Maximal ideals generated by irreducible elements
Sancho
what is (respectively) the most general type of ring R in which the
following statements hold?
1. f is an irreducible element of R ==> (f) is a maximal ideal of R
2. (f) is a maximal ideal of R an f is nonzero ==> f is an irreducible
element of R
I can prove both statements in a PID.
Thanks,
Sancho
Chip Eastham
Hi, Sancho:
I think (1) and (2) would be true vacuously
for rings R in which (a) irreducible elements
do not exist and (b) no maximal ideal is
principal. For example, suppose we start by
localizing Z at (2), i.e. introducing all the
rational fractions with odd denominators, and
then extending this local ring "integrally"
by adjoining all the roots 2^(1/k), k >= 2.
I haven't worked out the details, but I think
this will satisfy (a) and (b).
If you go in the direction of assuming a
supply of irreducible elements, probably you
will quickly get to having a UFD, and from
there to having a PID.
Some references that may be helpful:
[Going up and going down (prime ideals) -- Wikipedia]
http://en.wikipedia.org/wiki/Going_up_and_going_down
[Unique factorization domain -- Wikipedia]
http://en.wikipedia.org/wiki/Unique_factorization_domain
regards. chip
Hagen
(only lazyness)
(1) implies that every irreducible element is prime.
Moreover if R is noetherian, then (1) implies that
dim(R) = 1 (Principal Ideal Theorem). The localizations
at maximal ideals are then discrete valuation domains,
so that R is integrally closed and thus a Dedekind domain.
Altogether we (seem to) get that a noetherian domain
with property (1) is a PID.
If R is non-noetherian there can be to few irreducibles
to say anything interesting. For example: no irreducibles
at all.
H
Arturo Magidin
In any domain (commutative ring with identity and no zero divisors), f
is irreducible if and only if (f) is nonzero and maximal among proper
principal ideals of R (in particular, if (f) is maximal, then f is
irreducible; but (f) may be maximal among proper principal ideals and
yet not be maximal. And in particular, if R is a PID, them maximal
among proper principal ideals is equivalent to maximal).
If f is irreducible, and (a) is a proper principal ideal such that (f)
is contained in (a), then f = ax for some x, hence either a is a unit
or x is a unit; but a is not a unit since (a) is proper, so x is a
unit. Therefore, a is in (f), so (a) is contained in (f) and (a)=(f).
Conversely, if (f) is nonzero and maximal among proper principal
ideals, and f = xy, then (f) is contained in (x), so either (x)=(f) or
(x) = R. If (x)=R, then x is a unit. If (x)=(f), then x = fz for some
z, so f = xy = fzy. Since R is a domain, 1 = zy, so y is a unit. Thus,
f is irreducible.
--
Arturo Magidin
Bill Dubuque
SIMPLER since (a)>(b) <=> a|b (i.e. contains = divides)
(b) has no proper container (i.e. (b) is maximal)
<=> b has no proper divisor (i.e. b is irreducible)
The same is true in any domain where contains = divides for ideals,
e.g. Dedekind domains (e.g. number rings).
--Bill Dubuque
Bill Dubuque
>
> (1) implies that every irreducible element is prime.
>
> Moreover if R is noetherian, then (1) implies that
> dim(R) = 1 (Principal Ideal Theorem). The localizations
> at maximal ideals are then discrete valuation domains,
> so that R is integrally closed and thus a Dedekind domain.
> Altogether we (seem to) get that a noetherian domain
> with property (1) is a PID.
SIMPLER and more generally, avoiding PIT sledgehammer:
Domain D has ACCP & prime atoms => UFD D. Non0 max ideals
P in A UFD contain a non0 prime p (e.g. some prime factor
of any non0 elt of P), thus P = (p) by maximality of (p).
Now easily: UFD & max ideals principal => PID, cf. [1].
[1] http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2008;task=show_msg;msg=2902.0002.0001.0002
[2] http://google.com/groups?selm=y8z7i7sfcc1.fsf%40nestle.csail.mit.edu
Bill Dubuque
> On Feb 11, 5:22 am, Sancho <sanchopanch...@web.de> wrote:
>>
>> what is (respectively) the most general type of ring R in which the
>> following statements hold?
>>
>> 1. f is an irreducible element of R ==> (f) is a maximal ideal of R
>> 2. (f) is a maximal ideal of R an f is nonzero ==> f is an irreducible
>> element of R
>>
>> I can prove both statements in a PID.
>
> for rings R in which (a) irreducible elements
> do not exist and (b) no maximal ideal is
> principal. For example, suppose we start by
> localizing Z at (2), i.e. introducing all the
> rational fractions with odd denominators, and
> then extending this local ring "integrally"
> by adjoining all the roots 2^(1/k), k >= 2.
> I haven't worked out the details, but I think
> this will satisfy (a) and (b).
(2) is true in every ring. A standard example of
a ring with no atoms is the ring of all algebraic
integers: every elt splits r = r^(1/2) r^(1/2).
> If you go in the direction of assuming a
> supply of irreducible elements, probably you
> will quickly get to having a UFD, and from
> there to having a PID.
No, one requires a sufficient finiteness condition
e.g. Noetherian or ACCP -- see my other post here.
(1) implies AP (Atom => Prime). Generally one has
UFD <=> AP & atomic
atomic := nonzero nonunits factorize into atoms.
Atomic <= ACCP (Ascending Chain Condition on
Principal ideals), but not conversely.
> Some references that may be helpful:
>
> [Going up and going down (prime ideals) -- Wikipedia]
> http://en.wikipedia.org/wiki/Going_up_and_going_down
Why do you think that GU/GD is relevant?
--Bill Dubuque
Chip Eastham
Hi, Bill:
The thesis of my post was that (1) and (2)
might both be vacuously true without
providing anything nice in the way of a UFD
much less PID. I gave a construction,
localizing Z at a prime (say 2Z) and adding
enough roots to destroy all hope of finite
generation.
The point about going up/going down here is
that since the extension is integral over
the local ring, the spectrum of the extension
consists solely of primes lying over the
unique prime ideals of the local ring,
namely the zero ideal and the maximal ideal
generated by 2. So any maximal ideal of the
overring will contain 2, and by primality
all the roots thereof.
regards, chip