Mixing
Is there a “pseudo-UFD” without unity?
It is known that every euclidean domain has a product identity element. Does the same happens to the UFDs?
I tried to find a counter example playing with the powers of $2$ with
strange operations, but I get stuck. Any idea?
Important details
By domain I mean, a commutative ring (maybe without unity) with no zero divisors.
In a domain without unity, we can define the divisibility relation ($a$ divides $b$ if and only if there exists $c$ so that $ac=b$), also the irreducibility concept ($anot=0$ is reducible if and only if there exists $b,cnot=0$ such that $a=bc$, so, an irreducible element is an element which is not reducible), and the "associates" relation ($a$ and $b$ are associates if and only if $a|b$ and $b|a$).
So we can define something like an UFD, which we call a "pseudo-UFD", like this
Let $A$ be a commutative ring without unity and without zero divisor, we will call it "pseudo-UFD" if
- For each $ain A$, there exists irreducible elements $p_1,dots,p_r$ (not necessarily different) so that $a = p_1cdots p_r$.
- If $a=q_1cdots q_s$ (being $q_1,dots,q_s$ irreducible elements not necessarily different), then $r=s$ and, maybe after reordering, $p_i|q_i$ and $q_i|p_i$ for every $iin{1,dots,r}$.
As @rschwieb pointed out, in an unitless ring without zero divisors, the associates relation is empty, so the second item of the definition will change to
- If $a=q_1cdots q_s$ (being $q_1,dots,q_s$ irreducible elements not necessarily different), then $r=s$ and, maybe after reordering, $p_i=q_i$ for every $iin{1,dots,r}$.
So, is there a "pseudo-UFD" without unity?
abstract-algebra
asked Nov 21 '18 at 15:22
Álvaro G. Tenorio
13710
add a comment |
It is known that every euclidean domain has a product identity element. Does the same happens to the UFDs?
I tried to find a counter example playing with the powers of $2$ with
strange operations, but I get stuck. Any idea?
Important details
By domain I mean, a commutative ring (maybe without unity) with no zero divisors.
In a domain without unity, we can define the divisibility relation ($a$ divides $b$ if and only if there exists $c$ so that $ac=b$), also the irreducibility concept ($anot=0$ is reducible if and only if there exists $b,cnot=0$ such that $a=bc$, so, an irreducible element is an element which is not reducible), and the "associates" relation ($a$ and $b$ are associates if and only if $a|b$ and $b|a$).
So we can define something like an UFD, which we call a "pseudo-UFD", like this
Let $A$ be a commutative ring without unity and without zero divisor, we will call it "pseudo-UFD" if
- For each $ain A$, there exists irreducible elements $p_1,dots,p_r$ (not necessarily different) so that $a = p_1cdots p_r$.
- If $a=q_1cdots q_s$ (being $q_1,dots,q_s$ irreducible elements not necessarily different), then $r=s$ and, maybe after reordering, $p_i|q_i$ and $q_i|p_i$ for every $iin{1,dots,r}$.
As @rschwieb pointed out, in an unitless ring without zero divisors, the associates relation is empty, so the second item of the definition will change to
- If $a=q_1cdots q_s$ (being $q_1,dots,q_s$ irreducible elements not necessarily different), then $r=s$ and, maybe after reordering, $p_i=q_i$ for every $iin{1,dots,r}$.
So, is there a "pseudo-UFD" without unity?
abstract-algebra
asked Nov 21 '18 at 15:22
Álvaro G. Tenorio
13710
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 21 '18 at 17:42
add a comment |
It is known that every euclidean domain has a product identity element. Does the same happens to the UFDs?
I tried to find a counter example playing with the powers of $2$ with
strange operations, but I get stuck. Any idea?
Important details
By domain I mean, a commutative ring (maybe without unity) with no zero divisors.
In a domain without unity, we can define the divisibility relation ($a$ divides $b$ if and only if there exists $c$ so that $ac=b$), also the irreducibility concept ($anot=0$ is reducible if and only if there exists $b,cnot=0$ such that $a=bc$, so, an irreducible element is an element which is not reducible), and the "associates" relation ($a$ and $b$ are associates if and only if $a|b$ and $b|a$).
So we can define something like an UFD, which we call a "pseudo-UFD", like this
Let $A$ be a commutative ring without unity and without zero divisor, we will call it "pseudo-UFD" if
- For each $ain A$, there exists irreducible elements $p_1,dots,p_r$ (not necessarily different) so that $a = p_1cdots p_r$.
- If $a=q_1cdots q_s$ (being $q_1,dots,q_s$ irreducible elements not necessarily different), then $r=s$ and, maybe after reordering, $p_i|q_i$ and $q_i|p_i$ for every $iin{1,dots,r}$.
As @rschwieb pointed out, in an unitless ring without zero divisors, the associates relation is empty, so the second item of the definition will change to
- If $a=q_1cdots q_s$ (being $q_1,dots,q_s$ irreducible elements not necessarily different), then $r=s$ and, maybe after reordering, $p_i=q_i$ for every $iin{1,dots,r}$.
So, is there a "pseudo-UFD" without unity?
abstract-algebra
asked Nov 21 '18 at 15:22
Álvaro G. Tenorio
13710
It is known that every euclidean domain has a product identity element. Does the same happens to the UFDs?
I tried to find a counter example playing with the powers of $2$ with
strange operations, but I get stuck. Any idea?
Important details
By domain I mean, a commutative ring (maybe without unity) with no zero divisors.
In a domain without unity, we can define the divisibility relation ($a$ divides $b$ if and only if there exists $c$ so that $ac=b$), also the irreducibility concept ($anot=0$ is reducible if and only if there exists $b,cnot=0$ such that $a=bc$, so, an irreducible element is an element which is not reducible), and the "associates" relation ($a$ and $b$ are associates if and only if $a|b$ and $b|a$).
So we can define something like an UFD, which we call a "pseudo-UFD", like this
Let $A$ be a commutative ring without unity and without zero divisor, we will call it "pseudo-UFD" if
- For each $ain A$, there exists irreducible elements $p_1,dots,p_r$ (not necessarily different) so that $a = p_1cdots p_r$.
- If $a=q_1cdots q_s$ (being $q_1,dots,q_s$ irreducible elements not necessarily different), then $r=s$ and, maybe after reordering, $p_i|q_i$ and $q_i|p_i$ for every $iin{1,dots,r}$.
As @rschwieb pointed out, in an unitless ring without zero divisors, the associates relation is empty, so the second item of the definition will change to
- If $a=q_1cdots q_s$ (being $q_1,dots,q_s$ irreducible elements not necessarily different), then $r=s$ and, maybe after reordering, $p_i=q_i$ for every $iin{1,dots,r}$.
So, is there a "pseudo-UFD" without unity?
abstract-algebra
abstract-algebra
asked Nov 21 '18 at 15:22
Álvaro G. Tenorio
13710
asked Nov 21 '18 at 15:22
Álvaro G. Tenorio
13710
edited Nov 21 '18 at 17:41
asked Nov 21 '18 at 15:22
Álvaro G. Tenorio
13710
asked Nov 21 '18 at 15:22
Álvaro G. Tenorio
13710
asked Nov 21 '18 at 15:22
Álvaro G. Tenorio
13710
13710
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 21 '18 at 17:42
add a comment |
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 21 '18 at 17:42
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 21 '18 at 17:42
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 21 '18 at 17:42
add a comment |
2 Answers
2
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Personally, I would hope that any proposed definition would permit something like $2mathbb Z$ to be a pseudo-UFD.
The main problem is to define what "uniqueness of factorization" means, and that is what you've attempted with the criterion mentioned. Requiring two factorizations into irreducibles to be the same length is a must. The main barrier is to deal with associate elements without mentioning units. The way you have things now, we have to allow $pm x$ to be non-associate to one another, but this is unattractive.
One can patch this artificially by saying that $pm x$ are associates to one another, and then perhaps we do get $2mathbb Z$ to be a pseudo-UFD. But unfortunately, this convention does not extend the normal definition of UFDs, since lots of UFDs have many more associates!
So, one can see here the challenges posed by talking about factorization in domains without identity. Perhaps it would be most fruitful to explore which domains without identity can be embedded in domains such that the only units are $pm 1$, so that the ad-hoc definition of associates mentioned above works.
answered Nov 21 '18 at 17:58
rschwieb
105k12100244
add a comment |
Definition of a UFD is: $R$ is an integral domain ($R$ is a commutative ring having unity and no zero-divisors) such that each $ain R setminus (R^* cup {0})$ factors into a product of irreducible elements.Definition clears your doubt
answered Nov 21 '18 at 15:31
John Nash
7418
The definiton of domain I knew does not include the unity condition. And, as you can define irreducibility in rings without unity, the question is still alive, doesn' t it?
– Álvaro G. Tenorio
Nov 21 '18 at 15:40
add a comment |
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2 Answers
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2 Answers
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Personally, I would hope that any proposed definition would permit something like $2mathbb Z$ to be a pseudo-UFD.
The main problem is to define what "uniqueness of factorization" means, and that is what you've attempted with the criterion mentioned. Requiring two factorizations into irreducibles to be the same length is a must. The main barrier is to deal with associate elements without mentioning units. The way you have things now, we have to allow $pm x$ to be non-associate to one another, but this is unattractive.
One can patch this artificially by saying that $pm x$ are associates to one another, and then perhaps we do get $2mathbb Z$ to be a pseudo-UFD. But unfortunately, this convention does not extend the normal definition of UFDs, since lots of UFDs have many more associates!
So, one can see here the challenges posed by talking about factorization in domains without identity. Perhaps it would be most fruitful to explore which domains without identity can be embedded in domains such that the only units are $pm 1$, so that the ad-hoc definition of associates mentioned above works.
answered Nov 21 '18 at 17:58
rschwieb
105k12100244
add a comment |
Personally, I would hope that any proposed definition would permit something like $2mathbb Z$ to be a pseudo-UFD.
The main problem is to define what "uniqueness of factorization" means, and that is what you've attempted with the criterion mentioned. Requiring two factorizations into irreducibles to be the same length is a must. The main barrier is to deal with associate elements without mentioning units. The way you have things now, we have to allow $pm x$ to be non-associate to one another, but this is unattractive.
One can patch this artificially by saying that $pm x$ are associates to one another, and then perhaps we do get $2mathbb Z$ to be a pseudo-UFD. But unfortunately, this convention does not extend the normal definition of UFDs, since lots of UFDs have many more associates!
So, one can see here the challenges posed by talking about factorization in domains without identity. Perhaps it would be most fruitful to explore which domains without identity can be embedded in domains such that the only units are $pm 1$, so that the ad-hoc definition of associates mentioned above works.
answered Nov 21 '18 at 17:58
rschwieb
105k12100244
add a comment |
Personally, I would hope that any proposed definition would permit something like $2mathbb Z$ to be a pseudo-UFD.
The main problem is to define what "uniqueness of factorization" means, and that is what you've attempted with the criterion mentioned. Requiring two factorizations into irreducibles to be the same length is a must. The main barrier is to deal with associate elements without mentioning units. The way you have things now, we have to allow $pm x$ to be non-associate to one another, but this is unattractive.
One can patch this artificially by saying that $pm x$ are associates to one another, and then perhaps we do get $2mathbb Z$ to be a pseudo-UFD. But unfortunately, this convention does not extend the normal definition of UFDs, since lots of UFDs have many more associates!
So, one can see here the challenges posed by talking about factorization in domains without identity. Perhaps it would be most fruitful to explore which domains without identity can be embedded in domains such that the only units are $pm 1$, so that the ad-hoc definition of associates mentioned above works.
answered Nov 21 '18 at 17:58
rschwieb
105k12100244
Personally, I would hope that any proposed definition would permit something like $2mathbb Z$ to be a pseudo-UFD.
The main problem is to define what "uniqueness of factorization" means, and that is what you've attempted with the criterion mentioned. Requiring two factorizations into irreducibles to be the same length is a must. The main barrier is to deal with associate elements without mentioning units. The way you have things now, we have to allow $pm x$ to be non-associate to one another, but this is unattractive.
One can patch this artificially by saying that $pm x$ are associates to one another, and then perhaps we do get $2mathbb Z$ to be a pseudo-UFD. But unfortunately, this convention does not extend the normal definition of UFDs, since lots of UFDs have many more associates!
So, one can see here the challenges posed by talking about factorization in domains without identity. Perhaps it would be most fruitful to explore which domains without identity can be embedded in domains such that the only units are $pm 1$, so that the ad-hoc definition of associates mentioned above works.
answered Nov 21 '18 at 17:58
rschwieb
105k12100244
answered Nov 21 '18 at 17:58
rschwieb
105k12100244
answered Nov 21 '18 at 17:58
rschwieb
105k12100244
answered Nov 21 '18 at 17:58
rschwieb
105k12100244
105k12100244
add a comment |
add a comment |
Definition of a UFD is: $R$ is an integral domain ($R$ is a commutative ring having unity and no zero-divisors) such that each $ain R setminus (R^* cup {0})$ factors into a product of irreducible elements.Definition clears your doubt
answered Nov 21 '18 at 15:31
John Nash
7418
The definiton of domain I knew does not include the unity condition. And, as you can define irreducibility in rings without unity, the question is still alive, doesn' t it?
– Álvaro G. Tenorio
Nov 21 '18 at 15:40
add a comment |
Definition of a UFD is: $R$ is an integral domain ($R$ is a commutative ring having unity and no zero-divisors) such that each $ain R setminus (R^* cup {0})$ factors into a product of irreducible elements.Definition clears your doubt
answered Nov 21 '18 at 15:31
John Nash
7418
The definiton of domain I knew does not include the unity condition. And, as you can define irreducibility in rings without unity, the question is still alive, doesn' t it?
– Álvaro G. Tenorio
Nov 21 '18 at 15:40
add a comment |
Definition of a UFD is: $R$ is an integral domain ($R$ is a commutative ring having unity and no zero-divisors) such that each $ain R setminus (R^* cup {0})$ factors into a product of irreducible elements.Definition clears your doubt
answered Nov 21 '18 at 15:31
John Nash
7418
Definition of a UFD is: $R$ is an integral domain ($R$ is a commutative ring having unity and no zero-divisors) such that each $ain R setminus (R^* cup {0})$ factors into a product of irreducible elements.Definition clears your doubt
answered Nov 21 '18 at 15:31
John Nash
7418
edited Nov 21 '18 at 15:36
answered Nov 21 '18 at 15:31
John Nash
7418
answered Nov 21 '18 at 15:31
John Nash
7418
answered Nov 21 '18 at 15:31
John Nash
7418
7418
The definiton of domain I knew does not include the unity condition. And, as you can define irreducibility in rings without unity, the question is still alive, doesn' t it?
– Álvaro G. Tenorio
Nov 21 '18 at 15:40
add a comment |
The definiton of domain I knew does not include the unity condition. And, as you can define irreducibility in rings without unity, the question is still alive, doesn' t it?
– Álvaro G. Tenorio
Nov 21 '18 at 15:40
The definiton of domain I knew does not include the unity condition. And, as you can define irreducibility in rings without unity, the question is still alive, doesn' t it?
– Álvaro G. Tenorio
Nov 21 '18 at 15:40
The definiton of domain I knew does not include the unity condition. And, as you can define irreducibility in rings without unity, the question is still alive, doesn' t it?
– Álvaro G. Tenorio
Nov 21 '18 at 15:40
add a comment |
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Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 21 '18 at 17:42