factorization in noetherian domains
quasi
An element x in R will be called irreducible if
(1) x is not zero and not a unit
(2) x=ab implies a or b is a unit
Equivalently, x is irreducible if (x) is a maximal element of the set
of a nonzero proper ideals of R.
An element p in R is called prime if (p) is a nonzero prime ideal.
Clearly every prime is irreducible but the converse is not automatic.
In fact,
R is a unique factorization domain) (UFD) iff every irreducible is
prime.
Any noetherian domain which is not a field has at least one
irreducible element. In fact, for such a ring, every nonzero nonunit
element is either irreducible or else has at least 2 irreducible
factors (not necessarily distinct).
Question (1): In a noetherian domain, can a nonzero element have
infinitely many irreducible factors? In other words, can a nonzero
element be a member of infinitely many distinct maximally principal
ideals?
Question (2): If a noetherian domain is not a field, must there be at
least one element which is prime?
Next, we consider complete factorization without uniqueness.
Let R be an integral domain.
Call R a finite factorization domain (FFD) if every nonzero nonunit
element of R is either irreducible or a product of irreducibles.
FFD is a much weaker concept than UFD, since for example, the ring of
algebraic integers in a finite extension of Q is an FFD (I think), but
not necessarily a UFD.
There are various possible strengthenings of the concept of FFD.
For example:
FFD-1: The complete factorizations of a given element must have
bounded length (depending on the element).
FFD-2: The total number of distinct complete factorizations of a given
element must be finite:
But to start, let's stay with just FFD.
Let R be an FFD..
Question (3): Is every noetherian domain an FFD?
Disclaimer: I'm not very knowledgeable about this stuff -- I just know
the basics (up to perhaps the lower graduate level), so the new term
FFD which I coined above might already have some other standard name.
If so, please advise.
In any case, the questions I posed above were just some questions that
I had asked myself but couldn't answer, so I thought I'd share them.
quasi
mzafr...@usa.net
I am a bit tired. Have had a rough day with knowledeius and his crowd.
Here is a reference. There are more if you like but this will do for
now.
[1]. Anderson, Anderson, Zafrullah, Factorization in integral domains,
J. Pure Appl. Algebra 69(1990),1-19.
quasi wrote:
> Let R be an integral domain.
>
> An element x in R will be called irreducible if
>
> (1) x is not zero and not a unit
>
> (2) x=ab implies a or b is a unit
>
> Equivalently, x is irreducible if (x) is a maximal element of the set
> of a nonzero proper ideals of R.
An irreducible element is also called an atom.
> An element p in R is called prime if (p) is a nonzero prime ideal.
> Clearly every prime is irreducible but the converse is not automatic.
> In fact,
>
> R is a unique factorization domain) (UFD) iff every irreducible is
> prime.
You need to add: And every nonzero nonunit of R is a (finite) product
of irreducible elements.
Here are some examples in which every irreducible element is a prime
but the domain is not a UFD.
1. A discrete rank 2 valuation domain (V,M) has up to associates only
one irreducible element and that is the generator of the maximal
ideal. Let p be the prime them M=pV. Since V is of rank 2 M properly
contains a nonzero prime ideal Q. Now for every nonzero x in Q we have
p^{n}|x for every n. But in a UFD every nonzero nonunit can be epressed
as a finite product of finite powers of primes.
2. The ring of all algebraic integers A is another example. Because A
has no irreducible elements, "Every irreducible element is a prime"
holds vacuously. What is interesting is that in each polynomial ring
A[X_1,X_2 ,...,X_n] "Every irreducible element is a prime" holds but
not vacuously.
>
> Any noetherian domain which is not a field has at least one
> irreducible element. In fact, for such a ring, every nonzero nonunit
> element is either irreducible or else has at least 2 irreducible
> factors (not necessarily distinct).
Noetherian -->ACC on ideals --->ACC on principal ideals --> exery
nonzero nonunit is expressible as a finite product of irreducible
elements.
>
> Question (1): In a noetherian domain, can a nonzero element have
> infinitely many irreducible factors? In other words, can a nonzero
> element be a member of infinitely many distinct maximally principal
> ideals?
Yes. The example that I have is a bit kinky. Let D = R+XC[X] where R is
the set of real numbers and C that of complex numbers. D is the set of
polynomials f(X) over C such that f(0) is a real number. It can be
shown that D is Noetherian but each member of {(r+i)X: r in R} is a
factor of X^{2} and each of (r+i)X can be shown to be irreducible. You
can look up [1] for the ref. on D being Noetherian.
>
> Question (2): If a noetherian domain is not a field, must there be at
> least one element which is prime?
No. Take a one dimensional local Noetherian domain R with maximal ideal
M. If R contains an irreducible element p that is a prime then pR is a
prime ideal and obviously pR is contained in M and M is of height one.
Forcing M=pR. So if you require a one dimensional local Noetherian
domain to have at least one principal prime then you are asking it to
be a UFD.
>
> Next, we consider complete factorization without uniqueness.
I hope by complete factorization you mean x = a_1 a_2 ...a_n where a_i
are all irreducible elements.
>
> Let R be an integral domain.
>
> Call R a finite factorization domain (FFD) if every nonzero nonunit
> element of R is either irreducible or a product of irreducibles.
We call such a domain an atomic domain.
>
> FFD is a much weaker concept than UFD, since for example, the ring of
> algebraic integers in a finite extension of Q is an FFD (I think), but
> not necessarily a UFD.
True.
>
> There are various possible strengthenings of the concept of FFD.
>
> For example:
>
> FFD-1: The complete factorizations of a given element must have
> bounded length (depending on the element).
If by this you mean for every nonzero x there is an n =n(x) such that
for each atomic factorization x = a_1 a_2 ...a_r we must have r<=n Such
a domain is called a BFD (Bounded Factorization Domain). A description
can be found in [1].
>
> FFD-2: The total number of distinct complete factorizations of a given
> element must be finite:
These are known as FFD's (Finite factorization domains) in [1]. If I
understaqnd you correctly.
>
> But to start, let's stay with just FFD.
>
> Let R be an FFD..
>
> Question (3): Is every noetherian domain an FFD?
Every Noetherian domain is atomic i.e. every nonzero nonunit is
expressible as a product of finitely many irreducible elements.
>
> Disclaimer: I'm not very knowledgeable about this stuff -- I just know
> the basics (up to perhaps the lower graduate level), so the new term
> FFD which I coined above might already have some other standard name.
> If so, please advise.
>
> In any case, the questions I posed above were just some questions that
> I had asked myself but couldn't answer, so I thought I'd share them.
>
> quasi
This is the best I could do under my current state. If there are any
questions feel free.
Muhammad
quasi
Thank you, Muhammad.
I'm very low on time for the next week, so I won't have time to review
your reply carefully until next weekend.
But thanks again -- I appreciate the reply.
quasi