Is there a “pseudo-UFD” without unity?











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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?










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  • Comments are not for extended discussion; this conversation has been moved to chat.
    – Aloizio Macedo
    Nov 21 at 17:42















up vote
1
down vote

favorite












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?










share|cite|improve this question
























  • Comments are not for extended discussion; this conversation has been moved to chat.
    – Aloizio Macedo
    Nov 21 at 17:42













up vote
1
down vote

favorite









up vote
1
down vote

favorite











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?










share|cite|improve this question















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






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edited Nov 21 at 17:41

























asked Nov 21 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 at 17:42


















  • Comments are not for extended discussion; this conversation has been moved to chat.
    – Aloizio Macedo
    Nov 21 at 17:42
















Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo
Nov 21 at 17:42




Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo
Nov 21 at 17:42










2 Answers
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2
down vote



accepted










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.






share|cite|improve this answer




























    up vote
    -2
    down vote













    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






    share|cite|improve this answer























    • 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 at 15:40











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    2 Answers
    2






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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    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.






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      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.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 at 17:58









        rschwieb

        104k1299242




        104k1299242






















            up vote
            -2
            down vote













            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






            share|cite|improve this answer























            • 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 at 15:40















            up vote
            -2
            down vote













            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






            share|cite|improve this answer























            • 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 at 15:40













            up vote
            -2
            down vote










            up vote
            -2
            down vote









            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






            share|cite|improve this answer














            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







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 21 at 15:36

























            answered Nov 21 at 15:31









            John Nash

            6818




            6818












            • 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 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 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 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 at 15:40


















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