-- Maximal ideals in Z[x] and Z[x, y]
ballade1
Mariano Suárez-Alvarez
> What are the maximal ideals of Z[x]? Of Z[x, y] ?
Let I be a maximal ideal in Z[x]. In particular, it is
prime, so its preimage I' under the inclusion Z --> Z[x]
is also prime. Then either I' = 0 or there exists a prime
number p such that I' = (p).
Consider the second case. Then I contains the principal
ideal J = p Z[x] of Z[x], and I is the preimage under
the quotient map Z[x] --> Z[x] / J of a maximal ideal
of the ring Z[x] / J. Now it is quite obvious that Z[x] / J
is isomorphic to Z_p[x], with Z_p the field with p elements.
You should know by now what are the maximal ideals of Z_p[x],
so you only have to pull them back to Z[x].
Suppose now I' = 0, so that I intersects Z trivially.
Then the ideal K = (I, 2) is strictly larger than I,
and this is absurd for I is maximal. This means that
this case cannot occur.
-- m
KMF
For the case of Z[x], it is a Euclidean Ring under
standard division of polynomials, so it is a PID, so
there aren't many choices.
Gerry Myerson
<25712419.4090.12402837...@nitrogen.mathforum.org>,
KMF <tee...@hotmail.com> wrote:
So, what's the generator of the ideal generated by x and 2?
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
KMF
> <25712419.4090.1240283793942.JavaMail.jakarta@nitrogen
> .mathforum.org>,
> KMF <tee...@hotmail.com> wrote:
>
> > > What are the maximal ideals of Z[x]? Of Z[x, y] ?
> >
> > For the case of Z[x], it is a Euclidean Ring
> under
> > standard division of polynomials, so it is a PID,
> so
> > there aren't many choices.
>
> So, what's the generator of the ideal generated by x
> and 2?
>
O.K. Ballade did not bother stating that Z is the
integers in here, but let's assume this is what it
stands for. Then I think that for R[x] to be an
Euclidean Domain, we need for R to actually be a field.
I don't know if the same would work for F[x_1,..,x_n]
for n>1.
Steve Dalton
R[x] is a PID (Euclidean) <=> R is a field, no?
Steve
victor_me...@yahoo.co.uk
<mariano.suarezalva...@gmail.com> wrote:
> On Apr 20, 8:10 pm, ballade1 <balla...@yahoo.com> wrote:
>
> > What are the maximal ideals of Z[x]? Of Z[x, y] ?
>
> Let I be a maximal ideal in Z[x]. In particular, it is
> prime, so its preimage I' under the inclusion Z --> Z[x]
> is also prime. Then either I' = 0 or there exists a prime
> number p such that I' = (p).
>
> Then the ideal K = (I, 2) is strictly larger than I,
> and this is absurd for I is maximal.
No, since K could equal the whole of Z[X].
As an example, let I = <1 + 2X>. Obviously
I contains no integers save for zero, but I + <2> contains
1 and so is equal to Z[X]. You need a more subtle
argument to prove that an ideal like I is not maximal.
Steve Dalton
Well no maximal ideal in Z[x] can miss Z-{0} completely as then the quotient would contain an isomorphic copy of the rationals.
Steve
Hagen
So?
victor is refering to your argument:
> Then the ideal K = (I, 2) is strictly larger
> than I,
> and this is absurd for I is maximal.
This is only 'absurd' if K is a proper ideal.
But K could be the whole ring, in which
case you can conclude nothing about I.
Here is an example that demonstrates the
problem: replace Z by R:=Z_(2) the localization
of Z at the prime ideal generated by 2.
Consider the polynomial ring R[X].
Here the principal ideal (2X-1) is a prime ideal
and its intersection with R is 0 due to degree-
reasons. However: R[X]/(2X-1)=R[1/2]=Q,
so (2X-1) is maximal.
H
ge...@math.mq.edu.au
> R[x] is a PID (Euclidean) <=> R is a field, no?
I guess. Let R be a commutative ring with unity,
and suppose it has a non-zero non-unit, r. Then
I can't see the ideal generated by x and r being
principal.
victor_me...@yahoo.co.uk
What about R = Z/6Z and r = 2?
Steve Dalton
But this isn't my argument. I was only pointing out an alternative proof, that's all.
>
> > Then the ideal K = (I, 2) is strictly larger
> > than I,
> > and this is absurd for I is maximal.
>
> This is only 'absurd' if K is a proper ideal.
> But K could be the whole ring, in which
> case you can conclude nothing about I.
>
> Here is an example that demonstrates the
> problem: replace Z by R:=Z_(2) the localization
> of Z at the prime ideal generated by 2.
> Consider the polynomial ring R[X].
> Here the principal ideal (2X-1) is a prime ideal
> and its intersection with R is 0 due to degree-
> reasons. However: R[X]/(2X-1)=R[1/2]=Q,
> so (2X-1) is maximal.
>
> H
Steve
Gerry Myerson
<3a733216-73af-42d6...@q2g2000vbr.googlegroups.com>,
"victor_me...@yahoo.co.uk" <victor_me...@yahoo.co.uk>
wrote:
I must be dense - I don't see it - what's the generator
of the ideal generated by x and 2?
Derek Holt
wrote:
> In article
> <3a733216-73af-42d6-b003-40030186b...@q2g2000vbr.googlegroups.com>,
> "victor_meldrew_...@yahoo.co.uk" <victor_meldrew_...@yahoo.co.uk>
>
> wrote:
> > On 21 Apr, 13:52, ge...@math.mq.edu.au wrote:
> > > On Apr 21, 4:02 pm, Steve Dalton <sdal...@math.sunysb.edu> wrote:
>
> > > > R[x] is a PID (Euclidean) <=> R is a field, no?
>
> > > I guess. Let R be a commutative ring with unity,
> > > and suppose it has a non-zero non-unit, r. Then
> > > I can't see the ideal generated by x and r being
> > > principal.
>
> > What about R = Z/6Z and r = 2?
>
> I must be dense - I don't see it - what's the generator
> of the ideal generated by x and 2?
Try 2+3x.
I don't know whether all ideals of (Z/6Z)[x] are principal. Maybe
Victor can tell us?
But none of this affects the the claim that R[x] is a PID only if R
is a field, because if R[x] is a PID then R has to be a domain, and so
has no zero divisors. In that case, I believe you are correct in
saying that the ideal generated by a non-unit r in R and x cannot be
principal.
For let s be a generator of this ideal. If deg(s)=0, then r|s, and so
x cannot be a multiple of s. If deg(s)>0 then deg(us)>0 for all u in
R, so r is not a multiple of s.
Derek Holt.
victor_me...@yahoo.co.uk
> On 22 Apr, 00:53, Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
> wrote:
>
>
>
> > In article
> > <3a733216-73af-42d6-b003-40030186b...@q2g2000vbr.googlegroups.com>,
> > "victor_meldrew_...@yahoo.co.uk" <victor_meldrew_...@yahoo.co.uk>
>
> > wrote:
> > > On 21 Apr, 13:52, ge...@math.mq.edu.au wrote:
> > > > On Apr 21, 4:02 pm, Steve Dalton <sdal...@math.sunysb.edu> wrote:
>
> > > > > R[x] is a PID (Euclidean) <=> R is a field, no?
>
> > > > I guess. Let R be a commutative ring with unity,
> > > > and suppose it has a non-zero non-unit, r. Then
> > > > I can't see the ideal generated by x and r being
> > > > principal.
>
> > > What about R = Z/6Z and r = 2?
>
> > I must be dense - I don't see it - what's the generator
> > of the ideal generated by x and 2?
>
> Try 2+3x.
>
> I don't know whether all ideals of (Z/6Z)[x] are principal. Maybe
> Victor can tell us?
They certainly are: (Z/6Z)[x] is isomorphic to the Cartesian product
of (Z/2Z)[x] and (Z/3Z)[x] each of which are PIDs.
Hagen
There should be a result like:
R[X] is a principal ideal ring iff ???.
Does anyone of you know it?
+ R must be noetherian, but that probably
follows by faithfully flat descent or something the like.
+ dim(R)=0 also is necessary.
* A proposal for ??? could be: R is artinian
and reduced.
H
Gerry Myerson
<3ab0697f-efc2-402f...@3g2000yqk.googlegroups.com>,
Derek Holt <ma...@warwick.ac.uk> wrote:
> On 22 Apr, 00:53, Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
> wrote:
> > In article
> > <3a733216-73af-42d6-b003-40030186b...@q2g2000vbr.googlegroups.com>,
> > "victor_meldrew_...@yahoo.co.uk" <victor_meldrew_...@yahoo.co.uk>
> >
> > wrote:
> > > On 21 Apr, 13:52, ge...@math.mq.edu.au wrote:
> > > > On Apr 21, 4:02 pm, Steve Dalton <sdal...@math.sunysb.edu> wrote:
> >
> > > > > R[x] is a PID (Euclidean) <=> R is a field, no?
> >
> > > > I guess. Let R be a commutative ring with unity,
> > > > and suppose it has a non-zero non-unit, r. Then
> > > > I can't see the ideal generated by x and r being
> > > > principal.
> >
> > > What about R = Z/6Z and r = 2?
> >
> > I must be dense - I don't see it - what's the generator
> > of the ideal generated by x and 2?
>
> Try 2+3x.
OK, I will. 2 + 3 x times 4 is 2. (2 + 3 x) (3 + 2 x) = x. Excellent!
Bill Dubuque
Below I append one of my earlier posts on this topic.
As I mentioned here in thread below, generally one has
http://google.com/groups?threadm=y8zzlh3cdar.fsf%40nestle.csail.mit.edu
THEOREM TFAE for D a Krull domain (e.g. a UFD or Dedekind domain)
(i) D[x] has a max ideal M with M/\D = 0
(ii) D is a PID with finitely many primes
From: Bill Dubuque <w...@nestle.csail.mit.edu>; Date: 17 Dec 2005
URL: http://google.com/groups?threadm=y8zy82k2dni.fsf%40nestle.csail.mit.edu
matt <m...@hotmail.com> wrote: (paraphrased)
>
> Is it true that Z[x]/M a field => M/\Z > 0
M/\Z = 0 would imply Q = Z[1/n], n in Z, see the proof below,
direction (=>). The basic idea is that one can reduce to an
integral extension Z[1/n] < Z[x]/M where n is the lead coef of
any poly in M. But integral extensions can't introduce new units,
so since an integral extension of Z[1/n] is a field it must be
the case that Z[1/n] is already a field. See my prior post below
for details and references. See also other posts in that thread.
From: Bill Dubuque <w...@nestle.csail.mit.edu>; Date: 22 Mar 2005
URL: http://google.com/groups?threadm=y8z7jjzdivj.fsf%40nestle.csail.mit.edu
William Elliot <ma...@privacy.net> wrote:
>
> I think I'll have to take your word about it,
> that for maximal ideal M of Z[x], M /\ Z /= {0}.
The proof is simple, using nothing deeper than the fact that
(R-)algebraic integers are closed under sum & product, namely:
THEOREM For PID R, the poly ring R[x] has a maximal ideal M
with M/\R = 0 <=> R has finitely many nonassociate primes.
PROOF (<=) Let r = product of all nonassoc. primes p in R.
Then F = R[1/r] contains the inverse of every prime p, e.g.
1/p = (r/p)/r, so F is the fraction field of R. Let M be the
kernel of the hom h: R[x]->F which evaluates f(x) at x = 1/r.
Its image is the whole field R[1/r] = F, so M is a maximal.
Further M/\R = 0 since h restricts to the identity map on R.
(=>) Let w = x + M = image of x in the field F = R[x]/M.
M/\R = 0 => R is a subring of F and F = R[w], so by PROP.
below R[1/r] is a field for some nonzero r in R. Hence if
p is a prime in R then 1/p = s/r^n, for some s in R, n in N,
and ps = r^n => p divides r; i.e. every prime p of R occurs
among the prime factors of r, which is a finite set, since
r has a unique prime factorization (a PID is a UFD). QED
PROP. Suppose field F = R[w] for a subring R of F and w in F.
Then R[1/r] is a field for some nonzero r in R.
PROOF w is algebraic over R (else w is transcendental over R,
R[w] = R[x] is a poly ring, not a field, since 1/x not in R[x]).
Let p be a minimum poly of w over R and let r be its lead coef.
Field R[w] is integral over its subring S = R[1/r] since w is
integral over S, satisfying poly p with lead coef r unit in S.
Therefore the lemma below implies that R[1/r] is a field. QED
LEMMA Subring R of field F is a field if F is integral over R
PROOF Since field F contains R, F contains fraction field of R.
Therefore 1/r is integral over R for every nonzero r in R, i.e.
n n-1 n-2
(1/r) = a (1/r) + b (1/r) + ... + e, a,b,..,e in R
Times r^(n-1) this shows 1/r in R[r] = R, so R is a field. QED
These results are special cases of results at the basis of
an elegant abstract approach to the Hilbert Nullstellensatz
discovered circa 1950 by Goldman, Krull and Zariski. For more
on this look up the terms: G-domain, Hilbert ring, Jacobson ring.
See also D.J. Bernstein's page [1] which has some online links,
including expositions of an "elementary" proof by R. Munshi (1999).
Note that the theorem may be viewed as a generalization of
Euclid's theorem that Z has infinitely many primes, a view
that has sometimes been rediscovered by others apparently
not familiar with the above work, e.g. see [2].
--Bill Dubuque
[1] D.J. Bernstein: The Zariski-Goldman-Krull theorem
(the generalized Nullstellensatz)
http://cr.yp.to/zgk.html
[2] F. Zanello: When are There Infinitely Many Irreducible Elements
in a Principal Ideal Domain?
Amer. Math. Monthly 111 #2 (2004) 150-152.
http://arxiv.org/abs/math.AC/0411259
Bill Dubuque
>> On 22 Apr, 08:23, Derek Holt <ma...@warwick.ac.uk>> wrote:
>>> On 22 Apr, 00:53, Gerry Myerson>> <ge...@maths.mq.edi.ai.i2u4email> wrote:
>>
>> They certainly are: (Z/6Z)[x] is isomorphic to the
>> Cartesian product of (Z/2Z)[x] and (Z/3Z)[x] each of which are PIDs.
>
> There should be a result like:
>
> R[X] is a principal ideal ring iff ???.
>
> Does anyone of you know it?
>
> + R must be noetherian, but that probably
> follows by faithfully flat descent or something the like.
>
> + dim(R)=0 also is necessary.
>
> * A proposal for ??? could be: R is artinian and reduced.
More generally here is a semigroup version:
THEOREM TFAE for a semigroup ring R[S], with unitary ring R,
and nonzero torsion-free cancellative monoid S.
1) R[S] is a PIR (Principal Ideal Ring)
2) R[S] is a general ZPI-ring (i.e. a Dedekind _ring_, see below)
3) R[S] is a multiplication ring (i.e. I > J => I|J for ideals I,J)
4) R is a finite direct sum of fields, and
S is isomorphic to Z or N
A general ZPI-ring is a ring theoretic analog of a Dedekind domain
i.e. a ring where every ideal is a finite product of prime ideals.
A unitary ring R is a general ZPI-ring <=> R is a finite direct sum
of Dedekind domains and special primary rings (aka SPIR = special PIR)
i.e. local PIRs with nilpotent max ideals. ZPI comes from the German
phrase "Zerlegung in Primideale" = factorization in prime ideals.
The classical results on Dedekind domains were extended to rings
with zero divisors by S. Mori circa 1940, then later by K. Asano
and, more recently, by R. Gilmer. See Gilmer's book "Commutative
Semigroup Rings" sections 18 (and section 13 for the domain case).
See also the following MR's (not meant to be exhaustive).
------------------------------------------------------------------------------
49#5213 20M25 (13F05)
Gilmer, Robert; Parker, Tom
Semigroup rings as Prufer rings.
Duke Math. J. 41 (1974), 219--230.
------------------------------------------------------------------------------
Let RS be the semigroup ring of a torsion-free cancellative abelian
semigroup S with zero over a commutative ring R with identity. The
semigroup operation on S is written as addition. Such rings RS may be
essentially thought of as generalizations of polynomial rings. The authors
seek conditions on R and S under which the semigroup ring RS will have
a given ring-theoretic property. Some necessary and sufficient conditions are
found for the ring RS to fall into one of four classes of rings: Prufer
rings, Bezout rings, almost Dedekind rings, and general ZPI-rings. The
investigations are closely related to (but independent of) another paper of
the authors [Michigan Math. J. 21 (1974), 65--86; MR 49#7381].
In Sections 2 and 3, the case of Prufer rings is considered. A Prufer ring is
a commutative ring R with identity such that each finitely generated regular
ideal of R is invertible. In order to state the main result of these two
sections, we need some more definitions: Let Z, [Q] be the additive group of
integers [rationals], and let Z_0, [Q_0] be the additive semigroup of
nonnegative integers [nonnegative rationals]. Semigroups of the form G /\ Q_0,
where G is a subgroup of Q containing Z , are called Prufer sub-semigroups
of Q_0. A commutative ring with identity in which each finitely generated ideal
is principal is a Bezout ring. It is shown that if RS is a Prufer ring then R
is (von Neumann) regular. Further, the authors prove the equivalence of the
following three conditions: (1) RS is a Prufer ring; (2) RS is a Bezout ring;
(3) R is a regular ring, and to within isomorphism S is either a Prufer
subsemigroup of Q_0 or a subgroup of Q containing Z .
In Section 4 the authors deal with almost Dedekind rings (AD-rings).
Following M. D. Larsen [J. Reine Angew. Math. 245 (1970), 119--123;
MR 42#7662], an AD-ring is a Prufer ring in which regular prime ideals are
maximal and not idempotent. Some necessary and sufficient conditions for
RS to be an AD-ring are found in this section (Theorems 4.1 and 4.2).
In the final section (5), the notion of a general ZPI-ring is introduced.
These are commutative rings with identity in which each ideal is a
finite product of prime ideals. The following result (Theorem 5.1) is now
established. The semigroup ring RS is a general ZPI-ring if and only if R
is a finite direct sum of fields and S is isomorphic to Z_0 or to Z .
Reviewed by Uno Kaljulaid
------------------------------------------------------------------------------
82d:13019 13F20 (13F05)
Hardy, Bonnie R.; Shores, Thomas S.
Arithmetical semigroup rings.
Canad. J. Math. 32 (1980), no. 6, 1361--1371.
------------------------------------------------------------------------------
Throughout this paper, the ring R and the semigroup S are commutative with
identity; moreover, it is assumed that S is cancellative. An arithmetical
ring is a ring for which the ideals form a distributive lattice and a ZPI-ring
is one in which every ideal is a product of prime ideals.
The authors determine necessary and sufficient conditions on R and S that
the semigroup ring R[S] be arithmetical [respectively, semihereditary, a
ZPI-ring, a PIR (principal ideal ring)]. The main result is Theorem 3.6: Let
R and S be as above and G the group of quotients of S . Let \rho be a
congruence defined on S by x\rho y if and only if x=y+f for some
f in(S/\ tG) . Then R[S] is arithmetical if and only if one of the
following holds: (1) the torsion subgroup tG of G is a proper subsemigroup
of S , R[tG] is regular and the semigroup S/\rho of congruence classes of
\rho is isomorphic to an additive subgroup of Q or the positive cone of
such a group, (2) R is arithmetical and S=G is a torsion group such that
if its p -primary component G_p != 0 for some prime p = Char}(R/M),
where M is a *maximal* ideal of R , then G_p is cyclic or
quasicyclic and R_M is a field. Two other theorems, 4.1 and 4.2,
provide characterizations of R[S] that are ZPI-rings and PIRs.
This paper is closely related to the paper by R. Gilmer and T. Parker [Duke
Math. J. 41 (1974), 219--230; MR 49#5213], particularly the following
results (Corollary 3.1 and Corollary 5.1): If R and S are as above and
moreover S is torsion-free, then (a) R[S] is a Bezout ring if and only
if R[S] is a Prufer ring if and only if R is a (von Neumann) regular ring
and S is isomorphic to an additive subgroup of Q or the positive cone of
such a subgroup (the authors point out that each of the above statements is
also equivalent to another statement " R[S] is arithmetical"), and (b) R[S]
is a ZPI-ring if and only if R[S] is a PIR. Applying their theorems, the
authors give examples to show that the above results of Gilmer and Parker are
no longer true if the condition "S is torsion-free" is dropped.
Reviewed by Chin-Pi Lu
------------------------------------------------------------------------------
40 #1380 13.50
Wood, Craig A.
On general Z.P.I.-rings.
Pacific J. Math. 30 1969 837--846.
------------------------------------------------------------------------------
A general Z.P.I.-ring is a commutative ring R each ideal of which is a
finite product of prime ideals. Consider the cases (A) R has an identity,
(B) R has no identity, but has at least one proper prime ideal, and
(C) R has neither identity nor proper prime ideal. In each case the author
gives firstly a structure theorem for general Z.P.I.-rings and secondly
criteria for R to be a general Z.P.I.-ring. The structure theorems,
which have been given in a less clear form by S. Mori [J. Sci. Hiroshima
Univ. Ser. A 10 (1940), 117--136; MR 2, 121], are as follows. R is a
general Z.P.I.-ring if and only if R is a finite direct sum of
Dedekind domains and special P.I.R.'s in case (A), R = F (+) T in case (B)
and R = T in case (C), where F is a field and T is a ring without
identity and without non-zero ideals other than powers of T .
Reviewed by D. Kirby
------------------------------------------------------------------------------
13,313e 09.1X
Asano, Keizo
Uber kommutative Ringe, in denen jedes Ideal als Produkt von Primidealen
darstellbar ist. (German)
J. Math. Soc. Japan 3, (1951). 82--90.
------------------------------------------------------------------------------
Let a commutative ring R with identity element be called a Dedekind ring if
it is the direct sum of a finite number of Dedekind integral domains and of
rings having a nilpotent, principal, maximal ideal. Various conditions on
R are proved equivalent to its being Dedekind, among them the following: (1)
Every ideal in R is a product of prime ideals; (2) the zero ideal is a
product of prime ideals, and if a prime ideal P contains an ideal A, then
P is a factor of A. In the presence of the ascending chain condition, the
following are also equivalent: (3) For every maximal ideal M , there is no
ideal between M and M^2 ; (4) the lattice of ideals is distributive. These
results generalize known conditions for an integral domain to be Dedekind.
[Rings satisfying (1) have been studied by S. Mori, J. Sci. Hirosima
Univ. Ser. A. 10, 117--136 (1940); these Rev. 2, 121.]
Reviewed by I. S. Cohen
------------------------------------------------------------------------------
2,121a 09.1X
Mori, Shinziro
Allgemeine Z.P.I.-Ringe.
J. Sci. Hirosima Univ. Ser. A. 10 (1940). 117--136.
------------------------------------------------------------------------------
A commutative ring R is termed a general Z.P.I. ring if every ideal in R can be
expressed as a product of a finite number of prime ideals. Thus rings without
unit element for multiplication and rings with divisors of zero are included in
the class of rings considered by the author. As a main result the author proves
that a ring R is a Z.P.I. ring if and only if (1) every ideal of R has a finite
basis, (2) for every pair of maximal prime ideals P, P' (that is, R/P, R/P' are
fields != 0) there is no ideal Q with PP' < Q < P, (3) there is no ideal Q with
R^2 < Q < R. (The three conditions are independent.) This theorem essentially
depends on the fact that in a Z.P.I. ring P P_1 = P if P < P_1 and if R/P is
not a field. To prove the latter assertion it is necessary to investigate
the relationship between the ideal theory of R and R/P. Finally the author
formulates two theorems which are equivalent to his main theorem. For details
and the methods of proof see the original paper.
Reviewed by O. F. G. Schilling
Zbl Google Translation of http://www.emis.de/cgi-bin/Zarchive?an=0024.00801
K. Kubo (s. this. Zbl. 23,102) characterized those commutative rings, in which
every ideal from the whole ring and different from the zero-ideal ideal can be
represented uniquely as product of finitely many prime ideals. The uniqueness
idea is so sharply calm? that look for the occurrence of redundant (simply
omitable), from the total ring different prime ideal factors to be excluded is.
On this condition one (with more easily addition of the results won by Kubo
themselves) receives the main clause: A commutative ring with unique prime
ideal decomposition is either an integral domain, to which the well-known
Noether five axioms apply, or a "primary, detachable ring", i.e. a ring
with unit element, which contains only one prime ideal at ideals \\p and its
powers, whereby for a sufficiently large exponent \\p^n = (0) becomes.
-- By a Z.P.I. ring the author understands a commutative ring, which needs
to be neither zero-divisor free nor contain unit element, and in which each
ideal can be represented as product of finitely many prime ideals; Uniqueness
of the representation is not demanded in contrast to the work by K. Kubo.
The idea of the Z.P.I. ring is thus as far calm? as at all possible. As main
results are emphasized: All Z.P.I. rings are O-rings, thus rings with maximum
condition (divisor chain set). -- An O-ring with unit element is then and a
Z.P.I. ring only if with no maximum ring prime ideal \\p between \\p and \\p^2
a genuine intermediate ideal lies (the "Sono condition" characteristic of the
Japanese direction of the abstract ideal theory). -- The Z.P.I. rings with
unit element are nothing one but those already 1925 of W. Krull (S.-B.
Heidelberg. Akad. Wiss. 1925, 5. Abhandl.) in their structure exactly
described "multiplication rings with maximum condition".
-- A Ring \\R without unit element is then a Z.P.I. ring only if it possesses
a direct decmomposition \\R = \\F + \\m, whereby \\F represents (possibly only
from the nullelement existing) a field, while \\m is an O-ring without unit
element, which does not contain of (0) and \\m different prime ideal, and in
which between \\m and \\m^2 a genuine intermediate ideal does not lie.
Krull (Bonn).
--Bill Dubuque
Hagen
thanks a lot.
Although I am using multiplicative ideal theory frequently
I am usually quite ignorant when it comes to zero-divisors.
H
Bill Dubuque
>
> THEOREM TFAE for a semigroup ring R[S], with unitary ring R,
> and nonzero torsion-free cancellative monoid S.
>
> 1) R[S] is a PIR (Principal Ideal Ring)
>
>
> 3) R[S] is a multiplication ring (i.e. I>J => I|J for ideals I,J)
>
> 4) R is a finite direct sum of fields, and
> S is isomorphic to Z or N
THEOREM TFAE for a semigroup ring R[S], for domain R
and nonzero torsion-free cancellative monoid S.
0) R[S] is a Euclidean domain
1) R[S] is a PID
2) R[S] is a Dedekind domain
4) R is a field and S is isomorphic to Z or N
The polynomial ring R[x] is the special case S = N in R[S],
i.e. S = <x> = {1,x,x^2,...}. The case S = Z is its localization
at <x>, namely R[x,1/x], the ring of Laurent polynomials.
Recall that localizing a domain preserves these properties (and
many more: Euclidean, PID, UFD, valuation, Bezout, GCD, Dedekind,
Prufer, Krull, Noetherian, integrally closed, etc).
> A general ZPI-ring is a ring theoretic analog of a Dedekind domain
> i.e. a ring where every ideal is a finite product of prime ideals.
> A unitary ring R is a general ZPI-ring <=> R is a finite direct sum
> of Dedekind domains and special primary rings (aka SPIR = special PIR)
> i.e. local PIRs with nilpotent max ideals. ZPI comes from the German
> phrase "Zerlegung in Primideale" = factorization in prime ideals [...]
--Bill Dubuque