Suppose an ideal $q$ is $p$-primary, prove or disprove that if $(q:x)=q$, then $x$ is not in $p$












1












$begingroup$


Suppose $A$ is a commutative ring with unity and an ideal $q$ of $A$ is $p$-primary, i.e. $sqrt{q}=p$. It is known that for $x in A $, we have




  • if $x notin q$, then $(q:x)$ is $p$-primary.

  • if $xnotin p$, then $(q:x)=q$.


I was wondering whether the following is true:




if $(q:x)=q$, then $x$ is not in $p$.




Thank you for your time in advance.






commutative-algebra







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    1












    $begingroup$


    Suppose $A$ is a commutative ring with unity and an ideal $q$ of $A$ is $p$-primary, i.e. $sqrt{q}=p$. It is known that for $x in A $, we have




    • if $x notin q$, then $(q:x)$ is $p$-primary.

    • if $xnotin p$, then $(q:x)=q$.


    I was wondering whether the following is true:




    if $(q:x)=q$, then $x$ is not in $p$.




    Thank you for your time in advance.






    commutative-algebra







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Suppose $A$ is a commutative ring with unity and an ideal $q$ of $A$ is $p$-primary, i.e. $sqrt{q}=p$. It is known that for $x in A $, we have




      • if $x notin q$, then $(q:x)$ is $p$-primary.

      • if $xnotin p$, then $(q:x)=q$.


      I was wondering whether the following is true:




      if $(q:x)=q$, then $x$ is not in $p$.




      Thank you for your time in advance.






      commutative-algebra







      share|cite|improve this question









      $endgroup$




      Suppose $A$ is a commutative ring with unity and an ideal $q$ of $A$ is $p$-primary, i.e. $sqrt{q}=p$. It is known that for $x in A $, we have




      • if $x notin q$, then $(q:x)$ is $p$-primary.

      • if $xnotin p$, then $(q:x)=q$.


      I was wondering whether the following is true:




      if $(q:x)=q$, then $x$ is not in $p$.




      Thank you for your time in advance.






      commutative-algebra




      commutative-algebra






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      share|cite|improve this question










      asked Dec 2 '18 at 7:52









      Michael.LMichael.L

      877




      877






















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          $begingroup$

          Suppose $(q:x)=q$ and $x in p$, then we have
          $$x not in q quadtext{ and }quad x^n in q text{ for some $n>1$,}$$
          where the first statement is true since otherwise $(q:x)=(1)$.



          Choose $m$ to be the minimal positive integer such that $x^m in q$ and $x^{m-1} not in q$. Such $m$ exists since in the sequence $x,x^2,x^3,....$, we have $x not in q$ and $x^k in q$ for $kgeq n>1$.



          So by definition we have $x^{m-1} in (q:x)$, which contradict $(q:x)=q$.






          share|cite|improve this answer









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            $begingroup$

            Suppose $(q:x)=q$ and $x in p$, then we have
            $$x not in q quadtext{ and }quad x^n in q text{ for some $n>1$,}$$
            where the first statement is true since otherwise $(q:x)=(1)$.



            Choose $m$ to be the minimal positive integer such that $x^m in q$ and $x^{m-1} not in q$. Such $m$ exists since in the sequence $x,x^2,x^3,....$, we have $x not in q$ and $x^k in q$ for $kgeq n>1$.



            So by definition we have $x^{m-1} in (q:x)$, which contradict $(q:x)=q$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Suppose $(q:x)=q$ and $x in p$, then we have
              $$x not in q quadtext{ and }quad x^n in q text{ for some $n>1$,}$$
              where the first statement is true since otherwise $(q:x)=(1)$.



              Choose $m$ to be the minimal positive integer such that $x^m in q$ and $x^{m-1} not in q$. Such $m$ exists since in the sequence $x,x^2,x^3,....$, we have $x not in q$ and $x^k in q$ for $kgeq n>1$.



              So by definition we have $x^{m-1} in (q:x)$, which contradict $(q:x)=q$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Suppose $(q:x)=q$ and $x in p$, then we have
                $$x not in q quadtext{ and }quad x^n in q text{ for some $n>1$,}$$
                where the first statement is true since otherwise $(q:x)=(1)$.



                Choose $m$ to be the minimal positive integer such that $x^m in q$ and $x^{m-1} not in q$. Such $m$ exists since in the sequence $x,x^2,x^3,....$, we have $x not in q$ and $x^k in q$ for $kgeq n>1$.



                So by definition we have $x^{m-1} in (q:x)$, which contradict $(q:x)=q$.






                share|cite|improve this answer









                $endgroup$



                Suppose $(q:x)=q$ and $x in p$, then we have
                $$x not in q quadtext{ and }quad x^n in q text{ for some $n>1$,}$$
                where the first statement is true since otherwise $(q:x)=(1)$.



                Choose $m$ to be the minimal positive integer such that $x^m in q$ and $x^{m-1} not in q$. Such $m$ exists since in the sequence $x,x^2,x^3,....$, we have $x not in q$ and $x^k in q$ for $kgeq n>1$.



                So by definition we have $x^{m-1} in (q:x)$, which contradict $(q:x)=q$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 '18 at 11:37









                Michael.LMichael.L

                877




                877






























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