Definition of an irreducible element in an integral domain

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Discussion Overview

The discussion centers on the definition of an irreducible element in an integral domain, exploring its implications and the reasoning behind certain restrictions in the definition. Participants examine the distinctions between irreducible elements, units, and prime elements, as well as the necessity of working within integral domains rather than general rings.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Peter questions why the definition of irreducible elements is restricted to integral domains, suggesting that the absence of zero divisors is a key factor.
  • Another participant agrees with Peter's inquiry and mentions that irreducibility can be defined in commutative rings with identity, citing a number theory book.
  • There is a discussion on the rationale for excluding units from being considered irreducible, with one participant explaining that units complicate the definition and do not contribute to meaningful factorization.
  • Participants discuss the condition that for an irreducible element ##a##, if ##a = bc## then either ##b## or ##c## must be a unit, with the reasoning that this avoids genuine factorizations.
  • One participant notes that while the concept of prime and non-factorizable elements does not require an integral domain, the study of divisibility theory leads to integral domains due to their unique properties.
  • There is mention of the uniqueness of prime factorization in integral domains and the potential for non-factorizable elements that are not prime in certain integral domains.
  • Examples of rings where irreducibles are not prime are suggested, including the integers with algebraic irrational numbers adjoined.
  • Participants express interest in exercises to explore these concepts further, including proving properties of units and non-factorizables in specific rings.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of restricting definitions to integral domains and the implications of units in factorization. There is no consensus on whether irreducibility can be adequately defined outside of integral domains, and the discussion remains unresolved regarding the relationship between irreducible elements and prime elements in various contexts.

Contextual Notes

Participants acknowledge the complexity of definitions and the potential for counterexamples in different types of rings, particularly in distinguishing between irreducible and prime elements. The discussion highlights the importance of definitions and properties specific to integral domains versus general rings.

Who May Find This Useful

Mathematics students and researchers interested in abstract algebra, particularly those studying ring theory, divisibility, and the properties of integral domains.

  • #31
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.
 
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  • #32
Thanks fresh_42 ... will definitely try that ...

Peter
 
  • #33
Math Amateur said:
Thanks fresh_42 ... will definitely try that ...

Peter
I haven't done it, but it's all we have.
 
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  • #34
Just a clarifying question ... why did you multiply by the conjugate of p ... and not p itself ..

Peter
 
  • #35
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.
 
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  • #36
You can also calculate in ##R/(p)## and prove it has no zero divisors.
 
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  • #37
This can also be solved using the norm squared.
 
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  • #38
Thanks Lavinia ...
 
  • #39
fresh_42 said:
You can also calculate in ##R/(p)## and prove it has no zero divisors.

Thanks for the suggestion ...

Peter
 
  • #40
Math Amateur said:
=======================================================================================

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
Math Amateur said:
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
 
  • #41
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?
 
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  • #42
fresh_42 said:
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##.
 
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  • #43
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 ... :frown: ... )

Peter
 
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  • #44
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.
 
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  • #45
Math Amateur said:
Hi Lavinia, fresh_42

If ##5 \mid f^2## ... then since ##5## is prime, ##5 \mid f## ...
 
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  • #46

##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##
 
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  • #47
lavinia said:
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.
 
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  • #48
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
 
  • #49
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##.
 
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  • #50
Math Amateur said:
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 ... :frown: ... )

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
 
  • #51
Correct.
 
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  • #52
Thanks Lavinia ...

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

Peter
 
  • #53
Math Amateur said:
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.
 
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  • #54
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
 
  • #55
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.
 
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  • #56
fresh_42 said:
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 ... ... :smile:)
 
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  • #57
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.
 
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  • #58
fresh_42 said:
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
 
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  • #59
Yes it is. And you know something else about quotients with a prime ideal.
 
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  • #60
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
 

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