Prove or disprove this table is a field












3












$begingroup$


enter image description here



Any help will be appreciated on how to approach this and get started.






abstract-algebra discrete-mathematics







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$endgroup$








  • 4




    $begingroup$
    Hint: removal of $0$ should yield a group under multiplication.
    $endgroup$
    – Randall
    2 hours ago
















3












$begingroup$


enter image description here



Any help will be appreciated on how to approach this and get started.






abstract-algebra discrete-mathematics







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Hint: removal of $0$ should yield a group under multiplication.
    $endgroup$
    – Randall
    2 hours ago














3












3








3


1



$begingroup$


enter image description here



Any help will be appreciated on how to approach this and get started.






abstract-algebra discrete-mathematics







share|cite|improve this question









$endgroup$




enter image description here



Any help will be appreciated on how to approach this and get started.






abstract-algebra discrete-mathematics




abstract-algebra discrete-mathematics






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asked 2 hours ago









Wade KemmsiesWade Kemmsies

283




283








  • 4




    $begingroup$
    Hint: removal of $0$ should yield a group under multiplication.
    $endgroup$
    – Randall
    2 hours ago














  • 4




    $begingroup$
    Hint: removal of $0$ should yield a group under multiplication.
    $endgroup$
    – Randall
    2 hours ago








4




4




$begingroup$
Hint: removal of $0$ should yield a group under multiplication.
$endgroup$
– Randall
2 hours ago




$begingroup$
Hint: removal of $0$ should yield a group under multiplication.
$endgroup$
– Randall
2 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

A field has no non-zero divisors of zero.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Note from the multiplication table, the one for the $cdot$ operation, that



    $2 cdot 2 = 0; tag 1$



    thus $2$ is nilpotent; but a field $F$ can have no non-zero nilpotents (note $2 ne 0$ since $3 + 2 = 1 ne 3$).



    If $F$ is a field and $a in F$ were nilpotent, that is, if



    $a^k = 0, ; 2 le k in Bbb N, tag 2$



    then



    $a = a^{1 - k} a^k = a^{1 - k} = 0; tag 3$



    which shows a field has no non-zero nilpotent elements.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
      $endgroup$
      – J. W. Tanner
      4 mins ago













    Your Answer





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    2 Answers
    2






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    2 Answers
    2






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    active

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    active

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    2












    $begingroup$

    A field has no non-zero divisors of zero.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      A field has no non-zero divisors of zero.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        A field has no non-zero divisors of zero.






        share|cite|improve this answer









        $endgroup$



        A field has no non-zero divisors of zero.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        J. W. TannerJ. W. Tanner

        937




        937























            2












            $begingroup$

            Note from the multiplication table, the one for the $cdot$ operation, that



            $2 cdot 2 = 0; tag 1$



            thus $2$ is nilpotent; but a field $F$ can have no non-zero nilpotents (note $2 ne 0$ since $3 + 2 = 1 ne 3$).



            If $F$ is a field and $a in F$ were nilpotent, that is, if



            $a^k = 0, ; 2 le k in Bbb N, tag 2$



            then



            $a = a^{1 - k} a^k = a^{1 - k} = 0; tag 3$



            which shows a field has no non-zero nilpotent elements.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
              $endgroup$
              – J. W. Tanner
              4 mins ago


















            2












            $begingroup$

            Note from the multiplication table, the one for the $cdot$ operation, that



            $2 cdot 2 = 0; tag 1$



            thus $2$ is nilpotent; but a field $F$ can have no non-zero nilpotents (note $2 ne 0$ since $3 + 2 = 1 ne 3$).



            If $F$ is a field and $a in F$ were nilpotent, that is, if



            $a^k = 0, ; 2 le k in Bbb N, tag 2$



            then



            $a = a^{1 - k} a^k = a^{1 - k} = 0; tag 3$



            which shows a field has no non-zero nilpotent elements.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
              $endgroup$
              – J. W. Tanner
              4 mins ago
















            2












            2








            2





            $begingroup$

            Note from the multiplication table, the one for the $cdot$ operation, that



            $2 cdot 2 = 0; tag 1$



            thus $2$ is nilpotent; but a field $F$ can have no non-zero nilpotents (note $2 ne 0$ since $3 + 2 = 1 ne 3$).



            If $F$ is a field and $a in F$ were nilpotent, that is, if



            $a^k = 0, ; 2 le k in Bbb N, tag 2$



            then



            $a = a^{1 - k} a^k = a^{1 - k} = 0; tag 3$



            which shows a field has no non-zero nilpotent elements.






            share|cite|improve this answer











            $endgroup$



            Note from the multiplication table, the one for the $cdot$ operation, that



            $2 cdot 2 = 0; tag 1$



            thus $2$ is nilpotent; but a field $F$ can have no non-zero nilpotents (note $2 ne 0$ since $3 + 2 = 1 ne 3$).



            If $F$ is a field and $a in F$ were nilpotent, that is, if



            $a^k = 0, ; 2 le k in Bbb N, tag 2$



            then



            $a = a^{1 - k} a^k = a^{1 - k} = 0; tag 3$



            which shows a field has no non-zero nilpotent elements.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 hours ago

























            answered 2 hours ago









            Robert LewisRobert Lewis

            44.5k22964




            44.5k22964












            • $begingroup$
              Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
              $endgroup$
              – J. W. Tanner
              4 mins ago




















            • $begingroup$
              Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
              $endgroup$
              – J. W. Tanner
              4 mins ago


















            $begingroup$
            Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
            $endgroup$
            – J. W. Tanner
            4 mins ago






            $begingroup$
            Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
            $endgroup$
            – J. W. Tanner
            4 mins ago




















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