A negation of definition












1












$begingroup$


Let $R$ a ring. We know that an ideal $I$ of $R$ is said prime if for all $a,bin R$ $$abin IRightarrow ain Iquadtext{or}quad bin I.$$



When an ideal is not prime? That is, what is the negation of this definition formally?



EDIT



I understood thanks to your comments that an ideal is not prime if $exists a,b in R$ such that $anotin I$ and $bnotin I$. From here can I say that an ideal is not prime if for all $a,bin Rsetminus I$, $abin I$?



Thanks






abstract-algebra







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$endgroup$












  • $begingroup$
    Regarding edit! I afraid you can't
    $endgroup$
    – Cloud JR
    Nov 28 '18 at 19:00










  • $begingroup$
    Your edit is not quite right
    $endgroup$
    – Prince M
    Nov 29 '18 at 7:48
















1












$begingroup$


Let $R$ a ring. We know that an ideal $I$ of $R$ is said prime if for all $a,bin R$ $$abin IRightarrow ain Iquadtext{or}quad bin I.$$



When an ideal is not prime? That is, what is the negation of this definition formally?



EDIT



I understood thanks to your comments that an ideal is not prime if $exists a,b in R$ such that $anotin I$ and $bnotin I$. From here can I say that an ideal is not prime if for all $a,bin Rsetminus I$, $abin I$?



Thanks






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Regarding edit! I afraid you can't
    $endgroup$
    – Cloud JR
    Nov 28 '18 at 19:00










  • $begingroup$
    Your edit is not quite right
    $endgroup$
    – Prince M
    Nov 29 '18 at 7:48














1












1








1


0



$begingroup$


Let $R$ a ring. We know that an ideal $I$ of $R$ is said prime if for all $a,bin R$ $$abin IRightarrow ain Iquadtext{or}quad bin I.$$



When an ideal is not prime? That is, what is the negation of this definition formally?



EDIT



I understood thanks to your comments that an ideal is not prime if $exists a,b in R$ such that $anotin I$ and $bnotin I$. From here can I say that an ideal is not prime if for all $a,bin Rsetminus I$, $abin I$?



Thanks






abstract-algebra







share|cite|improve this question











$endgroup$




Let $R$ a ring. We know that an ideal $I$ of $R$ is said prime if for all $a,bin R$ $$abin IRightarrow ain Iquadtext{or}quad bin I.$$



When an ideal is not prime? That is, what is the negation of this definition formally?



EDIT



I understood thanks to your comments that an ideal is not prime if $exists a,b in R$ such that $anotin I$ and $bnotin I$. From here can I say that an ideal is not prime if for all $a,bin Rsetminus I$, $abin I$?



Thanks






abstract-algebra




abstract-algebra






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share|cite|improve this question













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share|cite|improve this question








edited Nov 28 '18 at 18:44







Jack J.

















asked Nov 28 '18 at 18:22









Jack J.Jack J.

4411419




4411419












  • $begingroup$
    Regarding edit! I afraid you can't
    $endgroup$
    – Cloud JR
    Nov 28 '18 at 19:00










  • $begingroup$
    Your edit is not quite right
    $endgroup$
    – Prince M
    Nov 29 '18 at 7:48


















  • $begingroup$
    Regarding edit! I afraid you can't
    $endgroup$
    – Cloud JR
    Nov 28 '18 at 19:00










  • $begingroup$
    Your edit is not quite right
    $endgroup$
    – Prince M
    Nov 29 '18 at 7:48
















$begingroup$
Regarding edit! I afraid you can't
$endgroup$
– Cloud JR
Nov 28 '18 at 19:00




$begingroup$
Regarding edit! I afraid you can't
$endgroup$
– Cloud JR
Nov 28 '18 at 19:00












$begingroup$
Your edit is not quite right
$endgroup$
– Prince M
Nov 29 '18 at 7:48




$begingroup$
Your edit is not quite right
$endgroup$
– Prince M
Nov 29 '18 at 7:48










2 Answers
2






active

oldest

votes


















3












$begingroup$

$exists a,bin R$ with $abin I, anotin I, bnotin I$



For example, $2cdot3in 6mathbb{Z}, 2notin 6mathbb{Z}, 3notin 6 mathbb{Z},$ so $6 mathbb{Z}$ is not a prime ideal of $mathbb{Z}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your ansewer. But it's corret said that $I$ is not prime if exist $a,bin R$ such that if $abin I$, then $anotin I$ and $bnotin I$, therefore if and only if for all $a,bin Rsetminus I$ $abin I$?
    $endgroup$
    – Jack J.
    Nov 28 '18 at 18:38










  • $begingroup$
    @JackJ. No, that's wrong.
    $endgroup$
    – egreg
    Nov 28 '18 at 18:46










  • $begingroup$
    and what is the correct formulation?
    $endgroup$
    – Jack J.
    Nov 28 '18 at 18:46










  • $begingroup$
    @Jack J , what he written is absolutely correct ...negation of if... Then 'is not' if ... Then not
    $endgroup$
    – Cloud JR
    Nov 28 '18 at 18:47










  • $begingroup$
    negation of $Pimplies Q$ is $P$ and $neg Q$. I added this in my answer too
    $endgroup$
    – Cloud JR
    Nov 28 '18 at 18:48



















2












$begingroup$

An Ideal $l$ is not prime if there exists $a,bin R$ such that



$abin I , anotin I $and $bnotin I$



(Note : negation of $Pimplies Q$ is $P$ and $neg Q$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    3












    $begingroup$

    $exists a,bin R$ with $abin I, anotin I, bnotin I$



    For example, $2cdot3in 6mathbb{Z}, 2notin 6mathbb{Z}, 3notin 6 mathbb{Z},$ so $6 mathbb{Z}$ is not a prime ideal of $mathbb{Z}.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for your ansewer. But it's corret said that $I$ is not prime if exist $a,bin R$ such that if $abin I$, then $anotin I$ and $bnotin I$, therefore if and only if for all $a,bin Rsetminus I$ $abin I$?
      $endgroup$
      – Jack J.
      Nov 28 '18 at 18:38










    • $begingroup$
      @JackJ. No, that's wrong.
      $endgroup$
      – egreg
      Nov 28 '18 at 18:46










    • $begingroup$
      and what is the correct formulation?
      $endgroup$
      – Jack J.
      Nov 28 '18 at 18:46










    • $begingroup$
      @Jack J , what he written is absolutely correct ...negation of if... Then 'is not' if ... Then not
      $endgroup$
      – Cloud JR
      Nov 28 '18 at 18:47










    • $begingroup$
      negation of $Pimplies Q$ is $P$ and $neg Q$. I added this in my answer too
      $endgroup$
      – Cloud JR
      Nov 28 '18 at 18:48
















    3












    $begingroup$

    $exists a,bin R$ with $abin I, anotin I, bnotin I$



    For example, $2cdot3in 6mathbb{Z}, 2notin 6mathbb{Z}, 3notin 6 mathbb{Z},$ so $6 mathbb{Z}$ is not a prime ideal of $mathbb{Z}.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for your ansewer. But it's corret said that $I$ is not prime if exist $a,bin R$ such that if $abin I$, then $anotin I$ and $bnotin I$, therefore if and only if for all $a,bin Rsetminus I$ $abin I$?
      $endgroup$
      – Jack J.
      Nov 28 '18 at 18:38










    • $begingroup$
      @JackJ. No, that's wrong.
      $endgroup$
      – egreg
      Nov 28 '18 at 18:46










    • $begingroup$
      and what is the correct formulation?
      $endgroup$
      – Jack J.
      Nov 28 '18 at 18:46










    • $begingroup$
      @Jack J , what he written is absolutely correct ...negation of if... Then 'is not' if ... Then not
      $endgroup$
      – Cloud JR
      Nov 28 '18 at 18:47










    • $begingroup$
      negation of $Pimplies Q$ is $P$ and $neg Q$. I added this in my answer too
      $endgroup$
      – Cloud JR
      Nov 28 '18 at 18:48














    3












    3








    3





    $begingroup$

    $exists a,bin R$ with $abin I, anotin I, bnotin I$



    For example, $2cdot3in 6mathbb{Z}, 2notin 6mathbb{Z}, 3notin 6 mathbb{Z},$ so $6 mathbb{Z}$ is not a prime ideal of $mathbb{Z}.$






    share|cite|improve this answer









    $endgroup$



    $exists a,bin R$ with $abin I, anotin I, bnotin I$



    For example, $2cdot3in 6mathbb{Z}, 2notin 6mathbb{Z}, 3notin 6 mathbb{Z},$ so $6 mathbb{Z}$ is not a prime ideal of $mathbb{Z}.$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 28 '18 at 18:26









    saulspatzsaulspatz

    14.1k21329




    14.1k21329












    • $begingroup$
      Thanks for your ansewer. But it's corret said that $I$ is not prime if exist $a,bin R$ such that if $abin I$, then $anotin I$ and $bnotin I$, therefore if and only if for all $a,bin Rsetminus I$ $abin I$?
      $endgroup$
      – Jack J.
      Nov 28 '18 at 18:38










    • $begingroup$
      @JackJ. No, that's wrong.
      $endgroup$
      – egreg
      Nov 28 '18 at 18:46










    • $begingroup$
      and what is the correct formulation?
      $endgroup$
      – Jack J.
      Nov 28 '18 at 18:46










    • $begingroup$
      @Jack J , what he written is absolutely correct ...negation of if... Then 'is not' if ... Then not
      $endgroup$
      – Cloud JR
      Nov 28 '18 at 18:47










    • $begingroup$
      negation of $Pimplies Q$ is $P$ and $neg Q$. I added this in my answer too
      $endgroup$
      – Cloud JR
      Nov 28 '18 at 18:48


















    • $begingroup$
      Thanks for your ansewer. But it's corret said that $I$ is not prime if exist $a,bin R$ such that if $abin I$, then $anotin I$ and $bnotin I$, therefore if and only if for all $a,bin Rsetminus I$ $abin I$?
      $endgroup$
      – Jack J.
      Nov 28 '18 at 18:38










    • $begingroup$
      @JackJ. No, that's wrong.
      $endgroup$
      – egreg
      Nov 28 '18 at 18:46










    • $begingroup$
      and what is the correct formulation?
      $endgroup$
      – Jack J.
      Nov 28 '18 at 18:46










    • $begingroup$
      @Jack J , what he written is absolutely correct ...negation of if... Then 'is not' if ... Then not
      $endgroup$
      – Cloud JR
      Nov 28 '18 at 18:47










    • $begingroup$
      negation of $Pimplies Q$ is $P$ and $neg Q$. I added this in my answer too
      $endgroup$
      – Cloud JR
      Nov 28 '18 at 18:48
















    $begingroup$
    Thanks for your ansewer. But it's corret said that $I$ is not prime if exist $a,bin R$ such that if $abin I$, then $anotin I$ and $bnotin I$, therefore if and only if for all $a,bin Rsetminus I$ $abin I$?
    $endgroup$
    – Jack J.
    Nov 28 '18 at 18:38




    $begingroup$
    Thanks for your ansewer. But it's corret said that $I$ is not prime if exist $a,bin R$ such that if $abin I$, then $anotin I$ and $bnotin I$, therefore if and only if for all $a,bin Rsetminus I$ $abin I$?
    $endgroup$
    – Jack J.
    Nov 28 '18 at 18:38












    $begingroup$
    @JackJ. No, that's wrong.
    $endgroup$
    – egreg
    Nov 28 '18 at 18:46




    $begingroup$
    @JackJ. No, that's wrong.
    $endgroup$
    – egreg
    Nov 28 '18 at 18:46












    $begingroup$
    and what is the correct formulation?
    $endgroup$
    – Jack J.
    Nov 28 '18 at 18:46




    $begingroup$
    and what is the correct formulation?
    $endgroup$
    – Jack J.
    Nov 28 '18 at 18:46












    $begingroup$
    @Jack J , what he written is absolutely correct ...negation of if... Then 'is not' if ... Then not
    $endgroup$
    – Cloud JR
    Nov 28 '18 at 18:47




    $begingroup$
    @Jack J , what he written is absolutely correct ...negation of if... Then 'is not' if ... Then not
    $endgroup$
    – Cloud JR
    Nov 28 '18 at 18:47












    $begingroup$
    negation of $Pimplies Q$ is $P$ and $neg Q$. I added this in my answer too
    $endgroup$
    – Cloud JR
    Nov 28 '18 at 18:48




    $begingroup$
    negation of $Pimplies Q$ is $P$ and $neg Q$. I added this in my answer too
    $endgroup$
    – Cloud JR
    Nov 28 '18 at 18:48











    2












    $begingroup$

    An Ideal $l$ is not prime if there exists $a,bin R$ such that



    $abin I , anotin I $and $bnotin I$



    (Note : negation of $Pimplies Q$ is $P$ and $neg Q$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      An Ideal $l$ is not prime if there exists $a,bin R$ such that



      $abin I , anotin I $and $bnotin I$



      (Note : negation of $Pimplies Q$ is $P$ and $neg Q$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        An Ideal $l$ is not prime if there exists $a,bin R$ such that



        $abin I , anotin I $and $bnotin I$



        (Note : negation of $Pimplies Q$ is $P$ and $neg Q$






        share|cite|improve this answer









        $endgroup$



        An Ideal $l$ is not prime if there exists $a,bin R$ such that



        $abin I , anotin I $and $bnotin I$



        (Note : negation of $Pimplies Q$ is $P$ and $neg Q$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 18:28









        Cloud JRCloud JR

        863517




        863517






























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