Algebra problems
Steve Mitchell
from the textbook. I have two that I have worked on the past couple of
days and cannot seem to get anywhere on them. Any hints and help would
be greatly appreciated.
Problem 1: Prove that the four-group (Z_2 x Z_2) is not isomorphic to
Z_4, but is isomorphic to the group of Euclidan symmetries of a
rectangle.
Problem 2: Prove that the set of all rational numbers with odd
denominators is a principal ideal domain.
ok, on the first problem, I can show that the four group is not
isomorphic to Z_4, but I cannot seem to show that it is to S4.
For Problem 2, I think that the ideal of Qodd is 2^m, but how to use
this to get a finished proof (if it is correct to begin with) I am not
sure about. Thanks for any help you can provide.
Sincerely,
Steve Mitchell
David C. Ullrich
wrote:
>I am studying for finals this week and am trying different problems
>from the textbook. I have two that I have worked on the past couple of
>days and cannot seem to get anywhere on them. Any hints and help would
>be greatly appreciated.
>
>Problem 1: Prove that the four-group (Z_2 x Z_2) is not isomorphic to
>Z_4, but is isomorphic to the group of Euclidan symmetries of a
>rectangle.
>
>Problem 2: Prove that the set of all rational numbers with odd
>denominators is a principal ideal domain.
>
>ok, on the first problem, I can show that the four group is not
>isomorphic to Z_4, but I cannot seem to show that it is to S4.
??? The group commonly known as S4 is not the same as the
group of symmetries of a rectangle.
>For Problem 2, I think that the ideal of Qodd is 2^m,
What do you mean by that statement?
>but how to use
>this to get a finished proof (if it is correct to begin with) I am not
>sure about. Thanks for any help you can provide.
>
>Sincerely,
>Steve Mitchell
************************
David C. Ullrich
Arturo Magidin
Steve Mitchell <ste...@yahoo.com> wrote:
>I am studying for finals this week and am trying different problems
>from the textbook. I have two that I have worked on the past couple of
>days and cannot seem to get anywhere on them. Any hints and help would
>be greatly appreciated.
>
>Problem 1: Prove that the four-group (Z_2 x Z_2) is not isomorphic to
>Z_4, but is isomorphic to the group of Euclidan symmetries of a
>rectangle.
>
>Problem 2: Prove that the set of all rational numbers with odd
>denominators is a principal ideal domain.
>
>ok, on the first problem, I can show that the four group is not
>isomorphic to Z_4, but I cannot seem to show that it is to S4.
S_4 is not "the Euclidean symmetries of a rectangle". S_4 would
allow you to leave two adjacent vertices fixed, and flip the remaining
two; that is not an operation you can do in Euclidean space and still
have a rectangle.
What are the "Euclidean symmetries of a rectangle"? Well, you can take
the rectangle and reflect it about a line that is parallel to the
short sides and bisects the long sides; or you can reflect it about a
line that is parallel to the long sides and bisects the short sides;
or you can reflect it about the intersection of these two lines;
or... is there anything else?
>For Problem 2, I think that the ideal of Qodd is 2^m, but how to use
>this to get a finished proof (if it is correct to begin with) I am not
>sure about. Thanks for any help you can provide.
What does "ideal of Qodd is 2^m" mean?
To do problem two you have to do the following things:
(1) Show that the set of all rational numbers with odd denominators
form a ring. For that, since you are using the operations from the
rationals, you just need to show that it is closed under additions
and multiplications, contains 0, and contains 1.
(2) That will show that it is a domain.
(3) Now show that any ideal is principal. I ->think<- that this is
what you were trying to say above. What will happen is that the
only ideals in this ring are the ideals of the form (0) or (2^m)
for m>0. How can you prove that?
One way that occurs to me is to show first that every ideal
contains some integer (easy); and then to show that any proper
ideal must only contain even integers (easy). Then show that if it
is not the zero ideal, it must contain some power of 2 (again,
easy). And then show that it is in fact equal to (2^m), where m is
the smallest positive integer such that 2^m lies in the ideal.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Steve Mitchell
<ull...@math.okstate.edu> wrote:
>??? The group commonly known as S4 is not the same as the
>group of symmetries of a rectangle.
Oops. Sorry. I guess I had S4 in my head from a previous problem. I
meant, symmetries of a rectangle V = {I, H, V, R}. Where I is the
identity, H is the horizontal reflection about the axis of symmetry, V
is the vertical reflection, and R is a rotation of 180 degrees.
William Elliot
Newsgroups: sci.math
Subject: Re: Algebra problems
Steve Mitchell <ste...@yahoo.com> wrote:
>>Prove that the set of all rational numbers with odd
>>denominators is a principal ideal domain.
>(1) Show that the set of all rational numbers with odd denominators
>form a ring. For that, since you are using the operations from
>the rationals, you just need to show that it is closed under
>additions and multiplications, contains 0, and contains 1.
>(2) That will show that it is a domain.
>(3) Now show that any ideal is principal. What will happen is that
>the only ideals in this ring are the ideals of the form (0) or (2^m)
>for m>0. How can you prove that?
>One way that occurs to me is to show first that every ideal
>contains some integer (easy); and then to show that any proper
>ideal must only contain even integers (easy). Then show that if it
>is not the zero ideal, it must contain some power of 2 (again,
>easy). And then show that it is in fact equal to (2^m), where m is
>the smallest positive integer such that 2^m lies in the ideal.
non-trivial ideal I ==> some m >= 0 with I = (2^m)
some k => 0, odd p,q with 2^k p/q in I; 2^k in I
let m = min{ k | 2^k in I }; 2^m in I; (2^m) subset I
If r in I: some k >= 0, p,q with r = 2^k p/q in I: 2^k in I
2^m | 2^k; 2^k in (2^m); r in (2^m)
I subset (2^m); I = (2^m); (m = 0 ==> I = whole ring)
-- exercise, generalization
R = { p/q | p in Z, q not in 3Z } is PID
I non-trivial ideal R ==> some m >= 0 with I = (3^m)
prime n ==> R = { p/q | p in Z, q not in nZ } is PID
I non-trivial ideal R ==> some m >= 0 with I = (n^m)
-- still a PID (principal ideal domain)?
R = { p/q | p in Z, q in 3Z + 1 }
What do the ideals look like?
--
> what I accept as reality
From the real to the complex are many imaginary steps,
but always i is there, there's no getting around that.
----
David C. Ullrich
wrote:
>On Sun, 09 May 2004 14:08:59 -0500, David C. Ullrich
And now this should make it possible to show that the group
is isomorphic to Z_2 x Z_2...
I'm still curious what you meant by "the ideal of Qodd is 2^m".
************************
David C. Ullrich
Arturo Magidin
William Elliot <ma...@privacy.net> wrote:
[.snip.]
>-- exercise, generalization
>R = { p/q | p in Z, q not in 3Z } is PID
>I non-trivial ideal R ==> some m >= 0 with I = (3^m)
>
>prime n ==> R = { p/q | p in Z, q not in nZ } is PID
>I non-trivial ideal R ==> some m >= 0 with I = (n^m)
Better generalization: show that a localization of a PID is a
PID. (You are localizing at (3)).
>-- still a PID (principal ideal domain)?
>R = { p/q | p in Z, q in 3Z + 1 }
>
>What do the ideals look like?
Either they don't look, or this is the localization at (3) again.
Given your inimitable telegraphic style, it is hard to see just what
it is you are defining as R: is it all rationals that MAY be written
as quotients in which the denominator is congruent to 1 mod 3, or just
the ones which, ->in reduced form<- have denominators congruent to 1
mod 3? Or is it the ring generated by these elements?
1/10 is in R; so is 5/1. But their product is 1/2 which is not in the
form given. So if you meant "in reduced form", then what you gave was
not a ring.
If, on the other hand, you meant "that could be written" in a quotient
like that, then every integer not divisible by 3 is a unit: for if n
is not divisible by 3, then either n = 1 (mod 3) or n=2 (mod 3). If
n=1 mod(3), then 1/n is in your ring. If n=2, then n^2 = 1 (mod 3), so
1/n^2 is in the ring, so is n/1, and therefore so is 1/n. In either
case, 1/n is in the ring. So R contains the localization at 3. It is
not hard to verify now that it is exactly the localization at (3).
William Elliot
William Elliot <ma...@privacy.net> wrote:
>>R = { p/q | p in Z, q not in 3Z } is PID
>>I non-trivial ideal R ==> some m >= 0 with I = (3^m)
>
>>prime n ==> R = { p/q | p in Z, q not in nZ } is PID
>>I non-trivial ideal R ==> some m >= 0 with I = (n^m)
>Better generalization: show that a localization
>of a PID is a PID. (You are localizing at (3)).
What ever that means. Anything similar to local ring?
Let's see. (3) ideal R and (3^n) subset (3) for all n in N.
Ok, R is a local ring. How's it local? A ring in the small?
Like R is locally compact, ie compact in the small.
But just a nanosec, big chuncks of R are compact.
Ok, R - (1/googleplex)Z is compact in the small.
>>-- still a PID (principal ideal domain)?
>>R = { p/q | p in Z, q in 3Z + 1 }
>
>>What do the ideals look like?
>1/10 is in R; so is 5/1. But their product is 1/2 which is not in
>the form given. So if you meant "in reduced form", then what you gave
>was not a ring.
No, I didn't require p,q to be coprime.
>If, on the other hand, you meant "that could be written" in a
>quotient like that,
1/(3k+2) = -1/(-3k-2) = -1/(-3(k+1) + 1) in R =
{ p/q | p in Z, q in 3Z + 1 } = { p/q | p in Z, q not in 3Z }
= blithereens, nothing new
>then every integer not divisible by 3 is a unit:
The non-units are { 3p/q | p in Z, q not in 3Z }.
R = Q - Z[1/3] is 2/3 field.
>So R contains the localization at 3. It is not hard to
>verify now that it is exactly the localization at (3).
----
Arturo Magidin
William Elliot <ma...@privacy.net> wrote:
>From: Arturo Magidin <mag...@math.berkeley.edu>
>Newsgroups: sci.math
>Subject: Re: Algebra problems
>
>William Elliot <ma...@privacy.net> wrote:
>
> >>R = { p/q | p in Z, q not in 3Z } is PID
> >>I non-trivial ideal R ==> some m >= 0 with I = (3^m)
>>
> >>prime n ==> R = { p/q | p in Z, q not in nZ } is PID
> >>I non-trivial ideal R ==> some m >= 0 with I = (n^m)
>
> >Better generalization: show that a localization
> >of a PID is a PID. (You are localizing at (3)).
>What ever that means.
Rather ironic coming from you...
> Anything similar to local ring?
What does that mean?
What should be similar? How should it be similar? What is it that
you asking about?
Let R be a commutative ring with 1. A "multiplicative subset" S of R
is a submonoid of the multiplicative monoid of R.
The "localization of R at S", denoted S^{-1}R, is constructed as
follows:
Take the set of all pairs (r,s), with r in R and s in S.
Define an equivalence relation on RxS as follows:
(r,s) ~ (u,t) if and only if there exists s' in S such that
s'(tr-su)=0.
Let S^{-1}R be the quotient of RxS modulo this equivalence relation.
Define a ring structure on S^{-1}R by defining addition to be as
follows (use [r,s] to denote the class of (r,s)):
[r,s] + [u,t] = [rt+us,st]
and define multiplication by
[r,s]*[u,t] = [ru,st].
Then S^{-1}R is a ring; there is a natural map i from R to S^{-1}R be
mapping r to [r,1]. S^{-1}R has the following universal property:
given any ring homomorphism f:R->T such that f(s) is a unit in T for
each s in S, there exists a unique morphism g:S^{-1}R->T such that
gi= f.
R is not always embedded into S^{-1}R; for example, if S contains a
zero divisor v, with vw=0, w<>0, then w maps to [w,1], but [w,1]=[0,1],
because v*(1w - 1*0) = 0, so w maps to zero.
But if R is a domain and S does not contain 0 (and in many other
circumstances as well), then there is a one-to-one inclusion
preserving correspondence between the ideals of R which are disjoint
from S, and the ideals of S^{-1}R.
It is called a "localization" because one of its most common
applications are to construct local rings: if P is any prime ideal,
then the set S = R-P is a multiplicative subset. The ring S^{-1}R has
S^{-1}P as its (unique) maximal ideal.
The ring {r/s in Q: gcd(r,s)=1, s not divisible by p}, where p is a
fixed prime, is the localization of R at the set S of all
non-multiples of p.
>Let's see. (3) ideal R and (3^n) subset (3) for all n in N.
>Ok, R is a local ring. How's it local? A ring in the small?
What does "A ring in the small" mean?
As far as I am aware, the terminology "local ring" is derived from the
fact that these rings arise when looking at 'local information' in
algebraic geometry.
>Like R is locally compact, ie compact in the small.
>But just a nanosec, big chuncks of R are compact.
>Ok, R - (1/googleplex)Z is compact in the small.
Sigh. Maybe I don't speak English as well as I thought... I have no
idea what any of the above means.
> >>-- still a PID (principal ideal domain)?
> >>R = { p/q | p in Z, q in 3Z + 1 }
>>
> >>What do the ideals look like?
>
> >1/10 is in R; so is 5/1. But their product is 1/2 which is not in
> >the form given. So if you meant "in reduced form", then what you gave
> >was not a ring.
>No, I didn't require p,q to be coprime.
Then you just have Z localized at (3): S^{-1}Z with S = Z-(3).
Arturo Magidin
William Elliot <ma...@privacy.net> wrote:
>From: Arturo Magidin <mag...@math.berkeley.edu>
>Newsgroups: sci.math
>Subject: Re: Algebra problems
>
>William Elliot <ma...@privacy.net> wrote:
>
> >>R = { p/q | p in Z, q not in 3Z } is PID
> >>I non-trivial ideal R ==> some m >= 0 with I = (3^m)
>>
> >>prime n ==> R = { p/q | p in Z, q not in nZ } is PID
> >>I non-trivial ideal R ==> some m >= 0 with I = (n^m)
>
> >Better generalization: show that a localization
> >of a PID is a PID. (You are localizing at (3)).
>What ever that means. Anything similar to local ring?
>Let's see. (3) ideal R and (3^n) subset (3) for all n in N.
>Ok, R is a local ring. How's it local? A ring in the small?
From the introduction to Nagata's "Local Rings", Interscience Tracts
in Pure and Applied Mathematics No. 13 (John Wiley and Sons, 1962)
"The history of the theory of local rings begins with Krull's paper
[Dimensiontheorie in Stellenringen', J. Reine Angew. Math. 179
(1938) pp. 204-226]. Here he defined a 'Stellenring' as a
Noetherian ring with only one maximal ideal (a Noetherian ring is a
commutative ring with units which satisfies the maximum condition
for ideals). The name 'Stellenring' was chosen because such rings
are often associated with points on algebraic and analytic
varieties. Chevalley [in 'On the theory of local rings',
Ann. Math. 44 (1943) pp. 690-708] renamed them 'local rings' since
a ring associated with a point on a variety gives local properties
of the variety."
> >If, on the other hand, you meant "that could be written" in a
> >quotient like that,
>1/(3k+2) = -1/(-3k-2) = -1/(-3(k+1) + 1) in R =
>{ p/q | p in Z, q in 3Z + 1 } = { p/q | p in Z, q not in 3Z }
>= blithereens, nothing new
Here's an exercise for you: prove that any subring of Q which contains
is obtained by localizing Z at some multiplicative subset S.
Bill Dubuque
>William Elliot <ma...@privacy.net> wrote:
>>
>> prime n ==> R = { p/q | p in Z, q not in nZ } is PID
>> I non-trivial ideal R ==> some m >= 0 with I = (n^m)
>
> Better generalization: show that a localization of a PID is a PID.
> (You are localizing at (3))
The fraction ring construction D S^-1 with denominator submonoid S < D*
preserves many properties of domains D, e.g. Euclidean, PID, UFD, valuation,
Bezout, GCD, Dedekind, Prufer, Krull, Noetherian, integrally closed, etc.
Recall that localization at a prime ideal P is simply the special case
where the denominator monoid S = D \ P is the complement of P within D.
In the present case of PIDs, such preservation follows immediately
from this obvious LEMMA: every ideal I in any fraction ring D S^-1
is the extension of its contraction to D, i.e. I = (I /\ D) S^-1.
Thus when D is a PID, I /\ D = i D is principal and therefore so
too is its extension to D S^-1, viz. I = (I /\ D) S^-1 = i D S^-1.
>> -- still a PID (principal ideal domain)?
>> R = { p/q | p in Z, q in 3Z + 1 }
>>
>> What do the ideals look like?
>
> Either they don't look, or this is the localization at (3) again.
>
> Given your inimitable telegraphic style, it is hard to see just what
> it is you are defining as R: is it all rationals that MAY be written
> as quotients in which the denominator is congruent to 1 mod 3, or just
> the ones which, ->in reduced form<- have denominators congruent to 1
> mod 3? Or is it the ring generated by these elements?
>
> 1/10 is in R; so is 5/1. But their product is 1/2 which is not in the
> form given. So if you meant "in reduced form", then what you gave was
> not a ring.
His ring is simply R = Z S^-1 with denominator monoid S = 3 Z + 1
R is a well-defined ring (of fractions) since S is a submonoid of Z*.
> If, on the other hand, you meant "that could be written" in a quotient
> like that, then every integer not divisible by 3 is a unit: for if n
> is not divisible by 3, then either n = 1 (mod 3) or n=2 (mod 3).
> If n=1 mod(3), then 1/n is in your ring. If n=2, then n^2 = 1 (mod 3),
> so 1/n^2 is in the ring, so is n/1, and therefore so is 1/n. In either
> case, 1/n is in the ring. So R contains the localization at 3. It is
> not hard to verify now that it is exactly the localization at (3).
Simpler: 1/(3n-1) = -1/(-3n+1) is in R. By your proof we may
assume that the monoid S is root-closed: n^k in S => n in S, k>0.
More generally, without changing the fraction ring F = D S^-1,
we may enlarge a denominator monoid S by adjoining all factors
of elements of S, yielding what is called the saturation of S.
Indeed: xy in S => 1/xy in F => y(1/xy) = 1/x in F
Hence suppose S is a saturated monoid: xy in S <=> x,y in S.
Since S contains 1 it contains all units u (= factors of 1)
therefore S is also closed under multiplication by units u.
So (u = -1) we may assume that S is closed under negation,
which yields the special case above: 3n-1 = -(-3n+1).
General fraction rings may be viewed as localizations
at a _set_ of primes, namely it is very easy to prove
THEOREM The following are equivalent for a subset S in a ring R
(1) S is a saturated multiplicatively closed set
(2) the complement of S is a union of prime ideals in R
-Bill Dubuque
Arturo Magidin
Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
[.snip.]
>> If, on the other hand, you meant "that could be written" in a quotient
>> like that, then every integer not divisible by 3 is a unit: for if n
>> is not divisible by 3, then either n = 1 (mod 3) or n=2 (mod 3).
>> If n=1 mod(3), then 1/n is in your ring. If n=2, then n^2 = 1 (mod 3),
>> so 1/n^2 is in the ring, so is n/1, and therefore so is 1/n. In either
>> case, 1/n is in the ring. So R contains the localization at 3. It is
>> not hard to verify now that it is exactly the localization at (3).
>
>Simpler:
I had the impression that most of your interactions with me involve me
writing a proof, only to be followed up by a post telling me how to do
it "Better", "Simpler", or "Easier." Made me think of the good/bad old
days when I was starting my research, and after a week of working hard
at something, my advisor would either explain how to do it or why it
cannot be done after giving it 1 minute's thought...
But then I had some free time right now; going through google, it
looks like only about 1/4-th of your replies do that...
>General fraction rings may be viewed as localizations
>at a _set_ of primes, namely it is very easy to prove
>
>THEOREM The following are equivalent for a subset S in a ring R
>(1) S is a saturated multiplicatively closed set
>(2) the complement of S is a union of prime ideals in R
And if R is a domain, then any ring between R and its field of
fractions is of the form S^{-1}R for some multiplicative subset of
R. If T is the ring, then let S = { t in R: t is a unit in T}.
Bill Dubuque
>
> And if R is a domain, then any ring between R and its field of
> fractions is of the form S^{-1}R for some multiplicative subset
> of R. If T is the ring, then let S = { t in R: t is a unit in T}.
You may wish to revise the above after consulting the following reviews
------------------------------------------------------------------------------
28#111 13.80 (16.00)
Davis, Edward D.
Overrings of commutative rings. II. Integrally closed overrings.
Trans. Amer. Math. Soc. 110 1964 196--212.
------------------------------------------------------------------------------
Part I appeared in same Trans. 104 (1962), 52--61 [MR 25#3061]. A ring in
this article is a commutative ring with unit, and an overring of a ring R
is a ring between R and the total quotient ring of R. A P-ring is a ring
such that every overring is integrally closed (in its total quotient ring).
A Q-ring is such that every overring is a ring of quotients of the ring.
In S.1, the author considers the case of integral domains and proves that
(1) a P-domain is nothing but a Prufer ring, (2) a Noetherian integral
domain is a Q-domain if and only if it is a Dedekind domain such that its
ideal class group is a torsion group, and (3) let D be an integrally closed
integral domain with field of quotients F, L an algebraic extension field
of F and let E be the integral closure of D in L, then if E/\G is a
Q-domain for every finite extension field G of F , then E is a Q-domain
and furthermore every integrally closed subring of L containing D is a
Q-domain. In S.2, the author gives a characterization of P-rings, which
asserts that a ring R is a P-ring if and only if R_{S(M)} is a
quasi-valuation ring for every maximal ideal M such that M contains at
least one non-zero-divisor; here S(M) denotes the set of non-zero-divisors
which are not contained in M . The author proves also in this section a
certain analogue of integral domains of Krull dimension 1 for the case of
rings with zero-divisors. In the last section, the author gives a certain
characterization of Noetherian P-rings (called D-rings), proves an
analogue of (2) above, and also proves that for a Noetherian ring R in which
0 has no imbedded component, R is a D-ring if and only if it is a direct
sum of Dedekind domains and rings whose non-units are all zero-divisors.
Reviewed by M. Nagata
------------------------------------------------------------------------------
28#3051 13.15 (16.00)
Gilmer, Robert; Ohm, Jack
Integral domains with quotient overrings.
Math. Ann. 153 1964 97--103.
------------------------------------------------------------------------------
An integral domain D with field of quotients K is said to have the
QR-property if every ring D' between D and K is a ring of quotients of D.
The article studies integral domains with the QR-property. Starting with some
simple properties of integral domains with the QR-property, the authors prove
the following theorem. Let D be an integral domain with the QR-property, and
let P be its arbitrary prime ideal. Then D_P is a valuation ring. If A is a
finitely generated ideal of D, then A is invertible and some power of A is
contained in a principal ideal.
As for the converse, the authors prove that if every valuation ring of K
containing D is a ring of quotients of D and if every prime ideal of D is the
radical of a principal ideal, then D has the QR-property. As a particular case,
it follows that a Noetherian integral domain D has the QR-property if and only
if it is a Dedekind domain in which the ideal class group is a torsion group.
Some properties of intermediate rings for such rings are observed.
{Reviewer's note: Though the authors added that the result on Noetherian
integral domains with the QR-property was obtained independently also by
E. D. Davis, the reviewer is afraid that this special case may be known by
some other people. Substantially the same result is also contained in the
paper of Goldman reviewed below [#3052].}
Reviewed by M. Nagata
------------------------------------------------------------------------------
28#3052 13.15 (16.00)
Goldman, Oscar
On a special class of Dedekind domains.
Topology 3 1964 suppl. 1, 113--118.
------------------------------------------------------------------------------
The main result is the following. Let R be a Dedekind domain which has the
following two properties: (F1) R/P is a finite field for every maximal ideal P,
and (F2) the group U(R) of units of R is finitely generated. Let K be the
field of quotients of R and let X be a transcendental element over K.
Then there is a Dedekind domain S such that (1) R[X] <= S <= K(X),
(2) S satisfies (F1), and (3) U(R) = U(S).
Some consequences of this theorem are also proved.
Reviewed by M. Nagata
------------------------------------------------------------------------------
31#5880
Richman, Fred
Generalized quotient rings.
Proc. Amer. Math. Soc. 16 1965 794--799.
13.80 (16.00)
------------------------------------------------------------------------------
Let A be an integral domain with the quotient field K , and consider an
intermediate ring A < B < K . The author calls an intermediate ring B a
"generalized quotient ring of A" if B is A-flat. (For example, if B = S^{-1} A
for some multiplicative subset S of A, then B is A-flat.) One of the
main theorems reads as follows. Let A be an integral domain with the quotient
field K. For an intermediate ring A < B < K, the following statements are
equivalent: (i) B is a generalized quotient ring of A; (ii) (A:Ab)B = B
for all b in B; (iii) B_M = A_{M/\A} for all maximal ideals M of B. The
author then studies the correspondence of ideals in A and in its generalized
quotient ring, and it is also shown that if B is a generalized quotient ring
of A, then the integral closure of B is a generalized quotient ring of the
integral closure of A. Finally, the author gives a new characterization of
Prufer rings; namely, an integral domain A is a Prufer ring if and only if
every extension of A in its quotient field is flat.
Reviewed by D. S. Rim
------------------------------------------------------------------------------
36#6397 13.15
Gilmer, Robert; Heinzer, William J.
Intersections of quotient rings of an integral domain.
J. Math. Kyoto Univ. 7 1967 133--150.
------------------------------------------------------------------------------
The following definitions are employed. Let D be an integral domain, K its
quotient field. Any integral domain between D and K is called an overring
of D. If N is a multiplicative system contained in D-{0} then the set
D_N = {d/n: d in D,n in N} is called a quotient ring of D. If M is a
maximal ideal in D then D_M is defined as the quotient ring of D with
respect to the complement of M . D is a Prufer domain if D_M is a
valuation ring for every maximal ideal M of D.
A domain is said to have the QR-property if every overring is a quotient ring
of it. It is said to have the QQR-property if every overring is the
intersection of quotient rings of it.
The authors investigate the question of how the following properties are
interrelated: a ring is a Prufer domain; a domain has the QQR property.
E. D. Davis [Trans. Amer. Math. Soc. 110 (1964), 247--254; MR 28#111]
showed that any Prufer domain has the QQR property and that if a domain has
the QQR property and if each prime ideal of it is of finite rank then it is
a Prufer domain. Here the authors show that a domain is Prufer if it satisfies
the ascending chain condition for prime ideals and if each of its valuation
overrings is the intersection of quotient rings of it. If the second of these
conditions is satisfied, then the integral closure of the domain is a Prufer
domain. If the domain has the QQR property then its integral closure is Prufer.
Similar results are obtained concerning rings possessing a unique minimal
overring. We quote one more result. A prime ideal P of a domain is called
unbranched if it is the only P-primary ideal. It is shown that if a domain
has the QQR property and if each of its maximal ideals is branched it is a
Prufer domain. Several instructive examples are given, in particular a domain
is constructed which has the QQR property but is not Prufer.
Reviewed by C. F. Moppert
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42#244 13.15
Heinzer, William
Quotient overrings of integral domains.
Mathematika 17 1970 139--148.
------------------------------------------------------------------------------
The results of this paper are closely related to those of several other papers,
includings ones by the reviewer and J. Ohm [Math. Ann. 153 (1964), 97-103; MR
28#3051], E. D. Davis [Trans. Amer. Math. Soc. 110 (1964), 196-212; MR 28#111],
R. L. Pendleton [Bull. Amer. Math. Soc. 72 (1966), 499-500; MR 32#7578] and
the reviewer and the author [J. Math. Kyoto Univ. 7 (1967), 133-150; MR 36
#6397]. Let D be an integral domain with identity with quotient field K ;
an overring of D is defined to be a subring of K containing D . If each
overring of D is a quotient ring of D , D is said to be a QR-domain.
A QR-domain is necessarily a Prufer domain, and conversely, a Prufer domain
with torsion class group is a QR-domain [the reviewer and J. Ohm, loc. cit.].
In the first section of this paper, the author gives an example of a
one-dimensional QR-domain of finite character whose class group is not a
torsion group, thereby answering a question that has been open for several
years. The construction of this example is quite clever, and is of independent
interest in itself.
Let S be a family of nonempty subsets of D such that S is closed under
multiplication. Define D_S to be the set of elements t in K such
that tA <= D for some A in S; D_S is called a generalized quotient
ring of D . Quotient rings, ideal transforms, flat overrings, and any
intersection of localizations of D are generalized quotient rings of D.
In the second section of this paper, the author calls D a GQR-domain if
each overring of D is a generalized quotient ring of D. A Prufer domain
is a GQR-domain (more generally, the so-called QQR-domains of the reviewer
and the author [loc. cit.] are GQR-domains), and the author proves, conversely,
that an integrally closed GQR-domain is a Prufer domain. In fact, he proves
that if ~D is the integral closure of the GQR-domain D, and if ~D
is a finite D-module, then ~D is a Prufer domain. The paper concludes
with an example of a GQR-domain J such that J_P is not a GQR-domain
for some prime ideal P of J .
{Reviewer's remark: In Proposition 1.2, the author could say, more
explicitly, that {V_\b} is the set of non-trivial valuation overrings
of D; this is true because each V_\b is of rank one.}
Reviewed by Robert Gilmer
------------------------------------------------------------------------------
45#5123 13B99
Wajnryb, Bronislaw; Zaks, Abraham
On the flat overrings of an integral domain.
Glasgow Math. J. 12 (1971), 162--165.
------------------------------------------------------------------------------
The main result of this paper is the following theorem. Let D be a Krull
domain and let X be the set of minimal prime ideals of D; the following
conditions are equivalent: (a) The divisor class group of D is a torsion
group; (b) for each subset Y of X, /\ {D_P|P in Y} is a quotient ring of D.
This theorem generalizes the following result, due to E. D. Davis [Trans.
Amer. Math. Soc. 110 (1964), 196-212; MR 28#111], the reviewer and J. Ohm
[Math. Ann. 153 (1964), 97--103; MR 28#3051] and O. Goldman [Topology 3 (1964),
suppl. 1, 113--118; MR 28#3052]. If J is a Dedekind domain, then each overring
of J is a quotient ring of J if and only if the class group of J is a torsion
group.
Reviewed by Robert Gilmer
------------------------------------------------------------------------------
48#3946 13B20
Davis, Edward D.
Overrings of commutative rings. III. Normal pairs.
Trans. Amer. Math. Soc. 182 (1973), 175--185.
------------------------------------------------------------------------------
Assume that R and S are commutative rings with a common identity and that
R is a subring of S . Then (R,S) is called a normal pair if each
intermediate subring T is integrally closed in S ; (R,S) is a QR-pair
if each such T is a quotient ring of R . The author restricts to the case
where R and S are integral domains although, as he notes, some of the
results extend to the case where S has zero divisors.
If D is an integral domain with identity and with quotient field K , then
the author has previously shown [same Trans. 110 (1964), 196--212;
MR 28#111] that (D,K) is a normal pair if and only if D is a Prufer
domain. The author [op. cit.], J. Ohm and the reviewer [Math. Ann. 153
(1964), 97--103; MR 28#3051], and R. Pendleton [Bull. Amer. Math. Soc.
72 (1966), 499--500; MR 32#7578] have considered the problem of determining
conditions under which (D,K) is a QR-pair; in particular, D is a Prufer domain
and if D is Noetherian, (D,K) is a QR-pair if and only if D is a Dedekind
domain with torsion class group. The results of this paper are therefore
generalizations of the case in which the second member of the pair is the
quotient field of the first member.
Assume that D is a quasi-local domain and J is an overring of D . Then
(D,J) is a normal pair if and only if x or x^{-1} is in D for each x
in J ; equivalently, (D,J) is normal if and only if there is a prime ideal
P of D such that J = D_P, PJ = J, and D/P is a valuation ring. Using the
fact that normality of a pair is preserved under quotient ring formation, the
author then applies the preceding result to the global case. Thus, if D' is
an integral domain with identity and J' is the composite of the family
{J_\a} of overrings of D' such that (D',J_\a) is normal, then (D',J')
is also normal. Moreover, J' = /\_\l D_{P_\l}' , where {P_\l} is the family of
prime ideals of D' such that D_{P_\l}' is trivial, in the sense that there is
no proper overring T of D_{P_\l}' such that (D_{P_\l}',T) is normal. The
pair (D',D") is normal, where D" is an overring of D' , if and only if each
intermediate ring is flat as a D'-module; this extends a result of F. Richman
[Proc. Amer. Math. Soc. 16 (1965), 794--799; MR 31#5880] from the case
where D" is the quotient field of D' .
If J is an overring of the domain D with identity, then (D,J) is normal
if (D,J) is a QR-pair. Extending a result of Pendleton [op. cit.] the
author proves that (D,J) is a QR-pair if and only if the radical of each
finitely generated ideal of D that extends to the unit ideal of J is the
radical of a principal ideal. Moreover, if J is the composite of the family
{J_\a} of overrings of D such that (D,J_\a) is a QR-pair, then (D,J)
is also a QR-pair.
In S.4, the author considers normal pairs and QR -pairs (D,J) , where D
is a member of a class of domains that includes the class of Noetherian
domains.
{Reviewer's remarks: In the proof of Theorem 5, the symbols A_x, A_y, and
A_{xy} stand for A[x^{-1}], and A[y^{-1}], and A[x^{-1}y^{-1}].}
Reviewed by Robert Gilmer
------------------------------------------------------------------------------
55#12718
Zaks, Abraham
Dedekind k -subalgebras of k(x).
Comm. Algebra 5 (1977), no. 4, 347--364.
------------------------------------------------------------------------------
Let k < k(x) be fields with x transcendental over k . Any valuation ring
V of k(x) containing k either contains x and thus is of the form
k[x]_{(p)} , p an irreducible element of k[x] , or else is the
(1/x) -adic valuation ring V_oo . The "minimal" integrally closed
domains R > k having quotient field k(x) are obtained by interesecting all
but one of these valuation rings and thus are either k[x] or a ring S of
the form k[x,1/p] /\ V_oo . Such an S is a 1-dim Krull domain and
hence Dedekind. Moreover, since any element z of k[x,1/p] may be written
as z = (a_0 + a_1 p +...+ a_t p^t)/p^s, a_i in k[x] and deg a_i < deg p = m,
and since the elements of V_oo are just the quotients of polynomials for
which deg denominator >= deg numerator, it follows that
S = k[1/p,x/p,...,x^{m-1}/p]. The author proves here that S has cyclic
class group (of order m) and hence by a classical result that every overring
of S is a localization of S (with respect to a multiplicative system).
Thus, the Dedekind domains R > k having quotient field k(x) may be
described as either localizations of k[x] or localizations of rings S of
the above form. The author also determines conditions under which such an R
is a PID.
Reviewed by J. Ohm
------------------------------------------------------------------------------
Arturo Magidin
Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>Arturo Magidin <mag...@math.berkeley.edu> wrote:
>>
>> And if R is a domain, then any ring between R and its field of
>> fractions is of the form S^{-1}R for some multiplicative subset
>> of R. If T is the ring, then let S = { t in R: t is a unit in T}.
>
>You may wish to revise the above after consulting the following reviews
Sigh. I (obviously too) briefly considered whether some weird
properties might cause problems...
William Elliot
William Elliot <ma...@privacy.net> wrote:
>From: Arturo Magidin <mag...@math.berkeley.edu>
>Newsgroups: sci.math
>Subject: Re: Algebra problems
>
>William Elliot <ma...@privacy.net> wrote:
>
> >>R = { p/q | p in Z, q not in 3Z } is PID
> >>I non-trivial ideal R ==> some m >= 0 with I = (3^m)
>>
> >>prime n ==> R = { p/q | p in Z, q not in nZ } is PID
> >>I non-trivial ideal R ==> some m >= 0 with I = (n^m)
>
>>Better generalization: show that a localization
>>of a PID is a PID. (You are localizing at (3)).
>Here's an exercise for you: prove that any subring of Q which
>contains is obtained by localizing Z at some multiplicative subset
>S.
Here's what came from locally sleeping upon the notions.
In the context of communative rings with identity:
A ring R is local ring when R contains a unique maximal ideal M.
An ideal I localizes a ring R when R\I = set of R's units.
If R is local ring with unique maximal ideal M, then M localizes R.
If a not in M and if (a) proper ideal, then by Zorn, some
maximal ideal I containing a. But as R is local, I = M.
Thus (a) not proper ideal, 1 in (a), a is unit.
If I localizes R, then I is maximal and unique, R local ring.
If a not in I, then a unit, hence ideal <a,I> = R. Thus I max.
If J proper ideal, then J subset I for if some j in J\I then
j is unit and J not proper and were J maximal ideal, then J = I.
Thus for prime p, R = Q - Z[1/p] is local ring
localized by (p) the unique maximal ideal of R
----