A ring without identity: deny the hypothesis












0












$begingroup$


Let $R$ a ring without identity. So, $$(Rquadtext{does not have the identity})Rightarrow (nexists;ein R;text{such that};ea=a;forall ain R)[text{It's correct?}].$$



Now, if I found an element $ein R$ and an element $ain R$ such that $ea=a$, can I conclude that this contradicts the hypothesis according to which $R$ not have identity?






abstract-algebra ring-theory







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  • $begingroup$
    For $e$ to be an identity we need $ea = a$ for all $a$, not just one single $a$. So you can‘t conclude what you said.
    $endgroup$
    – Lukas Kofler
    Dec 3 '18 at 14:25
















0












$begingroup$


Let $R$ a ring without identity. So, $$(Rquadtext{does not have the identity})Rightarrow (nexists;ein R;text{such that};ea=a;forall ain R)[text{It's correct?}].$$



Now, if I found an element $ein R$ and an element $ain R$ such that $ea=a$, can I conclude that this contradicts the hypothesis according to which $R$ not have identity?






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    For $e$ to be an identity we need $ea = a$ for all $a$, not just one single $a$. So you can‘t conclude what you said.
    $endgroup$
    – Lukas Kofler
    Dec 3 '18 at 14:25














0












0








0





$begingroup$


Let $R$ a ring without identity. So, $$(Rquadtext{does not have the identity})Rightarrow (nexists;ein R;text{such that};ea=a;forall ain R)[text{It's correct?}].$$



Now, if I found an element $ein R$ and an element $ain R$ such that $ea=a$, can I conclude that this contradicts the hypothesis according to which $R$ not have identity?






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$




Let $R$ a ring without identity. So, $$(Rquadtext{does not have the identity})Rightarrow (nexists;ein R;text{such that};ea=a;forall ain R)[text{It's correct?}].$$



Now, if I found an element $ein R$ and an element $ain R$ such that $ea=a$, can I conclude that this contradicts the hypothesis according to which $R$ not have identity?






abstract-algebra ring-theory




abstract-algebra ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 14:39









José Carlos Santos

157k22126227




157k22126227










asked Dec 3 '18 at 14:23









Jack J.Jack J.

4292419




4292419












  • $begingroup$
    For $e$ to be an identity we need $ea = a$ for all $a$, not just one single $a$. So you can‘t conclude what you said.
    $endgroup$
    – Lukas Kofler
    Dec 3 '18 at 14:25


















  • $begingroup$
    For $e$ to be an identity we need $ea = a$ for all $a$, not just one single $a$. So you can‘t conclude what you said.
    $endgroup$
    – Lukas Kofler
    Dec 3 '18 at 14:25
















$begingroup$
For $e$ to be an identity we need $ea = a$ for all $a$, not just one single $a$. So you can‘t conclude what you said.
$endgroup$
– Lukas Kofler
Dec 3 '18 at 14:25




$begingroup$
For $e$ to be an identity we need $ea = a$ for all $a$, not just one single $a$. So you can‘t conclude what you said.
$endgroup$
– Lukas Kofler
Dec 3 '18 at 14:25










2 Answers
2






active

oldest

votes


















2












$begingroup$

First of all, we write quantifiers before statements they are referring. The notation



$$notexists ein R text{ such that }ea=aforall ain R$$



is invalid, simply because it is unclear. We don't know if it means



$$forall ain R:notexists ein R text{ such that }ea=a$$
or
$$notexists ein R: forall ain R: text{ such that }ea=a.$$



The two statements above are not equivalent, and are in fact very much different. The second statement is much stronger.




Second of all, the negation of the statement "$R$ has an identity" is the negation of the statement



$$exists ein R: forall ain R: ae=e$$



and the negation of this statement is the second of the two statements above.



Your argument that if you find some $e,a$ such that $ea=a$, is not enough to prove that $R$ has no identity, since even in rings without an identity, we can set $a=e=0$ and have $ea=a$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for this. You said what I wanted to, and probably better.
    $endgroup$
    – Lubin
    Dec 3 '18 at 21:24



















2












$begingroup$

No, you cannot. In any ring (with or without identity) such elements exist: just take $e=a=0$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

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    active

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    active

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    2












    $begingroup$

    First of all, we write quantifiers before statements they are referring. The notation



    $$notexists ein R text{ such that }ea=aforall ain R$$



    is invalid, simply because it is unclear. We don't know if it means



    $$forall ain R:notexists ein R text{ such that }ea=a$$
    or
    $$notexists ein R: forall ain R: text{ such that }ea=a.$$



    The two statements above are not equivalent, and are in fact very much different. The second statement is much stronger.




    Second of all, the negation of the statement "$R$ has an identity" is the negation of the statement



    $$exists ein R: forall ain R: ae=e$$



    and the negation of this statement is the second of the two statements above.



    Your argument that if you find some $e,a$ such that $ea=a$, is not enough to prove that $R$ has no identity, since even in rings without an identity, we can set $a=e=0$ and have $ea=a$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for this. You said what I wanted to, and probably better.
      $endgroup$
      – Lubin
      Dec 3 '18 at 21:24
















    2












    $begingroup$

    First of all, we write quantifiers before statements they are referring. The notation



    $$notexists ein R text{ such that }ea=aforall ain R$$



    is invalid, simply because it is unclear. We don't know if it means



    $$forall ain R:notexists ein R text{ such that }ea=a$$
    or
    $$notexists ein R: forall ain R: text{ such that }ea=a.$$



    The two statements above are not equivalent, and are in fact very much different. The second statement is much stronger.




    Second of all, the negation of the statement "$R$ has an identity" is the negation of the statement



    $$exists ein R: forall ain R: ae=e$$



    and the negation of this statement is the second of the two statements above.



    Your argument that if you find some $e,a$ such that $ea=a$, is not enough to prove that $R$ has no identity, since even in rings without an identity, we can set $a=e=0$ and have $ea=a$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for this. You said what I wanted to, and probably better.
      $endgroup$
      – Lubin
      Dec 3 '18 at 21:24














    2












    2








    2





    $begingroup$

    First of all, we write quantifiers before statements they are referring. The notation



    $$notexists ein R text{ such that }ea=aforall ain R$$



    is invalid, simply because it is unclear. We don't know if it means



    $$forall ain R:notexists ein R text{ such that }ea=a$$
    or
    $$notexists ein R: forall ain R: text{ such that }ea=a.$$



    The two statements above are not equivalent, and are in fact very much different. The second statement is much stronger.




    Second of all, the negation of the statement "$R$ has an identity" is the negation of the statement



    $$exists ein R: forall ain R: ae=e$$



    and the negation of this statement is the second of the two statements above.



    Your argument that if you find some $e,a$ such that $ea=a$, is not enough to prove that $R$ has no identity, since even in rings without an identity, we can set $a=e=0$ and have $ea=a$.






    share|cite|improve this answer









    $endgroup$



    First of all, we write quantifiers before statements they are referring. The notation



    $$notexists ein R text{ such that }ea=aforall ain R$$



    is invalid, simply because it is unclear. We don't know if it means



    $$forall ain R:notexists ein R text{ such that }ea=a$$
    or
    $$notexists ein R: forall ain R: text{ such that }ea=a.$$



    The two statements above are not equivalent, and are in fact very much different. The second statement is much stronger.




    Second of all, the negation of the statement "$R$ has an identity" is the negation of the statement



    $$exists ein R: forall ain R: ae=e$$



    and the negation of this statement is the second of the two statements above.



    Your argument that if you find some $e,a$ such that $ea=a$, is not enough to prove that $R$ has no identity, since even in rings without an identity, we can set $a=e=0$ and have $ea=a$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 3 '18 at 14:34









    5xum5xum

    90.3k394161




    90.3k394161












    • $begingroup$
      Thanks for this. You said what I wanted to, and probably better.
      $endgroup$
      – Lubin
      Dec 3 '18 at 21:24


















    • $begingroup$
      Thanks for this. You said what I wanted to, and probably better.
      $endgroup$
      – Lubin
      Dec 3 '18 at 21:24
















    $begingroup$
    Thanks for this. You said what I wanted to, and probably better.
    $endgroup$
    – Lubin
    Dec 3 '18 at 21:24




    $begingroup$
    Thanks for this. You said what I wanted to, and probably better.
    $endgroup$
    – Lubin
    Dec 3 '18 at 21:24











    2












    $begingroup$

    No, you cannot. In any ring (with or without identity) such elements exist: just take $e=a=0$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      No, you cannot. In any ring (with or without identity) such elements exist: just take $e=a=0$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        No, you cannot. In any ring (with or without identity) such elements exist: just take $e=a=0$.






        share|cite|improve this answer









        $endgroup$



        No, you cannot. In any ring (with or without identity) such elements exist: just take $e=a=0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 14:25









        José Carlos SantosJosé Carlos Santos

        157k22126227




        157k22126227






























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