $R/R(s)$ has order $p$.

- Show that any ring of prime order that has an identity element is an integral domain.

- Conclude that $s$ is prime in $R$

Remarks:

- For $\sqrt -5$ the proof is easy to see by inspection because any multiple $(a + b\sqrt -5)\sqrt -5$ of $\sqrt -5$ is of the form $-5b + a\sqrt -5$ from which it is clear that the complex part can be arbitrary and that there are 5 equivalences classes in $R/R(\sqrt -5)$ and they are represented by the elements of the form $n + a\sqrt -5$ where $n$ is between $0$ and $4$.

- It is important that a ring of prime order has an identity element in order to conclude that it is an integral domain. Without an identity the trivial ring $st = 0$ for all $s$ and $t$ works. One concludes that for each prime $p$ there are exactly two rings, the trivial ring and $Z/Z(p)$.

- It would be interesting to see a direct proof rather than one that uses ideals and quotient rings.

- Does this proof depend upon $\sqrt -5$? Would it work for say for $\sqrt -3$ or $\sqrt -29$?

 

[Math Amateur]

Thanks Lavinia... wIll get onto this exercise shortly ...

Thanks once again for the learning experiences you are making possible ...!

Peter

 

[Math Amateur]

Hi Lavinia,

Sorry for the slowness in responding ...

I am traveling on the mainland having left my island state of Tasmania...

Will focus on your exercise when I return to my home state in several days time ...

Sorry for the delay ...

Peter

 

[Math Amateur]

Here is another problem in the ring $R=Z[\sqrt -5]$

Show that if the normed squared $N(s)$ is a prime in the integers then $s$ is a prime in $R$.

Here is a sequence of steps for one proof.

- Let $p = N(s)$ be a prime in the integers. Show that the ring $R/R(p)$ the quotient of $R$ by the principal ideal generated by $p$ is a finite ring of order $p^2$. (For instance, in the case of $\sqrt -5$ the quotient ring has order 25.)

- Show that the principal ideal generated by the coset of $s$ in $R/R(p)$ is a non-zero proper subset of $R/R(p)$. Use LaGrange's Theorem for subgroups of a finite group to conclude that this principal ideal has order $p$.

- Conclude that $R/R(s)$ has order $p$.

- Show that any ring of prime order that has an identity element is an integral domain.

- Conclude that $s$ is prime in $R$

Remarks:

- For $\sqrt -5$ the proof is easy to see by inspection because any multiple $(a + b\sqrt -5)\sqrt -5$ of $\sqrt -5$ is of the form $-5b + a\sqrt -5$ from which it is clear that the complex part can be arbitrary and that there are 5 equivalences classes in $R/R(\sqrt -5)$ and they are represented by the elements of the form $n + a\sqrt -5$ where $n$ is between $0$ and $4$.

- It is important that a ring of prime order has an identity element in order to conclude that it is an integral domain. Without an identity the trivial ring $st = 0$ for all $s$ and $t$ works. One concludes that for each prime $p$ there are exactly two rings, the trivial ring and $Z/Z(p)$.

- It would be interesting to see a direct proof rather than one that uses ideals and quotient rings.

- Does this proof depend upon $\sqrt -5$? Would it work for say for $\sqrt -3$ or $\sqrt -29$?

Hi Lavinia, fresh_42

Now working on Step 1 ... but need some help to get going ...Trying to show the following:Let $p = N(s)$ be a prime in the integers. Show that the ring $R/<p>$ the quotient of $R$ by the principal ideal generated by $p$ is a finite ring of order $p^2$.So let $s = u + v \sqrt{5} i$

Then $N(s) = u^2 + 5 v^2 = p$ where $p \in \mathbb{Z}$ and $p$ prime ... ...

Now, consider $<p> = \{ (a + b \sqrt{5} i ) p \ | \ a, b, p \in \mathbb{Z}, p$ is prime $\}$ ... BUT ... where do we go from here ... ?

... ... we do know that $R/ <p>$ is an integral domain since $<p>$ is a prime ideal ... but how do we use this ... ?Can you help ...

Peter

 

[lavinia]

Hi Lavinia, fresh_42

Now working on Step 1 ... but need some help to get going ...Trying to show the following:Let $p = N(s)$ be a prime in the integers. Show that the ring $R/<p>$ the quotient of $R$ by the principal ideal generated by $p$ is a finite ring of order $p^2$.So let $s = u + v \sqrt{5} i$

Then $N(s) = u^2 + 5 v^2 = p$ where $p \in \mathbb{Z}$ and $p$ prime ... ...

Now, consider $<p> = \{ (a + b \sqrt{5} i ) p \ | \ a, b, p \in \mathbb{Z}, p$ is prime $\}$... BUT ... where do we go from here ... ?

... ... we do know that $R/ <p>$ is an integral domain since $<p>$ is a prime ideal ... but how do we use this ... ?Can you help ...

Peter

As an abelian group under addition (not multiplication) $R$ is the same as $Z×Z$. It is a free abelian group on two generators. Mod p how many residue classes are there?

(For perfect rigor you want to prove that $R$ actually is isomorphic to $Z×Z$ as an abelian group.)

 

[Math Amateur]

As an abelian group under addition (not multiplication) $R$ is the same as $Z×Z$. It is a free abelian group on two generators. Mod p how many residue classes are there?

(For perfect rigor you want to prove that $R$ actually is isomorphic to $Z×Z$ as an abelian group.)

-------------------------------------------------------------------------------------------------------------------------------------------------------Hi Lavinia, fresh_42

Thanks for the hint, Lavinia ... will proceed as far as I can ...

... ...

To show that $R \cong \mathbb{Z} \times \mathbb{Z}$ ... ...Let $r, t \in R$ where $r = a + b \sqrt{5} i$ and $t = c + d \sqrt{5} i$

Define $\phi$ as follows ...

$\phi \ : \ R \longrightarrow \mathbb{Z} \times \mathbb{Z}$ is defined such that:

$\phi (r) = \phi (a + b \sqrt{5} i ) = (a,b)$

then

$\phi (r + s) = \phi ( (a + b \sqrt{5} i) + ( c + d \sqrt{5} i )$

$= \phi ( ( a+c ) + (b + d) \sqrt{5} i )$

$= (a+c, b + d) = (a, b) + (c,d)$

$= \phi ( r) + \phi (t)$... so $\phi$ is an additive group homomorphism ... ... clearly it is also injective and surjective ...

... so $\phi$ is an isomorphism between $R$ and $\mathbb{Z} \times \mathbb{Z}$ viewed as additive abelian groups ...

... that is $R \ \cong \ \mathbb{Z} \times \mathbb{Z}$

-------------------------------------------------------------------------------------------------------------------------------------------

Now we have (see previous post ... )

$p = p + 0. \sqrt{5} i = u^2 + 5 v^2$ ( see previous post where $N(s) = p = u^2 + 5 v^2$ )... ... now we have $R/ <p> \ \cong \ \mathbb{Z} \times \mathbb{Z}/ < \phi(p) >$ ... ... (BUT ... is this the case ... most unsure ...? !)... ... now $\phi (p) = (p,0)$ ...... ... so consider $\mathbb{Z} \times \mathbb{Z} / < \phi (p) > \ = \ \mathbb{Z} \times \mathbb{Z} / < (p, 0 ) >$

BUT...

... this seems to imply there are $p$ residue classes ... namely $(0,0) , (1,0) , (2,0) , \ ... \ ... \ (p-1, 0)$ ...

... seems like something is wrong ...
Can you help further ... seems like I should be working with $\mathbb{Z} \times \mathbb{Z} / < (p, p ) >$ ... ... but why ... ?

I am also very unsure of what $R \cong \mathbb{Z} \times \mathbb{Z}$ implies for the relationship between $R/ <p>$ and $\mathbb{Z} \times \mathbb{Z} / < \phi (p) >$ ...Hope you can help ...

Peter

 

[lavinia]

-------------------------------------------------------------------------------------------------------------------------------------------------------Hi Lavinia, fresh_42

Thanks for the hint, Lavinia ... will proceed as far as I can ...

... ...

To show that $R \cong \mathbb{Z} \times \mathbb{Z}$ ... ...Let $r, t \in R$ where $r = a + b \sqrt{5} i$ and $t = c + d \sqrt{5} i$

Define $\phi$ as follows ...

$\phi \ : \ R \longrightarrow \mathbb{Z} \times \mathbb{Z}$ is defined such that:

$\phi (r) = \phi (a + b \sqrt{5} i ) = (a,b)$

then

$\phi (r + s) = \phi ( (a + b \sqrt{5} i) + ( c + d \sqrt{5} i )$

$= \phi ( ( a+c ) + (b + d) \sqrt{5} i )$

$= (a+c, b + d) = (a, b) + (c,d)$

$= \phi ( r) + \phi (t)$... so $\phi$ is an additive group homomorphism ... ... clearly it is also injective and surjective ...

... so $\phi$ is an isomorphism between $R$ and $\mathbb{Z} \times \mathbb{Z}$ viewed as additive abelian groups ...

... that is $R \ \cong \ \mathbb{Z} \times \mathbb{Z}$

-------------------------------------------------------------------------------------------------------------------------------------------

Now we have (see previous post ... )

$p = p + 0. \sqrt{5} i = u^2 + 5 v^2$ ( see previous post where $N(s) = p = u^2 + 5 v^2$ )... ... now we have $R/ <p> \ \cong \ \mathbb{Z} \times \mathbb{Z}/ < \phi(p) >$ ... ... (BUT ... is this the case ... most unsure ...? !)... ... now $\phi (p) = (p,0)$ ...... ... so consider $\mathbb{Z} \times \mathbb{Z} / < \phi (p) > \ = \ \mathbb{Z} \times \mathbb{Z} / < (p, 0 ) >$

BUT...

... this seems to imply there are $p$ residue classes ... namely $(0,0) , (1,0) , (2,0) , \ ... \ ... \ (p-1, 0)$ ...

... seems like something is wrong ...
Can you help further ... seems like I should be working with $\mathbb{Z} \times \mathbb{Z} / < (p, p ) >$ ... ... but why ... ?

I am also very unsure of what $R \cong \mathbb{Z} \times \mathbb{Z}$ implies for the relationship between $R/ <p>$ and $\mathbb{Z} \times \mathbb{Z} / < \phi (p) >$ ...Hope you can help ...

Peter

What you have done is absolutely correct. Multiplication by $p$ in $R$ though is $p⋅(a +b\sqrt -5) = pa +pb\sqrt -5$ so the principal ideal $R(p)$ is all residue classes modulo multiples of $p$ in either coordinate. So the principal ideal generated by $p$ maps to $Z/Z(p)×Z/Z(p)$

It may have been confusing the way I stated the hint. Apologies for that.

Notice for instance that if $b=0$ one get s the residue classes mod $p$ of the real part and if $a=0$ one gets the residue classes mod $p$ of the complex part.

 

[Math Amateur]

What you have done is absolutely correct. Multiplication by $p$ in $R$ though is $p⋅(a +b\sqrt -5) = pa +pb\sqrt -5$ so the principal ideal $R(p)$ is all residue classes modulo multiples of $p$ in either coordinate. So the principal ideal generated by $p$ maps to $Z/Z(p)×Z/Z(p)$

It may have been confusing the way I stated the hint. Apologies for that.

Notice for instance that if $b=0$ one get s the residue classes mod $p$ of the real part and if $a=0$ one gets the residue classes mod $p$ of the complex part.

----------------------------------------------------------------------------------------------------------------------------------------Hi Lavinia, fresh_42

Thanks for the reassurance, Lavinia ... but could you expand on or clarify an important point ... as follows ...

You write: "the principal ideal generated by $p$ maps to $Z/Z(p)×Z/Z(p)$"

I am still struggling to see exactly why this follows ...

It appears that $R \cong \mathbb{Z} \times \mathbb{Z}$ ... ... implies that $R / <p> \ \cong \ \mathbb{Z} / <p> \times \mathbb{Z} / <p>$ ... ...But why, exactly and how, exactly does this follow ...Can you help ...

Peter---------------------------------------------------------------------------------------------------------------------------------------

*** EDIT ***

Just reflecting and thinking that maybe you did explain the above ... when you wrote:

" ... ... Multiplication by $p$ in $R$ though is $p⋅(a +b\sqrt -5) = pa +pb\sqrt -5$ so the principal ideal $R(p)$ is all residue classes modulo multiples of $p$ in either coordinate. ... ... "This seems to make sense to me ... but I wish I could be more sure of what exactly is going on ...

Peter

 

[fresh_42]

You have defined

$\phi \ : \ R \longrightarrow \mathbb{Z} \times \mathbb{Z}$ is defined such that:
$\phi(r)=\phi(a+b\sqrt{5}i)=(a,b)$

and $R(p)=p \cdot R$

$p⋅(a +b\sqrt{-5}) = pa +pb\sqrt{-5}$

So what is $\phi (R(p)) = \{\phi(p⋅(a +b\sqrt{-5}))\,\vert \,a,b \in \mathbb{Z}\}\,$?
Now you can build the quotients and get $\phi : R/pR \cong (\mathbb{Z} \times \mathbb{Z})/\phi(pR)$, because you factored out isomorphic ideals on both sides. All what's left is to formally prove
$$
(\mathbb{Z} \times \mathbb{Z})/\phi(pR) = \mathbb{Z}_p \times \mathbb{Z}_p
$$
I would set up the surjection $\mathbb{Z} \times \mathbb{Z} \twoheadrightarrow \mathbb{Z}_p \times \mathbb{Z}_p$ and compute its kernel.
(Remember that $R/\ker \varphi \cong \varphi(R)$.)

You can do similar if the first isomorphism is the source of your difficulties, i.e.
$$
R \cong_\varphi S \; \wedge I \unlhd R \; \Longrightarrow \; R/I \cong S/\varphi(I)
$$

 

[Math Amateur]

What you have done is absolutely correct. Multiplication by $p$ in $R$ though is $p⋅(a +b\sqrt -5) = pa +pb\sqrt -5$ so the principal ideal $R(p)$ is all residue classes modulo multiples of $p$ in either coordinate. So the principal ideal generated by $p$ maps to $Z/Z(p)×Z/Z(p)$

It may have been confusing the way I stated the hint. Apologies for that.

Notice for instance that if $b=0$ one get s the residue classes mod $p$ of the real part and if $a=0$ one gets the residue classes mod $p$ of the complex part.

You have defined

and $R(p)=p \cdot R$

So what is $\phi (R(p)) = \{\phi(p⋅(a +b\sqrt{-5}))\,\vert \,a,b \in \mathbb{Z}\}\,$?
Now you can build the quotients and get $\phi : R/pR \cong (\mathbb{Z} \times \mathbb{Z})/\phi(pR)$, because you factored out isomorphic ideals on both sides. All what's left is to formally prove
$$
(\mathbb{Z} \times \mathbb{Z})/\phi(pR) = \mathbb{Z}_p \times \mathbb{Z}_p
$$
I would set up the surjection $\mathbb{Z} \times \mathbb{Z} \twoheadrightarrow \mathbb{Z}_p \times \mathbb{Z}_p$ and compute its kernel.
(Remember that $R/\ker \varphi \cong \varphi(R)$.)

You can do similar if the first isomorphism is the source of your difficulties, i.e.
$$
R \cong_\varphi S \; \wedge I \unlhd R \; \Longrightarrow \; R/I \cong S/\varphi(I)
$$

-----------------------------------------------------------------------------------------------------------------------------------------

Hi Lavinia, fresh_42

Thanks for all your help ...

Just now reflecting on the last couple of posts ... especially the last one by fresh_42 ...

BUT ... ... Another issue that is bothering me is that $\phi$ is an isomorphism of abelian groups and not an isomorphism of rings ...

Yet we are using $\phi$ in the last few posts reasoning about rings modulus ideals ... how can this be correct ...? ... ...

Maybe we are only using aspects of the isomorphism that are established by the group isomorphism ...

Can you comment ...

Peter

 

[fresh_42]

You're right. But as far as the homomorphisms are concerned, it remains true what I've written applied on groups, simply replace "ideal" be "normal subgroup" or (as in this case of Abelian groups), simply by subgroups. As long as it is about the question of finiteness, it is sufficient to only consider the additive group of a ring. (Where should additional elements come from or even more important, how could they be outside of the ring?) So if you don't claim the rings to be isomorphic as rings, only the additive groups, everything is fine for counting elements.

 

[Math Amateur]

You have defined

and $R(p)=p \cdot R$

So what is $\phi (R(p)) = \{\phi(p⋅(a +b\sqrt{-5}))\,\vert \,a,b \in \mathbb{Z}\}\,$?
Now you can build the quotients and get $\phi : R/pR \cong (\mathbb{Z} \times \mathbb{Z})/\phi(pR)$, because you factored out isomorphic ideals on both sides. All what's left is to formally prove
$$
(\mathbb{Z} \times \mathbb{Z})/\phi(pR) = \mathbb{Z}_p \times \mathbb{Z}_p
$$
I would set up the surjection $\mathbb{Z} \times \mathbb{Z} \twoheadrightarrow \mathbb{Z}_p \times \mathbb{Z}_p$ and compute its kernel.
(Remember that $R/\ker \varphi \cong \varphi(R)$.)

You can do similar if the first isomorphism is the source of your difficulties, i.e.
$$
R \cong_\varphi S \; \wedge I \unlhd R \; \Longrightarrow \; R/I \cong S/\varphi(I)
$$

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Hi Lavinia, fresh_42

Based on previous posts we can say ... (details omitted)

$R/ pR \cong \mathbb{Z} \times \mathbb{Z} / \phi (pR) \cong \mathbb{Z}_p \times \mathbb{Z}_p$

So ... $R/pR$ has finite order equal to $p^2$ ...

BUT ... Lavinia writes the following:Let $p=N(s)$ be a prime in the integers. Show that the ring $R/R(p)$, the quotient of $R$ by the principal ideal generated by $p$ is a finite ring of order $p^2$. (For instance, in the case of $√−5$ the quotient ring has order $25$.)BUT ... how do we get $p^2 = 25$ in the case of $R = \mathbb{Z} [ \sqrt{-5} ]$ ?All we have is that $N(s) = p$ ...

Peter

 

[Math Amateur]

Hi Lavinia, fresh_42

Having great difficulty getting started on showing that the principal ideal generated by the coset of $s$ is a proper subset of $R/ pR$ ... ...

Indeed ... what is a useful expression for the principal ideal generated by the coset of $s$?

Peter

 

[fresh_42]

$R/ pR \cong \mathbb{Z} \times \mathbb{Z} / \phi (pR) \cong \mathbb{Z}_p \times \mathbb{Z}_p$

So ... $R/pR$ has finite order equal to $p^2$ ...

BUT ... Lavinia writes the following:

Let $p=N(s)$ be a prime in the integers. Show that the ring $R/R(p)$, the quotient of $R$ by the principal ideal generated by $p$ is a finite ring of order $p^2$. (For instance, in the case of $√−5$ the quotient ring has order $25$.)BUT ... how do we get $p^2 = 25$ in the case of $R = \mathbb{Z} [ \sqrt{-5} ]$ ?

You have $|R/pR|=|\mathbb{Z}_p \times \mathbb{Z}_p|=p^2$ and thus $|R/5R| = 25.$

All we have is that $N(s) = p$ ...

We have $N(s) = p$ is a prime in $\mathbb{Z}$, so e.g. $p=5$. I'm not sure, but I think the $5$ as $p$ and the $5$ in $\sqrt{-5}$ is only a coincidence, maybe not the best choice of an example. What you have is that $(R/pR \, , \,+) \cong ((\mathbb{Z}_p \times \mathbb{Z}_p) \, , \,+)$ for the additive groups and any prime, even for $p=5$. But as you correctly observed, this is not true for the ring structure, at least not by componentwise multiplication in $\mathbb{Z}_p \times \mathbb{Z}_p$. If we define another multiplication that mimics the one in $R/pR$ things might be different. But then we cannot (should not) write it as a direct product.

The next step @lavinia talked about was to show that $\langle \pi(s) \rangle \subseteq R/pR$ generates a principal ideal of order $p$ in $R/pR$, where $\pi : R \rightarrow R/pR$ is the projection map and $N(s)=p$.

As every ideal of $R_p := R/pR$ defines a subgroup of the additive group of $R/pR$ you have only three possible values for the number of elements in $\langle \pi(s) \rangle$.

 

[fresh_42]

Hi Lavinia, fresh_42

Having great difficulty getting started on showing that the principal ideal generated by the coset of $s$ is a proper subset of $R/ pR$ ... ...

Indeed ... what is a useful expression for the principal ideal generated by the coset of $s$?

Peter

Remember that $N(s)=p$. So can you rule out that $\langle \pi(s) \rangle$ contains a unit? Can it be the entire ring?

 

[Math Amateur]

Hi Lavinia, fresh_42

Am traveling in Queensland ... be back in a few days ...

Will post on my return...

Apologies for the delay ...

Peter

 

[lavinia]

Hi Lavinia, fresh_42

Am traveling in Queensland ... be back in a few days ...

Will post on my return...

Apologies for the delay ...

Peter

Have a good trip. Have you been to Darwin?

 

[Math Amateur]

Have a good trip. Have you been to Darwin?

I went through Brisbane to Toowoomba ...

I have never been to Darwin ... indeed it is the only Australian state capital I have not visited ...

Peter

 

[Math Amateur]

Here is another problem in the ring $R=Z[\sqrt -5]$

Show that if the normed squared $N(s)$ is a prime in the integers then $s$ is a prime in $R$.

Here is a sequence of steps for one proof.

- Let $p = N(s)$ be a prime in the integers. Show that the ring $R/R(p)$ the quotient of $R$ by the principal ideal generated by $p$ is a finite ring of order $p^2$. (For instance, in the case of $\sqrt -5$ the quotient ring has order 25.)

- Show that the principal ideal generated by the coset of $s$ in $R/R(p)$ is a non-zero proper subset of $R/R(p)$. Use LaGrange's Theorem for subgroups of a finite group to conclude that this principal ideal has order $p$.

- Conclude that $R/R(s)$ has order $p$.

- Show that any ring of prime order that has an identity element is an integral domain.

- Conclude that $s$ is prime in $R$

Remarks:

- For $\sqrt -5$ the proof is easy to see by inspection because any multiple $(a + b\sqrt -5)\sqrt -5$ of $\sqrt -5$ is of the form $-5b + a\sqrt -5$ from which it is clear that the complex part can be arbitrary and that there are 5 equivalences classes in $R/R(\sqrt -5)$ and they are represented by the elements of the form $n + a\sqrt -5$ where $n$ is between $0$ and $4$.

- It is important that a ring of prime order has an identity element in order to conclude that it is an integral domain. Without an identity the trivial ring $st = 0$ for all $s$ and $t$ works. One concludes that for each prime $p$ there are exactly two rings, the trivial ring and $Z/Z(p)$.

- It would be interesting to see a direct proof rather than one that uses ideals and quotient rings.

- Does this proof depend upon $\sqrt -5$? Would it work for say for $\sqrt -3$ or $\sqrt -29$?

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

Hi Lavinia, fresh_42

Now ... to show Lavinia's step 2 ... that is ...

Show that the principal ideal generated by the coset of $s$ in $R/Rp$ is a non-zero proper subset of $R/R(p)$.
Use LaGrange's Theorem for subgroups of a finite group to conclude that this principal ideal has order $p$.
We are considering the ring $R=Z[\sqrt -5] = \{ a + b \sqrt{ 5} i \ | \ a, b \in \mathbb{Z} \}$

and are also considering a specific element $s \in R$ where $s = u + v \sqrt{ 5} i$ and

$N(s) = u^2 + 5 v^2 = p$ where $p \in \mathbb{Z}$ and $p$ is prime ... ... Then ... the coset of $s$ in $R/Rp$ is ...

$s + I = s + Rp = \{ s + i \ | \ i \in Rp \}$

$= \{ s + ( a + b \sqrt{ 5} i ) p \}$

$= \{ (u + v \sqrt{ 5} i) + ( pa + pb \sqrt{ 5} i )$

$= \{ ( u + pa) + (v + pb) \sqrt{ 5} i \}$

But ... now we have to form the principal ideal generated by the coset of $s$ ...

I am assuming that this is $( s + I ) R$ ... ... is that correct?

Then it would be

$(s + I) R = \{ [( u + pa) + (v + pb) \sqrt{ 5} i ] (x + y \sqrt{ 5} i ) \}$BUT ... how do we proceed from here ... can you help ?Peter