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Udo Goebel
science forum beginner
Joined: 21 May 2006
Posts: 6
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 Posted: Sun May 21, 2006 11:45 am Post subject:
Definition for UFD
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Hello,
perhaps a stupid remark, but I'm just reading Ribenboim's "Classical
Theory of Algebraic Numbers" once again and right on the first page he
comes up with a definition for unique factorization domain which already
made me scratching my head on the first reading:
"The domain A is said to be a unique factorization domain when the
following statement is true in A:
Let S be a set of irreducible elements of A such that every irreducible
element of A is associated with one and only one element of S. Then
every element x of [the field of quotients] K, x != 0, may be written
uniquely in the form
x = u * Product_(p in S) p^v_p(x)
where u is a unit of A and v_p(x) in Z, and only finitely many v_p not 0."
He then concludes: "It follows that x in A iff v_p(x) >= 0 for all p."
1. Is the definition really equivalent to the usual one, where K is not
mentioned and the representation for x follows from the usual definition?
2. Equivalently: Why does it follow that x in A iff v_p >= 0 for all p?
Udo |
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Lukas-Fabian Moser
science forum beginner
Joined: 08 Apr 2005
Posts: 4
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| Posted: Sun May 21, 2006 9:15 pm Post subject:
Re: Definition for UFD
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Hello,
On 21 May 2006 07:45:04 -0400, Udo Goebel <udo_goebel@gmx.net> wrote:
[A a domain such that:]
| Quote: | Let S be a set of irreducible elements of A such that every irreducible
element of A is associated with one and only one element of S. Then
every element x of [the field of quotients] K, x != 0, may be written
uniquely in the form
x = u * Product_(p in S) p^v_p(x)
where u is a unit of A and v_p(x) in Z, and only finitely many v_p not 0."
2. Equivalently: Why does it follow that x in A iff v_p >= 0 for all p?
|
The "if" direction is trivial. As for the "only if" direction, assume
that x = Product_(p in S) p^(v_p) lies in A. We prove that v_p >= 0
for all p by induction on the number t of all p such that v_p does not
vanish. The case t = 0 is empty.
Now for t > 0, assume for one irreducible element p_0 we have v_(p_0)
= -k < 0. Then by the induction hypothesis, applied to y := x * p_0^k,
we see that all other v_p are non-negative. On the other hand, p_0
divides y in A, so p_0 must divide one of the other irreducible
elements, which is absurd. (We need the induction hypothesis to be
able to split y into a product of factors *in A*; I could imagine that
you can get around induction by multipying with several irreducible
elements at a time.)
Gr٤e, Lukas |
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Udo Goebel
science forum beginner
Joined: 21 May 2006
Posts: 6
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| Posted: Fri May 26, 2006 12:30 am Post subject:
Re: Definition for UFD
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Hello,
thanks for taking the time:
Lukas-Fabian Moser schrieb:
| Quote: | [...] 2. Equivalently: Why does it follow that x in A iff v_p >= 0 for all p?
The "if" direction is trivial.
True enough. |
| Quote: | As for the "only if" direction, assume
that x = Product_(p in S) p^(v_p) lies in A. We prove that v_p >= 0
for all p by induction on the number t of all p such that v_p does not
vanish. The case t = 0 is empty.
Now for t > 0, assume for one irreducible element p_0 we have v_(p_0)
= -k < 0. Then by the induction hypothesis, applied to y := x * p_0^k,
we see that all other v_p are non-negative. On the other hand, p_0
divides y in A, so p_0 must divide one of the other irreducible
elements, which is absurd. [...]
Well, this is exactly the place I had difficulties with trying various |
proofs. How can you assert that p_0 must divide one of the irreducibles?
This is certainly the case if A is a UFD (in the traditional
definition), but here we simply don't know.
Regards
Udo |
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Ryan Reich
science forum Guru Wannabe
Joined: 21 May 2005
Posts: 120
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| Posted: Tue May 30, 2006 3:00 am Post subject:
Re: Definition for UFD
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One 21 May 2006 07:45:04 -0400, Udo Goebel <udo_goebel@gmx.net> wrote:
| Quote: |
Hello,
perhaps a stupid remark, but I'm just reading Ribenboim's "Classical
Theory of Algebraic Numbers" once again and right on the first page he
comes up with a definition for unique factorization domain which already
made me scratching my head on the first reading:
"The domain A is said to be a unique factorization domain when the
following statement is true in A:
Let S be a set of irreducible elements of A such that every irreducible
element of A is associated with one and only one element of S. Then
every element x of [the field of quotients] K, x != 0, may be written
uniquely in the form
x = u * Product_(p in S) p^v_p(x)
where u is a unit of A and v_p(x) in Z, and only finitely many v_p not 0."
He then concludes: "It follows that x in A iff v_p(x) >= 0 for all p."
1. Is the definition really equivalent to the usual one, where K is not
mentioned and the representation for x follows from the usual definition?
2. Equivalently: Why does it follow that x in A iff v_p >= 0 for all p?
Udo
|
For such an obvious solution this took me an embarrassingly long time.
Let x = u * Product_{p in S} p^{v_p(x)} be some element of A,
let y = u * Product_{v_p(x) > 0} p^{v_p(x)} be its "numerator", and
let z = Product_{v_p(x) < 0} p^{-v_p(x)} be its "denominator".
We may assume that y and z are relatively prime; thus, none of the prime
factors of z are also prime factors of y. However, the equation
xz = y
(where all elements are in A) contradicts this. It follows that z = 1, so
every element of A has positive order for every irreducible.
--
Ryan Reich
ryan.reich@gmail.com
sci.math.research |
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Ryan Reich
science forum Guru Wannabe
Joined: 21 May 2005
Posts: 120
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| Posted: Tue May 30, 2006 1:30 pm Post subject:
Re: Definition for UFD
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One 21 May 2006 07:45:04 -0400, Udo Goebel <udo_goebel@gmx.net> wrote:
| Quote: |
Hello,
perhaps a stupid remark, but I'm just reading Ribenboim's "Classical
Theory of Algebraic Numbers" once again and right on the first page he
comes up with a definition for unique factorization domain which already
made me scratching my head on the first reading:
"The domain A is said to be a unique factorization domain when the
following statement is true in A:
Let S be a set of irreducible elements of A such that every irreducible
element of A is associated with one and only one element of S. Then
every element x of [the field of quotients] K, x != 0, may be written
uniquely in the form
x = u * Product_(p in S) p^v_p(x)
where u is a unit of A and v_p(x) in Z, and only finitely many v_p not 0."
He then concludes: "It follows that x in A iff v_p(x) >= 0 for all p."
1. Is the definition really equivalent to the usual one, where K is not
mentioned and the representation for x follows from the usual definition?
2. Equivalently: Why does it follow that x in A iff v_p >= 0 for all p?
Udo
|
Ah, forget my last reply, I mixed up unique factorization in A with unique
factorization in K. I shouldn't have used the word "embarrassing" :)
The correct answer is a slight modification. First note that A is contained
in the intersection of the localizations A_(p) at each prime ideal generated
by the elements of S (obviously, by definition of localization). Second, A
contains the set R of all elements of K with positive exponents for elements
of S. We can express R as the intersection of the sets R_p, the ring of
integers for the valuation v_p (consisting of all elements x of K with
v_p(x) >= 0).
I claim that A_(p) < R_p; indeed, this is the very definition, since A_(p)
consists of fractions of elements in A whose denominator does not contain p.
That is, certain elements of K with non-negative valuation at p. It follows
that we have the inclusions
A < Intersection_{p in S} A_(p) < R < A
so A = R. QED
--
Ryan Reich
ryan.reich@gmail.com
sci.math.research |
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Lukas-Fabian Moser
science forum beginner
Joined: 08 Apr 2005
Posts: 4
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| Posted: Tue May 30, 2006 1:30 pm Post subject:
Re: Definition for UFD
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Ryan Reich wrote:
| Quote: | Let x = u * Product_{p in S} p^{v_p(x)} be some element of A,
let y = u * Product_{v_p(x) > 0} p^{v_p(x)} be its "numerator", and
let z = Product_{v_p(x) < 0} p^{-v_p(x)} be its "denominator".
We may assume that y and z are relatively prime; thus, none of the prime
factors of z are also prime factors of y.
|
It's a bit dangerous to speak of "prime factors" before you know that A
is factorial... Anyway, we can assume that the unique representations
for y,z have no common irreducible factor.
| Quote: | However, the equation
xz = y
(where all elements are in A) contradicts this.
|
Can you explain why?
Lukas |
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Hagen
science forum beginner
Joined: 28 Apr 2005
Posts: 42
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| Posted: Tue May 30, 2006 1:30 pm Post subject:
Re: Definition for UFD
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The usual definition: the domain A is called a UFD
iff every element x of A can be written as a product
of prime elements.
Remark: the uniqueness of the factorization up to
associated primes follows.
As for the equivalence with Ribenboim's definition:
the implication --> follows from the fact, that in
a UFD the set of primes coincides with the set of
irreducibles, so that a set S as in Ribenboim's definition
consists of a full set of representatives for classes
of associated primes.
For the implication <-- one only has to show that every
irreducible element of A is prime, which is routine
using the uniqueness statement in Ribenboim's definition.
H |
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Ryan Reich
science forum Guru Wannabe
Joined: 21 May 2005
Posts: 120
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| Posted: Tue May 30, 2006 3:00 pm Post subject:
Re: Definition for UFD
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One Tue, 30 May 2006 13:30:05 +0000 (UTC), Ryan Reich <ryan.reich@gmail.com> wrote:
| Quote: | One 21 May 2006 07:45:04 -0400, Udo Goebel <udo_goebel@gmx.net> wrote:
Hello,
perhaps a stupid remark, but I'm just reading Ribenboim's "Classical
Theory of Algebraic Numbers" once again and right on the first page he
comes up with a definition for unique factorization domain which already
made me scratching my head on the first reading:
"The domain A is said to be a unique factorization domain when the
following statement is true in A:
Let S be a set of irreducible elements of A such that every irreducible
element of A is associated with one and only one element of S. Then
every element x of [the field of quotients] K, x != 0, may be written
uniquely in the form
x = u * Product_(p in S) p^v_p(x)
where u is a unit of A and v_p(x) in Z, and only finitely many v_p not 0."
He then concludes: "It follows that x in A iff v_p(x) >= 0 for all p."
1. Is the definition really equivalent to the usual one, where K is not
mentioned and the representation for x follows from the usual definition?
2. Equivalently: Why does it follow that x in A iff v_p >= 0 for all p?
Udo
Ah, forget my last reply, I mixed up unique factorization in A with unique
factorization in K. I shouldn't have used the word "embarrassing" :)
The correct answer is a slight modification. First note that A is contained
in the intersection of the localizations A_(p) at each prime ideal generated
by the elements of S (obviously, by definition of localization). Second, A
contains the set R of all elements of K with positive exponents for elements
of S. We can express R as the intersection of the sets R_p, the ring of
integers for the valuation v_p (consisting of all elements x of K with
v_p(x) >= 0).
I claim that A_(p) < R_p; indeed, this is the very definition, since A_(p)
consists of fractions of elements in A whose denominator does not contain p.
That is, certain elements of K with non-negative valuation at p. It follows
that we have the inclusions
|
Well, at least I'm converging on a proof. Unfortunately this line is
circular. To fix it, first note that in A_(p) we invert all the other primes
of S, since they are irreducible and not associate to p, hence not in (p).
Then say x in A_(p) had v_p(x) < 0, so x^{-1} has v_p(x^{-1}) > 0. This means
that it is in A_(p), since it is a product of inverses of primes not equal to
p, and p. This also means that it is in the maximal ideal, since it is
obviously a multiple of p. But this is impossible, since its inverse is also
in A_(p).
| Quote: | A < Intersection_{p in S} A_(p) < R < A
so A = R. QED
|
--
Ryan Reich
ryan.reich@gmail.com
sci.math.research |
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Ryan Reich
science forum Guru Wannabe
Joined: 21 May 2005
Posts: 120
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| Posted: Tue May 30, 2006 3:00 pm Post subject:
Re: Definition for UFD
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One Tue, 30 May 2006 13:30:06 +0000 (UTC), Lukas-Fabian Moser <lfm@gmx.de> wrote:
| Quote: | Ryan Reich wrote:
Let x = u * Product_{p in S} p^{v_p(x)} be some element of A,
let y = u * Product_{v_p(x) > 0} p^{v_p(x)} be its "numerator", and
let z = Product_{v_p(x) < 0} p^{-v_p(x)} be its "denominator".
We may assume that y and z are relatively prime; thus, none of the prime
factors of z are also prime factors of y.
It's a bit dangerous to speak of "prime factors" before you know that A
is factorial... Anyway, we can assume that the unique representations
for y,z have no common irreducible factor.
However, the equation
xz = y
(where all elements are in A) contradicts this.
Can you explain why?
Lukas
|
Sorry, this post was wrong. I've written (two) corrections, the second of
which may not have appeared yet.
--
Ryan Reich
ryan.reich@gmail.com
sci.math.research |
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Udo Goebel
science forum beginner
Joined: 21 May 2006
Posts: 6
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| Posted: Tue May 30, 2006 8:00 pm Post subject:
Re: Definition for UFD
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Thanks all,
the simple key is to remark that the usual proof for the equivalence of
irreducibles and primes goes through with Ribenboim's definition as
Hagen said. After that Lukas' proof goes through.
At least not too stupid to think 5 minutes about it  |
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Udo Goebel
science forum beginner
Joined: 21 May 2006
Posts: 6
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| Posted: Thu Jun 01, 2006 9:30 pm Post subject:
Re: Definition for UFD
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Perhaps I was a bit too careless in accepting Hagen's explanation. Lukas
pointed out to me that contrary to what Hagen claims the usual proof
for the equivalence of irreducible and primes does not go through with
Ribenboim's definition. Explicitly:
Let s be an irreducible in A and assume s divides a*b, both in A. Then s
*q = a*b for some q in A. And unique factorization according to
Ribenboim gives
q = r_1 ... r_n / t_1 ... t_n'
a = x_1 ... x_k / y_1 ... y_k'
b = u_1 ... u_l / v_1 ... v_l'
where factors can appear multiple times. Now s must be associated to
some x_i or y_j say x_1. Then we have
s x_2 ... x_k / y_1 ... y_k' = a
Now we don't know if x_2 ... x_k / y_1 ... y_k' is in A, so we cannot
conclude s divides a.
Perhaps Ribenboim is simply wrong with his definition? |
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Edgar Fuß
science forum beginner
Joined: 28 Nov 2005
Posts: 2
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| Posted: Sat Jun 03, 2006 2:30 pm Post subject:
Re: Definition for UFD
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Am I missing the point? It appears to me that the only non-obvious
thing to show is that if A has Ribenboim's property, then for x in A,
all the exponents are non-negative. But since x is in A, there is SOME
decomposition into irreducibles (with non-negative exponents, of
course), so if one of the exponents of the decomposition in K were
negative, we would have two different decompositions of x (in K),
which contradicts A having Ribenboim's property.
There is proably something stupidly wrong with this I will notice
immediately as soon as I have submitted my posting. |
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Pete Klimek
science forum beginner
Joined: 15 May 2005
Posts: 7
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| Posted: Sun Jun 04, 2006 2:30 pm Post subject:
Re: Definition for UFD
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"Udo Goebel" <udo_goebel@gmx.net> wrote in message
news:e5nm8v$tle$1@news.ks.uiuc.edu...
| Quote: | Perhaps I was a bit too careless in accepting Hagen's explanation. Lukas
pointed out to me that contrary to what Hagen claims the usual proof
for the equivalence of irreducible and primes does not go through with
Ribenboim's definition. Explicitly:
Let s be an irreducible in A and assume s divides a*b, both in A. Then s
*q = a*b for some q in A. And unique factorization according to
Ribenboim gives
q = r_1 ... r_n / t_1 ... t_n'
a = x_1 ... x_k / y_1 ... y_k'
b = u_1 ... u_l / v_1 ... v_l'
where factors can appear multiple times. Now s must be associated to
some x_i or y_j say x_1. Then we have
s x_2 ... x_k / y_1 ... y_k' = a
Now we don't know if x_2 ... x_k / y_1 ... y_k' is in A, so we cannot
conclude s divides a.
Perhaps Ribenboim is simply wrong with his definition?
|
Here is my attempt to show, by example, that Ribenboim's definition of UFD
differs from the usual one. Comments and criticisms are welcome.
Let R = {f(X) in Q[X]: f(0), f(1) are in Z}, where Q denotes the rational
numbers and Z the rational integers. Then R is an integral domain.
We claim that R is not a UFD (in the usual sense), but that its quotient
field K satisfies Ribenboim's UFD condition.
To prove the first part of the claim, suppose the R is a UFD. We make the
following observations:
(1) The only units in R are 1 and -1.
(2) The reducible elements of Z (i.e., the rational primes) remain
irreducible in R and hence are primes in R.
(3) In a UFD (such as R, by assumption) with a finite set of units, any
nonzero, nonunit element is divisible by only finitely many primes.
(These observations all require proof, but it seems to me that they all are
straightforward.)
Now let f(X) = X * (X - 1), which obviously lies in R.
For any rational prime p, Let g_p(X) = f(X)/p, all of which also obviously
lie in R. Then
f(X) = p * g_p(X)
Thus, f(X) is divisible by infinitely many primes. This contradicts
observation (3) and so R is not a UFD.
To prove the second part of the claim, we'll show that the quotient field K
of R is Q(X) (i.e., all rational functions of X with coefficients in Q).
Since
Q(X) clearly satisfies Ribenboim's UFD condition, this will complete the
proof.
Obviously K is contained in Q(X), so it suffices to show that Q(X) is
contained in K.
Let F(X)/G(X), with G(X) <> 0, be any rational function with coefficients in
Q.
Write
F(X) = r_n * X^n + ... + r_1 * X + r_0, with r_i rational numbers,
and
G(X) = s_m * X^m + ... + s_1 * X + s_0, with s_i rational numbers.
Let
r_i = a_i/b_i, with a_i, b_i integers and b_i <> 0
and
s_i = c_i/d_i, with c_i, d_i integers and d_i <> 0.
Define
B = b_n * ... * b_1 * b_0
and
D = d_m * ... * d_1 * d_0
Then both B and D are <> 0.
Define
F_1(X) = B * D * F(X)
and
G_1(X) = B * D * G(X)
It may be seen that F_1(0), F_1(1), G_1(0) and G_1(1) are all integers.
Therefore F_1(X) and G_1(X) are in R and F_1(X)/G_1(X) is in K. Since
F_1(X)/G_1(X) = F(X)/G(X)
we see that any rational function with coefficients in Q is a member of K.
Therefore, R is an integral domain that satisfies Ribenboim's UFD criterion,
but is not a UFD in the usual sense.
Acknowledgments: I benefited from the paper "RESTRICTED ELASTICITY AND RINGS
OF INTEGER-VALUED POLYNOMIALS DETERMINED BY FINITE SUBSETS" by SCOTT T.
CHAPMAN AND WILLIAM W. SMITH. It may be found here:
http://www.trinity.edu/schapman/RevFinalfactint.pdf. |
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William C Waterhouse
science forum beginner
Joined: 04 May 2005
Posts: 17
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| Posted: Wed Jun 07, 2006 3:30 pm Post subject:
Re: Definition for UFD
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In article <e5uqpe$2uo$1@news.ks.uiuc.edu>, "Pete Klimek" <Peter_delete_Klimek@comcast_delete_.net> writes:
| Quote: | ...
Here is my attempt to show, by example, that Ribenboim's definition of UFD
differs from the usual one. Comments and criticisms are welcome.
Let R = {f(X) in Q[X]: f(0), f(1) are in Z}, where Q denotes the rational
numbers and Z the rational integers. Then R is an integral domain.
We claim that R is not a UFD (in the usual sense), but that its quotient
field K satisfies Ribenboim's UFD condition.
To prove the first part of the claim...
|
That's fine.
| Quote: | ...
To prove the second part of the claim, we'll show that the quotient field K
of R is Q(X) (i.e., all rational functions of X with coefficients in Q).
Since
Q(X) clearly satisfies Ribenboim's UFD condition, this will complete the
proof.
...
|
No, this part doesn't work. It works if we write Ribenboim's condition
using the irreducibles in Z[X], which of course is a UFD. But there are
other irreducibles in R. For instance, clearly X+1 and X-2 are irreducible
in R. Their product is X^2 - 2X - 2. But that has value 2 at X=0 and X=1,
so (1/2)X^2 - X - 1 is in R and is easily seen to be another irreducible.
William C. Waterhouse
Penn State |
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Hagen
science forum beginner
Joined: 28 Apr 2005
Posts: 42
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| Posted: Wed Jun 07, 2006 3:30 pm Post subject:
Re: Definition for UFD
|
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| Quote: | Am I missing the point? It appears to me that the
only non-obvious
thing to show is that if A has Ribenboim's property,
then for x in A,
all the exponents are non-negative. But since x is in
A, there is SOME
decomposition into irreducibles (with non-negative
exponents, of
course),
|
No. This is true if R is noetherian. Otherwise it is
possible that a non-unit can be divided by an element
infinitely often.
So Ribenboim's definition is equivalent to the usual
one if R is noetherian.
H |
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